Elementary Differential Equations. solutions

INSTRUCTOR’S SOLUTIONS MANUAL LEE JOHNSON Virginia Tech

JEREMY BOURDON Virginia Tech

ELEMENTARY DIFFERENTIAL EQUATIONS WERNER KOHLER Virginia Tech

LEE JOHNSON Virginia Tech

Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors. Copyright © 2004 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston MA 02116 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN

0-201-77034-2

1 2 3 4 5 6 QEP 06 05 04 03

CONTENTS Chapter 1:

Introduction to Differential Equations

1

Chapter 2:

First Order Linear Differential Equations

5

Chapter 3:

First Order Nonlinear Differential Equations

26

Chapter 4:

Second Order Linear Differential Equations

51

Chapter 5:

Higher Order Linear Differential Equations

101

Chapter 6:

First Order Linear Systems

115

Chapter 7:

Laplace Transforms

178

Chapter 8:

Nonlinear Systems

214

Chapter 9:

Numerical Methods

250

Chapter 10:

Series Solutions of Linear Differential Equations

268

Chapter 1 Introduction to Differential Equations Section 1.1 1.

This D.E. is of order two because the highest derivative in the equation is y ¢¢ .

2.

Order is 1.

3.

This D.E. is of order one because the highest derivative in the equation is y ¢ . (Note:

( y ¢ ) 3 π y ¢¢¢ ) 4.

Order is 3.

5 (a).

y = Ce t . Differentiating gives us y ¢ = Ce t ◊ 2 t = 2 ty . Therefore, y ¢ - 2 ty = 0 for

2

2

any value of C . 2

5 (b). Substituting into the differential equation yields y (1) = Ce1 = Ce . Using the initial condition, y (1) = 2 = Ce . Solving for C , we find C = 2e -1 .

6.

y ¢¢¢ = 2.

t3 t2 y ¢¢ = 2 t + c1, y = t + c1t + c 2 , y = + c1 + c 2 t + c 3 . 3 2 1

Order = 3 7 (a).

2

3 arbitrary constants

y = C1 sin 2 t + C2 cos 2 t . Differentiating gives us y ¢ = 2C1 cos 2 t - 2C2 sin 2 t and y ¢¢ = -4 C1 sin 2 t - 4 C2 cos 2 t = -4 (C1 sin 2 t + C2 cos 2 t) = -4 y . Therefore, y ¢¢ + 4 y = -4 y + 4 y = 0 and thus y ( t) = C1 sin 2 t + C2 cos 2 t is a solution of the D.E. y ¢¢ + 4 y = 0 .

7 (b).

y ( p4 ) = C1 (1) + C2 (0) = C1 = 3 and y ¢ ( p4 ) = 2C1 (0) - 2C2 (1) = -2C2 = -2 fi C2 = 1.

8.

y = 2e -4 t .

y ¢ + ky = -8e -4 t + 2 ke -4 t = 2( k - 4 )e -4 t = 0

\ k = 4.

y (0) = 2 = y 0 .

9.

\ k = 4, y 0 = 2.

y = ct -1 . Differentiating gives us y ¢ = -ct -2 . Thus y ¢ + y 2 = -ct -2 + c 2 t -2 = (c 2 - c ) t -2 = 0 . Solving this for c , we find that c 2 - c = c (c - 1) = 0 . Therefore, c = 0,1.

10.

y = -e - t + sin t

y ¢ + y = g( t), y (0) = y 0 .

y ¢ = e - t + cos t

y ¢ + y = e - t + cos t - e - t + sin t = g \ g( t) = cos t + sin t, y (0) = -1 = y 0

2 • Chapter 1 Introduction to Differential Equations

11.

y = t r . Differentiating gives us y ¢ = rt r -1 and y ¢¢ = r( r - 1) t r - 2 . Thus t 2 y ¢¢ - 2 ty ¢ + 2 y = r( r - 1) t r - 2 rt r + 2 t r = [ r( r - 1) - 2 r + 2]t r = 0 . Solving this for r , we find that r( r - 1) - 2 r + 2 = r 2 - 3r + 2 = ( r - 2)( r - 1) = 0 . Therefore, r = 1, 2 .

12.

y = c1e 2 t + c 2e -2 t . y ¢ = 2c1e 2 t - 2c 2e -2 t , y ¢¢ = 4 c1e 2 t + 4 c 2e -2 t = 4 y \ y ¢¢ - 4 y = 0.

13.

From (12), y = C1e 2 t + C2e -2 t , which we differentiate to get y ¢ = 2C1e 2 t - 2C2e -2 t . Using the initial conditions, y(0) = 2 and y ¢ (0) = 0 , we have two equations containing C1 and C2 : C1 + C2 = 2 and 2C1 - 2C2 = 0 . Solving these simultaneous equations gives us C1 = C2 = 1.

Thus, the solution to the initial value problem is y = e 2 t + e -2 t = 2 cosh(2 t) .

y ( t) = e 2 t .

14.

y (0) = c1 + c 2 = 1, 2c1 - 2c 2 = 2 \ c1 = 1, c 2 = 0

15.

From (12), y ( t) = C1e 2 t + C2e -2 t . Using the initial condition y(0) = 3, we find that C1 + C2 = 3. From the initial condition lim y ( t) = 0 and the equation for y ( t) given to us in (12), we can t Æ•

conclude that C1 = 0 (if C1 π 0 , then lim = ±•). Therefore, C2 = 3 and y ( t) = 3e -2 t . tÆ•

lim y ( t) = 0 fi c 2 = 0 \ c1 = 10 & y ( t) = 10e 2 t .

16.

c1 + c 2 = 10

17.

From the graph, we can see that y ¢ = -1 and that y(1) = 1. Thus m = y ¢ - 1 = -1 - 1 = -2 and

t Æ-•

y 0 = y (1) = 1.

18.

y ¢ = mt fi y = Also c = -1.

19.

m 2 t + c. 2

From graph, y = -1 only at t = 0 \ t0 = 0.

From graph y (1) = -0.5 \ -

1 m = - 1 fi m = 1. 2 2

We know that this is a freefall problem, so we can begin with the generic equation for freefall

g situations: y ( t) = - t 2 + v 0 t + y 0 . The object is released from rest, so v 0 = 0 . The impact time 2 corresponds to the time at which y = 0 , so we are left with the following equation for the

g impact time t : 0 = - t 2 + y 0 . Solving this for t yields t = 2 of impact: v = y ¢ = - gt + v 0 = - gt = - 2 gy 0 . 20.

x ¢¢ = a

x ¢ = at + v 0 , v 0 = x 0 = 0 fi x =

at 2 + 0. 2

Ê 64 ˆ 88 = a(8) fi a = 11 ft / sec 2 . At t = 8, x = 11Á ˜ = 352 ft. Ë 2¯

2 y0 . For the velocity at the time g

Chapter 1 Introduction to Differential Equations • 3

21.

4 Ê pt ˆ Ê pt ˆ a = y ¢¢ = 32 - e sinÁ ˜ . Integrating gives us y ¢ = -32 t - e cosÁ ˜ + C . The object is Ë 4¯ Ë 4¯ p

4 4 dropped from rest, so y ¢ (0) = 0 = - e + C . Solving for C yields C = e , and putting this p p value back into the equation for y ¢ and simplifying gives us y ¢ = -32 t +

4 Ê Ê pt ˆ ˆ e Á1 - cosÁ ˜ ˜ . Ë 4 ¯¯ p Ë

2

4 Ê 4ˆ Ê pt ˆ Integrating again gives us y = -16 t + et - Á ˜ e sinÁ ˜ + C ¢ . Since the object is dropped Ë ¯ Ë 4¯ p p 2

from a height of 252 ft. (at t = 0 ), y (0) = C ¢ = 252 and thus 2

4 Ê 4ˆ Ê pt ˆ y = -16 t + et - Á ˜ e sinÁ ˜ + 252 . Finally, since y( 4 ) = 0 , Ëp ¯ Ë 4¯ p 2

2

4e p Ê 4ˆ y( 4 ) = 0 = -16 ◊ 4 + ◊ 4 - Á ˜ e sin(p ) + 252 . Solving for e yields e = . Ëp ¯ p 4 2

Section 1.2 1 (a). The equation is autonomous because y ¢ depends only on y . 1 (b). Setting y ¢ = 0 , we have 0 = - y + 1. Solving this for y yields the equilibrium solution: y = 1. 2 (a). not autonomous 2 (b). no equilibrium solutions, isoclines are t = const ant. 3 (a). The equation is autonomous because y ¢ depends only on y . 3 (b). Setting y ¢ = 0 , we have 0 = sin y . Solving this for y yields the equilibrium solutions: y = ± np . 4 (a). autonomous 4 (b).

y ( y - 1) = 0, y = 0, 1.

5 (a). The equation is autonomous because y ¢ does not depend explicitly on t . 5 (b). There are no equilibrium solutions because there are no points at which y ¢ = 0 . 6 (a). not autonomous 6 (b).

y = 0 is equilibrium solution, isoclines are hyperbolas.

7 (a).

c = -1: Setting c = -1 gives us - y + 1 = -1 which, solved for y , reads y = 2 . This is the isocline for c = -1.

c = 0 : Setting c = 0 gives us - y + 1 = 0 which, solved for y , reads y = 1. This is the isocline for c = 0 .

c = 1: Setting c = 1 gives us - y + 1 = 1 which, solved for y , reads y = 0 , the isocline for c = 1.

4 • Chapter 1 Introduction to Differential Equations

8 (a).

- y + t = -1 fi y = t + 1 -y + t = 0 fi y = t -y + t = 1 fi y = t -1

9 (a).

c = -1: Setting c = -1 gives us y 2 - t 2 = -1 which can be simplified to t 2 - y 2 = 1 (a hyperbola). This is the isocline for c = -1.

c = 0 : Setting c = 0 gives us y 2 - t 2 = 0 which can be simplified to y = ± t . This is the isocline for c = 0 .

c = 1: Setting c = 1 gives us y 2 - t 2 = 1 (a hyperbola). This is the isocline for c = 1. 10.

f (0) = f (2) = 0

y ¢ = y (2 - y )

y ¢ > 0 for 0 < y < 2, y ¢ < 0 for - • < y < 0 and 2 < y < • .

11.

One example that would fit these criteria is y ¢ = -( y - 1) 2 . For this autonomous D.E., y ¢ = 0 at y = 1 and y ¢ < 0 for -• < y < 1 and 1 < y < • .

12.

y ¢ = 1.

13.

One example that would fit these criteria is y ¢ = sin(2py ) . For this autonomous D.E., y ¢ = 0 at

y= 14.

c.

15.

f.

16.

a.

17.

b.

18.

d.

19.

e.

n . 2

Chapter 2 First Order Linear Differential Equations Section 2.1 1.

This equation is linear because it can be written in the form y ¢ + p( t) y = g( t) . It is nonhomogeneous because when it is put in this form, g( t) π 0 .

2.

nonlinear

3.

This equation is nonlinear because it cannot be written in the form y ¢ + p( t) y = g( t) .

4.

nonlinear

5.

This equation is nonlinear because it cannot be written in the form y ¢ + p( t) y = g( t) .

6.

linear, homogeneous

7.

This equation is nonlinear because it can be written in the form y ¢ + p( t) y = g( t) .

8.

nonlinear

9.

This equation is linear because it cannot be written in the form y ¢ + p( t) y = g( t) . It is nonhomogeneous because when it is put in this form, g( t) π 0 .

10.

linear, homogeneous

11 (a). Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since

t and sin( t) are t +1 2

both continuous for all t and -2 is on this interval. 11 (b). Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since

t and sin( t) are t +1 2

both continuous for all t and 0 is on this interval. 11 (c). Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since both continuous for all t and p is on this interval. 12 (a). 2 < t < • 12 (b). -2 < t < 2 12 (c). -2 < t < 2 12 (d). -• < t < -2

t and sin( t) are t +1 2

6 • Chapter 2 First Order Linear Differential Equations

13 (a). For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π 3. Therefore, Theorem 2.1 guarantees a unique solution for ( 3, •) , the largest interval that includes t = 5 . 13 (b). For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π 3. Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that

3 includes t = - . 2 13 (c). For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π 3. Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that includes t = 0 . 13 (d). For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π 3. Therefore, Theorem 2.1 guarantees a unique solution for (-•, -2) , the largest interval that includes t = -5 . 13 (e). For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π 3. Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that

3 includes t = . 2 14.

ln t + t -1 t-2

2

ln t t+1

=

t-2

undefined at t = 0, 2 .

14 (a). 2 < t < • . 14 (b). 0 < t < 2 . 14 (c). -• < t < 0 . 14 (d). -• < t < 0 . 15.

2

2

y ( t) = 3e t . Differentiating gives us y ¢ = 3e t (2 t) = 2 ty . Substituting these values into the given equation yields 2 ty + p( t) y = 0 . Solving this for p( t) , we find that p( t) = -2 t . Putting t = 0 into the equation for y gives us y 0 = 3 .

16(a). y = Ct r

y ¢ = Crt r -1

2 ty ¢ - 6 y = 0

\ 2Crt r - 6Ct r = (2 r - 6)Ct r = 0 fi (2 r - 6) y = 0 fi 2 r - 6 = 0 fi r = 3 y (-2) = C (-2) r = 8 fi C π 0 \ C (-2) 3 = 8 fi C = -1 16 (b). -• < t < 0 since p( t) = 16 (c). y ( t) = - t 3 , - • < t < • .

-3 t

Chapter 2 First Order Linear Differential Equations • 7

17.

y ( t) = 0 satisfies all of these conditions.

Section 2.2 1 (a). First, we will integrate p( t) = 3 to find P ( t) = 3t . The general solution, then, is

y ( t) = Ce - P ( t ) = Ce -3 t . 1 (b).

y (0) = C = -3 . Therefore, the solution to the initial value problem is y = -3e -3 t .

2 (a).

y¢ -

2 (b).

y (-1) = Ce - 2 = 2, C = 2e

1 y=0 2

t

t

(e - 2 y )¢ = 0, y = Ce 2 . 1

1

y ( t) = 2e

2

( t +1 ) 2

3 (a). We can rewrite this equation into the conventional form: y ¢ - 2 ty = 0 . Then we will integrate 2

p( t) = -2 t to find P ( t) = - t 2 . The general solution, then, is y ( t) = Ce - P ( t ) = Ce t .

3 (b).

y (1) = Ce = 3 . Solving for C yields C = 3e -1 . Therefore, the solution to the initial value 2

problem is y ( t) = 3e -1e t = 3e( t 4 (a).

ty ¢ - 4 y = 0 fi y ¢ -

4 y = 0. t

1 4 -4 4 y ¢ - 5 y = ( t y )¢ = 0 t t 4 (b).

2

-1)

.

4

Ú - t dt = -4 ln t = - ln(t

4

) \ m=

1 t4

y = Ct 4 .

y (1) = C = 1 \ y ( t) = t 4 .

5 (a). We can rewrite this equation into the conventional form: y ¢ +

p( t) =

4 y = 0 . Then we will integrate t

4 to find P ( t) = 4 ln t = ln t 4 . The general solution, then, is t -4

4

y ( t) = Ce - P ( t ) = Ce - ln t = Ce ln t = Ct -4 . 5 (b).

y (1) = C = 1. Therefore, the solution to the initial value problem is y ( t) = t -4 .

6 (a).

m = exp( t - cos t) \ y ( t) = Ce -( t - cos t ) .

6 (b).

p Êp ˆ y Á ˜ = Ce - 2 = 1 Ë 2¯

C=e

p

2

p

y = e 2e -( t - cos t ) = e

p

2 - t + cos t

.

7 (a). First, we will integrate p( t) = -2 cos(2 t) to find P ( t) = - sin(2 t) . The general solution, then, is

y ( t) = Ce - P ( t ) = Ce sin( 2 t ) . 7 (b).

y (p ) = C = -2 . Therefore, the solution to the initial value problem is y ( t) = -2e sin( 2 t ) .

8 (a).

(( t 2 + 1) y )¢ = 0

y=

C . t +1 2

8 • Chapter 2 First Order Linear Differential Equations

8 (b).

y (0) = C = 3 \ y ( t) =

3 . t +1 2

9 (a). We can rewrite this equation into the conventional form: y ¢ - 3( t 2 + 1) y = 0 . Then we will integrate p( t) = -3( t 2 + 1) to find P ( t) = - t 3 - 3t . The general solution, then, is y ( t) = Ce - P ( t ) = Ce t 9 (b).

3

+ 3t

.

y (1) = Ce 4 = 4 . Solving for C yields C = 4 e -4 . Therefore, the solution to the initial value problem is y ( t) = 4 e t

3

+ 3t - 4

10 (a). y ¢ + e - t y = 0 \

Úe

-t

10 (b). y (0) = Ce1 = 2

C = 2e -1

.

dt = -e - t

-t

(-e e y )¢ = 0

y ( t) = 2e e

-t

-1

-t

y = Ce e .

.

11 (a). #2 11 (b). #3 11 (c). #1 12.

y ( t) = y 0e -at

4 = y 0e -a , 1 = y 0e -3a 3

and y 0 = e 3a = e 2 13.

ln 4

1 Divide: 4 = e 2a fi a = ln 4 = ln 2 2

= e ln( 8 ) = 8. \ y ( t) = 8e -(ln 2 )t .

First, we should put the equation into our conventional form: y ¢ -

p( t) = -

a y = 0 . Integrating t

a gives us P ( t) = -a ln t = ln t -a . The general solution, then, is t

y ( t) = Ce - P ( t ) = Ce

- ln t -a

= Ce

ln t a

= Cta . Using the general solution and the point (2,1) , we can

solve for C in terms of a : y (2) = 1 = C ◊ 2a ; C = 2 -a . We can then substitute this value for C into the general solution at the point ( 4, 4 ) : y( 4 ) = 4 = 2 -a ◊ 4a = 4 -a / 2 ◊ 4a = 4a / 2 . Setting the exponents equal to each other yields 1 =

y 0 = y (1) = 2 -2 ◊ 12 =

a ;a = 2 . Finally, solving for y 0 , 2

1 . 4

14.

z¢ = 2 z, z = y + 2 \ z(0) = -1 + 2 = 1 fi z = e 2 t = y + 2 \ y = -2 + e 2 t

15.

Putting this equation into a form more like #14, we have y ¢ = -2 ty + 6 t = -2 t( y - 3) . We will then let z = y - 3 (and z¢ = y ¢ , accordingly). Substituting into our modified original equation yields an equation for z( t) : z¢ = -2 tz , or put in a more conventional form, z¢ + 2 tz = 0 . Using the same substitution for the initial condition yields z(0) = 4 - 3 = 1. Integrating p( t) = 2 t gives 2

us P ( t) = t 2 . The general solution is then z( t) = Ce - t . Our initial condition requires that C = 1,

Chapter 2 First Order Linear Differential Equations • 9 2

2

so the solution for z( t) is z( t) = e - t . In terms of y ( t) , this solution reads y - 3 = e - t . Solved 2

for y ( t) , this solution is y ( t) = e - t + 3. 16 (a).

dB = - kB, B(0) = - A* dc

16 (b). B(c ) = - A*e - kc = A(c ) - A* \ A(c ) = A* (1 - e - kc )

No. A(c ) ≠ A* as c ≠ •

16 (c). 0.95 A* = A* (1 - e - kc ) fi - 0.05 = -e - kc fi - kc = ln( 1 20) = - ln(20)

\ c 0.95 = 17.

1 ln(20) . k

Solving the equation y ¢ + cy = 0 with our method yields the general solution y ( t) = y 0e - ct . Looking at the graph, we can see that y (0) = 2 = y 0 and y (-0.4 ) = 3 = y 0e - c( -0.4 ) = 2e 0.4 c . 5 Ê 3ˆ Solving for c gives us c = lnÁ ˜ ª 1.01 . 2 Ë 2¯

18.

y ¢ = Ce - ct

y (1) = Ce - c = y 0 fi C = y 0e c \ y = y 0e - c( t -1)

y (1) = y 0 = -1

c ª-

y (0.3) ª -

1 1 Ê 1ˆ \ - = -e - c( -0.7 ) = 0.7c ª lnÁ ˜ Ë 2¯ 2 2

1 ln(2) = -0.990 \ c = -1. 0.7

19 (a). The general solution to this D.E. is y ( t) = y 0e - t , which can be rewritten as ln( y ) = - t + c . Thus, this D.E. corresponds to graph #2 with y 0 = y (0) = e ln( y( 0 )) = e 2 . 19 (b). The general solution to this D.E. is y ( t) = y 0e t sin 4 t , which can be rewritten as ln( y ) = t sin 4 t + c . Thus, this D.E. corresponds to graph #1 with y 0 = y (0) = e ln( y( 0 )) = 1.

19 (c). The general solution to this D.E. is y ( t) = y 0e

-t 2 2

t2 , which can be rewritten as ln( y ) = - + c . 2

Thus, this D.E. corresponds to graph #4 with y 0 = y (0) = e ln( y( 0 )) = e . 19 (d). The general solution to this D.E. is y ( t) = y 0e t - sin 4 t , which can be rewritten as ln( y ) = t - sin 4 t + c . Thus, this D.E. corresponds to graph #3 with y 0 = y (0) = e ln( y( 0 )) = 1.

20.

ln y ( t) =

1 3-1 t d t + 1 = + 1 \ p( t) = ln( y ( t)) = 2 4-0 2 dt

21 (a). Integrating p( t) = t n gives us P ( t) = y ( t ) = y 0e - t

n+1

n +1

y0 = e .

t n +1 . Thus the solution to this initial value problem is n +1

which can be rewritten as ln y = ln y 0 -

t n +1 . n +1

10 • Chapter 2 First Order Linear Differential Equations

Substituting values from the table gives us the necessary equations to solve for y 0 and n . First,

-

1 1 2 n +1 and -4 = ln y 0 can be combined to solve for n : = ln y 0 4 n +1 n +1

1 1 1 15 2 n +1 - 1 , so n = 3. - = ln y 0 - by substitution, and therefore y 0 = 1. 4- = = 4 4 4 4 n +1 21 (b). y ( t) = y 0e -

t n+1

n+1

= 1◊ e -

t4

4

-1

fi y (-1) = e 4 .

Section 2.3 1.

For this D.E., p( t) = 2 . Integrating gives us P ( t) = 2 t . An integrating factor is, then, m ( t) = e 2 t . Multiplying the D.E. by m ( t) , we obtain e 2 t y ¢ + 2e 2 t y = (e 2 t y )¢ = e 2 t . Integrating both sides

1 1 yields e 2 t y = e 2 t + C . Therefore, the general solution is y ( t) = + Ce -2 t . 2 2 2.

y ¢ + 2 y = e - t fi (e 2 t y )¢ = e t fi e 2 t y = e t + C fi y = e - t + Ce -2 t .

3.

For this D.E., p( t) = 2 . Integrating gives us P ( t) = 2 t . An integrating factor is, then, m ( t) = e 2 t . Multiplying the D.E. by m ( t) , we obtain e 2 t y ¢ + 2e 2 t y = (e 2 t y )¢ = 1. Integrating both sides yields

e 2 t y = t + C . Therefore, the general solution is y ( t) = te -2 t + Ce -2 t . 2

2 2 1 2 1 fi e t y = e t + C fi y = + Ce - t . 2 2

2

4.

y ¢ + 2 ty = t fi (e t y )¢ = te t

5.

Putting this equation into the conventional form gives us y ¢ +

2 2 y = t . For this D.E., p( t) = . t t 2

Integrating gives us P ( t) = 2 ln t . An integrating factor is, then, m ( t) = e ln t = t 2 . Multiplying the D.E. by m ( t) , we obtain t 2 y ¢ + 2 ty = ( t 2 y )¢ = t 3 . Integrating both sides yields

t2 y = 6.

1 4 1 t + C . Therefore, the general solution is y ( t) = t 2 + Ct -2 . 4 4

( t 2 + 4 ) y ¢ + 2 ty = t 2 ( t 2 + 4 ) fi y ¢ + 2

2

2

4

\ (( t + 4 ) y )¢ = t ( t + 4 ) = t + 4 t 7.

2

2 2t y = t 2 , m = e ln( t + 4 ) = t 2 + 4 t +4

2

t5 4 t3 fi (t + 4) y = + +C 5 3 2

t5

y=

3

5

+ 4t 3 + C . (t 2 + 4)

For this D.E., p( t) = 1. Integrating gives us P ( t) = t . An integrating factor is, then, m ( t) = e t . Multiplying the D.E. by m ( t) , we obtain e t y ¢ + e t y = (e t y )¢ = te t . Integrating both sides yields

e t y = te t - e t + C . Therefore, the general solution is y ( t) = t - 1 + Ce - t .

Chapter 2 First Order Linear Differential Equations • 11

8.

y ¢ + 2 y = cos 3t fi (e 2 t y )¢ = e 2 t cos 3t u = e 2t

dv = cos 3tdt

du = 2e 2 t dt

1 v = sin 3t 3

u = e 2t

dv = sin 3tdt 2t

du = 2e dt

1 v = - cos 3t 3

e 2t 2 sin 3t - Ú e 2 t sin 3tdt 3 3

e 2t 2 2t Ú e sin 3tdt = - 3 cos 3t + 3 Ú e cos 3tdt 2t

\ I=

4 2 ¸ e 2t e 2t 2 Ï e 2t sin 3t - Ì- cos 3t + I ˝ fi I (1 + ) = (sin 3t + 2 cos 3t) 9 3 3 ˛ 3Ó 3 3

\ I=

3 2t e (sin 3t + 2 cos 3t) 13

\ e 2t y = 9.

2t Ú e cos 3tdt =

3 3 2t e (sin 3t + 2 cos 3t) + C fi y = (sin 3t + 2 cos 3t) + Ce -2 t 13 13

For this D.E., p( t) = -3 . Integrating gives us P ( t) = -3t . An integrating factor is, then,

m ( t) = e -3 t . Multiplying the D.E. by m ( t) , we obtain e -3 t y ¢ - 3e -3 t y = (e -3 t y )¢ = 6e -3 t . Integrating both sides yields e -3 t y = -2e -3 t + C . Solving for y gives us y = -2 + Ce 3 t , and with our initial condition, y (0) = 1 = -2 + C . Solving for C yields C = 3 , and thus our final solution is y = -2 + 3e 3 t . 10.

y ¢ - 2 y = e 3 t , y (0) = 3.

(e -2 t y )¢ = e t fi e -2 t y = e t + C fi y = e 3 t + Ce 2 t

y (0) = 1 + C = 3 fi C = 2, y = e 3 t + 2e 2 t . 11.

Putting this D.E. in the conventional form, we have y ¢ + Integrating gives us P ( t) =

3 3 1 y = e t . For this D.E., p( t) = . 2 2 2

3 t 3 t . An integrating factor is, then, m ( t) = e 2 . Multiplying the D.E. 2

3 3 t t 3 3t 1 5t by m ( t) , we obtain e 2 y ¢ + e 2 y = (e 2 y )¢ = e 2 . Integrating both sides yields 2 2 3 - t 1 52 t 1 t e y = e + C . Solving for y gives us y = e + Ce 2 , and with our initial condition, 5 5 3 t 2

y (0) = 0 =

1 1 1 1 -3t + C . Solving for C yields C = - , and thus our final solution is y = e t - e 2 . 5 5 5 5

12 • Chapter 2 First Order Linear Differential Equations

12.

y ¢ + y = 1 + 2e - t cos(2 t), y (p 2) = 0 \ (e t y )¢ = e t + 2 cos 2 t e t y = e t + sin 2 t + C fi y = 1 + e - t sin 2 t + Ce - t p

p

p

y (p 2) = 1 + Ce - 2 = 0 fi C = -e 2 ; y = 1 + e - t sin 2 t - e -( t - 2 ) . 13.

Putting this D.E. in the conventional form, we have y ¢ +

p( t) =

cos( t) 3 y = - cos( t) . For this D.E., 2 2

sin( t ) cos( t) sin( t) . Integrating gives us P ( t) = . An integrating factor is, then, m ( t) = e 2 . 2 2

Multiplying the D.E. by m ( t) , we obtain e Integrating both sides yields e

sin( t ) 2

y = -3e

sin( t ) 2

sin( t ) 2

y¢ +

sin( t ) cos( t) sin(2 t ) 3 cos( t) sin(2 t ) . e y = (e 2 y )¢ = e 2 2

+ C . Solving for y gives us y = -3 + Ce

-

sin( t ) 2

,

and with our initial condition, y (0) = -4 = -3 + C . Solving for C yields C = -1 , and thus our final solution is y = -3 - e 14.

-

sin( t ) 2

.

y ¢ + 2 y = e - t + t + 1, y (-1) = e, (e 2 t y )¢ = e t + te 2 t + e 2 t t 1 1 1 1 ye 2 t = e t + te 2 t - e 2 t + e 2 t + C fi y = e - t + + + Ce -2 t 2 4 2 2 4 y (-1) = e -

1 1 1 + + Ce 2 = e fi C = e -2 2 4 4

\ y = e-t + 15.

t 1 1 -2( t +1) . + + e 2 4 4

Putting this D.E. in the conventional form, we have y ¢ +

3 3 1 y = 1 + . For this D.E., p( t) = . t t t 3

Integrating gives us P ( t) = 3 ln( t) . An integrating factor is, then, m ( t) = e 3 ln( t ) = e ln( t ) = t 3 . Multiplying the D.E. by m ( t) , we obtain t 3 y ¢ + 3t 2 y = ( t 3 y )¢ = t 3 + t 2 . Integrating both sides yields t 3 y =

t 1 1 4 1 3 t + t + C . Solving for y gives us y = + + Ct -3 , and with our initial 4 3 4 3

condition, y (-1) = is y =

1 1 1 1 = - + - C . Solving for C yields C = - , and thus our final solution 3 4 3 4

t 1 1 -3 + - t . The t -interval on which this solution exists is -• < t < 0 . 4 3 4

Chapter 2 First Order Linear Differential Equations • 13

16.

y¢ +

4 y = a t, m = t 4 t

t6 at 2 t y ¢ + 4 t y = a t = ( t y )¢ fi t y = a + C fi y = + Ct -4 6 6 4

3

5

4

4

1 a 1 a t2 y (1) = - = + C fi C = - - ∫ 0 fi a = -2, y = - . 3 6 3 6 3 17.

Multiplying both sides of the equation by the integrating factor, m ( t) = e 2 t , we have

e 2 t y = e 2 t (Ce -2 t + t + 1) = e 2 t ( t + 1) + C . Differentiating gives us (e 2 t y )¢ = e 2 t (1) + 2e 2 t ( t + 1) = e 2 t (2 t + 3) . Therefore, (e 2 t y )¢ = (m ( t) y )¢ = m ( t) ◊ g( t) = e 2 t (2 t + 3) fi g( t) = 2 t + 3 and

m ( t) = e 2 t = e P ( t ) fi P ( t) = 2 t fi p( t) = 2 . 2

2

2

2

18.

2 tCe t + pCe t = 0 fi p( t) = -2 t . Substituting, (Ce t + 2)¢ - 2 t(Ce t + 2) = -4 t fi g( t) = -4 t .

19.

Multiplying both sides of the equation by the integrating factor, m ( t) = t , we have

ty = t(Ct -1 + 1) = t + C . Differentiating gives us ( ty )¢ = 1. Therefore, ( ty )¢ = (m ( t) y )¢ = m ( t) ◊ g( t) = 1 = ( t)( t -1 ) fi g( t) = t -1 and 1 m ( t) = t = e P ( t ) fi P ( t) = ln t fi p( t) = = t -1 . t 20.

(e - t + t - 1)¢ + (e - t + t - 1) = t fi g( t) = t, y 0 = 0 .

21.

y ( t) = -2e - t + e t + sin t fi y 0 = y (0) = -2 + 1 + 0 = -1 . If y ( t) = -2e - t + e t + sin t , then y ¢ = 2e - t + e t + cos t . Substituting in y ¢ + y = g( t) , (2e - t + e t + cos t) + (-2e - t + e t + sin t) = 2e t + cos t + sin t = g( t) .

22.

y ¢ + (1 + cos t) y = 1 + cos t, y (0) = 3, m = e t + sin t . (e t + sin t y )¢ = (1 + cos t)e t + sin t = (e t + sin t )¢ fi e t + sin t y = e t + sin t + C fi y = 1 + Ce -( t + sin t ) . y (0) = 1 + C = 3 fi C = 2 \ y = 1 + 2e -( t + sin t ) and lim y ( t) = 1. t Æ•

23.

Putting this D.E. in the conventional form, we have y ¢ + 2 y = e - t - 2 . For this D.E., p( t) = 2 . An integrating factor is, then, m ( t) = e 2 t . Multiplying the D.E. by m ( t) , we obtain

e 2 t y ¢ + 2e 2 t y = (e 2 t y )¢ = e t - 2e 2 t . Integrating both sides yields e 2 t y = e t - e 2 t + C . Solving for y gives us y = e - t - 1 + Ce -2 t , and with our initial condition, y (0) = -2 = 1 - 1 + C . Solving for C yields C = -2 , and thus our final solution is y = e - t - 1 - 2e -2 t . Therefore, lim y ( t) = -1. t Æ•

14 • Chapter 2 First Order Linear Differential Equations

24.

On [1, 2]:

1 y ¢ + y = 3t, y (1) = 1. An integrating factor is m ( t) = t . Multiplying the D.E. by m ( t) , we t obtain ( ty )¢ = 3t 2 fi ty = t 3 + C fi y = t 2 + Ct -1, y (1) = 1 + C = 1 fi C = 0 . Therefore, the solution for 1 £ t £ 2 is y = t 2 and y(2) = 4 . On [2, 3]:

1 y ¢ + y = 0, y (2) = 4 . An integrating factor is m ( t) = t . Multiplying the D.E. by m ( t) , we t obtain ( ty )¢ = 0 fi ty = C fi y = Ct -1, y (2) =

C = 4 fi C = 8 . Therefore, the solution for 2

8 2 £ t £ 3 is y = . t 25.

On [0,p ] : y ¢ + (sin t) y = sin t, y (0) = 3 . An integrating factor is m ( t) = e - cos t . Multiplying the D.E. by

m ( t) , we obtain e - cos t y ¢ + e - cos t (sin t) y = (e - cos t y )¢ = (sin t)e - cos t . Integrating both sides yields e - cos t y = e - cos t + C . Solving for y gives us y = 1 + Ce cos t , and with our initial condition, y (0) = 3 = 1 + Ce fi C = 2e -1 . Therefore, the solution for 0 £ t £ p is y = 1 + 2e cos t -1 and y (p ) = 1 + 2e -2 . On [p , 2p ]:

y ¢ + (sin t) y = - sin t, y (p ) = 1 + 2e -2 . Multiplying the D.E. by m ( t) = e - cos t , we obtain e - cos t y ¢ + e - cos t (sin t) y = (e - cos t y )¢ = (- sin t)e - cos t . Integrating both sides yields e - cos t y = -e - cos t + C . Solving for y gives us y = -1 + Ce cos t , and with our initial condition, y (p ) = 1 + 2e -2 = -1 + Ce -1 fi C = 2e1 + 2e -1 . Therefore, the solution for p £ t £ 2p is y = -1 + 2e cos t +1 + 2e cos t -1 . 26.

On [0,1]: y ¢ = 2, y (0) = 1. y = 2 t + C, y (0) = C = 1 fi C = 1.

Therefore, the solution for 0 £ t £ 1 is y = 2 t + 1 and y(1) = 3.

1 On [1, 2]: y ¢ + y = 2, y (1) = 3. An integrating factor is m ( t) = t . Multiplying the D.E. by t

m ( t) , we obtain ( ty )¢ = 2 t fi ty = t 2 + C fi y = t + Ct -1, y (1) = 1 + C = 3 fi C = 2 . Therefore, 2 the solution for 1 £ t £ 2 is y = t + . t

Chapter 2 First Order Linear Differential Equations • 15

27.

On [0,1]: y ¢ + (2 t - 1) y = 0, y (0) = 3 . An integrating factor is m ( t) = e t

we obtain e t

2

-t

y¢ + et

2

-t

(2 t - 1) y = (e t

2

-t

2

-t

. Multiplying the D.E. by m ( t) ,

y )¢ = 0 . Integrating both sides yields e t

2

-t

y = C.

2

Solving for y gives us y = Ce t - t , and with our initial condition, y (0) = 3 = C . Therefore, the 2

solution for 0 £ t £ 1 is y = 3e t - t and y(1) = 3. On [1, 3]: y ¢ + (0) y = y ¢ = 0, y (1) = 3 . Integrating gives us y = C = 3 . Therefore, the solution for 1 £ t £ 3

is y = 3 and y( 3) = 3. On [ 3, 4 ]:

y ¢ + (- 1t ) y = 0, y ( 3) = 3 . An integrating factor is m ( t) = e - ln t = 1t . Multiplying the D.E. by m ( t) , we obtain 1t y ¢ -

1 t2

y = ( 1t y )¢ = 0 . Integrating both sides yields 1t y = C . Solving for y gives us

y = Ct , and with our initial condition, y ( 3) = 3 = C ( 3) fi C = 1. Therefore, the solution for

3 £ t £ 4 is y = t . 28.

y ( t) = t{Si( t) - Si(1) + 3}

Section 2.4 1.

P ( t) = A0e rt = 5000e.05 t . Thus, P ( 30) = 5000e.05◊30 = 22408.45 .

2.

r P2 ( t) = (1 + ) 2 t A0 2

P2 ( 30) = (1.025) 60 ◊ 5000

\ ln P2 ( 30) = 60 ln(1.025) + ln 5000 = 9.999

3 (a).

P2 ( 30) ª 21999

P1 ( t) = (1 + r) t A0 = (1.06) t A0 . Setting P1 ( t) = 2 A0 yields 2 = 1.06 t , and solving for t gives us t ª 11.9 years.

3 (b).

r P2 ( t) = (1 + ) 2 t A0 = (1.03) 2 t A0 . Setting P2 ( t) = 2 A0 yields 2 = 1.032 t , and solving for t gives us 2

t ª 11.72 years. 3 (c).

P ( t) = A0e rt = A0e.06 t . Setting P ( t) = 2 A0 yields 2 = e.06 t , and solving for t gives us t ª 11.55 years.

4.

With r = .05

P ( t) = e.05 t A0

P (10) = e 0.5 A0

With unknown r, P (8) = e r 8 A0 = e 0.5 A0 \ 8 r = 0.5 fi r =

1 16

ª 0.0625

(6.25%)

16 • Chapter 2 First Order Linear Differential Equations

5 (a).

PB ¢ = (0.04 + 0.004 t) PB ; PB (0) = A0 .

5 (b).

PB = A0e.04 t +.002 t . This can be verified easily through differentiation.

2

5 (c). For Plan A, PA ( t) = A0e.06 t . To find the time t at which Plan B “catches up” with Plan A, let us 2

set PA ( t) = PB ( t) : A0e.06 t = A0e.04 t +.002 t . Dividing by A0 and taking the natural logarithm of both sides yields .06 t = .04 t + .002 t 2 , and solving for t gives us t = 0 (the time of the initial investment) and t = 10 years (the time at which Plan B “catches up”).

P (10) = 1000e.05( 4 )+.07( 6 ) = 1000e.2 +.42 = 1858.93

6.

After 4 yrs, P ( 4 ) = 1000e.05( 4 ),

7.

We can simplify this problem by considering the two deposits separately and then adding the principals of each deposit together at a time of twelve years. We have, then, 1000e12 r + 1000e 6 r = 4000 . Introducing a new variable x ∫ e 6 r , we have x 2 + x - 4 = 0 .

Solving this with the quadratic formula yields one positive value of x : x ª 1.5616 = e 6 r . Solving for r yields r ª 0.0743. 8.

1 Ê 11ˆ 11, 000, 000 = 10, 000, 000e 5 k . Solving for k yields k = lnÁ ˜ . 5 Ë 10 ¯ 6 1 ln 11 ( 30 ) ln 11 P ( 30) = 10, 000, 000e 5 ( 10 ) = 10, 000, 000e ( 10 ) = 17, 715, 610 .

ln 2 ln 2 =5 ª 36.36 days. 11 k ln 10

9.

2 = e kt , and thus t =

10.

1 1.3 = e 2 k fi k = ln(1.3). 2

11.

1 80, 000 = 100, 000e 6 k . Solving for k yields k = ln(.8) . Using this value for k , we have 6

3 = e kt fi t =

ln 3 2 ln( 3) = ª 8.375 wks. k ln(1.3)

(80, 000 + 50, 000)e ln( 0.8 ) = 130, 000 ◊ 0.8 = 104, 000 . 12 (a). P ¢ = kP + M , P (0) = P0

e - kt P = -

M M M - kt e + C fi P = - + Ce kt , P0 = - + C k k k

\ P ( t) = 12 (b). P0 = -

P ¢ - kP = M , (e - kt P )¢ = Me - kt

M . k

M M + ( P0 + )e kt k k P0 and P must be nonnegative fi -

M ≥ 0. k

net growth rate k < 0 and vice versa. 12 (c). Set kP + M = 0 fi P = -

M . k

P ( t) = P0 = -

M in this case. k

If net immigration rate M > 0,

Chapter 2 First Order Linear Differential Equations • 17

13 (a). For Strategy I, we have M I = kP0 . For Strategy II, we have M II = P0 (e k - 1) . 13 (b). The net profit for each strategy would equal ( M )( profit fish ) , and so the profit for Strategy I is, then: PrI = 500, 000(.3172)(.75) = 118, 950 , and the profit for Strategy II is: PrII = 500, 000(e.3172 - 1)(0.6) ª 111, 983 . Strategy I would be more profitable for the farm. 14 (a). P1 (1) = -

M M M M + ( P0 + )e k , P1 (2) = P1 (1)e k = - e k + ( P0 + )e 2 k k k k k

P2 (1) = P0e k , P2 (2) = 14 (b). P1 (2) - P2 (2) = -

=

M k (e - 1) 2 . k

M M + ( P0e k + )e k k k

M M M M M k e + P0e 2 k + e 2 k + - P0e 2 k - e k = (e 2 k - 2e k + 1) k k k k k Since M > 0, P1 (2) > P2 (2) if k > 0 and P1 (2) < P2 (2) if k < 0 .

14 (c). If k > 0 , introduce the immigrants as early as possible. If k < 0 , introduce as late as possible. 15 (a). From the general solution of the radioactive decay equation, Q( t) = Ce - kt , we can use the data given to find C and k . Q(1) = Ce - k = 100 and Q( 4 ) = Ce -4 k = 30 , so combining these equations, we find that e -3 k =

1 Ê 10 ˆ 3 and therefore, k = lnÁ ˜ ª 0.4013 . Using this value of k 3 Ë 3¯ 10

with the t = 1 data, we find that C = Q0 = 149.4 mg . C = Q0 , since the exponential falls off the expression for Q at t = 0 . 15 (b). t =

ln 2 ª 1.727 months. k

15 (c). 0.01 = e - kt . Solving for t , we have t = 16 (a). t =

ln 2 ln 2 = 5730 fi k = . k 5730

t = ln( 103 ) ◊

ln(0.01) ª 11.475 months. k

0.3 = e - kt fi t =

Ê ln( 10 ) ˆ t = Á 3 ˜ t ª 9953 yr. ln 2 Ë ln 2 ¯

16 (b). From (a) t =

ln( 103 ) ln( 103 ) ln( 103 ) (t + 30) (t - 30) £ t £ t \ ln 2 ln 2 ln 2

or 9901 £ t £ 10005 yrs. 16 (c).

- ln(0.3) k

ln 2 Q(60, 000) -60, 000( 5730 ) = e -60,000 k = e ª 2.83(10 -5 ) . Q(0)

18 • Chapter 2 First Order Linear Differential Equations

17.

Q¢ = - kQ + M . Writing this D.E. in the conventional form, we have Q¢ + kQ = M . For this

D.E., p( t) = k and P ( t) = kt , which yields an integrating factor of m ( t) = e kt . Thus,

e kt Q¢ + ke kt Q = (e kt Q)¢ = e kt M . Integrating both sides gives us e kt Q = e kt

M M + Ce - kt . Q0 = + C , so our equation for Q in terms of Q0 now reads k k

we have Q = Q( t) =

k=

M + C . Solving for Q , k

M Mˆ M Ê + Á Q0 - ˜ e - kt = 50e - kt + (1 - e - kt ) . Setting Q(2) = 100 and substituting k k¯ k Ë

M ln 2 ln 2 M = ª 0.231 , we have 100 = 50e -2 k + (1 - e -2 k ) = 31.5 + (0.37) . Solving for 3 k t .231

M , we find M = 42.78 (mg/yr.). 18.

t=

ln 2 = 8 days. k

30 = Q0e 19.

3

fi Q0 = 30e 8

ln 2

- ln 2 tt

ª 38.9mg

0.99Q0 = Q0e - kt . Solving for t in terms of k , we have t=

20.

- 38 ln 2

Q( t) = Q0e - kt = Q0e

1 Ê 100 ˆ t Ê 100 ˆ 9 9 lnÁ lnÁ ˜= ˜ = 4 ◊ 10 ◊ 0.0145 ª 0.058 ◊ 10 =58 million years. Ë ¯ Ë ¯ 99 99 k ln 2

Contact angle is 180 - 30 + 45 = 195∞ or q 2 - q1 = 3.403 rad.

\ T2 = e 0.3( 3.403 ) (100) ª 277.6 lb. 21.

The contact angle, q 2 - q1 = 2p + 2p + p = 5p . T2 = e 0.1(q 2 -q 1 ) (100 ◊ 9.8) = e 0.1◊5p (980) ª 4714 N.

22.

Contact –: 90∞ + a + a + 90∞ = 240∞ where sin a =

q 2 - q1 =

23.

a 1 = fi a = 30∞ 2a 2

4 2p for T2 p for T3 and 3 3

T2 = 100e.2(

2p

T3 = 100e.2(

4p

3

)

ª 152 lb.

)

ª 231 lb.

3

The angle, a , is marked at various places on the diagram below. A right angle occurs at each of the dots.

Chapter 2 First Order Linear Differential Equations • 19

To determine the angle a , part of the diagram is shown here with the radii of the circles marked.

sin a =

2a a and sin a = fi y = 2x . x y

In the text, we are given that x + y = 5 a . Therefore, x + 2 x = 3 x = 5 a fi x =

5a a a \ sin a = = 5 a = .6 fi a ª .6435 radians . 3 x 3

The corresponding contact angles and belt tensions are: For T2 :

p 2

+ a ª 2.214 radians fi T2 = T1e m ( angle ) = 100e(.2 )( 2.214 ) ª 155.7 lb.

For T3 : ( p2 + a ) + 2a =

p 2

+ 3a ª 3.501 radians

fi T3 = T1e m ( angle ) = 100e(.2 )( 3.501) ª 201.4 lb. For T4 : ( p2 + 3a ) + a =

p 2

+ 4a ª 4.145 radians

fi T4 = T1e m ( angle ) = 100e(.2 )( 4.415 ) ª 229.1 lb. 24.

Contact –: 2p + 2p + 2p +

p 19p = 3 3

1 19p F = e 0.4 ( 3) ª 953.5 lb. 3

Section 2.5 1 (a). To begin, Q(0) = 0 and Q¢ = (0.2)( 3) -

Q ( 3) . Putting the second equation in the conventional 100

form, we have Q¢ + 0.03Q = 0.6 . Multiplying both sides of this equation by the integrating factor m ( t) = e 0.03 t gives us (e 0.03 t Q)¢ = 0.6e 0.03 t . Integrating both sides yields

e 0.03 t Q = 0.6 ◊

100 0.03 t e + C = 20e 0.03 t + C . Solving for Q , we have Q = 20 + Ce -0.03 t . 3

20 • Chapter 2 First Order Linear Differential Equations

Q(0) = 0 = 20 + C , so C = -20 . With this value for C , our final equation for Q is

Q = 20(1 - e -0.03 t ) . Thus, Q(10) = 20(1 - e -0.3 ) ª 5.18 lb. 1 (b). lim Q( t) = 20lb and the limiting concentration is 0.2 lb/gal. t Æ•

2.

V = 100( 70)(20) = 140, 000 m 3 . 0.01Q0 = Q0e

- rv 30

fi -

Q¢ = 0 -

v r 1 = ln(0.01) fi r = ln(100) . v 30 30

3 140, 000 ln(100) ª 21, 491 m min . 30

r=

Q -r t r fi Q = Q0e v v

3 (a). To begin, Q(0) = 5 and Q¢ = 0.25 r -

r 1 ln(100) = 0.1535 = v 30

( ª 15.4%) .

Q r . Putting the second equation in the conventional 200

form, we have Q¢ + 0.005 rQ = 0.25 r . Multiplying both sides of this equation by the integrating factor m ( t) = e 0.005 rt gives us (e 0.005 rt Q)¢ = 0.25 re 0.005 rt . Integrating both sides yields

e 0.005 rt Q = 0.25(200)e 0.005 rt + C = 50e 0.005 rt + C . Solving for Q , we have Q = 50 + Ce -0.005 rt . Q(0) = 5 = 50 + C , so C = -45 . With this value for C , our equation for Q now reads

Q = 50 - 45e

-0.005 rt

. We know that Q(20) = 30 = 50 - 45e

-

20 r 200

, and solving for r yields

Ê (50 - 30) ˆ Ê 9ˆ r = lnÁ ˜ (-10) = 10 lnÁ ˜ ª 8.11gal/min. Ë 45 ¯ Ë 4¯

3 (b). This would be impossible, since Q( t) < 50 lb for all 0 £ t < • . 4 (a).

Q¢ = (10 te Q¢ = -

Qe

t

50

-t

50

)(100) -

Q (100) 5000

Q(0) = 0

1 t t Q + 1000 te - 50 fi (Qe 50 )¢ = 1000 t 50

= 500 t 2 + C fi Q = 500 t 2e -

(

4 (b). Q¢ = 500 2 t -

t2 50

)e

- t 50

t

50

t

+ Ce - 50 .

Q(0) = C = 0.

\ Q( t) = 500 t 2e -

t

50

oz.

= 0 fi t 2 = 100 t fi t = 100 min.,

Q(100) 500(100) 2 -2 = e = 1000e -2 ª 135.3 oz gal 5000 5000 4 (c). Plot c ( t) vs t . Yes. 5 (a). To begin, Q(0) = 10 , V (0) = 100 , and V ( t) = 100 + t . Since the tank has a capacity of 700 gallons, 100 + t = 700 . Solving for t yields t = 600 minutes.

Chapter 2 First Order Linear Differential Equations • 21

5 (b). Q¢ = (0.5)( 3) -

Q 2 3 (2) . Putting this in the conventional form, we have Q¢ + Q= . 100 + t 100 + t 2

Multiplying both sides of the equation by the integrating factor m ( t) = e 2 ln(100 + t ) = (100 + t) 2

3 (100 + t) 3 gives us ((100 + t) 2 Q)¢ = (100 + t) 2 . Integrating both sides yields (100 + t) 2 Q = + C, 2 2 and solving for Q , we have Q =

100 + t C C + , and solving for C 2 . Q(0) = 10 = 50 + 2 (100 + t) 100 2

yields C = -40(100) 2 = -400, 000 . Substituting this value of C back into our equation for Q gives us our final equation for Q , 100 + t 400, 000 400 400, 000 = 197.5 lb. The 2 . V ( t) = 400 at t = 300 , so Q( 300) = 2 (100 + t) 2 ( 400) 2

Q( t) =

197.5 lb/gal. 400 700 400, 000 349.2 Q(600) = ª .4988 lb/gal. 2 ª 349.2 lb. The concentration, then, is 2 ( 700) 700

concentration, then, is 5 (c). 6 (a).

Q¢ = a

Q Q (15) (15) 500 500

6 (b). Q(180) = 0.01Q0

Q¢ =

-(1 - a ) (15)Q 500

Q = Q0e -.03(1-a )t

.01 = e -.03(1-a )(180 ) fi e -5.4 (1-a ) = .01 5.4 (1 - a ) = ln(100) fi 1 - a = 0.8528 fi a = 0.1472 .

7 (a).

Ê QA ˆ QA (0) = 1000 , QB (0) = 0 , QA ¢ = 0 - 1000Á ˜ , and Ë 500, 000 ¯ Ê QB ˆ Ê QA ˆ QB ¢ = 1000Á ˜ - 1000Á ˜. Ë 500, 000 ¯ Ë 200, 000 ¯

7 (b). Putting the equation for QA ¢ into the conventional form, we have QA ¢ = QA = 1000e QB ¢ + (QB e

-

t 500

1 Q . Thus, 500 A

. Putting the equation for QB ¢ into the conventional form, we have

t t 1 QB = 2e 500 . Multiplying both sides by the integrating factor m ( t) = e 200 yields 200

t 200

)¢ = 2e

1 ˆ Ê 1 tÁ ˜ Ë 200 500 ¯

= 2e

3t 1000

. Integrating both sides gives us QB e

t 200

3t 2000 1000 = + C , and e 3

22 • Chapter 2 First Order Linear Differential Equations t t 2000 2000 2000 - 500 . Substituting + C , so C = e + Ce 200 . QB (0) = 0 = 3 3 3

solving for QB , QB =

t t ˆ Ê 2000 ˆ Ê - 500 200 -e ˜ this value back into our equation, we have QB = Á ˜ Áe Ë 3 ¯Ë ¯ t t t t ˆ + 1 - 200 500 Ê 2000 ˆ Ê 1 - 500 e + e ˜ . Since e 500 200 = 7 (c). Setting QB ¢ = 0 , we have 0 = Á , ˜ ÁË 3 ¯ Ë 500 200 200 ¯

1000 Ê 5 ˆ 3 Ê 5ˆ lnÁ ˜ ª 305.4 hours. t = lnÁ ˜ , and thus t = Ë 2¯ Ë 2¯ 3 1000

7 (d). Here, we want to determine tA such that QA ( tA ) =

1 lb and tB such that QB ( t) £ 0.2 lb where 2

t £ tB . This can be solved via plotting: tA ª 3800 hours and tB ª 4056 hours. Therefore,

t ª 4056 hours. 8 (a).

ri = r0 = 3 + sin t fi V = constant.

8 (b). Expect lim Q( t) = .5(200) = 100 lb. t Æ•

The tank is being “flushed out”, albeit in a pulsating manner. 8 (c).

Q¢ = .5( 3 + sin t) Q¢ + Qe

Q ( 3 + sin t), Q(0) = 10 200

( 3 t - cos t ) ( 3 t - cos t ) 1 3 + sin t 1 200 200 Q = ( 3 + sin t) fi (Qe )¢ = ( 3 + sin t)e 2 200 2

( 3 t - cos t ) 200

= 100e 3 t - cos t + C fi Q = 100 + Ce

Q(0) = 10 = 100 + Ce - ( 3 t - cos t +1 )

1

200

fi C = -90e

-1

200

- ( 3 t - cos t ) 200

fi Q( t) = 100 - 90e

- ( 3 t - cos t +1 ) 200

.

8(d).

lim e

9.

f ( t) = 3 + sin t . Therefore, t = Ú ( 3 + sin s) ds = [ 3s - cos s]t0 = 3t - cos t + 1. Now,

200

t Æ•

= 0 fi lim Q( t) = 100 lb. t Æ•

t

0

1 dQ Q and Q(0) = 10 . Putting the first equation into the conventional form, we = 0.5 200 dt have

t 1 dQ Q = 0.5 , and multiplying both sides by the integrating factor m ( t) = e 200 gives + dt 200

t t t Ê t ˆ¢ us Á e 200 Q˜ = 0.5e 200 . Integrating both sides yields e 200 Q = 100e 200 + C , and solving for Q , Ë ¯

Q = 100 + Ce

-

t 200

. Now, Q(t = 0) = 10 = 100 + C , and therefore, C = -90 .

Chapter 2 First Order Linear Differential Equations • 23

Substituting this back into our equation for Q yields Q = 100 - 90e reads Q = 100 - 90e

-

3 t - cos t +1 200

-

t 200

, which in terms of t

.

10 (a). No limit since we do not expect concentration to stabilize. 10 (b). Q¢ = .2(1 + sin t)( 3) 10 (c). Q¢ +

Q ( 3), Q(0) = 10 200

3 Q = 0.6(1 + sin t) 200

at at Ú e sin tdt = e

(e

3

(- cos t + a sin t) (1 + a 2 )

200 t

3 Q ¢ = 0.6e 200 t (1 + sin t).

)

e

3

200 t

Ï 200 3 200 t e Q = 0.6Ì e + Ó 3

3

200 t

3 (- cos t + 200 sin t) ¸ ˝+C 2 1 + ( 3 200) ˛

3 Ï 200 (- cos t + 200 sin t) ¸ -0.6 cos t + 0.009 sin t -3 -3 t + Q( t) = 0.6Ì + Ce 200 t ˝ + Ce 200 = 40 + 2 3 1 + ( 200) 1.000225 Ó 3 ˛

Q(0) = 10 = 40 Q( t) = 40 - 30e

-3

0.6 0.6 + C fi C = -30 + 1.000225 1.000225 200 t

Ê 0.6(e +Á Ë

-3

200 t

- cos t) + 0.009 sin t) ˆ ˜ 1.000225 ¯

10 (d).

11 (a). First, Q = Q0e - kt for the radioactive material. To find k from the half-life of the material,

ln2 1 . Thus for the decay of the radioactive material Q0 = Q0e -18 k . Solving for k , we have k = 18 2 alone, we have Q( t) = 5e

-

ln 2 t 18

with t measured in hours. Now, for the lake, we know that Q

varies both with decay and with the water flow. Accordingly, we will begin with the relationship Q( t + Dt) - Q( t) ª - kQ( t) Dt -

Q( t) rDt . V

24 • Chapter 2 First Order Linear Differential Equations

Using a form of the definition of the derivative and solving for Q© , we have

60, 000 ˆ rˆ Ê ln 2 Ê + = -Á k + ˜ Q = -Á Q© ˜ Q ª -0.0885Q . We know that Q0 = Q(0) = 5 lb, so our Ë Ë 18 1, 200, 000 ¯ V¯ final equation for Q reads Q( t) = 5e -0.0885 t . 11 (b). Here, (0.0001)(5) = 5e -0.0885 t . Thus, t = 104.07 hours. 12.

q ¢ = k ( S - q ) , S = 72 , q (0) = 350 , q (10) = 290

q ¢ + kq = kS fi (e ktq )¢ = ke kt S fi e ktq = e kt S + C fi q = S + Ce - kt q (0) = q 0 = S + C fi C = q 0 - S fi q = S + (q 0 - S )e - kt Ê 278 ˆ 290 = 72 + ( 350 - 72)e - k(10 ) fi 218 = 278e -10 k , 10 k = lnÁ ˜ Ë 218 ¯ k=

1 Ê 278 ˆ 48 - kt lnÁ fi e - kt = ˜ ; 120 = 72 + ( 350 - 72)e 10 Ë 218 ¯ 278

1 Ê 48 ˆ 10 ln( 278 10(1.756) 48 ) t = - lnÁ = ª 72.2 min. ˜= 278 0.243 k Ë 278 ¯ ln( 218 ) 13.

To begin, q = S + (q 0 - S )e - kt . With our substitutions for the time the food was in the oven, this equation reads 120 = 350 + ( 40 - 350)e -10 k . Solving for k , we have k=-

1 Ê 350 - 120 ˆ lnÁ ˜ ª .02985 . The temperature of the food after 20 minutes in the oven is, 10 Ë 350 - 40 ¯

then, q (20) = 350 + ( 40 - 350)e -20 k = 350 - ( 310)(0.550) ª 179.5 degrees. Finally, the food is cooled at room temperature, so q ( t) = 110 = 72 + (179.5 - 72)e -0.02985 t . Solving for t yields t ª-

14.

1 Ê 110 - 72 ˆ lnÁ ˜ ª 34.8 minutes. 0.02985 Ë 179.5 - 72 ¯

q = S + (q 0 - S )e - kt ; 170 = 212 + ( 72 - 212)e - k 5 1 Ê 140 ˆ 1 Ê 10 ˆ -1 k = lnÁ ˜ = lnÁ ˜ min. 5 Ë 42 ¯ 5 Ë 3 ¯ t ÏÔ Ê ˆ ¸Ô 1 Ê Q( t) ˆ P ¢ = rÁ1 ( ) q s ds ˜ P , P = P0 expÌrÁ t ˜˝ Ú Ë 140 ¯ ÔÓ Ë 140 0 ¯ Ô˛

q ( t) = 212 + ( 72 - 212)e - kt = 212 - 140e - kt t

Ú qds = 212t 0

140 1 - e - kt ) ( k

Ï Ê 140 1 È ˘ˆ ¸ 2120 1 - e -10 k ) ˙˜ ˝ \ 0.01 = expÌrÁ10 ( Í 140 Î k ˚¯ ˛ Ó Ë

Chapter 2 First Order Linear Differential Equations • 25

212 1 1 Ï Ê ˆ¸ = expÌrÁ10 + (1 - e -10 k )˜ ˝ ¯˛ 14 k 100 Ó Ë Ê ˆ 5 - ln(100) = rÁ10 - 15.143 + 10 (1 - .09)˜ = r(-5.143 + 3.78) ln( 3 ) Ë ¯

-4.6052 ª r(-1.363) fi r ª 3.379 min -1 15.

For the first cup, q1 = 72 + ( 34 - 72)e - kt . Thus, with the proper substitutions, 53 = 72 - 38e - kt1 . e - kt1 , then, is equal to

19 . For the second cup, q 2 = 34 + ( 72 - 34 )e - kt . With the proper 38

substitutions, we have 53 = 34 + 38e - kt 2 . e - kt 2 , then, is equal to

19 . Thus, the two times are 38

equal. 16.

q = S + (q 0 - S )e - kt

For casserole, 45 = 72 + ( 40 - 72)e - k 2

1 Ê 32 ˆ -27 = -32e - k 2 , k = lnÁ ˜ 2 Ë 27 ¯ S ( t) = 72 + 228(1 - e -at )

1 - e -2a =

S (2) = 150 = 72 + 228(1 - e -a 2 )

78 150 1 Ê 228 ˆ fi e -2a = fi a = lnÁ ˜ 228 228 2 Ë 150 ¯

q ¢ = k ( S ( t) - q ) fi q ¢ + kq = kS ( t) fi (e ktq )¢ = ke kt S ( t) = ke kt ( 72 + 228 - 228e -at ) = ke kt ( 300 - 228e -at ) e ktq = 300e kt -

228 k ( k -a )t 228 k -at e + Ce - kt e + C fi q = 300 k -a k -a

q (0) = 45 = 300 q (8) = 300 -

228 k 228 k +C fi C= - 255 k -a k -a

228 k -8a Ê 228 k ˆ e +Á - 255˜ e -8 k Ëk -a ¯ k -a 4

4

4 Ê 150 ˆ Ê 27 ˆ -8 k -2 k 4 e -8a = (e -2a ) = Á ˜ ª .1873, e = (e ) = Á ˜ = .5068 Ë 228 ¯ Ë 32 ¯

1 Ê 32 ˆ k = lnÁ ˜ ª 0.08495 2 Ë 27 ¯

k - a ª -0.1244

1 Ê 228 ˆ a = lnÁ ˜ ª 0.2094 2 Ë 150 ¯

k = -0.682843 k -a

q (8) = 300 - 228(-0.682843)(.1873379) + (228(-.682843) - 255)(.5068216) = 300 + 29.166 - 208.14565 = 121.02∞

Chapter 3 First Order Nonlinear Differential Equations Section 3.1 1 1 1 (a). Solving for y ¢ , we have y ¢ = (1 - 2 t cos y ) . Thus, f ( t, y ) = (1 - 2 t cos y ) . 3 3 1 (b).

2 ∂f 1 ∂f = (0 + 2 t sin y ) = t sin y . f and are continuous in the entire ty plane. 3 ∂y 3 ∂y

1 (c). The largest open rectangle is the entire ty plane, since f and

∂f are continuous in the entire ∂y

ty plane.

1 (1 - 2 cos y ) . 3t

2 (a).

f ( t, y ) =

2 (b).

2 ∂f ∂f = sin y . f and are continuous when t < 0, t > 0 . ∂y 3t ∂y

2 (c).

R = {( t, y ): t > 0, -• < y < •} .

3 (a). Solving for y ¢ , we have y ¢ = 3 (b).

2t 2t . 2 . Thus, f ( t, y ) = 1+ y2 1+ y

4 ty ∂f ∂f = (-2 t)(-1)(1 + y 2 ) -2 (2 y ) = are continuous in the entire ty plane. 2 2 . f and (1 + y ) ∂y ∂y

3 (c). The largest open rectangle is the entire ty plane, since f and

∂f are continuous in the entire ∂y

ty plane. -2 t . 1+ y3

4 (a).

f ( t, y ) =

4 (b).

∂f 6 ty 2 ∂f = are continuous everywhere in the ty -plane except on the line 3 2 . f and ∂y (1 + y ) ∂y y = -1.

4 (c).

R = {( t, y ):-• < t < •, y > -1} . 1

1

5 (a). Solving for y ¢ , we have y ¢ = tan t - ty 3 . Thus, f ( t, y ) = tan t - ty 3 .

Chapter 3 First Order Nonlinear Differential Equations • 27

5 (b).

1 -2 ∂f ∂f 1ˆ Ê = - ty 3 . f and are continuous except on the lines t = Á n + ˜ p (where n is an Ë 3 ∂y ∂y 2¯ integer) and y = 0 .

p p ¸ Ï 5 (c). The largest open rectangle is R = Ì( t, y ):- < t < , 0 < y < •˝ . 2 2 ˛ Ó t2 - e-y . y2 - 9

6 (a).

f ( t, y ) =

6 (b).

∂f ( y 2 + 2 y - 9)e - y - 2 t 2 y ∂f = . f and are continuous everywhere in the ty -plane except 2 2 ∂y ( y - 9) ∂y y = ±3 .

6 (c).

R = {( t, y ):-• < t < •, -3 < y < 3} .

7 (a). Solving for y ¢ , we have y ¢ = 7 (b).

2 + tan t 2 + tan t . Thus, f ( t, y ) = . cos y cos y

∂f ∂f -2 = (2 + tan t)(-1)(cos y ) (- sin y ) = (2 + tan t) sec y tan y . f and are continuous except on ∂y ∂y 1ˆ 1ˆ Ê Ê the lines t = Á n + ˜ p (where n is an integer) and y = Á m + ˜ p (where m is an integer). Ë Ë 2¯ 2¯

p¸ p p p Ï 7 (c). The largest open rectangle is R = Ì( t, y ):- < t < , - < y < ˝ . 2 2 2 2˛ Ó

2 + tan y . cos 2 t

8 (a).

f ( t, y ) =

8 (b).

∂f sec 2 y ∂f = . f and are continuous except where tan y is not defined and cos2 t = 0 , or ∂y cos 2 t ∂y where y = ( n + 12 )p , n = ..., -2, -1, 0,1, 2,..., and t = ( m +

8 (c).

3p 5p p p¸ Ï ,- < y < ˝ . R = Ì( t, y ):
9.

One possible example is y ¢ =

10 (a). f ( t, y ) =

)

1 p 2 2

, m = ..., -2, -1, 0,1, 2,...

1 with ( t0 , y 0 ) = (2, 0) . t( t - 4 )( y + 1)( y - 2)

y 2 ∂f 2 y ∂f = 2 . f and are continuous except where t = 0 . 2 , t ∂y t ∂y

R = {( t, y ):0 < t < •, -• < y < •} . 10 (b). No contradiction. If the hypotheses are not satisfied, “bad things need not happen”. 11.

y ( t) =

2 1 - ( t - 1)

, so y (0) =

2 = 2. 2

28 • Chapter 3 First Order Nonlinear Differential Equations

12.

3

3

y ( t) = ( 4 + ( t - t0 )) 2 , so y (0) = ( 4 - t0 ) 2 = 1 fi t0 = 3.

13 (a). z1 ( t) = y ( t + 2), so z1 (-5) = y (-3) = 2 . 13 (b). z2 ( t) = y ( t - 2), so z2 ( 3) = y (1) = 0 .

Section 3.2 1 (a). Antidifferentiation gives us

(-2) 2 2

+ cos

y2 + cos t = C . From the initial condition, we have 2

p = C = 2 . Then we have y 2 = 4 - 2 cos t, y = - 4 - 2 cos t . 2

1 (b). -• < t < • 2 (a).

y 2 y ¢ = 1, so

5 8 y3 - t = C . From the initial condition, we have - 1 = = C . Then we have 3 3 3 1

y 3 = 3t + 5 fi y = ( 3t + 5) 3 . 2 (b). -• < t < • 3 (a).

( y + 1) y ¢ + 1 = 0, have

so

y2 + y + t = C . From the initial condition, we have 0 + 0 + 1 = C . Then we 2

-2 ± 4 - 8( t - 1) y2 . Since y (1) = 0 , we only + y + t = 1 fi y 2 + 2 y + 2( t - 1) = 0, y = 2 2

want the plus sign. Finally, y = 3 (b). -• < t £ 4 (a).

-2 + 4 - 8( t - 1) = -1 + 3 - 2 t . 2

3 2

y -2 y ¢ - 2 t = 0, so - y -1 - t 2 = C . From the initial condition, we have 1 - 0 = C . Then we have - y -1 = t 2 + 1 fi y =

-1 . 1 + t2

4 (b). -• < t < • 5 (a).

y -3 y ¢ - t = 0, so

y -2 + t 2 =

1 , y= 4

1 1 5 (b). - < t < 2 2

1 y -2 t 2 - = C . From the initial condition, we have C = - . Then we have 8 -2 2

1 1 - t2 4

=

2 1 - 4 t2

.

Chapter 3 First Order Nonlinear Differential Equations • 29

6 (a).

e - y y ¢ + ( t - sin t) = 0, so - e -y +

(

t2 2

-1 + 1 = 0 = C . Then we have e -y =

)

+ cos t = C . From the initial condition, we have t2 2

(

+ cos t fi y = - ln

t2 2

)

+ cos t .

6 (b). -• < t < • 7 (a).

1 p -1 . Then we have 2 y ¢ - 1 = 0, so tan y - t = C . From the initial condition, we have C = 1+ y 2 tan -1 y = t -

p Ê , y = tanÁ t Ë 2

pˆ ˜. 2¯

7 (b). 0 < t < p 8 (a).

(cos y ) y ¢ + t -2 = 0, so sin y - t -1 = C . From the initial condition, we have 0 - (-1) = 1 = C . Then we have sin y = 1 + t -1 fi y = sin -1 (1 + t -1 ) .

8 (b). -• < t < 9 (a).

1 2

1 y¢ - t = 0 . 1- y2

1 1 -1 -1 1 1 y + 1 t2 2 2 = = = + - = C. By partial fractions, , and so ln 1 - y 2 y 2 - 1 ( y - 1)( y + 1) y - 1 y + 1 2 y -1 2 -

From the initial condition, we have

1 ln 3 = C . Then we have 2 2

3e t - 1 1 Ê y + 1ˆ y +1 2 2 , and solving for y yields . y = = - t = ln 3 fi ln Á t ln 2 ˜ 3 Ë y - 1¯ y -1 3e t + 1 9 (b). -• < t < • 10 (a). 3 y 2 y ¢ + 2 t - 1 = 0, so y 3 + t 2 - t = C . From the initial condition, we have -1 + 1 - (-1) = 1 = C . 1

Then we have y 3 = 1 + t - t 2 fi y = (1 + t - t 2 ) 3 . 10 (b). -• < t < • 11 (a). e y y ¢ - e t = 0, so e y - e t = C . From the initial condition, we have C = e - 1. Then we have e y - e t = e - 1, y = ln(e t + e - 1) .

11 (b). -• < t < • 12 (a). yy ¢ - t = 0, so

y2 t 2 - = C . From the initial condition, we have 2 - 0 = C . Then we have 2 2

y2 t 2 - = 2 fi y = - 4 + t2 . 2 2

30 • Chapter 3 First Order Nonlinear Differential Equations

12 (b). -• < t < • 13 (a). sec 2 y ( y ¢ ) + e - t = 0, so tany - e - t = C . From the initial condition, we have C = 1 - 1 = 0 . Then we have tan y = e - t , y = tan -1 (e - t ) . 13 (b). -• < t < • 14 (a). (2 y - sin y )( y ¢ ) + ( t - sin t) = 0, so y 2 + cos y +

t2 + cos t = C . From the initial condition, we 2

have 0 + 1 + 0 + 1 = 2 = C . Then we have y 2 + cos y = 2 -

t2 - cos t . There is no explicit 2

solution. 15 (a). ( y + 1)e y ¢ + ( t - 2) = 0, so ye + y

y

(t - 2) 2 2

1 = C . From the initial condition, we have C = 2e 2 + . 2

2

Then we have ye y = 2e 2 + 16.

1 ( t - 2) . There is no explicit solution. 2 2

1 1 1 1 -1 -3 -1 y = ( 4 + t) 2 , so y ¢ = - ( 4 + t) 2 = - y 3 fi y ¢ + y 3 = 0, y (0) = 4 2 = . Therefore, 2 2 2 2 1 1 a = , n = 3, y 0 = . 2 2 2

17.

y=

-24 t 3 6 2 3 2 4 -2 3 3Ê yˆ , so y ( ) + t t = = t = t y . Then we have = 6 1 5 4 24 ¢ Á ˜ ( ) ( ) 2 4 Ë 6¯ 3 (5 + t ) (5 + t 4 )

6 2 2 = 1. y ¢ + t 3 y 2 = 0 , so a = , n = 3, y 0 = 5 +1 3 3 18.

y 3 + t 2 + sin y = 4 fi 3 y 2 y ¢ + 2 t + (cos y ) y ¢ = 0 fi ( 3 y 2 + cos y ) y ¢ + 2 t = 0 . When t = 2, y 03 + 4 + sin y 0 = 4 fi y 03 + sin y 0 = 0 fi y 0 = 0 fi y (2) = 0 .

19.

First, y ¢e y + ye y y ¢ + 2 t = cos t . Then (1 + y )e y y ¢ + (2 t - cos t) = 0 . At t0 = 0 , we have

y 0e y 0 + 0 = 0, so y 0 = 0 , and thus y (0) = 0 . 20.

y -2 y ¢ = 2 fi - y -1 = 2 t + C, - y 0-1 = C fi - y -1 = 2 t - y 0-1 fi y -1 = y 0-1 - 2 t fi y =

1 Require y 0-1 - 2( 4 ) = 0 fi y 0 = . 8 ÊK ˆ 21 (a). Á + 1˜ S ¢ + a = 0, so K ln S + S + at = C . From the initial condition, we have ËS ¯ K ln S0 + S0 = C, so K ln S + S = -at + K ln S0 + S0 .

1 . y - 2t -1 0

Chapter 3 First Order Nonlinear Differential Equations • 31

21 (b). When t = 0 , S (0) = S0 = 1, so C = K ◊ 0 + 1 = 1. Then we have K ln S + S = -at + 1. From the 1 Ê 3ˆ 3 Ê 3ˆ other conditions, we have K lnÁ ˜ + = -a + 1 fi Á ln ˜ K + a = and Ë 4¯ 4 Ë 4¯ 4 7 Ê 1ˆ 1 Ê 1ˆ K lnÁ ˜ + = -6a + 1 fi Á ln ˜ K + 6a = . Solving these simultaneous equations yields Ë 8¯ 8 Ë 8¯ 8

K ª 1.769 and a ª 0.759 . Ê 1ˆ 1 21 (c). K lnÁ ˜ + = -at + 1, so 1.769(-3.912) + 0.02 = -0.759 t + 1. Solving for t yields t ª 10.41. Ë 50 ¯ 50

22.

y ¢ = 1 + ( y + 1) 2 . Let u = y + 1, u¢ = 1 + u 2 , Then, y (0) = 0 fi u(0) = 1,

1 -1 2 u¢ = 1 fi tan ( u) = t + C . (1 + u )

p p Ê pˆ = 0 + C fi tan -1 ( u) = t + fi u = y + 1 = tanÁ t + ˜ . Ë 4 4 4¯

3p p Ê pˆ Therefore, y = tanÁ t + ˜ - 1,
23.

(

2

)

y ¢ = t ( y + 2) + 1 . Letting u = y + 2 , we have u¢ = t( u 2 + 1), so tan -1 u =

1 u¢ = t . Then u +1 2

t2 + C . From the initial condition, we have y (0) = -3 and u(0) = -1, so 2

Ê t2 pˆ p p t2 p -1 - = 0 + C, C = - , and tan u = - . In terms of y, this reads y = -2 + tanÁ - ˜ . 4 4 2 4 Ë 2 4¯ p t2 p p Setting - < - < and simplifying, we have 2 2 4 2 -

24.

p 3p < t2 < fit< 2 2

y ¢ = ( y + 1) 2 sin t.

3p 3p fi
3p . 2

y¢ -1 = - cos t + C . 2 = sin t fi ( y + 1) y +1

Then, y (0) = 0 fi -1 = -1 + C fi C = 0 fi

-1 = - cos t . y +1

Therefore, y + 1 = sec t fi y = sec t - 1. 25.

Q-3Q¢ + k = 0 , so

Q-2 + kt = C ¢ and Q-2 = 2 kt - C . From the implicit initial condition, we have -2

Q0 -2 = -C, so Q-2 = 2 kt + Q0 -2 . Solved for Q, we have Q( t) =

1 2 kt + Q0

-2

=

Q0 1 + 2 kQ0 2 t

.

32 • Chapter 3 First Order Nonlinear Differential Equations

Thus

Q0 1 , where t is the half-life of the reactant. Therefore, Q0 = 2 1 + 2 kQ0 2t

2 = 1 + 2 kQ0 2t , which, solved for t , gives t =

26.

3 . Thus the half-life depends upon Q0 . 2 kQ0 2

Q¢ = - kQ2 , Q(0) = Q0 ; Q-2Q¢ = - k fi -Q-1 = - kt + C, C = -Q0-1. Therefore, Q-1 = kt + Q0-1 fi Q = 0.4 Q0 =

1 Q0 , Q(10) = 0.4 Q0 . Then, -1 = 1 + kQ0 t kt + Q0

Q0 Q0 fi 0.4 + 4 kQ0 = 1 fi kQ0 = 0.15 and Q = . 1 + kQ0 (10) 1 + .15 t

Set Q = 0.25Q0 . Then, 0.25 =

1 fi t = 20 min. 1 + .15 t

27 (a). The equation is nonlinear and separable.

Ï y, y ≥ 0 27 (b). y = Ì . Thus Ó- y, y < 0

1 y¢ - 1 = 0 . y

Ï y (0)e t , y > 0 dy Ï ln y, y > 0 Ú y = ÌÓ- ln y, y < 0 fi y (t) = ÌÓy (0)e - t , y < 0 .

Since y(0) = 1 > 0 , the solution y ( t) = e t of y ¢ = y , y (0) = 1 will be identical to that of y ¢ = y, y (0) = 1 as long as y ( t) = e t ≥ 0 . This is true for all t, however, and so the two solution

curves agree. 27 (c). If y(0) = -1 < 0 , then the solution of y ¢ = y , y (0) = -1, is y ( t) = -e - t , but the solution of y ¢ = y, y (0) = -1, is y ( t) = -e t .

28.

y ¢ = - y 2 is graph c. y ¢ = y 3 is graph a. y ¢ = y ( 4 - y ) is graph b.

29.

This is a translation three units to the right of graph (a) in problem 28.

30.

Yes.

31.

y ¢ sin y + y (cos y ) y ¢ - 3 = 0 fi y ¢ =

1 y¢ - 1 = 0. f ( y) 3 = f ( y ) . The solution of sin y + y cos y

y ¢ = f ( y ) is H ( y ) = t + C , where H ( y ) = Ú

1 dy . The solution of y ¢ = t 2 f ( y ) is therefore f ( y)

H ( y) =

t3 + C2 . From the initial condition, we know that H (0) = 1 + C fi C = H (0) - 1 , and 3

H (0) =

1 2 + C2 . Thus C2 = C + . Now we have H ( y ) = t + C and y sin y - 3t + 3 = 0 , so 3 3

y sin y = 3( t - 1) fi 13 y sin y = t + (-1) fi H ( y ) = 13 y sin y, and C = -1.

Chapter 3 First Order Nonlinear Differential Equations • 33

Therefore, C2 = C +

2 1 2 = -1 + = - and the implicit solution of the initial value problem is 3 3 3

y sin y t 3 1 = - fi y sin y - t 3 + 1 = 0 . 3 3 3

Section 3.3 1.

M = 3t 2 - 2, N = y, M y = N t = 0 , so the equation is exact.

y2 ∂H =yfiH= + g( t) and ∂y 2

y2 3 ∂H = g¢ = 3t 2 - 2 fi g( t) = t 3 - 2 t + C fi + t - 2t = C . 2 ∂t From the initial condition, we have

(-2) 2 2

- 1 - 2(-1) = 2 - 1 + 2 = 3 = C , and thus

y2 3 + t - 2 t = 3 fi y = - 6 + 4 t - 2 t 3 (the minus sign can be checked by the initial condition). 2 2.

M = y + t 3 , N = t + y 3 , M y = N t = 1, so the equation is exact.

y4 dh dh ∂H t4 ∂H =t+ = N = t + y3 fi = y3 fi h = . = M = y + t 3 fi H = yt + + h ( y ) and dy dy ∂y 4 ∂t 4 Therefore, yt + 3.

t4 y 4 y4 t4 + yt + = 4 fi y 4 + 4 yt + t 4 = 16 . + = C, y (0) = -2 fi C = 4 and 4 4 4 4 -1

The equation is separable, and therefore it is exact. ( y 2 + 1) y ¢ = 3t 2 + 1 gives us

tan -1 y = t 3 + t + C , and from the initial condition we have

p Ê = C . Thus y = tanÁ t 3 + t + Ë 4

pˆ ˜. 4¯

4.

M = 3t 2 y, N = 6 t + y 3 , M y = 3t 2 , Nt = 6 . Therefore, the differential equation is not exact.

5.

M = e t + y + 3t 2 , N = e t + y + 2 y, M y = N t = e t + y , so the equation is exact. dh dh ∂H ∂H = et +y + = N = et +y + 2 y fi = 2y , = M = e t + y + 3t 2 fi H = e t + y + t 3 + h ( y ) and dy dy ∂y ∂t

and so h = y 2 + C . From the initial condition, we have 1 + 0 + 0 = C , and thus

e t + y + t 3 + y 2 = 1. 6.

2

M = y cos( ty ) + 1, N = t cos( ty ) + 2 ye y , M y = N t = cos( ty ) - ty sin( ty ) , so the equation is

exact.

∂H = M = y cos( ty ) + 1 fi H = sin( ty ) + t + h ( y ) ∂t

34 • Chapter 3 First Order Nonlinear Differential Equations

and

2 2 dh dh ∂H 2 = t cos( ty ) + = N = t cos( ty ) + 2 ye y fi = 2 ye y , and so h = e y . From the initial dy dy ∂y

2

condition, we have 0 + p + 1 = C , and thus sin( ty ) + t + e y = p + 1. 7.

M y = cos( t + y ) - y sin( t + y ) + 1, N t = cos( t + y ) - y sin( t + y ) + 1, so the equation is exact. t2 ∂H = M = y cos( t + y ) + y + t fi H = y sin( t + y ) + yt + + h ( y ) and ∂t 2 dh ∂H = sin( t + y ) + y cos( t + y ) + t + = sin( t + y ) + y cos( t + y ) + t + y . Thus dy ∂y

y2 dh = yfih= + C , and from the initial condition, we have dy 2 1 2

1 2

(-1) sin(1 - 1) + (-1)(1) + + = 0 = C , and thus y sin(t + y ) + yt + 8.

t2 y 2 + = 0. 2 2

M = a t 3 y n , N = t m y 2 , M y = na t 3 y n -1, N t = mt m -1 y 2 . Therefore,

m - 1 = 3 fi m = 4, n - 1 = 2 fi n = 3, 3a = 4 fi a =

4 . 3

9.

y3 N = t + y sin t, N t = M y = 2 t + y cos t . Thus M = 2 ty + cos t + m( t). 3

10.

M = t 2 + y 2 sin t, M y = 2 y sin t = N t fi N = -2 y cos t + n ( y ).

11.

0 + 1 + y 0 2 = 5 fi y 0 = ±2 . 3t 2 y + t 3 y ¢ + e t + 2 yy ¢ = 0 , so ( t 3 + 2 y ) y ¢ + ( 3t 2 y + e t ) = 0 and thus

2

2

2

M = 3t 2 y + e t and N = t 3 + 2 y . 12.

y = - t - 4 - t 2 fi y (0) = -2 = y 0 . Also, N t = a = M y fi exact.

y2 bt 2 Then, H y = y + at fi H = . + aty + g( t), H t = at + g¢ = ay + bt fi g¢ = bt fi g = 2 2 Therefore, H=

y2 bt 2 + aty + = C fi y 2 + 2 aty + (bt 2 - 2C ) = 0 2 2

fiy=

-2 at ± 4 a 2 t 2 - 4 (bt 2 - 2C ) = - at ± a 2 t 2 - bt 2 + 2C = - at ± 2C - (b - a 2 ) t 2 2

Choose the negative and a = 1, 2C = 4, b - a 2 = 1 fi b = 2 . 13.

2y 1 y¢ + = 0 fi ln( y 2 + 1) + ln( t + 1) = C . From the initial condition, we have ln2 = C . 2 y +1 t +1

Thus ( y 2 + 1)( t + 1) = 2 , and solving for y yields y =

1- t . 1+ t

Chapter 3 First Order Nonlinear Differential Equations • 35

14.

One example: ( y + 2 t) + (2 y + t) y ¢ = 0, M = y + 2 t, N = 2 y + t dh ∂H ∂H =t+ = 2 y + t . Therefore, = M = y + 2 t fi H = yt + t 2 + h ( y ), dy ∂y ∂t dh = 2 y fi h = y 2 fi yt + t 2 + y 2 = C . dy

Section 3.4 1 (a). The equation is both separable and exact. 1 (b). (i) y ¢ = y (2 - y ) fi y ¢ - 2 y = - y 2 fi 1 - n = -1 = m, v = y -1 fi y = v -1 , thus y ¢ = -v 2v ¢ and -v -2v ¢ = 2v -1 - v -2 or v ¢ + 2v = 1, v (0) = 1.

1 1 (ii) (e 2 t v )¢ = e 2 t fi e 2 t v = e 2 t + C or v = + Ce -2 t . From the initial condition, 2 2

1 1 1 + C = 1 fi C = , and so v = (1 + e -2 t ) . 2 2 2 (iii) y = v -1 = 1 (c).

2 . 1 + e -2 t

-• < t < •

2 (a). The equation is both separable and exact. 2 (b). (i) y ¢ = 2 ty - 2 ty 2 fi 1 - 2 = -1 = m, v = y -1 fi y = v -1 , thus -v -2v ¢ = 2 tv -1 - 2 tv -2 or v ¢ + 2 tv = 2 t, v (0) = -1. 2 2 2 2 2 ¢ (ii) e t v = 2 te t fi e t v = e t + C or v = 1 + Ce - t . From the initial condition,

( )

2

1 + C = -1 fi C = -2 , and so v = 1 - 2e - t . (iii) y = v -1 = 2 (c).

1 2 . 1 - 2e - t

- ln 2 < t < ln 2

3 (a). The equation is neither separable nor exact. 3 (b). (i) m = 1 - n = -1, v = y -1 fi y = v -1 , thus y ¢ = -v -2v ¢ = -v -1 + e t v -2 fi v ¢ = v - e t or

v ¢ - v = -e t , v (-1) = -1. (ii) (e - t v )¢ = -1 fi e - t v = - t + C or v = - te t + Ce t . From the initial condition,

e -1 + Ce -1 = -1 fi C = -(1 + e) , and so v = e t (- t - 1 - e) = -( t + 1)e t - e t +1. (iii) y = v -1 =

-1 . (t + 1)e t + e t +1

36 • Chapter 3 First Order Nonlinear Differential Equations

3 (c).

-(1 + e) < t < •

4 (a). The equation is both separable and exact. 1 1 1 -1 1 -1 -1 4 (b). (i) 1 - n = 2 = m, v = y 2 fi y = v 2 , thus y ¢ = v 2 v ¢ and v 2 v ¢ = v 2 + v 2 or 2 2

v ¢ = 2v + 2, v (0) = 1.

(ii) (e -2 t v )¢ = 2e -2 t fi e -2 t v = -e -2 t + C or v = -1 + Ce 2 t . From the initial condition,

-1 + C = 1 fi C = 2 , and so v = -1 + 2e 2 t . (iii) y = - -1 + 2e 2 t . 4 (c).

1 - ln 2 < t < • 2

5 (a). The equation is neither separable nor exact. 1 2 1 - 23 1 - 23 3 -3 3 5 (b). (i) m = 1 - n = 3, v = y fi y = v , thus y ¢ = v v ¢ . Then t ◊ v v ¢ + v = t v , and so 3 3 1 3

3

tv ¢ + 3v = 3t 3 , v (1) = 1. (ii) ( t 3v )¢ = 3t 5 fi t 3v = and so v =

1 1 t6 C 1 + C or v = t 3 + 3 . From the initial condition, + C = 1 fi C = , 2 2 t 2 2

1 3 -3 (t + t ) . 2 1

1 3

Ê1 ˆ3 (iii) y = v = Á ( t 3 + t -3 )˜ . Ë2 ¯

5 (c).

0< t<•

6 (a). The equation is neither separable nor exact. 3 1 2 3 2 3 1 3 1 6 (b). (i) m = 1 - n = , v = y 3 fi y = v 2 , thus y ¢ = v 2 v ¢ . Then v 2 v ¢ - v 2 = tv 2 , and so 3 2 2

2 2 v ¢ - v = t, v (0) = 4 . 3 3

2 -2t -2t ¢ -2t (ii) e 3 v = te 3 fi e 3 v = 3

( )

2 2 Ê 3 - 23 t 9 - 23 t ˆ 3 t Á - te - e ˜ + C or v = - t - + Ce 3 . From the initial Ë ¯ 3 2 4 2

11 3 Ê 3ˆ 11 2 t condition, - + C = 4 fi C = , and so v = -Á t + ˜ + e 3 . Ë 2¯ 2 2 2 3 2

Ê 11 Ê 3ˆ ˆ (iii) y = -Á e - Á t + ˜ ˜ . Ë 2¯ ¯ Ë2 2 3t

6 (c).

-• < t < •

Chapter 3 First Order Nonlinear Differential Equations • 37

7.

First, let u = e y . Then y = ln u and y ¢ = us

1 2 u¢ u¢ . Therefore, = 2 t -1 + fi u¢ - u = 1 which gives u u u t

1 2 1 -2 ¢ -2 fi t -2 u = - t -1 + C . Solving for u gives us 2 u¢ - 3 u = 2 . Then we have ( t u) = t t t t

u = - t + Ct 2 . From the initial condition, we have y (1) = 0 fi u(1) = 1 , and so

1 u = - t + 2 t 2 fi y = ln(2 t 2 - t), t > . 2 8.

First, let u = y + 1, u¢ = - u + tu -1, 1 - n = 3. Therefore, 1 1 1 -2 1 -2 -2 v = u 3 , u = v 3 , u¢ = v 3 v ¢ fi v 3 v ¢ + v 3 = tv 3 . Then, 3 3

v ¢ + 3v = 3t, y (0) = 1 fi u(0) = 2 fi v (0) = 8 and

1 v = Ce -3 t + at + b, a + 3( at + b) = 3t fi a = 1, b = - . Therefore, 3 25 1 1 v = Ce -3 t + t - , v (0) = C - = 8 fi C = . Then, 3 3 3 1 1 1ˆ 3 25 1 Ê 25 v = e -3 t + t - , y = u - 1 = v 3 - 1 = Á e -3 t + t - ˜ - 1, - • < t < • . Ë 3 3¯ 3 3

9.

y 0 = 3 by substitution. Differentiating yields

Ê -1 ˆ Ê ˆ 3 9 -3e - t t t 2 e + 3e - t Á 3 y¢ = ( ) = + 2˜ 2 2t ˜ = - y + e y . t Á 1 - 3t (1 - 3t)e Ë (1 - 3t) ¯ Ë (1 - 3t) e ¯

Thus q( t) = e t .

Section 3.5 1.

1 - n = -1, v = P -1, P = v -1 . Thus -v -2v ¢ - rv -1 = v = Ce - rt +

1 1 1 , v (0) = = C + . Solving for C yields C = P0 -1 - Pe -1 , so we have Pe P0 Pe

(

)

v = P -1 = P0 -1 - Pe -1 e - rt + Pe -1. Thus P =

2.

r -2 r v , or v ¢ + rv = , v (0) = P0 -1 . Then Pe Pe

-1 - rt

P0 e

P0 Pe 1 . = -1 - rt - Pe (e - 1) P0 - ( P0 - Pe )e - rt

Since P is measured in millions, P0 = 0.1, r = 0.1, Pe = 3 . Therefore, P=

0.3 0.1( 3) fi e -.1t ª .003831417 -0.1t , 0.9 Pe = 2.7 fi 2.7 = 0.1 + 2.9e -0.1t 0.1 - (0.1 - 3)e

fi t ª 55.65 years.

38 • Chapter 3 First Order Nonlinear Differential Equations

Ê M P2 Pˆ +P+ = 0 . Then 3 (a). Setting rÁ1 - ˜ P + M = 0 , we have r Pe Ë Pe ¯ Pe ± Pe 2 + 4 Pe M r M . This makes sense; migration would alter the P - Pe P - Pe =0fiP= r 2 2

equilibrium state. 3 (b).

2 1 1 1 1 M (2P - 1) 2 = 1 + 4 x fi ÊÁ P - ˆ˜ = x + , where x = . This is a parabola with vertex ÊÁ - , ˆ˜ . Ë ¯ Ë ¯

2

4

4 2

r

For x > 0 , there is one nonnegative equilibrium solution. Two such solutions exist for

-

1 < x £ 0. 4

1 3 (c). When x = - , the two nonnegative equilibrium solutions coalesce into a single equilibrium 4 1 value. There are no equilibrium solutions for x < - . This makes sense, since if the migration 4 out of the colony is too large relative to reproduction, equilibrium could not be achieved. 4.

Equilibrium values at 4 and - 2. P =

4 - (-2) = 6 = Pe 1 + 5.

P ¢ = (1 - P ) P -

Pe Ê 4M ˆ 1± 1+ fi 4 + (-2) = Pe fi Pe = 2 . Á Pe r ˜¯ 2Ë

4M 2M 2M M or 3 1 + =8= fi = 4. Pe r r r r

3 3ˆ Ê 1ˆ 1 3 Ê , P (0) = . Then P ¢ = - P 2 + P - = -Á P - ˜ Á P - ˜ . Then we have Ë 16 4¯ Ë 4¯ 2 16

1 P ¢ + 1 = 0 , and by partial fractions, we have 1ˆ Ê 3ˆ Ê ÁP - ˜ÁP - ˜ Ë 4¯ Ë 4¯ È Í -2 2 Ú ÍÍ Ê 1 ˆ + Ê ÁP - ˜ ÁP ÍÎ Ë 4¯ Ë

˘ P˙ ˙dP + t = C , and so 2 ln 3ˆ ˙ P˜ 4 ¯ ˙˚

Ê 2 Á -2 + Á 1 ÁPPË 4

ˆ ˜ P ¢ + 1 = 0 . Then 3˜ ˜ 4¯

3 4 + t = C. 1 4

From the initial condition, we have C = 0 , so

3 t P4 = e - 2 . But 1 < P < 3 fi P 1 4 4 P4

3 4

3 3 1 - 2t t -P + e = e 2 , and then P ( t) = 4 4 t . = -( P - 43 ) fi 4 1 P1+ e 2 4

Chapter 3 First Order Nonlinear Differential Equations • 39

2

6.

1ˆ 1ˆ 1 Ê Ê P ¢ = (1 - P ) P - , P0 = 1 fi P ¢ = -Á P 2 - P + ˜ = -Á P - ˜ which is separable. Ë Ë 4¯ 2¯ 4 1

1ˆ Ê 2 P ¢ + 1 = 0 fi -Á P - ˜ Ë 2¯ 1ˆ Ê ÁP - ˜ Ë 2¯

1ˆ Ê Therefore, Á P - ˜ Ë 2¯

7.

-1

Ê 1ˆ + t = C, - Á1 - ˜ Ë 2¯

-1

=2+tfiP=

-1

+ 0 = -2 = C

1 1 2 + 2t + = . 2 2+t 2+t

P ¢ = .5(1 + 2 sin(2 pt))(1 - P ) P, P (0) = P0 . Following the derivation in the chapter with r( t) , t

1Ê 1 ˆ we have R( t) = 0.5 Ú (1 + 2 sin(2 ps)) ds = Á s - cos(2 ps)˜ 0 ¯0 2Ë p t

P0 1Ê 1 1ˆ 1 Ê 1 ˆ with = Á t - cos(2 pt) + ˜ = Á t + [1 - cos(2 pt)]˜ . Therefore P = ¯ P0 - ( P0 - 1)e - R ( t ) 2Ë p p¯ 2 Ë p

1Ê 1 ˆ R( t) = Á t + [1 - cos(2 pt)]˜ . ¯ 2Ë p

40 • Chapter 3 First Order Nonlinear Differential Equations

8.

t

t = Ú r( s) ds fi 0

P (t ) =

9.

dP = (1 - P ) P . The solution procedure in the text leads to dt

P0 1Ê 1 ˆ Á t + [1 - cos(2pt)]˜ . -t . Substitute t = ¯ P0 - ( P0 - 1)e 2Ë p

P ¢ = r(1 - P ) P = rP - rP 2 fi v = P -1, P = v -1 , -v -2v ¢ = rv -1 - rv -2 fi v ¢ + rv = r, v (0) = P0 -1 . t

Letting R( t) = Ú r( s) ds , we have (e R v )¢ = re R fi e R v = e R + C fi v = 1 + Ce - R . v (0) = P0 -1 , so 0

(

)

C = P0 -1 - 1 and thus v = 1 + P0 -1 - 1 e - R . Finally, P = v -1 =

1

(

-1

)

1 + P0 - 1 e

-R

=

P0 P0 - ( P0 - 1)e - R

1Ê 1 ˆ with R( t) = t = Á t + [1 - cos(2 pt)]˜ . ¯ 2Ë p

10 (a). P ¢ = k ( N - P ) P with N and P in units of 100,000 and t in months. N = 5, k = 2e - t - 1. t

t

0

0

t = Ú (2e - s - 1) ds = (-2e - s - s) = (-2e - t - t + 2) = - t + 2(1 - e - t ) . dP = (5 - P ) P which is separable. dt P A B 1 1 1 1 1 = + , A = , B = fi (ln P - ln 5 - P ) = t + C = ln (5 - P ) P P 5 - P 5 5- P 5 5 5

Chapter 3 First Order Nonlinear Differential Equations • 41

From the initial condition, Therefore, P =

1 P P 1 Ê 1ˆ Ê 1ˆ lnÁ ˜ = 0 + C fi ln = 5t + lnÁ ˜ fi = e 5t Ë 4¯ 5- P 4 5 Ë 4¯ 5- P

5e 5t -t 5t , t = - t + 2(1 - e ) . 4+e

10 (b). From the plot, Pmax ª 2.7 (270, 000) . 10 (c). From the plot, t ª 1.8 months. 11 (a). ( A - B)¢ = - kAB + kAB = 0, A( t) - B( t) = A(0) - B(0) = 5 - 2 = 3 moles. 11 (b). B = A - 3, A¢ = - kA( A - 3) = k ( 3 - A) A, A(0) = 5 . 5◊ 3 15 Ê Aˆ 11 (c). A(1) = 4, A¢ = 3k Á1 - ˜ A . Using equation (5), A( t) = . -3 kt . Thus A( t) = Ë 5 - (5 - 3)e 3¯ 5 - 2e -3 kt We know that A(1) = 4 , so A( 4 ) =

15 Ê 5ˆ 5 - 2Á ˜ Ë 8¯

4

15 5 -3 k yields e -3 k = . Thus -3 k = 4 . Solving for e 5 - 2e 8

= 3.195 moles. B = A - 3 = 0.195 moles.

Section 3.6 1.

With v 0 = 0, v = -

k k k - tˆ - t - t mg Ê 1 mg 1 1 m m m e 1 . Setting gives us . Thus = v = 1 e e = , Á ˜ k Ë 2 k 2 2 ¯

m k t = ln 2, t = ln 2 . k m

2.

mv ¢ = - mg + kv 2 , v (0) = 0 fi v ¢ = - g +

k 2 k Ê 2 mg ˆ v ¢ k = v = Áv ˜ k ¯ v 2 - mg m m mË k

42 • Chapter 3 First Order Nonlinear Differential Equations

1 mg v2 k

A

=

mg k

v-

B

+ v+

mg k

fi A=

1 1 . Therefore, , B=mg mg 2 2 k k

v1 ln mg v+ 2 k

mg k = k t + C, v (0) = 0 fi C = 0 and - mg < v £ 0 . Then, k mg m k

mg k = mg v+ k

mg -v 2 k =e mg +v k

v-

kg t m

Ê -2 mg Á 1 - e fiv=k ÁÁ -2 Ë1 + e

3.

Ê 5280 ˆ 10 mi / hr = 10Á ˜ = 14.67 ft/sec. Then 14.67 = Ë 3600 ¯

4 (a).

m

ˆ ˜ = - mg tanhÊ kg tˆ . Á m ˜ kg ˜ k t Ë ¯ m ˜ ¯

kg t m

200 lb ◊ sec 2 . fi k ª .929 ft 2 k

dv 3000 -kt slug + kv = 0 fi v ( t) = v 0e m , m = dt 32

v ( 4 ) 50 3000 Ê 22 ˆ Ê 22 ˆ 128 - k ◊ 32 ◊ 4 k . Then, k = = = e 3000 fi lnÁ ˜ = lnÁ ˜ = 34.725 lb ◊ sec/ ft. Ë 5 ¯ 3000 v0 220 128 Ë 5 ¯

4 (b).

4

0

0

d = Ú v ( t) dt = v 0 Ú e =

5.

4

4

- mk t

mv 0 Ê m - k tˆ -4k dt = v 0 Á - e m ˜ = 1- e m Ë k ¯0 k

(

)

5280 ˆ Ê 1 ˆ Ê 170 ˆ 3000 Ê Á 220 ◊ ˜Á ˜Á ˜ ª 673 ft. 3600 ¯ Ë 34.725 ¯ Ë 220 ¯ 32 Ë

mv ¢ + kv 2 = 0 fi v -1 =

v¢ k k fi -v -1 = - t + C , C = -v 0 -1 . Then we have 2 = v m m

v0 k t + v 0 -1 fi v = . From the condition provided, we have k m 1 + v0t m

5 11 k v ( 4 ) 50 22 = 17 . Solving for k yields = = fi 4 v0 = 5 5 220 1 + 4 k v v0 m 0 22 m

k=

lb ◊ sec 2 17 m 17 3000 1 Ê Ê 5280 ˆ ˆ = ◊ ∏ Á 220Á . ˜ ˜ ª .247 Ë 3600 ¯ ¯ ft 2 5 4 v 0 5 32 4 Ë 4

4 dt dt Ê 20 ˆ Ê 17 ˆ = v0 Ú = v 0 Á ˜ lnÁ1 + t˜ For the distance traveled, d = Ú v ( t) dt = v 0 Ú 0 0 0 17 kv 0 Ë 17 ¯ Ë 20 ¯ 0 t 1+ 1+ t 20 m 4

Ê 5280 ˆ Ê 20 ˆ Ê 17 ˆ = 220Á ˜ Á ˜ lnÁ1 + ˜ = 562.4 ft. Ë 3600 ¯ Ë 17 ¯ Ë 5¯

4

Chapter 3 First Order Nonlinear Differential Equations • 43

6.

mv ¢ + kv = - mg, v (0) = v 0 fi v ( t) = -

mg ˆ - mk t mg Ê + Á v0 + ˜ e . Set k ¯ k Ë

Ê kv ˆ m Ê kv ˆ k mg mg Ê kv 0 ˆ - mk t m = fi tm = lnÁ1 + 0 ˜ fi tm = lnÁ1 + 0 ˜ . Á1 + ˜e k Ë mg ¯ mg ¯ m k k Ë mg ¯ Ë

v = 0:

tm

7.

tm

0

0

h = Ú v ( t) dt = Ú =-

8.

tm

È mg Ê È mg mg ˆ - mk t ˘ mg ˆ - mk t ˘ mÊ e dt v + + v t = + Á ˜ Á ˜e ˙ Í ˙ Í 0 k ¯ k ¯ kË 0 Î k Ë ˚ Î k ˚0

k - tm ˆ mg ˆ Ê mÊ mg tm + Á v 0 + ˜ Á1 - e m ˜ . k ¯Ë kË k ¯

mv ¢ = - mg fi v ¢ = - g, v (0) = v 0 . Therefore, v ( t) = v 0 - gt and y = - g

impact time is given by - g 9 (a).

t2 v + v 0 t, tm = 0 . The g 2

2v ti 2 g + v 0 ti = 0 fi - ti + v 0 = 0 fi ti = 0 = 2 tm . g 2 2

1 v ¢ = - g, v 0 = 0 fi v = - gt = y ¢ fi y = - gt 2 + y 0 . We want to find the time t at which y=7. 2 Thus 7 = -

32 2 t + 555 , and solving for t yields t ª 5.852 sec. At that time, 2

v = -32(5.852) ª -187.3 ft/sec. k k t Ê mk t ˆ ¢ t kv mg mk t m m 9 (b). mv ¢ + kv = - mg fi v ¢ + = - g, v 0 = 0 . Thus Á ve ˜ = - ge fi ve = e + C . From m k Ë ¯

the initial condition, we have C =

mg , and so k t

k t - tˆ m - mk sˆ mg Ê mg Ê mg Ê m Ê - mk t ˆ ˆ m v=t + Á e - 1˜ ˜ . Á1 - e ˜ fi y = y 0 + Ú0 v ( s) ds = y 0 Á s + e ˜ = y0 k k Ë k Ë kË k ÁË ¯ ¯0 ¯¯ 5

m=

1 8

16

32

mg =

=

41 m 41 = ª 5.56098 sec -1 . slug, so 8 16 32 0 . 0018 k ( )( )( ) 8 ◊ 16 ◊ 32

41 ª 0.3203125 lb , and so solving for t yields 8(16)

-t Ê È ˘ˆ 7 = 555 - 177.95139Á t - 5.56098 Í1 - e 5.56098 ˙˜ fi t = 7.08513 sec. Substitution gives us Ë Î ˚¯

v=

-7.08513 ˘ -0.3203125 È 5.56098 1 e Í ˙ ª -128.18 ft/sec. 0.0018 Î ˚

44 • Chapter 3 First Order Nonlinear Differential Equations

10.

mg = 180 lb. For 0 £ t £ 10, v ¢ = - g, v (0) = 0. For 10 < t £ 14, mv ¢ + kv = - mg, y (14 ) = 0.

For mg = 200,

5280 3600(200) 200 = 10 fik= = 13.63636364. 3600 5280(10) k

10 (a). v = - gt At t = 10, v = -320 ft / sec. 10 (b). Solve v ¢ + v ( t) = -

k v = - g, v (0) = -320 , for v( 4 ). m

mg ˆ - mk t 180 ˆ - 13.63180( 32 ) ( 4 ) mg Ê 180 Ê + Á v0 + + Á -320 + ˜ e fi v (4) = ˜e k ¯ 13.63¯ k Ë 13.63 Ë

= -13.2 - 306.8(.000061469) = -13.219 ft / sec (basically the terminal velocity). 4

mg ˆ - 4mk mg mÊ mg ˘Ê m ˆ - mk t ˆ Ê mg È 10 (c). h = - Ú v ( t) dt = Á t - Ív 0 + (4) + Á v0 + Á- ˜ e ˜ = ˜ e -1 ˙ 0 Ë k ¯0 k ¯ k kË k ˚Ë k ¯ Î

(

4

=

)

4 ( 13. 63 ) 32 mg ˆ 180 ˆ Ê 4 mg m Ê 4 (180) 180 Ê - 4k - Á v0 + ˜ 1- e m = Á -320 + ˜ Ë1 - e 180 ˆ¯ k ¯ 13.63¯ k kË 13.63 32(13.63) Ë

(

)

= 52.8 - 0.4125(-306.8)(0.99994 ) = 179.347 ft.

1 10 (d). hballoon = h + g(10) 2 = 179.347 + 1600 = 1779.347 ft. 2 11.

For the first situation, mv1¢ + kv1 = 0, v1 = v 0e

-

k t m

, m=

3000 , k = 25 . Then 32

-25◊ 32 t1 50 3000 22 = e 3000 fi t1 = ln ª 5.556 sec. 220 25( 32) 5

For the second situation, mv 2¢ + k (tanh t)v 2 = 0, v 2¢ + linear equation. Letting m = e

k ln(cosh t ) m

k tanh t(v 2 ) = 0 . This is a first order m

k

= (cosh t) m , we have

k ¢ k Ê ˆ Á v 2 (cosh t) m ˜ = 0 fi v 2 = C (cosh t) m . Ë ¯

From the initial condition, we have cosh(0) = 1 fi C = v 0 . Then m

3000

k Ê v ˆ k Ê 220 ˆ 32◊25 v2 = (cosh t) m fi cosh t2 = Á 0 ˜ = Á . ˜ Ë 50 ¯ v0 Ë v2 ¯

Ê 22 ˆ ln(cosh t2 ) = 3.75 lnÁ ˜ ª 5.55602 fi cosh t2 ª 258.79 , so t2 ª cosh -1 (258.79) ª 6.249 sec. Ë 5¯

This would be expected, since the size of the drag coefficient would be less for the second situation. Comparing the two values gives us t1 ª 0.89 t2 . These values do not seem appreciably

Chapter 3 First Order Nonlinear Differential Equations • 45

different. However, it can be shown that this difference in stopping time leads to a difference in stopping distance of approximately 110 ft. If this distance is important for a certain situation, then the idealization is not reasonable.

Section 3.7 1.

dv k dv k dv k k x3 = - x 2v fi v = - x 2v fi = - x2 fi v = + C . When x = 0 , v = v 0 . dx m dx m dt m m 3 1

m k x3 Ê m ˆ3 Therefore, v 0 = C , and so v = + v 0 and x f 3 = 3 v 0 fi x f = Á 3 v 0 ˜ . Ë k ¯ k m 3

2.

mv

dv k dv k dv = - kxv 2 fi = - xv fi + xv = 0 (first order linear). dx m dx m dx

2 2 d Ê mk x22 ˆ - kx - kx 2m 2m . Since v > 0, 0 £ x < •, x f = • . e v v Ce C v v v e 0 , = fi = = fi = 0 0 ¯ dx Ë

3.

v 2 k -x dv k - x dv v02 k -x - , and so mv = - ke fi v + e = 0 fi - e = C . Then C = dx m dx 2 m 2 m 1

È v02 k k -x ˘ k 2k È ˘2 v = 2Í - + e ˙ fi v = Ív 0 2 - 2 (1 - e - x ) ˙ . If v 0 2 ≥ , then v>0 for all nonnegative m m Î ˚ Î 2 m m ˚ 2

x and x f = • . If v 0 2 <

2k 2k -x , then we have v 0 2 = 1 - e f , which, solved for x f , yields m m

(

)

Ê mv 0 2 ˆ x f = - lnÁ1 ˜. 2k ¯ Ë 4.

mv

k dv kv dv kÊ 1 ˆ =fi =- Á ˜ fi v = - ln(1 + x ) + C, v 0 = C . Therefore, 1+ x m dx dx m Ë1 + x ¯

v = v0 5.

m

mv 0 mv 0 k ln(1 + x ) and = ln(1 + x f ) fi x f = e k - 1. k m

dv + kv 2 = 0, v (0) = v 0 , x (0) = 0 . We want to find v when x=d. dt

mv

k k - x - x dv k dv + kv 2 = 0 fi + v = 0 fi v = Ce m . From the initial condition, v = v 0e m , and so at dx m dx

x=d, v = v 0e

-

k d m

.

46 • Chapter 3 First Order Nonlinear Differential Equations

6.

m

dv k dv dv = - mg - kv 2 , v (0) = v 0 , x (0) = 0 fi mv = - mg - kv 2 fi = - v - gv -1 dy m dy dt

fi

dv k + v = - gv -1 (Bernoulli). dy m 1

1 - n = 2, u = v 2 fi v = u 2 ,

dv 1 - 12 du 1 - 1 du k 12 -1 = u fi u 2 + u = - gu 2 . Therefore, dy m dy dy 2 2

2 ky 2 ky mg mg 2mky d 2mky du 2 k e + C, C = v 02 + e u = -2 ge m fi e m u = u = -2 g, u = v 02 when y = 0. + k k dy dy m

( )

1

mg Ê 2 mg ˆ - 2mky È mg Ê 2 mg ˆ - 2mky ˘ 2 2 + Á v0 + + Á v0 + Therefore, u = ˜ e = v fi v = ͘ e ˙ . This equation is k ¯ k ¯ k Ë Î k Ë ˚ valid for 0 £ y £ h, where h = maximum height. È mgk ˘ m È kv 02 ˘ mg Ê 2 mg ˆ - 2mk h 2k . + Á v0 + = 0 fi - m h = ln Í 2 mg ˙ fi h = ln 1 + ˜e k ¯ k Ë 2 k ÍÎ mg ˙˚ Î v0 + k ˚

7.

With x measured as shown and v =

x

cosq =

(x

2

1 , we have - mv

+ h2)2

dx dv , we have - m = F cosq . Defining dt dt

dv = dx

v = 0 when x = D , so C = - F ( D + h 2

Fx

(x

1 2 2

)

1

2

fi -m

+ h2)2

1 v2 = F ( x 2 + h 2 ) 2 + C . We know that 2

1 1 2Ê 2 2 2ˆ 2 2 2 . Then we have v = Á F ( D + h ) - F ( x + h ) ˜ . mË ¯ 2

1

Ê 2F Ê D When x = , v = -Á D2 + h 2 Á 3 Ë m Ë

8.

P = Fv, m

ˆˆ 2 D 2 + h ˜˜ . 9 ¯¯ 2

dv P v3 P dv P P dv = F = fi mv = fi v2 = fi = x +C dx m dx v v dt 3 m

v3 P v3 P v3 P v13 P = x1 + C fi C = 1 - x1, 2 = x 2 + 1 - x1. Therefore, 3 m 3 m 3 m 3 m

m Ê v 23 v13 ˆ m 3 3000 v 2 - v13 ), m = x 2 - x1 = Á - ˜ = , P = 200(550) ft ◊ lb / sec ( P Ë 3 3 ¯ 3P 32 v 2 = 50

5280 2 3000 1 1 Ê 50(5280) ˆ ◊ ◊ ft / sec, v1 = v 2 fi Dx = Á ˜ 3600 5 32 3 200(550) Ë 3600 ¯

fi Dx = 112.04 (.936) ª 104.87 ft.

9 (a).

mv

dv + k 0 xv 2 = 0, v = v 0 when x = 0 . dx

3

Ê Ê 2ˆ 3ˆ Á1 - ÁË 3˜¯ ˜ Ë ¯

Chapter 3 First Order Nonlinear Differential Equations • 47

2

9 (b).

Ê k 0x dv k 0 + xv = 0 fi Á e 2 m dx m Ë

k x ˆ¢ - 0 v˜ = 0 fi v = v 0e 2 m . Setting x = d and v = 0.01v 0 , we have ¯ 2

k 0d 2 2m

2m k 0d 2 0.01v 0 = v 0e fi = ln100 . Solving for k 0 yields k 0 = 2 ln100 . d 2m GM dv k dv GmM e 10 (a). mv =+ kv 2 fi = v - 2 e v -1, v = 0 when r = Re + h . 2 r dr m dr r 1 dv 1 - 12 du k 12 GM e - 12 10 (b). Bernoulli equation: 1 - n = -1 fi n = 2, u = v 2 fi v = u 2 fi = u = u - 2 u r dr m dr 2 du 2k 2GM e . Therefore, ufi = r2 dr m -

(e u) - 2mk r

= 0-e

¢

= 2GM e

- 2mk ( R e )

u r = Re

e

- 2mk r

r2

fie

- 2GM e Ú

- 2mk ( R e + h )

u r = Re + h

Re + h

e

Re

-e

- 2mk ( R e )

- 2GM e Ú

u

Re + h

Re

r = Re

e

- 2mk r

r2

dr

- 2mk r

r2

dr . 1 2k

- r È ˘2 Re + h e m k dr (R ) Since u = v 2 , v = dr < 0, v impact = -e m e Í2GM e Ú ˙ . Let Re r2 dt ÍÎ ˙˚ 1

r = Re + s. Then v impact

11.

m

- 2k s ˘2 È h e m = - Í2GM e Ú ds˙ . 0 (Re + s)2 ˙˚ ÍÎ

dv GM e dv GM e m v2 G when . Thus v v v =fi = , = = M e + C , and from our r = R 0 e dr r2 dt r2 r 2

initial condition,

Ê1 1 ˆ v 2 v02 v02 G Me + C fi = = + GM e Á - ˜ . Since v = 0 when r = Re + h , Re 2 2 2 Ë r Re ¯

Ê 1 1 ˆ v 0 2 = 2GM e Á ˜ . Thus Ë Re Re + h ¯ 1

1

24 -11 È Ê 1 1 ˆ ˘ 2 È 2(6.673)(10 )(5.976)(10 ) Ê 1 1 ˆ ˘2 v 0 = Í2GM e Á Á ˜ ˙ ª 2044 m/sec. ˜˙ = Í Ë 6.371 6.591¯ ˙ 10 6 Ë Re Re + h ¯ ˚ Î ÎÍ ˚

12 (a). ml 2q ¢¢ = - mgl sin q = ml 2

ml 2w

dw dw fi ml 2w = - mgl sin q dq dt

dw = - mgl sin q and w = -w 0 when q = p . dq

2 2 2 w2 2 w0 2w 2 w0 12 (b). ml = mgl cosq + C, ml = - mgl + C fi ml - mgl cosq = ml + mgl . 2 2 2 2 2

48 • Chapter 3 First Order Nonlinear Differential Equations

12 (c). When q = 0, ml 2 fi w = w 02 +

13.

ml 2

w2 w2 4g Ê 2 ˆ - mgl = ml 2 0 + mgl fi w 2 = w 02 + 2 mglÁ 2 ˜ = w 02 + Ë ml ¯ 2 2 l

4g . l

w2 w2 = mgl cosq + C, w = w 0 when q = 0 . Therefore, C = ml 2 0 - mgl , and so 2 2

2 3p w2 2 w0 , so ml = mgl cosq + ml - mgl . We know that w = 0 when q = 4 2 2 2

-

mgl ml 2w 0 2 2 1 ˆ g Ê + - mgl = 0 fi w 0 2 = ˜ = 2 + 2 . Thus 2 mgl Á1 + Ë 2 ml 2¯ l 2

w0 =

(

g 2 + 2 = 16 2 + 2 ª 7.391 rad/sec. l

(

)

(

)

Section 3.8 Note: for exercises 1-5, h=0.1 1 (a).

y = t 2 - t + C, y (1) = C = 0 fi y = t 2 - t

1 (b).

y k +1 = y k + h (2 tk - 1)

1 (c).

y1 = 0.1, y 2 = 0.22, y 3 = 0.36

1 (d).

y (1.1) = 0.11, y (1.2) = 0.24, y (1.3) = 0.39

2 (a).

y = Ce - t , y (0) = C = 1 fi y = e - t

2 (b).

y k +1 = y k - hy k

2 (c).

y1 = 0.9, y 2 = 0.81, y 3 = 0.729

2 (d).

y (0.1) = 0.90484, y (0.2) = 0.81873, y (0.3) = 0.74082 -

t2 2

, y (0) = C = 1 fi y = e

-

t2 2

3 (a).

y = Ce

3 (b).

y k +1 = y k - h ( tk y k )

3 (c).

y1 = 1, y 2 = 0.99, y 3 = 0.9702

3 (d).

y (0.1) = 0.99501, y (0.2) = 0.98020, y (0.3) = 0.955997

4 (a).

y = Ce - t + t - 1, y (0) = C - 1 = 0 fi y = e - t + t - 1

4 (b).

y k +1 = y k + h (- y k + tk )

4 (c).

y1 = 0, y 2 = 0.01, y 3 = 0.029

4 (d).

y (0.1) = 0.0048374, y (0.2) = 0.01873075, y (0.3) = 0.040818

)

Chapter 3 First Order Nonlinear Differential Equations • 49

1 1- t

5 (a).

y -2 y ¢ = 1, - y -1 = t + C, C = -1 fi y =

5 (b).

y k +1 = y k + h y k 2

5 (c).

y1 = 1.1, y 2 = 1.221, y 3 = 1.3700841

5 (d).

y (0.1) = 1.111111, y (0.2) = 1.25, y (0.3) = 1.4285714

6.

y k +1 = y k + 0.1(a tk + b ) . From k=0, t0 = 0, y 0 = -1 .

( )

For k = 0, y1 = y 0 + 0.1(a t0 + b ) fi -0.9 = -1 + .1(0 + b ) fi 0.1 = .1b fi b = 1. For k = 1, y 2 = y1 + 0.1(a t1 + b ) fi -0.81 = -0.9 + .1(a (0.1) + 1) fi -0.01 = .01a fi a = -1. 7.

(

) For k = 0, y = y + .1( y For k = 1, y = y + .1( y

y k +1 = y k + 0.1 y k n + a . From k=0, t0 = 1, y 0 = 1. n

1

0

0

2

1

n 1

)

+ a fi 0.9 = 1 + .1(1n + a ) fi (1n + a ) = -1 fi a = -2 .

)

- 2 fi 0.781 = 0.9 + .1(.9 n - 2) fi (.9 n - 2) = -1.19

fi .9 n = .81 fi n = 2 .

8 (a). (i)Euler’s method will underestimate the exact solution. (ii)Euler’s method will overestimate the exact solution. (iii)Euler’s method will underestimate the exact solution. (iv)Euler’s method will overestimate the exact solution. 8 (b). Exercise 2: decreasing, concave up, underestimates Exercise 3: decreasing, concave down, overestimates Exercise 5: increasing, concave up, underestimates 8 (c). Euler’s method should initially underestimate (when solution curves are concave up) and then tend to “catch up” (when solution curves become concave down). 9.

y k +1 = y k + h ( tk y k + sin(2 ptk )), y 0 = 1, h = 0.01, k = 0,1,..., 99 .

10.

V (0) = 90, V ( t) = 90 + 5 t, V (T ) = 100 when T = 2 fi 0 £ t £ 2

Q dQ = 6(2 - cos(p t)) - 1 ◊ , Q(0) = 0 dt 90 + 5 t È Qk ˘ Qk +1 = Qk + h Í6(2 - cos(ptk )) , Q = 0, h = 0.01, k = 0,1, 2,,...,199 90 + 5 tk ˙˚ 0 Î Result: Q(2) = 23.7556...lb. 11.

È Ê 1 ˆ ˘ 1 Ê Pˆ P ¢ = 0.1Á1 - ˜ P + e - t , P (0) = . Pk +1 = Pk + h Í0.1Á1 - Pk ˜ Pk + e - t k ˙, P0 = 0.5 . With Ë 2 3¯ Î Ë 3 ¯ ˚ h = 0.01, k = 0,1,...,199, tk = 0.01k , P(2) = 1.502477 million.

50 • Chapter 3 First Order Nonlinear Differential Equations

12 (a). y = Ce t - 1, C = 1 fi y = e t - 1. 12 (b). y k +1 = y k + h ( y k + 1), y 0 = 0 . For y k(1), h = 0.02, k = 0,1,...49 For y k( 2 ), h = 0.01, k = 0,1,...99 . 13 (a). y ¢ - ly = 0, (e -lt y )¢ = C, y = Celt , y (0) = C = y 0 . Thus y = elt y 0 . 13 (b). y k +1 = y k + hly k = (1 + lh ) y k . Therefore 2

n

y1 = (1 + lh ) y 0 , y 2 = (1 + lh ) y1 = (1 + lh ) y 0 , y n = (1 + lh ) y 0 , n

n

aˆ Ê lt ˆ Ê 13 (c). y n = Á1 + ˜ y 0 . Since lim Á1 + ˜ = e a , the result follows. n Æ• Ë Ë n¯ n¯ 14 (a). y -2 y ¢ = 1, - y -1 = t + C, C = -1, y =

1 , -•< t <1 1- t

14 (b). y k +1 = y k + hy k2 , y 0 = 1, h = 0.1, k = 0,1,...11 14 (c). Numerical solution becomes worse as tk ≠ 1. The numerical solution gives the mistaken impression that the interval of existence extends to t ≥ 1.

Chapter 4 Second Order Linear Differential Equations Introduction 1 (a). The block’s weight is W b = p(0.5) 2 (2)(50) = 25 p lb. Therefore, we have

25 p = p(0.5) 2 (Y )(62.4 ) . Solving for Y yields Y = 1.60256 feet. 1 (b).

rl g W g 62.4 32 1 2 = = = 19.968 . Therefore, w = 4.469 , and thus y 0 = - , y© 0 = 0 , and w = rL W b L 50 2 4 1 our equation for y ( t) is y ( t) = - cos( 4.469 t) . 4

1 (c). The maximum depth to which the block sinks would be Y + y 0 = 1.603 + .0.25 = 1.853 feet. 2.

y ( t) =

y 0¢ y¢ sin wt. Y + 0 = 2 fi y 0¢ = w (2 - Y ) ª 1.77596 ft. w w

3 (a). We know that the frequency of oscillation is given by w 2 =

rl g . We can see that the density rL

for the second drum is greater than that of the first. Since all the other variables determining w would be the same for both drums, r2 > r1 means that w 2 < w1 , which in turn means that T2 > T1 . Thus the first drum bobs more rapidly.

3 (b). By the same rationale as in the first part, L2 > L1 means that w 2 < w1 , which in turn means that T2 > T1 . Thus the first drum bobs more rapidly.

4 (a).

y 0 = 0, T = 2

4 (b).

y 0¢ =

1 1 1 ft. y ( t) = sin w t, with w ◊ 2 = 2p fi w = p . \ y ( t) = sin p t, 2 2 2 2

4 (c). w 2 = p 2 =

W g Wl g 45p 3 Ê 3ˆ ft fi W b = 2l ; V = p Á ˜ ◊ 5 = Ë 2¯ Wb L 4 w L

9 Ê 45p ˆ 1 32 W b = 62.4 Á = 62.4 ◊ ◊ 8 ª 1430.1 lb. ˜ 2 Ë 4 ¯p 5 p

5 (a). Using the model provided, my ¢¢ = mg - rlVg . We can rewrite this equation with m in terms of

V and r : rVy ¢¢ = rVg - rlVg . Simplifying this equation and solving for y ¢¢ , we have Ê r - rl ˆ y ¢¢ = Á ˜ g . We need not restrict this equation to the motion of cylindrical objects. Ë r ¯

52 • Chapter 4 Second Order Linear Differential Equations

5 (b). Using these initial conditions, antidifferentiation yields the general solution

Ê r - rl ˆ t 2 y=Á ˜ g + y 0¢ t + y 0 . Ë r ¯ 2 6.

r - rl 30 - 62.4 = = -.51923 . r 62.4 \ 0 = -.51923 ◊ 32 ◊

t2 + 0 + 99 fi 8.30769 t 2 = 99 fi t = 11.9166... sec. 2

Section 4.1 1.

All the relevant functions are continuous everywhere, so Theorem 4.1 guarantees a unique solution for the interval (-•, •) .

p 3p .
2.

t0 = p . Since g( t) = tan t,

3.

Dividing the equation by e t yields the functions p( t) = 0 , q( t) =

1 t

2

e ( t - 1)

, and g( t) =

4 . te t

These functions are discontinuous at the points ±1 and 0 , and since t0 = -2 , the largest tinterval on which Theorem 4.1 guarantees a unique solution is (-•, -1) .

sin 2 t 2 , q( t) = , 0 < t < 3. 2 t t( t - 9)

4.

p( t) =

5.

y ¢¢ + y = 0 , y ( t0 ) = y 0 , y ¢ ( t0 ) = y 0¢ for any t0 , y 0 , y 0¢ .

6.

y ¢¢ +

1 y = 0, y ( t0 ) = y 0 , y ¢ ( t0 ) = y 0¢ , t0 > 3, any y 0 , y 0¢ (not both 0) . t- 3

7.

y ¢¢ +

y¢ y + = 0 , -1 < t0 < 5, y ( t0 ) = y 0 , y ¢ ( t0 ) = y 0¢ . y 0 and y 0¢ cannot both be zero. t - 5 t +1

8 (a).

y ¢¢ -

y¢ y + = 0, t0 = 1. 0 < t < •. t t2

8 (b).

t 2 (0) - t(1) + t = 0 .

8 (c). No. 9.

Theorem 4.1 guarantees a unique solution on the interval (-4, 4 ) , so it is not possible for the limit to hold.

10.

Theorem 4.1 guarantees a unique solution on the interval (-•, 3) , so it is not possible for the limit to hold.

Chapter 4 Second Order Linear Differential Equations • 53

11 (a). B 11 (b). D 11 (c). A 11 (d). C

Section 4.2 1 (a).

y1¢¢ = 4 e 2 t = 4 y1 and y 2¢¢ = 8e -2 t = 4 y 2 , so both equations are solutions.

1 (b). W =

e 2t 2e 2 t

2e -2 t = -8 π 0 , so yes, the functions do form a fundamental set of solutions. -4 e -2 t

1 (c). The general solution would be y = c1e 2 t + c 2 ◊ 2e -2 t . Differentiating yields y ¢ = 2c1e 2 t - 4 c 2e -2 t . y (0) = c1 + 2c 2 = 1 and y ¢ (0) = 2c1 - 4 c 2 = -2 , and solving these two simultaneous equations

1 yields c1 = 0 and c 2 = . Thus the unique solution for this initial value problem is y ( t) = e -2 t . 2 2 (a).

y1¢¢ = 2e t = y1 and y 2¢¢ = e - t + 3 = y 2 , so both equations are solutions.

2e t 2 (b). W = t 2e

e-t +3 = -4 e 3 π 0 , so yes, the functions do form a fundamental set of solutions. -t +3 -e

2 (c). The general solution would be y = c1 ◊ 2e t + c 2 ◊ e - t + 3 . Differentiating yields y ¢ = 2c1e t - c 2e - t + 3 .

y (-1) = 2c1e -1 + c 2e 4 = 1 and y ¢ (-1) = 2c1e -1 - c 2e 4 = 0 , and solving these two simultaneous equations yields c1 = y ( t) = 3 (a).

e 1 and c 2 = 4 . Thus the unique solution for this initial value problem is 4 2e

e t +1 e - t -1 . + 2 2

y1 is not a solution. y 2 is a solution.

4 (a). Both equations are solutions. 4 (b). W =

cos t sin t = 1 π 0 , so yes, the functions do form a fundamental set of solutions. - sin t cos t

4 (c). The general solution would be y = c1 cos t + c 2 sin t . Differentiating yields Êp ˆ Êp ˆ y ¢ = -c1 sin t + c 2 cos t . y Á ˜ = c 2 = 1 and y ¢ Á ˜ = -c1 = 1 fi c1 = -1. Thus the unique solution Ë 2¯ Ë 2¯

for this initial value problem is y ( t) = - cos t + sin t .

54 • Chapter 4 Second Order Linear Differential Equations

5 (a).

y1¢ = 2e 2 t = 2 y1 and y1¢¢ = 4 e 2 t = 4 y1 . Thus y1¢¢ - 4 y1¢ + 4 y1 = 4 y1 - 8 y1 + 4 y1 = 0 . y 2¢ = 2 te 2 t + e 2 t and y 2¢¢ = 4 te 2 t + 2e 2 t + 2e 2 t = 4 te 2 t + 4 e 2 t . Thus y 2¢¢ - 4 y 2¢ + 4 y 2 = 4 te 2 t + 4 e 2 t - 8 te 2 t - 4 e 2 t + 4 te 2 t = 0 , and so both equations are solutions.

e 2t 5 (b). W = 2 t 2e

te 2 t = e 4 t π 0 , so yes, the functions do form a fundamental set of solutions. 2t (2 t + 1)e

5 (c). The general solution would be y = c1e 2 t + c 2 te 2 t . Differentiating yields

y ¢ = 2c1e 2 t + 2c 2 te 2 t + c 2e 2 t . y (0) = c1 = 2 and y ¢ (0) = 2c1 + c 2 = 0 fi c 2 = -4 . Thus the unique solution for this initial value problem is y ( t) = 2e 2 t - 4 te 2 t . 6 (a). Both equations are solutions. 1 6 (b). W = 0

t

e2 t = 12 e 2 π 0 , so yes, the functions do form a fundamental set of solutions. 1 t2 2e t

t

6 (c). The general solution would be y = c1 + c 2e 2 . Differentiating yields y ¢ = 12 c 2e 2 .

y (2) = c1 + c 2e = 0 and y ¢ (2) = 12 c 2e = 2 fi c 2 = 4e -1 and c1 = -4 . Thus the unique solution for t -2

this initial value problem is y ( t) = -4 + 4 e 2 . 7 (a).

1 Êt y1¢ = cosÁ + 2 Ë2

pˆ 1 Êt ˜ and y1¢¢ = - sinÁ + 3¯ 4 Ë2

pˆ Êt ˜ . Thus 4 y1¢¢ + y1 = - sinÁ + Ë2 3¯

pˆ Êt ˜ + sinÁ + Ë2 3¯

pˆ ˜ = 0. 3¯

1 Ê t pˆ 1 Ê t pˆ Ê t pˆ Ê t pˆ y 2¢ = cosÁ - ˜ and y 2¢¢ = - sinÁ - ˜ . Thus 4 y 2¢¢ + y 2 = - sinÁ - ˜ + sinÁ - ˜ = 0 , Ë 2 3¯ Ë 2 3¯ 2 Ë 2 3¯ 4 Ë 2 3¯

and so both equations are solutions.

Ê t pˆ sinÁ + ˜ Ë 2 3¯ 7 (b). W = 1 Ê t pˆ cosÁ + ˜ 2 Ë 2 3¯

=

Ê t pˆ sinÁ - ˜ Ë 2 3¯ 1 Ê t pˆ cosÁ - ˜ 2 Ë 2 3¯

1 È Ê t pˆ Ê t pˆ ˘ Ê t pˆ Ê t pˆ sinÁ + ˜ cosÁ - ˜ - sinÁ - ˜ cosÁ + ˜ ˙ Í Ë 2 3¯ ˚ Ë 2 3¯ Ë 2 3¯ 2 Î Ë 2 3¯

1 Êt p t = sinÁ + - + 2 Ë2 3 2

p ˆ 1 Ê 2p ˆ ˜ = sinÁ ˜ π 0 , so yes, the two equations do form a fundamental set 3¯ 2 Ë 3 ¯

of solutions. Ê t pˆ Ê t pˆ 7 (c). The general solution would be y = c1 sinÁ + ˜ + c 2 sinÁ - ˜ . Differentiating yields Ë 2 3¯ Ë 2 3¯ 1 Êt y ¢ = c1 cosÁ + Ë2 2

pˆ 1 Êt ˜ + c 2 cosÁ Ë2 ¯ 3 2

pˆ ˜ . 3¯

Chapter 4 Second Order Linear Differential Equations • 55

Ê pˆ Ê pˆ Using the first initial condition, y (0) = c1 sinÁ ˜ + c 2 sinÁ - ˜ = 0 , and so c1 = c 2 . From the Ë 3¯ Ë 3¯ 1 1 Ê pˆ 1 Ê pˆ 1 second initial condition, y ¢ (0) = c1 cosÁ ˜ + c 2 cosÁ - ˜ = c1 + c 2 = 1 . Thus c1 = c 2 = 2 , Ë 3¯ 4 Ë 3¯ 2 2 4 Ê t pˆ Ê tˆ Ê pˆ Ê tˆ Ê t pˆ and the unique solution is y = 2 sinÁ + ˜ + 2 sinÁ - ˜ = 4 sinÁ ˜ cosÁ ˜ = 2 sinÁ ˜ . Ë 2 3¯ Ë 2¯ Ë 3¯ Ë 2¯ Ë 2 3¯

8 (a). Both equations are solutions. 8 (b). W =

2e t 2e t

e 2t = 2e 3 t π 0 , so yes, the functions do form a fundamental set of solutions. 2t 2e

8 (c). The general solution would be y = c1 ◊ 2e t + c 2 ◊ e 2 t . Differentiating yields y ¢ = 2c1e t + 2c 2e 2 t .

y (-1) = 2c1e -1 + c 2e -2 = 1 and y ¢ (-1) = 2c1e -1 + 2c 2e -2 = 0 , and solving these two simultaneous equations yields c1 = e and c 2 = -e 2 . Thus the unique solution for this initial value problem is

y ( t) = 2e t +1 - e 2( t +1) . 9 (a).

1 1 1 1 y1¢ = , y1¢¢ = - 2 , y 2¢ = , and y 2¢¢ = - 2 . Thus ty1¢¢ + y1¢ = ty 2¢¢ + y 2¢ = 0 , and so both t t t t equations are solutions.

ln t ln 3t 1 1 = (ln t - ln 3t) = t -1 (ln t - ln 3 - ln t) = - t -1 ln 3 π 0 on (0, •) , so yes, the two 9 (b). W = 1 t t t equations do form a fundamental set of solutions. 9 (c). The general solution would be y = c1 ln t + c 2 ln 3t = (c1 + c 2 )ln t + c 2 ln 3 . Differentiating yields

y¢ =

c1 + c 2 . From the first initial condition, y ( 3) = (c1 + c 2 )ln 3 + c 2 ln 3 = 0, so c1 + 2c 2 = 0 . t

From the second initial condition, y ¢ ( 3) =

c1 + c 2 = 3 . Thus c1 = 18 and c 2 = -9 , and so the 3

unique solution is y = 18 ln t - 9 ln 3t, 0 < t < • . 10 (a). Both equations are solutions. 10 (b). W =

ln t ln 3 - ln 3 = π 0 , so yes, the functions do form a fundamental set of solutions. 1 0 t t

10 (c). The general solution would be y = c1 ◊ ln t + c 2 ◊ ln 3 . Differentiating yields y ¢ =

c1 + 0. t

y (1) = 0 + c 2 ln 3 = 0 and y ¢ (1) = c1 = 3, and solving these two simultaneous equations yields c1 = 3 and c 2 = 0 . Thus the unique solution for this initial value problem is y ( t) = 3 ln t, 0 < t < • .

56 • Chapter 4 Second Order Linear Differential Equations

11 (a). y1¢ = 3t 2 , y1¢¢ = 6 t, y 2¢ = t -2 , and y 2¢¢ = -2 t -3 . Thus t 2 y1¢¢ - ty1¢ - 3 y1 = t 2 y 2¢¢ - ty 2¢ - 3 y 2 = 0 , and so both equations are solutions. 11 (b). W =

t3 3t 2

- t -1 = t + 3t = 4 t π 0 on (-•, 0) , so yes, the two equations do form a fundamental t -2

set of solutions. 11 (c). The general solution would be y = c1t 3 - c 2 t -1 . Differentiating yields y ¢ = 3c1t 2 + c 2 t -2 . From the first initial condition, y (-1) = -c1 + c 2 = 0 . From the second initial condition,

1 y ¢ (-1) = 3c1 + c 2 = -2 . Thus c1 = c 2 = - , and so the unique solution is 2 1 1 y = - t 3 + t -1, - • < t < 0 . 2 2 12 (a). Both equations are solutions. 12 (b). W =

e-t -e - t

2e1- t = 0 , Therefore, the Wronskian calculation does not establish that y1 and y 2 -2e1- t

form a fundamental set. 13 (a). y1¢ = 1, y1¢¢ = 0, y 2¢ = -1, and y 2¢¢ = 0 . Thus both equations are solutions. 13 (b). W =

t + 1 -t + 2 = - t - 1 + t - 2 = -3 π 0 , so yes, the two equations do form a fundamental set 1 -1

of solutions. 13 (c). The general solution would be y = c1 ( t + 1) + c 2 (2 - t) . Differentiating yields y ¢ = c1 - c 2 . From the first initial condition, y (1) = 2c1 + c 2 = 4 . From the second initial condition, y ¢ (1) = c1 - c 2 = -1 . Thus c1 = 1 and c 2 = 2 , and so the unique solution is y = t + 1 + 4 - 2t = - t + 5 .

14 (a). Both equations are solutions. 14 (b). W =

sin p t + cosp t sin p t - cosp t p cosp t - p sin p t p cosp t + p sin p t

= 2p sin p t cosp t + p (sin 2 p t + cos2 p t) - 2p sin p t cosp t + p (sin 2 p t + cos2 p t) , = 2p π 0 so yes, the functions do form a fundamental set of solutions. 14 (c). The general solution would be y = c1 (sin p t + cosp t) + c 2 (sin p t - cosp t) . Differentiating yields y ¢ = pc1 (cosp t - sin p t) + pc 2 (cosp t + sin p t) .

Ê 1ˆ Ê 1ˆ y Á ˜ = c1 + c 2 = 1 and y ¢ Á ˜ = -pc1 + pc 2 = 0 fi c1 = c 2 = 12 . Ë 2¯ Ë 2¯

Chapter 4 Second Order Linear Differential Equations • 57

Thus the unique solution for this initial value problem is

y = 12 (sin p t + cosp t) + 12 (sin p t - cosp t) = sin p t . 15 (a). y1 is not a solution. y 2 is not a solution. 1 1 Êp ˆ Êp ˆ 16 (b). y = sin(2 t) cosÁ ˜ + cos(2 t) sinÁ ˜ = (2 cos 2t) + (sin 2 t) . Ë 4¯ Ë 4¯ 2 2 2

Thus c1 =

1 2 2

1 . 2

and c 2 =

1 17 (a). y ¢ = 2 + 1 + ln 3t = ln 3t + 3, y ¢¢ = . Thus t 2 y ¢¢ - ty ¢ + y = t - 3t - t ln 3t + 2 t + t ln 3t = 0 . t 17(b). y = 2 t + t(ln t + ln 3) = t(2 + ln 3) + t ln t . Thus c1 = 2 + ln 3 and c 2 = 1. 18 (b). y = 2 cosh( 2t ) = e 2 + e t

-t

(

= 1 ◊ e - 2 + (- 12 ) ◊ -2e t

2

t

).

2

Thus c1 = 1 and c 2 = - 12 . 19.

Substituting y1 into the equation yields 9e 3 t + 3ae 3 t + be 3 t = 0; 3a + b = -9 . Substituting y 2 into the equation yields 9e -3 t - 3ae -3 t + be -3 t = 0; - 3a + b = -9 . Solving the two simultaneous equations gives us a = 0 and b = -9 .

20 (a). y1 ( t) = t, y1¢( t) = 1, y1¢¢( t) = 0, y 2 ( t) = e t , y 2¢ ( t) = y 2¢¢( t) = e t .

0 + p( t) ◊ 1 + q( t) ◊ t = 0, e t + p( t) ◊ e t + q( t) ◊ e t = 0 p + qt = 0, e t (1 + p + q) = 0 fi p + q = -1 \ ( t - 1)q = 1 fi q =

1 -t , p= t -1 t -1

20 (b). Both p and q continuous on (-•,1) and (1, •) . t et 20 (c). W = = ( t - 1)e t t 1 e

W π 0 on (-•,1) and (1, •)

20 (d). Yes, W π 0 on the two intervals on which p and q are both continuous. t

2

21.

sds - sds From Abel’s Theorem, we have W ( t) = W (1)e Ú1 fi W (2) = 4 e Ú1 = 4 e -( 4 -1) 2 = 4 e - 3 2 .

22.

Substitute, 4 e 2 t + 2ae 2 t + be 2 t = 0 fi 2a + b = -4 .

-

Also, W = e - t fi p( t) = a = 1. \ a = 1, b = -6 ( y ¢¢ + y ¢ - 6 y = 0) . 23. 24.

p( t) = 0 . From Abel’s Theorem, W ( t) = W ( t0 ), which is a constant. Therefore, W ( 4 ) = -3.

y ¢¢ + py ¢ + 3 y = 0, W = W ( t0 )e

\ p( t) =

d 2 (t ) = 2t . dt

-

t

Út0 pds

2

= e-t .

58 • Chapter 4 Second Order Linear Differential Equations

Section 4.3 1 (a). W (1) =

2 -1 2 -1

= 0 , so the two solutions do not form a fundamental set.

1 1 (b). Since - y1 and y 2 satisfy identical initial conditions, we can conclude from Theorem 4.1 that 2 1 1 - y1 ( t) ∫ y 2 ( t) . Therefore, y1 ( t) + y 2 ( t) = 0, - • < t < • . 2 2 2 (a). W (-2) =

1 0 2 1

= 1 π 0 , so the two solutions do form a fundamental set.

2 (b). Yes (Theorem 4.7) 3 (a). W (0) =

0 -1 = 1 π 0 , so the two solutions do form a fundamental set. 1 0

3 (b). From Theorem 4.7, the two solutions do form a linearly independent set of functions on

- • < t < •. 4 (a). W ( 3) =

0 1 = 0 , so the two solutions do not form a fundamental set. 0 2

4 (b). By Theorem 4.1, y1 ( t) ∫ 0 . Therefore, 1 ◊ y1 ( t) + 0 ◊ y 2 ( t) = 0 and y1 ( t), y 2 ( t) are linearly dependent. 5 (a).

y1¢¢ - 4 y1 = 4 e 2 t - 4 e 2 t = 0, y 2¢¢ - 4 y 2 = 4 e -2 t - 4 e -2 t = 0 , so yes, both are solutions to the

differential equation. 5 (b).

y1 (1) = e 2 , y1¢ (1) = 2e 2 , y 2 (1) = e -2 , y 2¢ (1) = -2e -2 .

e2 5 (c). W (1) = 2 2e

e -2 = -4 π 0 , so the solutions do form a fundamental set. -2e -2

(

)-e

= 0, 4 y 2¢¢ - y 2 = 4

(

) - (-2e ) = 0 ,

6 (a).

4 y1¢¢ - y1 = 4 14 e

6 (b).

y1 (-2) = e -1, y1¢ (-2) = 12 e -1, y 2 (-2) = -2e, y 2¢ (-2) = e .

e -1 6 (c). W (-2) = e -1 2 7 (a).

t

2

t

2

-1 2

e

-t

2

-t

2

-2e e

= 2 π 0 , so the solutions do form a fundamental set.

y1¢¢ + 9 y1 = -9 sin( 3( t - 1)) + 9 sin( 3( t - 1)) = 0, y 2¢¢ + 9 y 2 = -9(2 cos( 3( t - 1))) + 9(2 cos( 3( t - 1))) = 0 , so yes, both are solutions to the differential equation.

Chapter 4 Second Order Linear Differential Equations • 59

7 (b).

y1 (1) = 0, y1¢ (1) = 3, y 2 (1) = 2, y 2¢ (1) = 0 .

7 (c). W (1) = 8 (a).

0 2 3 0

= -6 π 0 , so the solutions do form a fundamental set.

y1¢ = e -2 t (-2 cos t - sin t), y1¢¢= e -2 t ( 3 cos t + 4 sin t) y1¢¢ + 4 y1¢ + 5 y1 = e -2 t ( 3 cos t + 4 sin t - 8 cos t - 4 sin t + 5 cos t) = 0

y 2¢ = e -2 t (-2 sin t + cos t), y 2¢¢ = e -2 t ( 3 sin t - 4 cos t) y 2¢¢ + 4 y 2¢ + 5 y 2 = e -2 t ( 3 sin t - 4 cos t - 8 sin t + 4 cos t + 5 sin t) = 0

8 (b).

y1 (0) = 1, y1¢ (0) = -2, y 2 (0) = 0, y 2¢ (0) = 1.

8 (c). W (0) = 9 (a).

1 0 = 1 π 0 , so the solutions do form a fundamental set. -2 1

y1¢¢ + 2 y1¢ - 3 y1 = 9e -3 t - 6e -3 t - 3e -3 t = 0, y 2¢¢ + 2 y 2¢ - 3 y 2 = 9e -3( t - 2 ) - 6e -3( t - 2 ) - 3e -3( t - 2 ) = 0 , so yes, both are solutions to the differential

equation. 9 (b).

y1 (2) = e -6 , y1¢ (2) = -3e -6 , y 2 (2) = 1, y 2¢ (2) = -3.

9 (c). W (2) =

e -6 -3e -6

1 = 0 , so the solutions do not form a fundamental set. -3

10 (a). y1¢¢ - 6 y1¢ + 9 y1 = e 3( t + 2 ) (9 - 18 + 9) = 0 y 2¢¢ - 6 y 2¢ + 9 y 2 = e 3( t + 2 ) (9 t + 6 - 6( 3t + 1) + 9 t)) = 0

10 (b). y1 (-2) = 1, y1¢ (-2) = 3, y 2 (-2) = -2, y 2¢ (-2) = -5 . 10 (c). W (-2) =

1 -2 3 -5

= 1 π 0 , so the solutions do form a fundamental set.

11.

[y

y 2 ] = [ y1

È 2 1˘ 2 1 y 2 ]Í = 3 π 0 , so {y1 ˙; Î-1 1˚ -1 1

y 2 } is a fundamental set.

12.

[y

y 2 ] = [ y1

È2 1˘ 2 1 y 2 ]Í = 0 , so {y1 ˙; Î-2 -1˚ -2 -1

y 2 } is not a fundamental set.

13.

[y

y 2 ] = [ y1

È0 2 ˘ 0 2 y 2 ]Í = -2 π 0 , so {y1 ˙; Î1 -1˚ 1 -1

14.

The set is linearly independent since one function is not a constant multiple of the other.

15.

1

1

1

y 2 } is a fundamental set.

f 2 = ln t 2 = 2 ln t = 2 f1 , so the set is linearly dependent (2 f1 - f 2 = 0) .

60 • Chapter 4 Second Order Linear Differential Equations

16.

The set is linearly independent since one function is not a constant multiple of the other.

17.

The set is linearly independent since one function is not a constant multiple of the other.

18.

Set k1 ◊ 2 + k2 ◊ t + k3 ◊ (- t 2 ) = 0 and evaluate at t = -1, 0,1.

È2 -1 -1˘ È k1 ˘ È0 ˘ 2 -1 -1 -1 -1 ˙Í ˙ Í ˙ Í . k = 2 0 0 0 2 0 0 = 2 = -4 π 0 . 2 ˙Í ˙ Í ˙ Í 1 -1 ÍÎ2 1 -1˙˚ ÍÎk3 ˙˚ ÍÎ0 ˙˚ 2 1 -1 Therefore, k1 = k2 = k3 = 0 and the set is linearly independent. 19.

f1 = 2, f 2 = sin 2 t, f 3 = 2 cos2 t . Therefore, 2 f 2 + f 3 - f1 = 0 on - 3 < t < 2 , so the set is linearly dependent.

20.

f1 = e t , f 2 = 2e - t , f 3 = sinh t . Therefore,

1 2

f1 -

1 4

f 2 - f 3 = 0 , so the set is linearly dependent.

21 (a). f1 = t, f 2 = 2 t = 2 f1, so the functions form a linearly dependent set. 21 (b). f1 = t, f 2 = - t = - f1, so the functions form a linearly dependent set. 21 (c). f1 = t, f 2 = t - 1. These functions form a linearly independent set since one function is not a constant multiple of the other. 22 (a). If f1 = cf 2 , then f1 - cf 2 = 0 . Therefore, the set is linearly dependent. 22 (b). Assume { f1, f 2 } is a linearly dependent set. Then k1 f1 ( t) + k2 f 2 ( t) = 0 with k1 and k2 not both zero. Assume that k1 π 0 . Then f1 ( t) = -

( ) f (t) on k2 k1

2

the domain. 23 (a). If f 2 = 3 f1 - 2 f 3 , then 3 f1 - f 2 - 2 f 3 = 0 on the domain. Therefore, the set is linearly dependent. 23 (b). Assume { f1, f 2 , f 3 } is a linearly dependent set. Then k1 f1 ( t) + k2 f 2 ( t) + k3 f 3 ( t) = 0 with k1, k2 , k3 not all zero. Assume, without loss of generality, that k1 π 0 . Then f1 ( t) = -

24.

( ) f (t) - ( ) f (t). k2 k1

2

k3 k1

3

Any set of functions containing the zero function is linearly dependent. Consider {0, f 2 , f 3 ,..., f n }. Then 1 ◊ 0 + 0 ◊ f 2 + 0 ◊ f 3 + ... + 0 ◊ f n = 0.

25.

Suppose that f 3 = a1 f1 + a2 f 2 and f 3 = b1 f1 + b2 f 2 . Then ( a1 - b1 ) f1 + ( a2 - b2 ) f 2 = 0 . Since the functions are linearly independent, a1 - b1 = a2 - b2 = 0; a1 = b1, a2 = b2 .

26.

On 0 < t < •, f 2 = t = t = f1 . \ f1 - f 2 = 0 and { f1, f 2 } is linearly dependent. On -• < t < •, let k1 f1 + k2 f 2 = k1t + k2 t = 0 and evaluate at t = ±1. Then k1 + k2 = 0, - k1 + k2 = 0 fi k1 = k2 = 0 and this is a linearly independent set.

Chapter 4 Second Order Linear Differential Equations • 61

27.

È 2 1˘ Í-1 1˙ Î ˚

28.

È-1 -3˘ Í2 1 ˙ Î ˚

Section 4.4 1 (a).

l2 + l - 2 = (l + 2)(l - 1) = 0 . Thus y = c1e -2 t + c 2e t .

1 (b).

y (0) = c1 + c 2 = 3, y ¢ (0) = -2c1 + c 2 = -3 . Solving these simultaneous equations yields c1 = 2 and c 2 = 1. Thus the unique solution to the initial value problem is y = 2e -2 t + e t .

1 (c).

lim y = • and lim y = • .

t Æ-•

t Æ•

2 (a).

1 t t l2 - 14 = 0 fi l = ± . Thus y = c1e - 2 + c 2e 2 . 2

2 (b).

y (2) = c1e -1 + c 2e1 = 1, y ¢ (2) = - 12 c1e -1 + 12 c 2e1 = 0 . Therefore, c 2e = c1e -1 = y = 12 e

2 (c).

- ( t -2 ) 2

e e -1 1 . Thus the unique solution to the initial value problem is fi c1 = and c 2 = 2 2 2

+ 12 e

( t -2 ) 2

.

lim y = • and lim y = • .

t Æ-•

t Æ•

3 (a).

l - 4 l + 3 = (l - 3)(l - 1) = 0 . Thus y = c1e 3 t + c 2e t .

3 (b).

y (0) = c1 + c 2 = -1, y ¢ (0) = 3c1 + c 2 = 1. Solving these simultaneous equations yields

2

c1 = 1 and c 2 = -2 . Thus the unique solution to the initial value problem is y = e 3 t - 2e t .

3 (c).

lim y = 0 and lim y = • .

t Æ-•

t Æ•

t

4 (a).

2l2 - 5l + 2 = (2l - 1)(l - 2) = 0 . Thus y = c1e 2 + c 2e 2 t .

4 (b).

y (0) = c1 + c 2 = -1, y ¢ (0) = 12 c1 + 2c 2 = -5 . Solving these simultaneous equations yields t

c1 = 2 and c 2 = -3 . Thus the unique solution to the initial value problem is y = 2e 2 - 3e 2 t .

4 (c).

lim y = 0 and lim y = -• .

t Æ-•

t Æ•

5 (a).

l2 - 1 = (l + 1)(l - 1) = 0 . Thus y = c1e - t + c 2e t .

5 (b).

y (0) = c1 + c 2 = 1, y ¢ (0) = -c1 + c 2 = -1. Solving these simultaneous equations yields c1 = 1 and c 2 = 0 . Thus the unique solution to the initial value problem is y = e - t .

5 (c). 6 (a).

lim y = • and lim y = 0 .

t Æ-•

t Æ•

l2 + 2l = l (l + 2) = 0 . Thus y = c1e -2 t + c 2 .

62 • Chapter 4 Second Order Linear Differential Equations

6 (b). 6 (c).

y (-1) = c1e 2 + c 2 = 0, y ¢ (-1) = -2c1e 2 = 2 . Therefore, c1 = -e -2 and c 2 = 1, and y = 1 - e -2( t -+1). lim y = -• and lim y = 1.

t Æ-•

t Æ•

7 (a).

l2 + 5l + 6 = (l + 2)(l + 3) = 0 . Thus y = c1e -3 t + c 2e -2 t .

7 (b).

y (0) = c1 + c 2 = 1, y ¢ (0) = -3c1 - 2c 2 = -1 . Solving these simultaneous equations yields c1 = -1 and c 2 = 2 . Thus the unique solution to the initial value problem is y = -e -3 t + 2e -2 t .

7 (c). 8 (a).

lim y = -• and lim y = 0.

t Æ-•

t Æ•

l - 5l + 6 = (l - 2)(l - 3) = 0 . Thus y = c1e 2 t + c 2e 3 t . 2

8 (b). c1 + c 2 = 1, 2c1 + 3c 2 = -1. Therefore, c1 = 4 and c 2 = -3 , and y = 4 e 2 t - 3e 3 t . 8 (c).

lim y = 0 and lim y = -• .

t Æ-•

t Æ•

9 (a).

l - 4 = (l + 2)(l - 2) = 0 . Thus y = c1e -2 t + c 2e 2 t .

9 (b).

y ( 3) = c1e -6 + c 2e 6 = 0, y ¢ ( 3) = -2c1e -6 + 2c 2e 6 = 0 . Solving these simultaneous equations yields

2

c1 = 0 and c 2 = 0 . Thus the unique solution to the initial value problem is y = 0 .

9 (c).

lim y = 0 and lim y = 0 .

t Æ-•

t Æ•

t

t

10 (a). 8l2 - 6l + 1 = ( 4 l - 1)(2l - 1) = 0 . Thus y = c1e 4 + c 2e 2 . 1

1

10 (b). c1e 4 + c 2e 2 = 4, y = 2e

( t -1 ) 4

+ 2e

1

1

1 4

c1e 4 + 12 c 2e 2 = 32 . Therefore, c1 = 2e

2

.

( t -1 )

-1

4

-1

and c 2 = 2e 2 , and

10 (c). lim y = 0 and lim y = • . t Æ-•

t Æ•

3

t

11 (a). 2l2 - 3l = (l )(2l - 3) = 0 . Thus y = c1 + c 2e 2 .

3 11 (b). y (-2) = c1 + c 2e -3 = 3, y ¢ (-2) = c 2e -3 = 0 . Solving these simultaneous equations yields 2 c1 = 3 and c 2 = 0 . Thus the unique solution to the initial value problem is y = 3.

11 (c). lim y = 3 and lim y = 3 . t Æ-•

t Æ•

12 (a). l - 6l + 8 = (l - 2)(l - 4 ) = 0 . Thus y = c1e 2 t + c 2e 4 t . 2

12 (b). c1e 2 + c 2e 4 = 2, 2c1e 2 + 4 c 2e 4 = -8 . Therefore, c1 = 8e -2 and c 2 = -6e -4 , and

y = 8e 2( t -1) - 6e 4 ( t -1). 12 (c). lim y = 0 and lim y = -• . t Æ-•

t Æ•

13 (a). l2 + 4 l + 2 = 0 . Thus l =

-4 ± 16 - 8 = -2 ± 2 and y = c1e( -2 2

2 )t

+ c 2e( -2 +

2 )t

.

Chapter 4 Second Order Linear Differential Equations • 63

13 (b). y (0) = c1 + c 2 = 0, y ¢ (0) = (-2 - 2 )c1 + (-2 + 2 )c 2 = 4 . Solving these simultaneous equations yields c1 = - 2 and c 2 = 2 . Thus the unique solution to the initial value problem is y = - 2e( -2 -

2 )t

+ 2e( -2 +

2 )t

.

13 (c). lim y = -• and lim y = 0. t Æ-•

t Æ•

14 (a). l2 - 4 l - 1 = 0 fi l =

4 ± 16 + 4 = 2 ± 5 . Thus y = c1e( 2 2

5 )t

+ c 2e ( 2 +

5 )t

.

14 (b). c1 + c 2 = 1, (2 - 5 )c1 + (2 + 5 )c 2 = 2 + 5 . Therefore, c1 = 0 and c 2 = 1, and y = e( 2 +

5 )t

.

14 (c). lim y = 0 and lim y = • . t Æ-•

t Æ•

15 (a). 2l2 - 1 = 0 . Thus l = ±

- t 1 and y = c1e 2

15 (b). y (0) = c1 + c 2 = -2, y ¢ (0) = -

t 2

+ c 2e

2

.

1 1 c1 + c 2 = 2 . Solving these simultaneous equations yields 2 2

c1 = -2 and c 2 = 0 . Thus the unique solution to the initial value problem is y = -2e

- t

2

.

15 (c). lim y = -• and lim y = 0. t Æ-•

16.

t Æ•

Since lim e -3 t = 0, y ( t) = c1e -3 t + 2, y (0) = c1 + 2 = 1 fi c1 = -1. Therefore, t Æ•

y ( t) = 2 - e -3 t , l2 + al + b = l (l + 3) fi a = 3, b = 0 and y ¢ (0) = y 0¢ = 3. 17 (a). W =

e-t -e - t

cel 2 t = (l 2 + 1)ce(l 2 -1)t = 4 e 2 t . Thus l 2 = 3 and c = 1. The second member of the l 2cel 2 t

fundamental set is then y 2 = e 3 t . 17 (b). l2 + al + b = (l + 1)(l - 3) . Therefore a = -2 and b = -3. 17 (c). The general solution is y = c1e - t + c 2e 3 t . Using the initial conditions, we have y (0) = c1 + c 2 = 3, y ¢ (0) = -c1 + 3c 2 = 5 . Solving these simultaneous equations yields c1 = 1 and c 2 = 2 . The unique solution to the initial value problem is y = e - t + 2e 3 t .

18 (a). l2 + 2l = 0 fi l = 0, -2 . Graph (C) since it is the only equation admitting a nonzero limit as

t Æ •. 1 1 18 (b). 6l2 - 5l + 1 = ( 3l - 1)(2l - 1) = 0 fi l = , . Graph (B) 3 2 18 (c). l2 - 1 = 0 fi l = 1, -1. Graph (A). 19.

Utilizing the hint given, we can make the substitution u( t) = y ¢ ( t) . The equation then becomes

u¢¢ - 5 u¢ + 6 u = 0 .

64 • Chapter 4 Second Order Linear Differential Equations

The characteristic equation of this new differential equation is l2 - 5l + 6 = (l - 3)(l - 2) = 0 . Thus u( t) = c1¢ e 2 t + c 2¢ e 3 t = y ¢ ( t) . Antidifferentiation gives us

1 1 y = c1¢ e 2 t + c 2¢ e 3 t + c 3 = c1e 2 t + c 2e 3 t + c 3 . 2 3 20 (a). mx ¢¢ + kx ¢ = 0, ml2 + kl = ml (l + 20 (b). x ¢ = - mk c1e x ( t) = -

mv 0 k

- mk t

e

- mk t

20 (c). lim x ( t) = x 0 + t Æ•

k m

k

) = 0 fi l = - m ,0 . Thus

x ( t) = c1e

- mk t

+ c2 .

\ c1 + c 2 = x 0 , - mk c1 = v 0 fi c1 = - mk v 0 , c 2 = x 0 + mk v 0 . + x0 + mv 0 k

mv 0 k

= x0 +

mv 0 k

(1 - e

- mk t

).

.

21 (a). l2 - W2 = (l + W)(l - W) = 0 . Thus r( t) = c1e -Wt + c 2eWt . From the initial conditions, r(0) = c1 + c 2 = r0 and r¢ (0) = -Wc1 + Wc 2 = r0¢ . Solving these simultaneous equations yields

c1 =

1 1 r0 - W-1r0¢) and c 2 = ( r0 + W-1r0¢) . Thus the unique solution is ( 2 2

r=

1 1 r0 - W-1r0¢)e -Wt + ( r0 + W-1r0¢)eWt = r0 cosh(Wt) + W-1r0¢ sinh(Wt) . When r0 = r0¢ = 0 , ( 2 2

r( t) = 0 and the particle remains at rest at the pivot.

30(2 p) 1 21 (b). r0 = 0, r0¢ = , l = 3, W = = p . We need to find t when r( t) = l = 3. With this condition, 60 5 1 1 ◊ sinh( pt) . Solved for t, t = 1.44705 seconds. p 5

we have r( t) = 3 =

k 22 (a). l + l - W2 = 0 . Thus l ± = m 2

-

k ± m

( k m)

2

+ 4 W2

and the general solution to the differential

2

equation is r = c1el - t + c 2el + t . From the initial conditions, we have c1 + c 2 = 0 and l -c1 + l +c 2 = r0¢ . Solving these simultaneous equations yields

c 2 = -c1 =

r0¢e

r0¢

( m) k

, and thus r( t) =

2

22 (b). m ª R 3 , k ª R 2 \

+ 4W

2

k 1 ª would decrease. m R

-

k t 2m

2 È1 Ê ˆ ˘ ◊ 2 sinh Í Á k m + 4 W2 ˜ t ˙ ¯ ˚ Î2 Ë 2 2 k m + 4W

( )

( )

Chapter 4 Second Order Linear Differential Equations • 65

22 (c). W = 20 rev min = 20(2p ) / 60 =

2p 3

rad

sec

k = 4 s-1 m

, r0¢ = 1,

2

Ê 4p 2 ˆ k Ê kˆ 2 = 5.79189, = 2, r0¢ = 1. Á ˜ + 4 W = 16 + 4 Á ˜ Ë m¯ 2m Ë 9 ¯ \ r ( t) =

1 ◊ e -2 t ◊ 2 sinh[2.89594 t] , r(2) = 1.03605cm. 5.79189

[

]

23 (a). r = c 2 el + t - el - t where l + > 0 and l - < 0 (from question 22). For the positive limit,

l+ =

W 1Ê k = Á- + 2 2Ë m

W+

k = m

2

2

ˆ + 4 W2 ˜ . Therefore ¯ 2

2

k Ê kˆ Ê kˆ 2 2 2 + 4 W , and W + 2 W + = Á ˜ Á ˜ + 4 W . Then m Ë m¯ Ë m¯ m

( ) k

( k m)

2

k k 3 W = 3W2 , and = W. m m 2 2

25 9 Ê kˆ 23 (b). Using the relation reached in part (a), Á ˜ + 4 W2 = W2 + 4 W2 = W2 . Therefore Ë m¯ 4 4

( k m)

2

1È k 5 + 4 W2 = W and l - = Í- 2Î m 2

( k m)

5 ˘ ˘ 1È 3 + 4 W2 ˙ = Í- W - W˙ = -2W . 2 ˚ ˚ 2Î 2

2

˘ r0¢ È W2 t 1È 3 5 ˘ W e - e -2Wt ˙ . l + = Í- W + W˙ = , and so r( t) = Í 5 Î 2Î 2 2 ˚ 2 ˚ W 2

Section 4.5 1 (a).

l2 + 2l + 1 = (l + 1) 2 = 0 . Thus y = c1e - t + c 2 te - t .

1 (b).

y ¢ = -c1e - t + c 2 (1 - t)e - t . From the initial conditions, we have c1e -1 + c 2e -1 = 1 and - c1e -1 + c 2 ◊ 0 = 0, and thus c1 = 0 and c 2 = e . The unique solution is then y ( t) = te -( t -1) .

1 (c).

lim y = -• and lim y = 0.

t Æ-•

t Æ•

t

t

2 (a).

9l2 - 6l + 1 = ( 3l - 1) 2 = 0 . Thus y = c1e 3 + c 2 te 3 .

2 (b).

y ¢ = 13 c1e 3 + c 2 (1 + 3t )e 3 . c1e + 3c 2e = -2 and 13 c1e + 2c 2e = - 53 fi c1 = e -1 and c 2 = - e -1 . The

t

t

unique solution is then y ( t) = e 2 (c). 3 (a).

( t -3 ) 3

- te

( t -3 ) 3

= (1 - t)e

lim y = 0 and lim y = -• .

t Æ-•

t Æ•

l2 + 6l + 9 = (l + 3) 2 = 0 . Thus y = c1e -3 t + c 2 te -3 t .

( t -3 ) 3

.

66 • Chapter 4 Second Order Linear Differential Equations

3 (b).

y ¢ = -3c1e -3 t + c 2 (1 - 3t)e -3 t . From the initial conditions, we have c1 = 2 and - 3c1 + c 2 = -2, thus c 2 = 4 . The unique solution is then y ( t) = (2 + 4 t)e -3 t .

3 (c).

lim y = -• and lim y = 0.

t Æ-•

t Æ•

4 (a).

25l2 + 20l + 4 = (5l + 2) 2 = 0 . Thus y = c1e

4 (b).

y¢ =

- 25 t -2 5 1

ce

+ c 2 (1 - 25t )e

- 25 t

c1e -2 + 5c 2e -2 = 4 e -2 and

+ c 2 te

- 25 t

.

.

-2 -2 5 1

c e + (1 - 2)c 2e -2 = - 35 e -2 fi c1 = -1 and c 2 = 1.

The unique solution is then y ( t) = ( t - 1)e 4 (c).

- 25 t

- 25 t

.

lim y = -• and lim y = 0.

t Æ-•

t Æ•

2

t 2

2

t 2

5 (a).

4 l - 4 l + 1 = (2l - 1) = 0 . Thus y = c1e + c 2 te .

5 (b).

t 2t 1 2t y ¢ = c1e + c 2 (1 + )e . From the initial conditions, we have 2 2 1 1 1 12 3 12 2 c1e + c 2e = -4 and c1e + c 2e = 0, and thus c1 = -6e and c 2 = 2e 2 . The unique 2 2 1 2

1 2

solution is then y ( t) = (-6 + 2 t)e 5 (c). 6 (a).

t -1 2

.

lim y = 0 and lim y = • .

t Æ-•

t Æ•

l2 - 4 l + 4 = (l - 2) 2 = 0 . Thus y = c1e 2 t + c 2 te 2 t .

6 (b). c1e -2 - c 2e -2 = 2 and 2c1e -2 + (1 - 2)c 2e -2 = 1 fi c1 = - e 2 and c 2 = -3e 2 The unique solution is then y ( t) = (-1 - 3t)e 2( t +1) . 6 (c).

lim y = 0 and lim y = -• .

t Æ-•

t Æ•

2

t 4

2

t 4

7 (a). 16l - 8l + 1 = ( 4 l - 1) = 0 . Thus y = c1e + c 2 te . 7 (b). From the initial conditions, we have y (0) = c1 = -4 and y 0¢ = t

the unique solution is y ( t) = (-4 + 4 t)e 4 . 7 (c). 8 (a).

lim y = 0 and lim y = • .

t Æ-•

t Æ•

l2 + 2 2l + 2 = (l + 2 ) 2 = 0 . Thus y = c1e -

+ c 2 te -

2t

2t

8 (b). c1 + 0 = 1 and - 2c1 + (1 - 0)c 2 = 0 fi c1 = 1 and c 2 = 2 The unique solution is then y ( t) = (1 + 2 t)e 8 (c).

lim y = -• and lim y = 0.

t Æ-•

t Æ•

2t

.

.

1 c + c = 3. Thus c 2 = 4 , and so 4 1 2

Chapter 4 Second Order Linear Differential Equations • 67 2

5

5

9 (a).

t t 5ˆ Ê l2 - 5l + 6.25 = Á l - ˜ = 0 . Thus y = c1e 2 + c 2 te 2 . Ë 2¯

9 (b).

5 5 5t Ê 5 ˆ t y ¢ = c1e 2 + c 2 Á1 + t˜ e 2 . From the initial conditions, we have Ë 2 ¯ 2

c1e -5 - 2c 2e -5 = 0 and 5

then y ( t) = (2 + t)e 2 9 (c).

5 -5 c e - 4 c 2e -5 = 1, and thus c1 = 2e 5 and c 2 = e 5 . The unique solution is 2 1

( t + 2)

.

lim y = 0 and lim y = • .

t Æ-•

t Æ•

10 (a). 3l2 + 2 3l + 1 = ( 3l + 1) 2 = 0 . Thus y = c1e 10 (b). y ¢ =

-1 3

c1e

-

1 t 3

+ c 2 (1 -

t 3

)e

-

1 t 3

-

1 t 3

. c1 + 0 = 2 3 and

The unique solution is then y ( t) = (2 3 + 5 t)e

-

1 t 3

+ c 2 te -1 3

-

1 t 3

.

c1 + (1 - 0)c 2 = 3 fi c1 = 2 3 and c 2 = 5 .

.

10 (c). lim y = -• and lim y = 0. t Æ-•

11.

t Æ•

2

l2 - 2al + a 2 = (l - a ) = 0 . Thus y = c1eat + c2 teat . From the initial conditions and the graph provided, y(0) = c1 = 0 and at the maximum, y ¢ = c2 (1 + at )eat = 0 . Solving for the t coordinate at the maximum gives us tmax = -

1 1 = 2, and thus a = - . 2 a

1 Solving for the y coordinate at the maximum gives us ymax = c2 Ê - ˆ e -1 = 2c2 e -1 = 8e -1 , and Ë a¯ t 2

1 thus c2 = 4 . Finally, the equation for y(t ) is y(t ) = 4te , and a = - , y0 = 0, y0¢ = 4. 2 -

12.

1 y ¢¢ = 0 \ a = 0, y = y 0¢ t + y 0 , y (0) = y 0 = 2, y ( 4 ) = ( y 0¢ )( 4 ) + 2 = 0 fi y 0¢ = - . 2 1 Therefore, y ( t) = - 12 t + 2, a = 0, y 0 = 2, y 0¢ = - . 2

13.

2

4l + 4l + 1 = (2l + 1) = 0 . Thus y = c1e 2

y(1) = c1e

-

1 2

+ c2 e

-

1 2

-

-

t 2

-

t 2

+ c2 te . From the first point given, we have

1 2

= e . From the second, we have y(2) = c1e -1 + 2c2 e -1 = 0 . Solving these

two simultaneous equations yields c1 = 2 and c2 = -1. Finally, we have y(t ) = 2e differentiation gives us y ¢ = -e

-

t 2

t

1 - (2 - t )e 2 . Thus y(0) = 2 and y ¢(0) = -2 . 2

-

t 2

-

t

- te 2 , and

68 • Chapter 4 Second Order Linear Differential Equations

14 (a). y 2 = e tu, y 2¢ = e tu + e tu ¢, y 2¢¢ = e t v + 2e t v ¢ + e tu ¢¢ . Therefore t(u ¢¢ + 2u ¢ + v ) - (2 t + 1)(u ¢ + v ) + ( t + 1)u = 0 fi tu ¢¢ + (2 t - 2 t - 1)u ¢ + ( t - 2 t - 1 + t + 1)v = tu ¢¢ - u ¢ = 0 ;

u = v ¢, u¢ - t -1u = 0, ( t -1u)¢ = 0, u = ct fi v = c et 14 (b). W = t e

14 (c). p( t) = -

t2 \ y 2 = t 2e t 2

t 2e t = 2 te 2 t π 0 on (-•, 0) and (0, •). ( t 2 + 2 t)e t

(2 t + 1) t +1 , q( t) = , continuous on (-•, 0) and (0, •). t t

15 (a). y2 = tu , y2¢ = u + tu ¢, y2¢¢ = tu ¢¢ + 2u ¢ . Therefore

t 2 ( tu ¢¢ + 2u ¢ ) - t(u + tu ¢ ) + tu = t 3u ¢¢ + t 2u ¢ = t 2 ( tv ¢¢ + v ¢ ) = 0 , and so

(tu ¢)¢ = 0, which means that u ¢ =

c . Antidifferentiation yields u = c ln t + c ¢ , and thus t

y2 = t ln t . 15 (b). W =

t t ln t = t π 0 on ( -•, 0) and (0, •). 1 ln t + 1

1 1 15 (c). p(t ) = - , q(t ) = 2 , continuous on ( -•, 0) and (0, •). t t 16 (a). y 2 = u sin t, y 2¢ = u ¢ sin t + u cos t, y 2¢¢ = v ¢¢ sin t + 2v ¢ cos t - v sin t . Therefore

(u ¢¢ sin t + 2u ¢ cos t - v sin t) - (2 cot t)(v ¢ sin t + v cos t) + (1 + 2 cot 2 t)sin v = 0 fi

u ¢¢ (sin t) + u ¢ (2 cos t - 2 cot t sin t) + v (- sin t - 2 cot t cos t + sin t + 2 cot 2 t sin t) = 0 fi v ¢¢ sin t = 0 fi v ¢¢ = 0, v = c1t + c 2 \ y 2 = t sin t .

16 (b). W =

sin t t sin t = sin 2 t π 0 on (np ,(n +1)p ),n = ...,-2,-1,0,1,2,.... cos t sin t + t cos t

16 (c). p( t) = -2 cot t, q( t) = 1 + 2 cot 2 t, continuous on same intervals. 17 (a). y 2 = ( t + 1) 2 u, y 2¢ = 2( t + 1)u + ( t + 1) 2 u ¢, y 2¢¢ = ( t + 1) 2 u ¢¢ + 4 ( t + 1)u ¢ + 2u . Therefore

[

]

( t + 1) 4 u ¢¢ + 4 ( t + 1) 3 u ¢ + 2( t + 1) 2 u - 4 2( t + 1) 2 u + ( t + 1) 3 u ¢ + 6( t + 1) 2 v

= ( t + 1) 4 u ¢¢ = 0 , and so u ¢¢ = 0. Antidifferentiation yields u = c1 (t + 1) + c2 , and thus y2 = (t + 1)3 . (t + 1)2 (t + 1)3 17 (b). W = = (t + 1)4 π 0 on ( -•, -1) and ( -1, •). 2 2(t + 1) 3(t + 1)

Chapter 4 Second Order Linear Differential Equations • 69

17 (c). p(t ) = -

4 6 , q(t ) = , continuous on ( -•, -1) and ( -1, •). t +1 (t + 1)2

2

2

2

18 (a). y 2 = e - t u, y 2¢ = -2 te - t u + e - t u ¢, . 2

2

2

2

2

y 2¢¢ = -2e - t v + 4 t 2e - t v - 2 te - t v ¢ - 2 te - t v ¢ + e - t u ¢¢ . Therefore, u ¢¢ - 4 tu ¢ + (-2 + 4 t 2 )v + 4 t(-2 tv + u ¢ ) + (2 + 4 t 2 )u = 0 fi v ¢¢ = 0 ; 2

\ v = c1t + c 2 , y 2 = te - t . 2

e-t 18 (b). W = 2 -2 te - t

2

e-t

2

2 te - t = e -2 t π 0 on (-•, •). 2 -t 2 - 2t e

18 (c). p( t) = 4 t, q( t) = 2 + 4 t 2 , continuous on (-•, •). 19 (a). y2 = (t - 2)2 u , y2¢ = 2(t - 2)u + (t - 2)2 u ¢, y2¢¢ = (t - 2)2 u ¢¢ + 4(t - 2)u ¢ + 2u . Therefore

(t - 2)4 u ¢¢ + 4(t - 2)3 u ¢ + 2(t - 2)2 u + 2(t - 2)2 u + (t - 2)3 u ¢ - 4(t - 2)2 u = 0 , and so

u ¢¢ +

c1 5 u ¢ = 0 . Thus ((t - 2)5 u ¢)¢ = 0 , and antidifferentiation yields u = + c2 , (t - 2 ) 4( t - 2 ) 4

and thus y2 = (t - 2)-2 . (t - 2 ) 2 (t - 2)-2 4 19 (b). W = =π 0 on ( -•, 2) and (2, •). -3 (t - 2 ) 2(t - 2) -2(t - 2) 19 (c). p(t ) =

1 4 , q(t ) = , continuous on ( -•, 2) and (2, •). t-2 (t - 2 ) 2

20 (a). y 2 = e tu, y 2¢ = e tu + e tu ¢, y 2¢¢ = e t v + 2e t v ¢ + e tu ¢¢ . Therefore n - 1ˆ Ê n - 1ˆ Ê u ¢¢ + 2u ¢ + v - Á 2 + ˜u = 0 fi ˜ (u ¢ + v ) + Á1 + Ë Ë t ¯ t ¯

u ¢¢ 20 (b). W =

n -1 v ¢ = 0 fi ( t -( n -1)v ¢ )¢ = 0 fi v ¢ = c1t n -1 fi v = t et et

c1 n

tn , \ y2 = tnet

t ne t = nt n -1e 2 t . ( nt n -1 + t n )e t

If n = 1, W π 0 on (-•, •). If n ≥ 2, W π 0 on (-•, 0),(0, •).

n - 1ˆ Ê n - 1ˆ Ê 20 (c). p( t) = -Á 2 + ˜ , continuous on same intervals. ˜ , q( t) = Á1 + Ë Ë t ¯ t ¯

70 • Chapter 4 Second Order Linear Differential Equations

Section 4.6 1 (a).

p p 2ei p 3 = 2Ê cos + i sin ˆ = 1 + i 3 Ë 3 3¯

-p -p 1 (b). -2 2e - i p 4 = -2 2 Ê cos + i sin ˆ = -2 + i 2 Ë 4 4¯ 1 (c). 1 (d).

(2 - i)e i 3 p 2 = (2 - i)(-i) = -1 - i2 . 1 2 2

ei 7 p 6 = 4

Ê 1 2p 2p 3ˆ = 4ei 2 p 3 = 4Ê cos + i sin ˆ = 4Á - + i ˜ = -2 + i 2 3 Ë 3 3¯ 2 ¯ Ë 2

1 (e).

(

2 (a).

2 cos( 2 t) + i2 sin( 2 t)

2 (b).

2 p

2 (c).

- 12 e 2 t [cos( t + p ) + i sin( t + p )] = 12 e 2 t cos t + i 12 e 2 t sin t .

2 (d).

3 3e 3 t cos( 3t) + i 3 3e 3 t sin( 3t)

2 (e).

-

3 (a).

l2 + 4 = 0 , and thus l = ±i2 .

3 (b).

y = c1 cos(2t ) + c2 sin(2t )

3 (c).

y ¢ = -2c1 sin(2t ) + 2c2 cos(2t ) . Using the initial conditions, we have

2ei p 6

)

1 Ê 7p 7p 3 1 + i sin ˆ = cos -i 2 2Ë 6 6¯ 4 2 4 2

e -2 t cos( 3t) - i p2 e -2 t sin( 3t)

2 4

(cos 3pt + i sin 3pt) = -

2 4

cos( 3pt) - i

2 4

sin( 3pt)

y( p 4) = c2 = -2 and y ¢( p 4) = -2c1 = 1 . Thus

1 1 c1 = - , c2 = -2, and y = - cos(2t ) - 2 sin(2t ) . 2 2 -2 ± 4 - 8 = -1 ± i . 2

4 (a).

l2 + 2l + 2 = 0 fi l =

4 (b).

y = c1e - t cos t + c 2e - t sin t

4 (c).

y ¢ = -c1e - t cos t - c1e - t sin t - c 2e - t sin t + c 2e - t cos t . y (0) = c1 = 3 and y ¢ (0) = -c1 + c 2 = -1. Thus c1 = 3, c 2 = 2, and y = 3e - t cos t + 2e - t sin t .

5 (a).

1 9l2 + 1 = 0 , and thus l = ±i . 3

5 (b).

t t y = c1 cosÊ ˆ + c2 sinÊ ˆ Ë 3¯ Ë 3¯

Chapter 4 Second Order Linear Differential Equations • 71

5 (c).

1 t 1 t y ¢ = - c1 sinÊ ˆ + c2 cosÊ ˆ . Using the initial conditions, we have Ë ¯ Ë 3 3 3 3¯

3 1 1 1 1 Ê 3ˆ c1 + c2 = 4 and y ¢( p 2) = - c1 Ê ˆ + c2 Á ˜ = 0 . Ë ¯ 2 2 3 2 3 Ë 2 ¯

y( p 2 ) =

Solving these simultaneous equations gives us t t c1 = 2 3, c2 = 2, and y = 2 3 cosÊ ˆ + 2 sinÊ ˆ . Ë 3¯ Ë 3¯ 2± 4-8 1 1 = ±i . 4 2 2

6 (a).

2l2 - 2l + 1 = 0 fi l =

6 (b).

y = c1e 2 cos( 2t ) + c 2e 2 sin( 2t )

6 (c).

y ¢ = 12 c1e 2 cos( 2t ) - 12 c1e 2 sin( 2t ) + 12 c 2e 2 sin( 2t ) + 12 c 2e 2 cos( 2t ) .

t

t

t

y (-p ) = -c 2e

t

- p2

t

t

-p

p

p

c1 = -3e 2 , c 2 = -e 2 , and y = -3e

t +p 2

cos( 2t ) - e

7 (a).

1 3 l2 + l + 1 = 0 , and thus l = - ± i . 2 2

7 (b).

y = c1e

7 (c).

Ê 3 ˆ 1 - 2t 3 - 2t Ê 3 ˆ y ¢ = - c1e cosÁ t˜ c1e sinÁ t˜ 2 Ë 2 ¯ 2 Ë 2 ¯

-

t 2

-p

= 1 and y ¢ (-p ) = - 12 c1e 2 (-1) + 12 c 2e 2 (-1) = -1 . Thus t +p 2

sin( 2t ) .

t Ê 3 ˆ Ê 3 ˆ cosÁ t ˜ + c2 e 2 sinÁ t˜ Ë 2 ¯ Ë 2 ¯

Ê 3 ˆ 1 - 2t Ê 3 ˆ 3 - 2t - c 2e sinÁ t˜ + c 2e cosÁ t˜ . 2 2 Ë 2 ¯ Ë 2 ¯ 1 3 Using the initial conditions, we have y(0) = c1 = -2 and y ¢(0) = - c1 + c2 = -2 . 2 2

Solving these simultaneous equations gives us

c1 = -2, c2 = -2 3, and y = -2e

-

t 2

t Ê 3 ˆ Ê 3 ˆ cosÁ t ˜ - 2 3e 2 sinÁ t˜ . Ë 2 ¯ Ë 2 ¯

-4 ± 16 - 20 = -2 ± i . 2

8 (a).

l2 + 4 l + 5 = 0 fi l =

8 (b).

y = c1e -2 t cos t + c 2e -2 t sin t

8 (c).

y ¢ = -2c1e -2 t cos t - c1e -2 t sin t - 2c 2e -2 t sin t + c 2e -2 t cos t . y ( p2 ) = c 2e -p =

1 2

and y ¢ ( p2 ) = -c1e -p - 2c 2e -p = -2 . Thus

-2 t - p c1 = ep , c 2 = 12 ep , and y = e ( 2 ) (cos t + 12 sin t) .

72 • Chapter 4 Second Order Linear Differential Equations

9 (a).

1 1 9l2 + 6l + 2 = 0 , and thus l = - ± i . 3 3

9 (b).

t t y = c1e cosÊ ˆ + c2 e 3 sinÊ ˆ Ë 3¯ Ë 3¯

9 (c).

t t t 1 - 3t Ê t ˆ 1 -3 Ê t ˆ 1 -3 Ê t ˆ 1 -3 Ê tˆ y ¢ = - c1e cosÁ ˜ - c1e sinÁ ˜ - c 2e sinÁ ˜ + c 2e cosÁ ˜ . Using the initial Ë ¯ Ë ¯ Ë ¯ Ë 3¯ 3 3 3 3 3 3 3

-

t 3

t

1 1 conditions, we have y ( 3p) = -c1e -p fi c1 = 0 and y ¢ ( 3p) = - c 2e -p = . Solving these 3 3 t

t simultaneous equations gives us c1 = 0, c2 = -e p , and y = -e 3 e p sinÊ ˆ . Ë 3¯

10 (a). l2 + 4p 2 = 0 fi l = ±2pi . 10 (b). y = c1 cos(2pt) + c 2 sin(2pt) 10 (c). y ¢ = -2pc1 sin(2pt) + 2pc 2 cos(2pt) .

y (1) = c1 = 2 and y ¢ (1) = 2pc 2 = 1 fi c 2 = Thus y = 2 cos(2pt) +

1 . 2p

1 sin(2pt) . 2p

11 (a). l2 - 2 2 l + 3 = 0 , and thus l = 2 ± i . 11 (b). y = c1e

2t

cos(t ) + c2 e

11 (c). y ¢ = 2c1e

2t

2t

sin(t )

cos(t ) - c1e

we have y(0) = c1 = -

2t

sin(t ) + 2c2 e

2t

sin(t ) + c2 e

2t

cos(t ) . Using the initial conditions,

1 and y ¢(0) = 2c1 + c2 = 2 . 2

Solving these simultaneous equations gives us 1 3 2 1 c1 = - , c2 = , and y = - e 2 2 2

2t

cos(t ) +

3 2 e 2

2t

sin(t ) .

p 12 (a). 9l2 + p 2 = 0 fi l = ± i . 3 12 (b). y = c1 cos( p3 t) + c 2 sin( p3 t)

p p 12 (c). y ¢ = - c1 sin( p3 t) + c 2 cos( p3 t) . 3 3 p y ( 3) = -c1 = 2 and y ¢ ( 3) = - c 2 = -p fi c1 = -2, c 2 = 3 . 3 Thus y = -2 cos( p3 t) + 3 sin( p3 t)

Chapter 4 Second Order Linear Differential Equations • 73

13.

l = ±i, so the characteristic equation must be l2 + 1 = 0 . Thus a = 0 and b = 1. y0 = y( p 4) =

14.

1 1 - 1 and y0¢ = y ¢( p 4) = + 1. 2 2

l = ±2i , so the characteristic equation must be l2 + 4 = 0 . Thus a = 0 and b = 4 . y 0 = y ( p 4 ) = 2 and y 0¢ = y ¢ ( p 4 ) = -2 .

15.

l = -2 ± i , so the characteristic equation must be l2 + 4l + 5 = 0 . Thus a = 4 and b = 5 . y0 = y(0) = 1 and y0¢ = y ¢(0) = -2 - 1 = -3 .

16.

l = 1 ± 2i , so the characteristic equation must be (l - 1) 2 + 4 = l2 - 2l + 5 = 0 . Thus a = -2 and b = 5 . y 0 = y ( p 6) =

17.

3 1 3 3 1 . and y 0¢ = y ¢ ( p 6) = - 2 2 2 2

l = ±ip , so the characteristic equation must be l2 + p 2 = 0 . Thus a = 0 and b = p 2 . y 0 = y (1 2) = -1 and y 0¢ = y ¢ (1 2) = - 3p .

18.

y = sin t + cos t, so a = 0 and b = 1 . R cosd = 1 and R sin d = 1 , so R = 2 and d =

Ê y = 2 cosÁ t Ë

19.

p . Thus 4

pˆ ˜. 4¯

y = cos pt - sin pt, so a = 0 and b = p . R cos d = 1 and R sin d = -1, so R = 2 and d =

7p . 4

7p Thus y = 2 cosÊ pt - ˆ . Ë 4¯ 20.

y = e t cos t + 3e t sin t, so a = 1 and b = 1. R cosd = 1 and R sin d = 3 , so R = 2 and d = Ê pˆ Thus y = 3e t cosÁ t - ˜ . Ë 3¯

21.

y = -e - t cos t + 3e - t sin t, so a = -1 and b = 1. R cos d = -1 and R sin d = 3 , so

R = 2 and d = 22.

2p 2p . Thus y = 2e - t cosÊ t - ˆ . Ë 3¯ 3

y = e -2 t cos 2 t - e -2 t sin 2 t, so a = -2 and b = 2 . R cosd = 1 and R sin d = -1, so R = 2 and d =

7p ˆ 7p Ê . Thus y = 2e -2 t cosÁ 2 t - ˜ . Ë 4¯ 4

23.

p ˆ p2 Ê y(t ) = 2 cos t , a = 0, b = , y0 = 2, y0¢ = 0 . Ë2 ¯ 4

24.

9 3 Êp ˆ Ê3 pˆ Êp ˆ y ( t) = cosÁ t - ˜ , a = 0, b = , y 0 = cosÁ ˜ , y 0¢ = sinÁ ˜ . Ë2 Ë 8¯ 4 2 Ë 8¯ 8¯

p . 3

74 • Chapter 4 Second Order Linear Differential Equations

25.

y( t ) =

1 5p 1 5p 5p cosÊ 2t - ˆ , a = 0, b = 4, y0 = cos , y0¢ = sin . Ë ¯ 2 6 2 6 6

26 (a). l2 + ml + w 2 = 0 . l = m

y = c1e

-2t

cosÊ w 2 Ë

-m ± m 2 - 4w 2 m i 4w 2 - m 2 =- ± 2 2 2 m - t tˆ + c 2e 2 sinÊ w 2 ¯ Ë

m2 4

m 26 (b). y (0) = c1 = 2, y ¢ (0) = - c1 + w 2 2 m

y=e

-2t

È Í2 cosÊ w 2 Ë Í Î

m 26 (c). a = - , b = w 2 2 R2 = 4 +

m2 w2 -

2

m 4

=

tˆ + ¯

m2 4

m2 4

m2 4

m w2 -

m2 4

m2 4

tˆ . ¯

m

c2 = 0 fi c2 =

sinÊ w 2 Ë

m2 4

w2 -

.

m2 4

˘ ˆ t ˙. ¯˙ ˚

.

4w 2

w2 -

2

m 4

2w

fiR=

2

m 4

2

w -

m

, tan d =

2

2 w -

2

m 4

=

m 4w 2 - m 2

.

1 1 1 1 27 (a). eat cos bt = e(a + ib ) t + e(a - ib ) t , eat sin bt = e(a + ib ) t - e(a - ib ) t . 2 2 2i 2i

] [

27 (b). eat cos bt , eat sin bt = e(

[

a + ib ) t

È- 1 1 Í 27 (c). det A = - , so A-1 = -2i Í 2i 1 2i ÍÎ 2

[e e(

(a + ib ) t

a + ib ) t

, e(

a - ib ) t

] = [e

at

, e(

-

a - ib ) t

È1 Í2 Í1 Í Î2

]

1 ˘ 2i ˙ . 1˙ - ˙ 2i ˚

1˘ 2i ˙ = È1 1 ˘ . 1 ˙ ÍÎi -i ˙˚ ˙ 2 ˚

È1 1 ˘ cos bt , eat sin bt Í ˙ , so Îi -i ˚

]

= eat cos bt + ieat sin bt and e(

a - ib ) t

= eat cos bt - ieat sin bt .

28.

From Abel’s Theorem, a = 0 \ y = c1 cos 3t + c 2 sin 3t .

29.

l2 + 4il + 5 = 0 , and thus l =

-4i ± -16 - 20 = i, - 5i . Thus y = c1eit + c2 e -5it . 2

Chapter 4 Second Order Linear Differential Equations • 75

Section 4.7 9.8(10) = 3266.67 N / m or 3.2667 N / mm . 0.030

2 (a).

ky = mg, k =

2 (b).

my ¢¢ + ky = 0, y ¢¢ + 326.67 y = 0, y (0) = 0.07, y ¢ (0) = 0 .

2 (c). w =

k m

= 18.0739..., y = c1 coswt + c 2 sin wt .

y (0) = 0.07, c 2w = 0 fi c 2 = 0 \ y = 0.07 cos(18.0739 t) .

3.

my ¢¢ + ky = 0, y(0) = 0, y ¢(0) = 2 . The general solution to this differential equation is y = c1 cos wt + c2 sin wt , where w =

k . From the initial conditions, we have m

y(0) = c1 = 0 and y ¢(0) = wc2 = 2, and thus c2 = We know that ymax = 0.2 = 4 (a).

T = 4 ( 45 -

4 (b).

f =

4 (c).

R = 3; the first maximum occurs at t=

3 4

)=2

2 , so w = 10 = w

2 2 . The unique solution is then y = sin wt . w w

k . Solving for k yields k = 2000 N/m. 20

s.

1 1 = Hz, w = 2pf = p rad / s. T 2

3 Ê5 -Á 4 Ë4

p 3ˆ 1 ˜ = \ w ( 14 ) - d = 0 fi d = , y = 3 cos(pt - p4 ) cm. ¯ 4 4 4 3p = 6.6643...cm / s. 2

4 (d).

y (0) = 3 cos( p4 ) = 2.1213...cm, y ¢ (0) = -3p sin(- p4 ) =

5 (a).

my ¢¢ + gy ¢ + ky = 0 with m = 10, g = 7, k = 100, y(0) = 0.5, y ¢(0) = 1. Thus with numerical

constants the initial value problem reads y ¢¢ + 0.7 y ¢ + 10 y = 0, y(0) = 0.5, y ¢(0) = 1. 5 (b). l2 + 0.7l + 10 = 0 , and thus l = -0.35 ± i3.14285 = -a ± ib . The general solution is then

y = c1e -0.35t cos 3.14285t + c2 e -0.35t sin 3.14285t . From the initial conditions, we have y(0) = c1 = 0.5 and y ¢(0) = -0.35c1 + 3.14285c2 = 1 . Solving these simultaneous equations

yields c1 = 0.5 and c2 = 0.37386 . Thus the unique solution to the initial value problem is

y = e -0.35t (0.5 cos(3.14285t ) + 0.37386 sin(3.14285t )) . lim y(t ) = 0 , which means that the t Æ•

damping dissipates the energy of the system, causing the motion to decrease. 6 (a).

-g ± g 2 - 4 mk . Critical damping: g 2 = 4 mk . my ¢¢ + gy ¢ + ky = 0, ml + gl + k = 0 fi l = 2m 2

76 • Chapter 4 Second Order Linear Differential Equations g

6 (b).

y = c1e

- 2m t

(

g

+ c 2 te

y ¢ ( t) = c 2 1 y ( tm ) = c 2

g 2m

- 2m t

, y (0) = c1 = 0, y ¢ (0) = c 2 = 4.

)

g

te

- 2m t

, y ¢ ( t) = 0 when tm =

2m . g

2m e m e 2 m -1 1 e = \4 = fi = . 2 2 g g g 16

m = 1 slug, g = 16e -1 ª 5.886...lb ◊ sec/ ft, k = 7 (a).

g2 = 8.6614...lb / ft. 4m

my ¢¢ + gy ¢ + ky = 0 with y(0) = y0 , y ¢(0) = 0 . ml2 + gl + k = 0 , and thus l = We can rewrite this as l1 =

-g ± g 2 - 4 mk . 2m

-g - g 2 - 4 mk -g + g 2 - 4 mk and l2 = . The general solution 2m 2m

to this initial value problem is y = c1e l1t + c2 e l 2 t . From the initial conditions, we have

y(0) = c1 + c2 = y0 and y ¢(0) = l1c1 + l2 c2 = 0 . Solving these simultaneous equations for c1 and c2 gives us c1 =

(l e y= 2

l1 t

- l1e l 2 t

l2 - l1

l2 y0 - l1 y0 and c1 = , and thus the unique solution is l2 - l1 l2 - l1

)y . 0

7 (b).

Ê g g 2 - 4 mk ˆ Ê g g 2 - 4 mk ˆ g2 g 2 - 4 mk + = l1l 2 = Á ˜Á ˜ 2 2m 2m 4m2 Ë 2m ¯ Ë 2m ¯ 4m k 4 mk k 2k m . Therefore, lim l 2 = 0 . = fi = =l1l 2 = l 2 2 g Æ• 4m m l1 g + g 2 - 4 mk

g Ê -g ˆ Ê g ˆ Then, since l1 + l 2 = 2Á ˜ = - , lim l1 = lim Á - - l 2 ˜ = -• . g Æ• Ë m Ë 2m ¯ ¯ m g Æ• 7 (c).

ÈÊ l 2el 1 t - l1el 2 t ˆ ˘ ÈÊ l 2el 1 t l1el 2 t ˆ ˘ lim y ( t) = lim ÍÁ ÍÁ ˜ y ˙ = y 0 glim ˜˙ g Æ• g Æ• Ë Æ• Ë l - l l 2 - l1 ¯ 0 ˚ l 2 - l1 ¯ ˚ 1 Î Î 2

È el 2 t ˘ È 1 ˘ = y 0 Í0 - lim l 2 = y 0 . As damping increases, the motion becomes suppressed; ˙ = y 0 ÍÎ 0 - 1 ˙˚ ÍÎ g Æ• l 1 - 1 ˙˚ the system “locks up” and tends to stay at its initial displacement. 8 (a).

y ¢¢ + gy ¢ + y = 0, y (0) = 1, y ¢ (0) = 0, l2 + gl + 1 = 0 fi l = 2 g crit - 4 = 0 fi g crit = 2.

-g ± g 2 - 4 , 2

Chapter 4 Second Order Linear Differential Equations • 77

8 (b).

The plots are consistent. For a fixed t, y is tending toward y 0 = 1 as g increases. 9.

For this problem, we must make m =

rg k g and l = . The volume of the drum is rL m m

2

Ê 5ˆ V = pÁ ˜ 8 = 50 p cubic feet. Therefore, Ë 2¯

rl weight of equiv. vol. of water 50 p(62.4 ) k rl g Ê 32 ˆ = = = 1.634 . = = 1.634 ◊ Á ˜ = 6.535 s-2 Ë 8¯ weight of drum 6000 r m r L and 10.

g = 0.1 s-1, m = 5 kg, k = 32.675 N m, g = 0.5 kg s . m

my ¢¢ + gy ¢ + ky = 0, y (0) = 0, y ¢ (0) = y 0¢ , ml2 + gl + k = 0 fi l =

10 (a). Underdamped: l = -a ± ib, a =

g 2m

-g ± g 2 - 4 mk . 2m

4 mk - g 2 2m

, b=

y = c1e -at cos bt + c 2e -at sin bt, y (0) = c1 = 0, y ¢ (0) = bc 2 = y 0¢ \ y ( t) =

y 0¢ b

e -at sin bt.

Critically damped: g

y = c1e

- 2m t

g

+ c 2 te

- 2m t

Overdamped: l = -

g

, y (0) = c1 = 0, y ¢ (0) = c 2 = y 0¢ \ y ( t) = y 0¢ te

g g 1 ± g 2 - 4 mk = ± 2m 2m 2m

g2 4m2

-

- 2m t

[

]

g

10 (b). As g Æ g crit , e

- 2m t

Æe

-

k mt

1 g

2

4m2

g

e -

k m

- 2m t

È Í- e ÍÎ

g2 4 m2

- mk t

k mt

.

+e

g2 4 m2

- mk t

y 0¢ = -c1. (l 2 - l1 )

˘ ˙ y 0¢ . ˙˚

sinh( xt) sin( xt) = t, lim = t. xØ 0 x≠ 0 x x

and use: lim

-

k m

y = c1el 1 t + c 2el 2 t , c1 + c 2 = 0, l1c1 + l 2c 2 = y 0¢ = (l 2 - l1 )c 2 \ c 2 =

1 y ( t) = el 1 t - el 2 t y 0¢ = l 2 - l1 2

= y 0¢ te

78 • Chapter 4 Second Order Linear Differential Equations

Section 4.8 1 (a).

y P¢ = 3, y P¢¢ = 0, 0 - 2( 3) - 3( 3t - 1) = -9 t - 6 + 3 = -9 t - 3.

1 (b). l2 - 2l - 3 = (l - 3)(l + 1) = 0 , and thus yC = c1e - t + c 2e 3 t . 1 (c).

y = c1e - t + c 2e 3 t + 3t - 1. From the initial conditions, we have y (0) = c1 + c 2 - 1 = 1 and y ¢ (0) = -c1 + 3c 2 + 3 = 3. Solving these simultaneous equations yields

c1 =

3 1 3 1 and c 2 = , and so y = e - t + e 3 t + 3t - 1. 2 2 2 2

2 (b).

yC = c1e - t + c 2e 3 t .

2 (c).

1 y = c1e - t + c 2e 3 t - e 2 t . 3 y (0) = c1 + c 2 - 13 = 1 and y ¢ (0) = -c1 + 3c 2 - 23 = 0 fi c1 =

5 1 and c 2 = , and so 6 2

y = 56 e - t + 12 e 3 t - 13 e 2 t . 3 (a).

y P¢ = 8e 4 t , y P¢¢ = 32e 4 t , e 4 t ( 32 - 8 - 2(2)) = 20e 4 t .

3 (b). l2 - l - 2 = (l - 2)(l + 1) = 0 , and thus yC = c1e - t + c 2e 2 t . 3 (c).

y = c1e - t + c 2e 2 t + 2e 4 t . From the initial conditions, we have y (0) = c1 + c 2 + 2 = 0 and y ¢ (0) = -c1 + 2c 2 + 8 = 1. Solving these simultaneous equations yields c1 = 1 and c 2 = -3 , and so y = e - t - 3e 2 t + 2e 4 t .

4 (b).

yC = c1e - t + c 2e 2 t .

4 (c).

y = c1e - t + c 2e 2 t - 5 . y (-1) = c1e1 + c 2e -2 - 5 = 0 and y ¢ (-1) = -c1e1 + 2c 2e -2 = 1 fi c1 = 3e -1 and c 2 = 2e 2 , and so y = 3e -( t +1) + 2e 2( t +1) - 5 .

5 (a).

2 + (2 t - 2) = 2 t .

5 (b). l2 + l = l (l + 1) = 0 , and thus yC = c1e - t + c 2 . 5 (c).

y = c1e - t + c 2 + t 2 - 2 t . From the initial conditions, we have y (1) = c1e -1 + c 2 - 1 = 1 and y ¢ (1) = -c1e -1 + 2 - 2 = -2 . Solving these simultaneous equations yields c1 = 2e and c 2 = 0 , and so y = 2e -( t -1) + t 2 - 2 t .

6 (b).

yC = c1e - t + c 2 .

6 (c).

y = c1e - t + c 2 - 2 te - t . y (0) = c1 + c 2 = 2 and y ¢ (0) = -c1 - 2 = 2 fi c1 = -4 and c 2 = 6 , and so y = -4 e - t + 6 - 2 te - t .

7 (a).

y P¢ = 2 - 2 sin 2 t, y P¢¢ = -4 cos 2 t, - 4 cos 2 t + (2 t + cos 2 t) = 2 t - 3 cos 2 t .

Chapter 4 Second Order Linear Differential Equations • 79

7 (b). l2 + 1 = 0 , and thus yC = c1 cos t + c 2 sin t . 7 (c).

y = c1 cos t + c 2 sin t + 2 t + cos 2 t . From the initial conditions, we have y (0) = c1 + 1 = 0 and y ¢ (0) = c 2 + 2 = 0 . Solving these simultaneous equations yields c1 = -1 and c 2 = -2 , and so y = - cos t - 2 sin t + 2 t + cos 2 t .

8 (b).

yC = c1 cos 2 t + c 2 sin 2 t .

8 (c).

y = c1 cos 2 t + c 2 sin 2 t + 2e t -p . y (p ) = c1 + 2 = 2 and y ¢ (p ) = 2c 2 + 2 = 0 fi c1 = 0 and c 2 = -1, and so y = - sin2 t + 2e t -p .

9 (a). 10 - 2(10( t + 1)) + 10( t + 1) 2 = 10 - 20 t - 20 + 10 t 2 + 20 t + 10 = 10 t 2 . 9 (b). l2 - 2l + 2 = 0, so l = 1 ± i , and thus yC = c1e t cos t + c 2e t sin t . 9 (c).

y = c1e t cos t + c 2e t sin t + 5( t + 1) 2 . From the initial conditions, we have y (0) = c1 + 5 = 0 and y ¢ (0) = c1 + c 2 + 10 = 0 . Solving these simultaneous equations yields c1 = -5 and c 2 = -5 , and so y = -5e t cos t - 5e t sin t + 5( t + 1) 2 .

10 (b). yC = c1e t cos t + c 2e t sin t . 10 (c). y = c1e t cos t + c 2e t sin t + 2 cos t + sin t . p

p

p

y ( p2 ) = c 2e 2 + 1 = 1 and y ¢ ( p2 ) = -c1e 2 + c 2e 2 - 2 = 0 fi c1 = -2e

y = -2e

t - p2

- p2

and c 2 = 0 , and so

cos t + 2 cos t + sin t .

Ê t2 ˆ t 1 t2 t 1 1 2 t 2 2 t t 11 (a). y P¢ = (2 t + t )e , y P¢¢ = (2 + 4 t + t )e , and Á + 2 t + 1˜ e - 2 ◊ (2 t + t )e + e = e t . 2 2 2 2 Ë2 ¯ 11 (b). l2 - 2l + 1 = (l - 1) 2 = 0 , and thus yC = c1e t + c 2 te t . t2 t 11 (c). y = c1e + c 2 te + e . From the initial conditions, we have 2 t

t

y (0) = c1 = -2 and y ¢ (0) = c1 + c 2 = 2 . Solving these simultaneous equations yields c1 = -2 and c 2 = 4 , and so y = -2e t + 4 te t +

t2 t e. 2

12 (b). yC = c1e t + c 2 te t . 12 (c). y = c1e t + c 2 te t + t 2 + 4 t + 10 + cos t . y (0) = c1 + 10 + 1 = 1 and y ¢ (0) = c1 + c 2 + 4 = 3 fi c1 = -10 and c 2 = 9 , and so

y = -10e t + 9 te t + t 2 + 4 t + 10 + cos t . 13.

First, y P = a1u + a2v . Now we have

y P¢¢ + p( t) y P¢ + q( t) y P = a1 ( u¢¢ + p( t) u¢ + q( t) u) + a2 (v ¢¢ + p( t)v ¢ + q( t)v ) = a1g1 + a2 g2

80 • Chapter 4 Second Order Linear Differential Equations

2 1 1 t 1 2 = 2e + 1 + [ 3t] . Thus y P = u1 + u3 . 3 2 2 2 3

[

]

14.

e t + 2t +

15.

4 e - t - 2 = 2 2e - t - t - 1 +

16.

1 1 1 1 1 1 1 1 cosh t = e t + e - t = 2e t + 1 + 2e - t - t - 1 + [ 3t] . Thus y P = u1 + u2 + u3 . 12 4 4 4 12 2 2 4

17.

Differentiation gives us y P¢ = 2e 2 t - 2 t, y P¢¢ = 4 e 2 t - 2 . From the given differential equation, we

[

]

2 2 [3t]. Thus y P = 2u2 + u3 . 3 3

[

]

[

]

have g( t) = 4 e 2 t - 2 + 2e 2 t - 2 t - e 2 t + t 2 = 5e 2 t + t 2 - 2 t - 2 . 18.

1 -1 1 -3 Differentiation gives us y P¢ = 3 + t 2 , y P¢¢ = - t 2 . From the given differential equation, we 2 4 1 -3 1 -1ˆ 1 -3 Ê -1 have g( t) = - t 2 - 2Á 3 + t 2 ˜ = -6 - t 2 - t 2 . Ë 4 2 ¯ 4

19.

Differentiation gives us y P¢ = 3, y P¢¢ = 0 . From the given differential equation, we have

g( t) = t ◊ 0 + e t ◊ 3 + 2 ◊ 3t = 3e t + 6 t . 20.

Differentiation gives us y P¢ = have g( t) = -

21.

1 1 , y P¢¢ = . From the given differential equation, we 1+ t (1 + t) 2

1 + ln(1 + t), t > -1. (1 + t) 2

Differentiation gives us y P¢ = - sin t, y P¢¢ = - cos t . From the given differential equation, we have

g( t) = - cos t - sin 2 t + 2 t cos t = (2 t - 1) cos t - sin 2 t . 22.

(l - 1)(l - 2) = l2 - 3l + 2 = 0 fi a = -3, b = 2 . Differentiation gives us y P¢ = -4 e -2 t , y P¢¢ = 8e -2 t , and so g( t) = 8e -2 t - 3(-4 e -2 t ) + 2(2e -2 t ) = 24 e -2 t .

23.

l = -1, 0 fi (l + 1)l = 0, so l2 + l = 0 and a = 1, b = 0 . Differentiation gives us y P¢ = 2 t, y P¢¢ = 2 , and so g( t) = 2 + 2 t .

24.

(l - 1) 2 = l2 - 2l + 1 = 0 fi a = -2, b = 1. Differentiation gives us y P¢ = ( t 2 + 2 t)e t , y P¢¢ = ( t 2 + 4 t + 2)e t , and so g( t) = ( t 2 + 4 t + 2)e t - 2( t 2 + 2 t)e t + t 2e t = 2e t .

25.

2

l = 1 ± i fi (l - 1) = -1, so l2 - 2l + 2 = 0 and a = -2, b = 2 . Differentiation gives us y P¢ = e t + cos t, y P¢¢ = e t - sin t , and so

[

] [

]

g( t) = e t - sin t - 2 e t + cos t + 2 e t + sin t = e t - 2 cos t + sin t .

26.

l2 + 4 = 0 fi a = 0, b = 4 . Differentiation gives us y P¢ = cos t, y P¢¢ = - sin t , and so g( t) = - sin t + 4 (-1 + sin t) = 3 sin t - 4 .

Chapter 4 Second Order Linear Differential Equations • 81

Section 4.9 1 (a).

l2 - 4 = 0 fi yC = c1e -2 t + c 2e 2 t

1 (b).

y P = A2 t 2 + A1t + A0 , y P¢ = 2 A2 t + A1, y P¢¢ = 2 A2 . 1 y P¢¢ - 4 y P = 2 A2 - 4 ( A2 t 2 + A1t + A0 ) = 4 t 2 fi A0 = - , A1 = 0, A2 = -1. 2 Therefore, y P = - t 2 -

1 2 1 2

1 (c).

y = c1e -2 t + c 2e 2 t - t 2 -

2 (a).

yC = c1e -2 t + c 2e 2 t

2 (b).

1 y P = - sin 2 t . 8

2 (c).

1 y = c1e -2 t + c 2e 2 t - sin 2 t . 8

3 (a).

l2 + 1 = 0 fi yC = c1 cos t + c 2 sin t

3 (b).

y P = Ae t , y P¢ = Ae t , y P¢¢ = Ae t . y P¢¢ + y P = 2 Ae t = 8e t fi A = 4 . Therefore, y P = 4 e t

3 (c).

y = c1 cos t + c 2 sin t + 4 e t

4 (a).

yC = c1 cos t + c 2 sin t

4 (b).

2 1 y P = - e t cos t + e t sin t . 5 5

4 (c).

2 1 y = c1 cos t + c 2 sin t - e t cos t + e t sin t . 5 5

5 (a).

(l - 2) 2 = 0 fi yC = c1e 2t + c 2 te 2t

5 (b).

y P = At 2e 2 t , y P¢ = (2 At 2 + 2 At)e 2 t , y P¢¢ = ( 4 At 2 + 8 At + 2 A)e 2 t . 1 t 2 2t y P¢¢ - 4 y P¢ + 4 y P = e fi A = . Therefore, y P = e 2 2 2t

t 2 2t e 2

5 (c).

y = c1e 2 t + c 2 te 2 t +

6 (a).

yC = c1e 2 t + c 2 te 2 t

6 (b).

1 y P = cos 2 t + 2 . 8

6 (c).

1 y = c1e 2 t + c 2 te 2 t + cos 2 t + 2 . 8

82 • Chapter 4 Second Order Linear Differential Equations

7 (a).

l2 + 2l + 2 = 0 fi l = -1 ± i fi yC = c1e - t cos t + c 2e - t sin t

7 (b).

y P = A3 t 3 + A2 t 2 + A1t + A0 , y P¢ = 3 A3 t 2 + 2 A2 t + A1, y P¢¢ = 6 A3 t + 2 A2 . 3 3 1 y P¢¢ + 2 y P¢ + 2 y P = t 3 fi A0 = 0, A1 = , A2 = - , A3 = . 2 2 2 Therefore, y P =

1 3 3 2 3 t - t + t 2 2 2

7 (c).

1 3 3 y = c1e - t cos t + c 2e - t sin t + t 3 - t 2 + t 2 2 2

8 (a).

yC = c1e 2 + c 2e 2 t

8 (b).

y P = - te t + e t .

8 (c).

y = c1e 2 + c 2e 2 t - te t + e t .

9 (a).

yC = c1e - t cos t + c 2e - t sin t

9 (b).

y P = A0e - t + A1 cos t + A2 sin t, y P¢ = - A0e - t - A1 sin t + A2 cos t,

t

t

1 2 y P¢¢ = A0e - t - A1 cos t - A2 sin t . y P¢¢ + 2 y P¢ + 2 y P = e - t + cos t fi A0 = 1, A1 = , A2 = 5 5 1 2 Therefore, y P = e - t + cos t + sin t 5 5 9 (c).

1 2 y = c1e - t cos t + c 2e - t sin t + e - t + cos t + sin t 5 5

10 (a). yC = c1e - t + c 2 10 (b). y P = 2 t 3 - 6 t 2 + 12 t . 10 (c). y = c1e - t + c 2 + 2 t 3 - 6 t 2 + 12 t . t 1 2 11 (a). 2l - 5l + 2 = (2l - 1)(l - 2) = 0 fi l = , 2 fi yC = c1e + c 2e 2 t 2 2

t t

11 (b). y P = A0 te 2 . Substituting into the differential equation yields y P = 2 te 2 . t 2

2t

11 (c). y = c1e + c 2e + 2 te

t 2

12 (a). yC = c1e - t + c 2

1 1 12 (b). y P = - cos t + sin t . 2 2 1 1 12 (c). y = c1e - t + c 2 - cos t + sin t . 2 2 2

2

t 3

13 (a). 9l - 6l + 1 = ( 3l - 1) = 0 fi yC = c1e + c 2 te

t 3

Chapter 4 Second Order Linear Differential Equations • 83 t

13 (b). y P = ( A1t 3 + A0 t 2 )e 3 . Substituting into the differential equation yields y P =

1 3 3t te 6

t t 1 t 13 (c). y = c1e 3 + c 2 te 3 + t 3e 3 6

14 (a). yC = c1e -2 t cos t + c 2e -2 t sin t 14 (b). y P = t -

5 e-t . + 4 2

14 (c). y = c1e -2 t cos t + c 2e -2 t sin t + t -

5 e-t . + 4 2

15 (a). l2 + 4 l + 5 = 0 fi l = -2 ± i fi yC = c1e -2 t cos t + c 2e -2 t sin t 15 (b). y P = Ae -2 t + B1 cos t + B2 sin t . Substituting into the differential equation yields

1 1 y P = 2e -2 t + cos t + sin t . 8 8 1 1 15 (c). y = c1e -2 t cos t + c 2e -2 t sin t + 2e -2 t + cos t + sin t 8 8 16 (a). yC = c1e - t + c 2e 3 t 16 (b). y P = Ae - t cos t + Be - t cos t + C2 t 2 + C1t + C0 + t( D1 + D0 )e 3 t

= Ae - t cos t + Be - t cos t + C2 t 2 + C1t + C0 + D1t 2e 3 t + D0 te 3 t 17 (a). yC = c1 cos 3t + c 2 sin 3t 17 (b). y P = t( A2 t 2 + A1t + A0 ) cos 3t + t( B2 t 2 + B1t + B0 ) sin 3t + C cos t + D sin t = ( A2 t 3 + A1t 2 + A0 t) cos 3t + ( B2 t 3 + B1t 2 + B0 t) sin 3t + + C cos t + D sin t

18 (a). yC = c1 + c 2e t 18 (b). y P = t( A2 t 2 + A1t + A0 ) + t( B2 t 2 + B1t + B0 )e t = A2 t 3 + A1t 2 + A0 t + ( B2 t 3 + B1t 2 + B0 t)e t 19 (a). l2 - 2l + 2 = (l - 1) 2 + 1 = 0 ; yC = c1e t cos t + c 2e t sin t 19 (b). y P = Ae - t cos 2 t + Be - t sin 2 t + C1t + C0 + e - t ( D1t + D0 ) cos t + e - t ( E1t + E 0 ) sin t 20 (a). yC = c1e t + c 2e - t 20 (b). y P = Ate t + Bte - t + Ce 2 t + De -2 t 21 (a). yC = c1 cos 2 t + c 2 sin 2 t 21 (b). Using sin(2 t) = 2 sin t cos t and cos(2(2 t)) = 2 cos2 2 t - 1,

1 1 1 sin t cos t + cos2 2 t = sin 2 t + + cos 4 t . Therefore, 2 2 2 y P = At cos 2 t + Bt sin 2 t + C + D cos 4 t + E sin 4 t

84 • Chapter 4 Second Order Linear Differential Equations

22 (a). yC = c1 sin 2 t + c 2 cos 2 t 22 (b). y P = Ae -2 t + B + Ce 2 t 23.

(l + 1)(l - 2) = l2 - l - 2 = 0 , so a = -1 and b = -2 . y ¢¢ - y ¢ - 2 y = 4 t , which leads to the general solution of y = c1e - t + c 2e 2 t - 2 t + 1.

24.

l (l + 1) = l2 + l = 0 , so a = 1 and b = 0 . y ¢¢ + y ¢ = t , which leads to the general solution of y = c1 + c 2e - t +

25.

t2 - t. 2

(l + 2)(l + 2) = l2 + 4l + 4 = 0 , so a = 4 and b = 4 . y ¢¢ + 4 y ¢ + 4 y = 5sin t , which leads to the general solution of y = c1e -2 t + c 2 te -2 t -

4 3 cos t + sin t . 5 5

26.

1 a = 0 and b = 1, y = c1 cos t + c 2 sin t + t - sin 2 t . 3

27.

l = -1 ± i2, so (l + 1) = -4 and thus l2 + 2l + 5 = 0 . Therefore. a = 2 and b = 5 .

2

y ¢¢ + 2 y ¢ + 5 y = 8e - t , which leads to the general solution of y = c1e - t cos 2 t + c 2e - t sin 2 t + 2e - t . 28.

Since y P = t( A1t + A0 ) + B0 te 3 t , we know that 1 and e 3 t are solutions and l2 - 3l = 0 , and thus

a = -3 and b = 0 . 29.

Since y P = A0 te 2 t + B0 te -2 t + C1t + C0 , we know that e 2 t and e -2 t are solutions of the homogeneous differential equation. This means that l = ±2 , so (l + 2)(l - 2) = l2 - 4 = 0 , and thus a = 0 and b = -4 .

30.

We know that cos 2 t and sin 2 t are solutions and l2 + 4 = 0 , and thus a = 0 and b = 4 .

31 (a). Graph C 31 (b). Graph E 31 (c). Graph A 31 (d). Graph B 31 (e). Graph D

32.

W b = 200 lb. The weight of an equivalent weight of water is W b = 8(62.4 ) = 499.2 lb.

mb =

499.2 r 200 Ê 32 ˆ = 2.496 fi w 2 = 2.496Á ˜ fi w = 39.936 . . Note: l = Ë 2¯ 200 p 32

32 (a). y ¢¢ + w 2 y =

10 sin wt, y (0) = 0, y ¢ (0) = 0 . mb

Chapter 4 Second Order Linear Differential Equations • 85

32 (b). y c = c1 coswt + c 2 sin wt, y p = At coswt + Bt sin wt

y ¢p = A coswt - wAt sin wt + B sin wt + wBt coswt = ( A + wBt)coswt + ( B - wAt)sin wt y ¢¢p = wB coswt - w ( A + wBt)sin wt - wA sin wt + w ( B - wAt)coswt y ¢¢p + w 2 y p = wB coswt - wA sin wt - wA sin wt + wB coswt =

10 sin wt . mb

Therefore, B = 0, A = -

10 5 =and 2wmb wmb

y = c1 coswt + c 2 sin wt -

5 5 5 t coswt, y (0) = c1 = 0, y ¢ (0) = wc 2 = 0 fi c2 = 2 . wmb wmb w mb

y=

5 5 sin wt t coswt, w ª 6.3195 sec -1, mb = 6.25 slug . w mb wmb 2

32 (c). Before being put into motion, the block is floating with a depth Y submerged, where 62.4 ( 4 )Y = 200 fi Y = 0.80128... ft. Therefore, the model is valid if -0.80128... £ y £ 2 - 0.80128 . From the graph, y = -0.80128 at t ª 7 sec.

33.

y P¢¢ + p( t) y P¢ + q( t) y P = ( u¢¢ + iv ¢¢ ) + p( t)( u¢ + iv ¢ ) + q( t)( u + iv ) = ( u¢¢ + p( t) u¢ + q( t) u) + i(v ¢¢ + p( t)v ¢ + q( t)v ) = g1 ( t) + ig2 ( t) . The real and imaginary parts must be equal on both sides of the equation, so u¢¢ + p( t) u¢ + q( t) u = g1 ( t) and v ¢¢ + p( t)v ¢ + q( t)v = g2 ( t) .

34 (a). y ¢¢ - y = e i 2 t , y p = Ae i 2 t , y ¢p = i2 Ae i 2 t , y ¢¢p = -4 Ae i 2 t

1 -4 Ae i 2 t - Ae i 2 t = e i 2 t fi A = - . 5 1 1 1 34 (b). Ae i 2 t = - (cos 2 t + i sin 2 t) fi u = - cos 2 t, v = - sin 2 t . 5 5 5 35 (a). y P¢ = iAe it , y P¢¢ = - Ae it . - Ae it + 2iAe it + Ae it = 2iAe it = e it , so A =

i 1 i = - and y P = - e it . 2i 2 2

86 • Chapter 4 Second Order Linear Differential Equations

35 (b). y P = -

i i 1 1 1 (cos t + i sin t) = sin t - cos t . Thus u = sin t and v = - cos t . For the real function, 2 2 2 2 2

1 Ê1 ˆ 1 u¢¢ + 2 u¢ + u = - sin t + 2Á cos t˜ + sin t = cos t . For the imaginary function, Ë ¯ 2 2 2 1 Ê1 ˆ 1 v ¢¢ + 2v ¢ + v = cos t + 2Á sin t˜ - cos t = sin t . Ë2 ¯ 2 2

36 (a). y P¢ = iAe it , y P¢¢ = - Ae it . - Ae it + 4 Ae it = e it fi A = 36 (b). y P =

1 1 and y P = e it . 3 3

1 1 1 (cos t + i sin t) fi u = cos t, v = sin t . 3 3 3

37 (a). y P¢ = A(1 + i2 t)e i 2 t , y P¢¢ = (i2 + i2 - 4 t) Ae i 2 t = (-4 t + i 4 ) Ae i 2 t . (-4 t + i 4 ) Ae i 2 t + 4 Ate i 2 t = e i 2 t , so

A=-

i i and y P = - te i 2 t . 4 4

i t t t Ê t ˆ 37 (b). y P = - t(cos 2 t + i sin 2 t) = sin 2 t + iÁ - cos 2 t˜ . Thus u = sin 2 t and v = - cos 2 t . For the Ë 4 ¯ 4 4 4 4 Êt ˆ real function, u¢¢ + 4 u = cos 2 t - t sin 2 t + 4 Á sin 2 t˜ = cos 2 t . For the imaginary function, Ë4 ¯ Êt ˆ v ¢¢ + 4 v = sin 2 t + t cos 2 t - 4 Á cos 2 t˜ = sin 2 t . Ë4 ¯

38 (a). y P¢ = -i2 Ae - i 2 t , y P¢¢ = -4 Ae - i 2 t . (-4 - i2) Ae - i 2 t = e - i 2 t fi A =

-( 4 - i2) -1 1 1 = = - + i and 4 + i2 20 5 10

1ˆ Ê 1 y p = Á - + i ˜ e - i2t . Ë 5 10 ¯ 1 1 1 1 1ˆ Ê 1 38 (b). y P = Á - + i ˜ (cos 2 t - i sin 2 t) fi u = - cos 2 t + sin 2 t, v = sin 2 t + cos 2 t . Ë 5 10 ¯ 10 5 10 5 2

39 (a). y P¢ = (1 + i) Ae(1+ i) t , y P¢¢ = (1 + i) Ae(1+ i) t = i2 Ae(1+ i) t . i2 Ae(1+ i) t + Ae(1+ i) t = e(1+ i) t , so

A=

1 1 - i2 Ê 1 2ˆ and y P = Á - i ˜ e(1+ i) t . = Ë 5 5¯ 1 + i2 5

2 2 Ê 1 2ˆ Ê1 ˆ Ê1 ˆ 39 (b). y P = Á - i ˜ e t (cos t + i sin t) = e t Á cos t + sin t˜ + ie t Á sin t - cos t˜ . Thus Ë 5 5¯ Ë5 ¯ Ë5 ¯ 5 5 2 2 Ê1 ˆ Ê1 ˆ u = e t Á cos t + sin t˜ and v = e t Á sin t - cos t˜ . For the real function, Ë5 ¯ Ë5 ¯ 5 5 2 2 Ê4 ˆ Ê1 ˆ u¢¢ + u = e t Á cos t - sin t˜ + e t Á cos t + sin t˜ = e t cos t . For the imaginary function, Ë5 ¯ Ë5 ¯ 5 5 4 1 Ê2 ˆ Ê 2 ˆ v ¢¢ + v = e t Á cos t + sin t˜ + e t Á - cos t + sin t˜ = e t sin t . Ë5 ¯ Ë 5 ¯ 5 5

Chapter 4 Second Order Linear Differential Equations • 87

Section 4.10 1 (a).

yC = c1 cos 2 t + c 2 sin 2 t

1 (b).

sin 2 t ˘È u1¢ ˘ È 0 ˘ È cos 2 t Í-2 sin 2 t 2 cos 2 t ˙Íu¢ ˙ = Í2 sin 2 t ˙ , so Î ˚Î 2 ˚ Î ˚

È u1¢ ˘ 1 È2 cos 2 t - sin 2 t ˘È 0 ˘ È - sin 2 2 t ˘ ˙. Íu¢ ˙ = 2 Í 2 sin 2 t cos 2 t ˙Í2 sin 2 t ˙ = Í Î 2˚ Î ˚Î ˚ Îsin 2 t cos 2 t ˚

sin 2 2 t t 1 . Thus Antidifferentiation gives us u1 = - + sin 4 t and u2 = 4 2 8

1 t 1 y P = - cos 2 t + sin 4 t cos 2 t + sin 3 2 t and 4 8 2 1 t 1 y = c1 cos 2 t + c 2 sin 2 t - cos 2 t + sin 4 t cos 2 t + sin 3 2 t . 4 8 2 1 (c).

y P = At sin 2 t + Bt cos 2 t , y P¢ = ( A - 2 Bt)sin 2 t + ( B + 2 At)cos 2 t , y P¢¢ = (-4 B - 4 At)sin 2 t + ( 4 A - 4 Bt)cos 2 t . -4 B sin 2 t + 4 A cos 2 t = 2 sin 2 t , and thus

1 1 A = 0, B = - , and y P = - t cos 2 t . Combining the particular solution with the 2 2 t complementary solution gives us y = C1 cos 2 t + C2 sin 2 t - cos 2 t . 2 To reconcile, y P = 18 sin 4 t cos 2 t + 14 sin 3 2 t = 14 sin 2 t cos2 2 t + 14 sin 3 2 t = 14 sin 2 t . Therefore, the

t solution in (b) can be written y = c1 cos 2 t + (c 2 + 1 4 )sin 2 t - cos 2 t . 2 2 (a).

yC = c1 cos t + c 2 sin t

È cos t sin t ˘È u1¢ ˘ È 0 ˘ 2 (b). Í ˙Í ˙ = Í ˙ , so Î- sin t cos t ˚Îu2¢ ˚ Îsec t ˚

sin t È u1¢ ˘ Ècos t - sin t ˘È 0 ˘ È- cos t˘ Íu¢ ˙ = Í sin t cos t ˙Ísec t ˙ = Í 1 ˙ . Antidifferentiation gives Î 2˚ Î ˚Î ˚ Î ˚

us u1 = ln cos t and u2 = t . Thus y P = (cos t)ln cos t + t sin t and

y = c1 cos t + c 2 sin t + (cos t)ln(cos t) + t sin t, since cos t > 0 for - p2 < t < p2 . 2 (c). Method of undetermined coefficients is not applicable. 3 (a).

yC = c1e t + c 2 t 2e t t

3 (b).

Èe Í t ÍÎe

2 t

˘È u1¢ ˘ È0 ˘ te , so = t ˙Í (t + 2t)e ˙˚Îu2¢ ˙˚ ÍÎ t ˙˚ 2

È u1¢ ˘ 1 È( t + 2 t)e Íu¢ ˙ = 2 te 2 t Í t Î 2˚ Î -e 2

t

È t2 -t ˘ - t e ˘È0 ˘ Í- 2 e ˙ ˙Í ˙ = Í ˙. e t ˚Î t ˚ Í 1 e - t ˙ Î 2 ˚ 2 t

Ê t2 ˆ 1 Antidifferentiation gives us u1 = Á + t + 1˜ e - t and u2 = - e - t . Thus 2 Ë2 ¯ Ê t2 ˆ 1 y P = Á + t + 1˜ e - t e t - e - t t 2e t = t + 1 and y = c1e t + c 2 t 2e t + t + 1. 2 Ë2 ¯

88 • Chapter 4 Second Order Linear Differential Equations

3 (c). The method of undetermined coefficients is not applicable. 4 (a).

yC = c1e - t + c 2e t

4 (b).

È e-t Í -t Î-e

e t ˘È u1¢ ˘ È 0 ˘ ˙Í ˙ = Í 1 ˙ , so e t ˚Îu2¢ ˚ Î1+ e t ˚

È u1¢ ˘ 1 È e t Íu¢ ˙ = 2 Í - t Î 2˚ Îe

t

-e t ˘È 0 ˘ È- 12 1+e e t ˘ ˙ . Antidifferentiation gives us -t ˙Í 1 ˙ = Í e - t ˚Î1+ e t ˚ ÍÎ 12 1e+ e t ˙˚

1 1 1 1 et e-t u1 = - ln(1 + e t ) and u2 = - e - t + ln(1 + e - t ) . Thus y P = ln(1 + e t ) - + ln(1 + e - t ) 2 2 2 2 2 2 1 et e-t t and y = c1e + c 2e ln(1 + e ) - + ln(1 + e - t ). 2 2 2 -t

t

4 (c). The method of undetermined coefficients is not applicable. 5 (a).

yC = c1e - t + c 2e t

5 (b).

È e-t Í -t Î-e

e t ˘È u1¢ ˘ È 0 ˘ ˙Í ˙ = Í t ˙ , so e t ˚Îu2¢ ˚ Îe ˚

È u1¢ ˘ 1 È e t Í u¢ ˙ = 2 Í - t Î 2˚ Îe

È 1 2t ˘ -e t ˘È 0 ˘ Í- 2 e ˙ ˙Í t ˙ = Í ˙ . Antidifferentiation gives us e - t ˚Îe ˚ Í 1 ˙ Î 2 ˚

t 1 t t 1 1 u1 = - e 2t and u2 = . Thus y P = - e t + e t and y = c1e - t + c 2e t - e t + e t . 4 2 4 2 4 2 5 (c).

y P = Ate t , and differentiation gives us y P¢ = A(1 + t)e t and y P¢¢ = A(2 + t)e t . Then we have t t 1 A(2 + t)e t - Ate t = e t , and so A = , y P = e t , and y = C1e - t + C2e t + e t . 2 2 2 t To reconcile, the solution in (b) can be written y = c1e - t + (c 2 - 1 4 )e t + e t . 2

6 (a).

y1 = t 2 . Use Reduction of Order to obtain y 2 ( t). y 2 = t 2v, y 2¢ = 2 tv + t 2v ¢, y 2¢¢ = 2v + 4 tv ¢ + t 2v ¢¢ Therefore,

2v + 4 tv ¢ + t 2v ¢¢ - 4 v - 2 tv ¢ + 2v = 0 fi t 2v ¢¢ + 2 tv ¢ = ( t 2v ¢ )¢ fi v ¢ =

k1 k1 + k2 . 2 fiv = t t

Using v = t -1, y 2 = t, y c = c1t 2 + c 2 t . 6 (b).

È t 2 t ˘È u1¢ ˘ È 0 ˘ Í ˙Í ˙ = Í t ˙ , so Î2 t 1˚Îu2¢ ˚ Î1+ t 2 ˚

1 È u1¢ ˘ 1 È 1 - t ˘ È 0 ˘ È 1+ t 2 ˘ = = Í ˙ Í ˙ . Antidifferentiation gives us Íu¢ ˙ t 2 ÍÎ-2 t t 2 ˙˚Î1+tt 2 ˚ Î1+-tt2 ˚ Î 2˚

t 1 u1 = tan -1 t and u2 = - ln(1 + t 2 ) . Thus y P = t 2 tan -1 t - ln(1 + t 2 ) and 2 2 t y = c1t 2 + c 2 t + t 2 tan -1 t - ln(1 + t 2 ) . 2 6 (c). The method of undetermined coefficients is not applicable. 7 (a).

yC = c1e t + c 2 te t

Chapter 4 Second Order Linear Differential Equations • 89

7 (b).

Èe t Í t Îe

˘È u1¢ ˘ È 0 ˘ È u1¢ ˘ 1 È( t + 1)e t , so Í ˙ = 2 t Í = t (t + 1)e t ˙˚ÍÎu2¢ ˙˚ ÍÎe t ˙˚ Îu2¢ ˚ e Î -e te t

yP = 7 (c).

- te t ˘È 0 ˘ È- t ˘ ˙Í t ˙ = Í ˙ . Integrating gives us e t ˚Îe ˚ Î 1 ˚

t2 t 2 t t2 t t2 e + t e = e and y = c1e t + c 2 te t + e t . 2 2 2

y P = At 2e t , and differentiation gives us y P¢ = A( t 2 + 2 t)e t and y P¢¢ = A( t 2 + 4 t + 2)e t . Then we have A( t 2 + 4 t + 2)e t - 2 A( t 2 + 2 t)e t + At 2e t = e t , and so 1 t2 t2 A = , y P = e t , and y = cƒ1e t + cƒ2 te t + e t . 2 2 2

8 (a).

yC = c1 cos 6 t + c 2 sin 6 t

sin 6 t ˘È u1¢ ˘ È 0 ˘ È cos 6 t 8 (b). Í ˙Í ˙ = Í 3 ˙ , so Î-6 sin 6 t 6 cos 6 t ˚Îu2¢ ˚ Îcsc 6 t ˚ Antidifferentiation gives us u1 =

yP =

3

È u1¢ ˘ 1 È6 cos 6 t - sin 6 t ˘È 0 ˘ È- csc6 6 t ˘ Íu¢ ˙ = 6 Í 6 sin 6 t cos 6 t ˙Ícsc 3 6 t ˙ = Í cos 6 t ˙ . Î 2˚ Î ˚Î ˚ ÍÎ 6 sin 3 6 t ˙˚

1 1 cot(6t) and u2 = - csc 2 (6 t) . Thus 36 72

1 1 1 1 cos(6t)cot(6t) - csc(6 t) and y = c1 cos 6 t + c 2 sin 6 t + cos(6t)cot(6t) - csc(6 t) . 36 72 36 72

8 (c). The method of undetermined coefficients is not applicable. 9 (a).

yC = c1 sin t + c 2 t sin t

t sin t ˘È u1¢ ˘ È 0 ˘ È sin t 9 (b). Í ˙Í ˙ = Í ˙ , so Îcos t sin t + t cos t ˚Îu2¢ ˚ Ît sin t ˚ Antidifferentiation gives us u1 = -

2 È u1¢ ˘ 1 Èsin t + t cos t - t sin t ˘È 0 ˘ È- t ˘ = Í ˙. Íu¢ ˙ = sin 2 t Í - cos t sin t ˙˚ÍÎt sin t ˙˚ Î t ˚ Î 2˚ Î

t3 t2 t3 t3 t3 , u2 = , y P = - sin t + sin t = sin t , and 3 2 3 2 6

t3 y = c1 sin t + c 2 t sin t + sin t . 6 9 (c). The method of undetermined coefficients is not applicable. 10 (a). yC = c1t + c 2 t ln | t | 2 Èt t ln t ˘È u1¢ ˘ È 0 ˘ È u1¢ ˘ 1 Èln t + 1 - t ln t ˘È 0 ˘ È- 1t (ln t ) ˘ 1 1 ˙ , so Í ˙ = Í ˙= Í 10 (b). Í ˙. ˙Í ˙ = Í ˙Í t ˚ÍÎ t ln t ˙˚ ÍÎ 1t ln t ˙˚ Îu2¢ ˚ t Î -1 Î1 ln t + 1˚Îu2¢ ˚ ÍÎ t ln t ˙˚

Antidifferentiation gives us u1 = -

yP = -

1 1 3 2 ln t ) and u2 = (ln t ) . Thus ( 3 2

t t t t 3 3 3 3 ln t ) + (ln t ) = (ln t ) and y = c1t + c 2 t ln t + (ln t ) . ( 3 2 6 6

10 (c). The method of undetermined coefficients is not applicable. 11 (a). yC = c1t + c 2e t

90 • Chapter 4 Second Order Linear Differential Equations

Èt e t ˘È u1¢ ˘ È 0 ˘ È u1¢ ˘ 1 11 (b). Í , so = = ˙ Í Í ˙ ˙ Í ˙ t t t Îu2¢ ˚ ( t - 1)e Î1 e ˚Îu2¢ ˚ Î( t - 1)e ˚

gives us u1 = -e t , u2 =

È e t -e t ˘È 0 ˘ È-e t ˘ Í ˙Í ˙ . Antidifferentiation t˙= Í Î-1 t ˚Î( t - 1)e ˚ Î t ˚

Ê t2 ˆ t2 t2 , y P = - te t + e t , and y = c1t + c 2e t + Á - t˜ e t . 2 2 Ë2 ¯

11 (c). The method of undetermined coefficients is not applicable. 2

12 (a). yC = c1e - t + c 2 te - t 2

È e-t 12 (b). Í -t 2 ÍÎ-2 te

2

2

˘È u1¢ ˘ È 0 ˘ te - t 2 ˙Í ˙ = Í 2 - t 2 ˙ , so (1 - 2 t 2 )e - t ˙˚Îu2¢ ˚ Ît e ˚

Antidifferentiation gives us u1 = y = c1e

-t 2

+ c 2 te

-t 2

È u1¢ ˘ 2 t 2 È(1 - 2 t 2 )e - t Íu¢ ˙ = e Í -t 2 ÍÎ 2 te Î 2˚

2

- te - t 2 e-t

2

˘È 0 ˘ È- t 3 ˘ ˙Í 2 - t 2 ˙ = Í 2 ˙ . ˙˚Ît e ˚ Î t ˚

t4 t3 t4 2 t4 2 t4 2 and u2 = . Thus y P = - e -t + e -t = e -t and 4 3 4 3 12

t 4 -t 2 + e . 12

12 (c). The method of undetermined coefficients is not applicable. 2

2

13 (a). y1 = ( t - 1) , and using reduction of order we have y 2 = ( t - 1) v . Differentiation yields 2

2

y 2¢ = 2( t - 1)v + ( t - 1) v ¢ and y 2¢¢ = 2v + 4 ( t - 1)v ¢ + ( t - 1) v ¢¢ . Then we have

(t - 1) 4 v ¢¢ + 4(t - 1) 3 v ¢ + 2(t - 1) 2 v - 4(t - 1)[2(t - 1)v + ( t - 1) 2 v ¢ ] + 6( t - 1) 2 v = 0 . Thus (t - 1) 4 v ¢¢ = 0 , and antidifferentiation of v ¢¢ = 0 gives us v = k1( t - 1) + k2 . Then y 2 = (t - 1) 3 , 2

and so yC = c1 ( t - 1) + c 2 ( t - 1)

3

È( t - 1) 2 ( t - 1) 3 ˘È u1¢ ˘ È 0 ˘ t ˙ , so 13 (b). Í ˙= Í 2 ˙Í ÍÎ2( t - 1) 3( t - 1) ˙˚Îu2¢ ˚ ÍÎ ( t - 1) 2 ˙˚ È u1¢ ˘ 1 4 Í u¢ ˙ = Î 2 ˚ ( t - 1) -1

u1 = ( t - 1) +

2

È 3( t - 1) Í ÍÎ-2( t - 1)

t ˘ È 0 ˘ È Í -( t - 1) ˘Í t ˙ (t - 1) 3 ˙˙ Í . Antidifferentiation gives us = ˙ (t - 1) 2 ˙˚ÍÎ (t - 1) 2 ˙˚ Í t 4 ˙ ÍÎ ( t - 1) ˙˚ 3

1 1 1 t 1 1 1 1 (t - 1) -2, u2 = - (t - 1) -2 - ( t - 1) -3, y P = (t - 1) + - (t - 1) - = - , 3 2 3 2 3 2 2 2 2

3

and y = c1 ( t - 1) + c 2 ( t - 1) +

t 1 - . 2 3

13 (c). The method of undetermined coefficients is not applicable.

Chapter 4 Second Order Linear Differential Equations • 91

14 (a). y1 = e t . Use Reduction of Order to obtain y 2 ( t). y 2 = e t v, y 2¢ = e t v + e t v ¢, y 2¢¢ = e t (v + 2v ¢ + v ¢¢ ) Therefore, v ¢¢ + 2v ¢ + v - (2 +

2 t

)(v + v ¢) + (1 + 2t )v = 0 fi v ¢¢ - 2t v ¢ = 0

t3 fi ( t v ¢ )¢ = 0 fi t v ¢ = k1 fi v = k1 + k2 . 3 -2

-2

Using y 2 = t 3e t , y c = c1e t + c 2 t 3e t . Èe t 14 (b). Í t Îe

˘È u1¢ ˘ È 0 ˘ t 3e t ˙ = Í t ˙ , so 3 2 t ˙Í ( t + 3t )e ˚Îu2¢ ˚ Îe ˚

- t 3e t ˘È 0 ˘ È- 3t ˘ ˙Í t ˙ = Í ˙ . e t ˚Îe ˚ Î 31t 2 ˚

È u1¢ ˘ e -2 t È( t 3 + 3t 2 )e t Íu¢ ˙ = 3t 2 Í -e t Î 2˚ Î

t2 t -1 t2 t t2 t t2 t Antidifferentiation gives us u1 = - and u2 = - . Thus y P = - e - e = - e and 6 3 6 3 2 y = c1e t + c 2 t 3e t -

t2 t e. 2

14 (c). The method of undetermined coefficients is not applicable.

15.

È y2g ˘ È y1 y 2 ˘È u1¢ ˘ È0 ˘ È u1¢ ˘ 1 È y 2¢ - y 2 ˘È0 ˘ Í- W ˙ Í y ¢ y ¢ ˙Íu¢ ˙ = Íg ˙ fi Íu¢ ˙ = W Í- y ¢ y ˙Íg ˙ = Í y g ˙ . Antidifferentiation yields 1 ˚Î ˚ Î 2˚ Î 1 2 ˚Î 2 ˚ Î 1 Î ˚ Í 1 ˙ Î W ˚ u1 = - Ú

t

0

t [ y ( t) y (l ) - y ( t) y (l )] t y (l ) g(l ) y 2 (l ) g(l ) 1 1 2 g(l ) dl dl , and so y P = Ú 2 dl and u2 = Ú 1 0 0 W (l ) W (l ) W (l )

and y = c1 y1 + c 2 y 2 + Ú

t

0

[ y (t) y (l ) - y (t) y (l )]g(l )dl . 2

1

1

2

W (l )

y (0) = c1 y1 (0) + c 2 y 2 (0) = y 0 and

y ¢ (0) = c1 y1¢ (0) + c 2 y 2¢ (0) = y 0¢ , since y P (0) = y P¢ (0) = 0 .

16.

For this problem, we have y1 = cos 2 t, y 2 = sin 2 t, and W =

cos 2 t sin 2 t = 2 . Then we -2 sin 2 t 2 cos 2 t

sin 2 t cos 2l - cos 2 t sin 2l 1 t g(l ) dl = Ú sin(2( t - l )) g(l ) dl . Since y = y P , 0 2 2 0

have y P = Ú

t

a = 0, b = 4, y 0 = 0, y 0¢ = 0 . 17.

e-t For this problem, we have y1 = e , y 2 = e , and W = - t -e -t

yP = Ú

t

0

[e e

t -l

]

t

et = 2 . Then we have et

t - e - t el g(l ) dl = Ú sinh( t - l ) g(l ) dl . Thus we can see that y = e - t + y P , and so 0 2

a = 0, b = -1, y 0 = 1, y 0¢ = -1.

92 • Chapter 4 Second Order Linear Differential Equations

18.

For this problem, we have y1 = 1, y 2 = t, and W =

1 t 0 1

= 1. Then we have

t

y P = Ú [ t - l ]g(l ) dl . Thus we can see that y = t + y P ( t) , and so a = 0, b = 0, y 0 = 0, y 0¢ = 1 . 0

Section 4.11 1 (a).

yC = c1 cosw 0 t + c 2 sin w 0 t

1 (b). Case i: w π w 0 . y P = A coswt + B sin wt , and differentiation yields

(

)

(

)

y P¢¢ = -w 2 A coswt - w 2 B sin wt . Then we have w 0 2 - w 2 A coswt + w 0 2 - w 2 B sin wt = F coswt , and thus B = 0 and A =

F F coswt . 2 2 . The particular solution is then y P = w0 - w w0 - w 2 2

Case ii: : w = w 0 . y P = At cosw 0 t + Bt sin w 0 t , and differentiation yields

y P¢¢ = w 0 B cosw 0 t - w 0 ( A + w 0 Bt) sin w 0 t - w 0 A sin w 0 t + w 0 ( B - w 0 At) cosw 0 t

(

)

(

)

= 2w 0 B - w 0 2 At cosw 0 t + -2w 0 A - w 0 2 Bt sin w 0 t . Then we have 2w 0 B cosw 0 t - 2w 0 A sin w 0 t = F cosw 0 t , and thus A = 0 and B =

then y P = 2 (a).

F . The particular solution is 2w 0

F t sin w 0 t . 2w 0

ky = mg, k =

10 ◊ 9.8 = 1000 N m . 0.098

2 (b). 10 y ¢¢ + 1000 y = 20 cos(10 t); y ¢¢ + 100 y = 2 cos(10 t), y (0) = 0, y ¢ (0) = 0 . y P = At cos(10 t) + Bt sin(10 t) and differentiation yields y P¢¢ = (20 B - 100 At)cos(10 t) + (-20 A - 100 Bt)sin(10 t) . Then we have

y ¢¢p + 100 y p = 20 B cos(10 t) - 20 A sin(10 t) = 2 cos(10 t) fi B = so y P =

1 , A = 0 , and 10

t t sin(10 t) , and y = c1 cos10 t + c 2 sin10 t + sin(10 t) . From the initial conditions, we 10 10

have y (0) = c1 = 0 and y ¢ (0) = 10c 2 = 0 . Thus the unique solution is y =

t sin(10 t) . 10

Chapter 4 Second Order Linear Differential Equations • 93

2 (c). There is no maximum excursion.

3 (a).

ky = mg, k =

10 ◊ 9.8 = 1000 N m . 0.098

3 (b). 10 y ¢¢ + 1000 y = 20e - t ; y ¢¢ + 100 y = 2e - t , y (0) = 0, y ¢ (0) = 0 .

y P = Ae - t , and differentiation yields y P¢¢ = Ae - t . Then we have Ae - t + 100 Ae - t = 2e - t , and so A=

2 -t 2 2 -t , yP = e . From the initial conditions, we e , and y = c1 cos10 t + c 2 sin10 t + 101 101 101

have y (0) = c1 +

2 1Ê 2 ˆ 2 2 and c 2 = Á = 0 and y ¢ (0) = 10c 2 = 0 , and thus c1 = ˜. 101 10 Ë 101¯ 101 101

Thus the unique solution is y = 3 (c).

y max ª 0.035m.

4 (a).

ky = mg, k =

1 2 Ê -t ˆ Á - cos10 t + sin10 t + e ˜ . ¯ 10 101 Ë

10 ◊ 9.8 = 1000 N m . 0.098

4 (b). 10 y ¢¢ + 1000 y = 20 cos(8 t); y ¢¢ + 100 y = 2 cos(8 t), y (0) = 0, y ¢ (0) = 0 . y P = A cos(8 t) + B sin(8 t) and differentiation yields y P¢¢ = -64 A cos(8 t) - 64 B sin(8 t) . Then we

have y ¢¢p + 100 y p = 36 A cos(8 t) + 36 B sin(8 t) = 2 cos(8 t) fi A = so y P =

1 1 cos(8 t) , and y = c1 cos10 t + c 2 sin10 t + cos(8 t) . From the initial conditions, we 18 18

have y (0) = c1 +

y=-

1 , B = 0 , and 18

1 1 = 0 and y ¢ (0) = 10c 2 = 0 fi c1 = - , c 2 = 0 . Thus the unique solution is 18 18

1 1 cos(10 t) + cos(8 t) . 18 18

94 • Chapter 4 Second Order Linear Differential Equations

4 (c).

y max ª 0.11m.

5 (a). See 3 (a) 5 (b). On 0 £ t £ p : y ¢¢ + 100 y = 2, y (0) = 0, y ¢ (0) = 0 . y P = A , and differentiation yields y P¢¢ = 0 . Then we have 0 + 100 A = 2 , and so A =

1 1 1 , y P = , and y = c1 cos10 t + c 2 sin10 t + . From 50 50 50

the initial conditions, we have y (0) = c1 +

c1 = -

1 = 0 and y ¢ (0) = 10c 2 = 0 , and thus 50

1 1 1 and c 2 = 0 . Thus the unique solution is y = - cos10 t . At 50 50 50

t = p, y ( p) =

1 1 10 - cos(10 p) = 0 and y ¢ ( p) = sin(10 p) = 0 . Then we have 50 50 50

y ¢¢ + 100 y = 0, y ( p) = 0, y ¢ ( p) = 0 for t > p . y = c1 cos10 t + c 2 sin10 t fi y (p ) = c1 = 0, y ¢ (p ) = 10c 2 = 0. Thus y = 0 for this region.

2 = 0.04m . 50

5 (c).

y max =

6.

È1 ˘ y = 0.1sin(pt)sin( 7pt) = 0.1Í (cos(6pt) - cos(8pt)) ˙ = 0.05[cos(6pt) - cos(8pt] Î2 ˚

Therefore, y c = A cos(6pt) + B sin(6pt) and

k = (6p ) 2 . If y p = A cos(8pt) + B sin(8pt) , then m

-(8p ) 2 [ A cos(8pt) + B sin(8pt)] + (6p ) 2 [ A cos(8pt) + B sin(8pt)] = Therefore, (-64 + 36)p 2 A = and

20 cos(8pt) . m

20 1 20 5 , (-64 + 36)p 2 B = 0 fi A = =- 2 2 28p m 7p m m

5 k = 36p 2 , - 2 = -0.05 fi m = 1.447...kg, k = 36p 2 m = 514.2857...N / m . 7p m m

Chapter 4 Second Order Linear Differential Equations • 95

7 (a).

2 y ¢¢ + 8 y ¢ + 80 y = 20 cos 8 t, y (0) = 0, y ¢ (0) = 0 . For the complementary solution, we have

l2 + 4 l + 40 = 0, and so l = -2 ± i6 . Thus yC = c1e -2 t cos 6 t + c 2e -2 t sin 6 t . For the particular solution, we have y P = A cos 8 t + B sin 8 t , and differentiation yields y P¢ = -8 A sin 8 t + 8 B cos 8 t and y P¢¢ = -64 A cos 8 t - 64 B sin 8 t . Then we have

-64 A cos 8 t - 64 B sin 8 t + 4 (-8 A sin 8 t + 8 B cos 8 t) + 40( A cos 8 t + B sin 8 t) = 10 cos 8 t . Solving for

A and B yields A = -

1 3 1 3 and B = . Thus y = c1e -2 t cos 6 t + c 2e -2 t sin 6 t - cos 8 t + sin 8 t . From 5 20 5 20

the initial conditions, we have y (0) = c1 simultaneous equations yields c1 =

y=

8 3 = 0 and y ¢ (0) = -2c1 + 6c 2 + = 0 . Solving these 5 20

3 13 and c 2 = - , so 20 60

1 3 13 3 -2 t e cos 6 t - e -2 t sin 6 t - cos 8 t + sin 8 t . 5 20 60 20

7 (b). For sufficiently large t, y ( t) ª -

1 3 cos 8 t + sin 8 t , and so the limit does not exist. This equation is 5 20

called the steady state solution. 8 (a).

yC = c1e -2 t cos 6 t + c 2e -2 t sin 6 t . For the particular solution, we have y P = Ae - t , y ¢¢p + 4 y ¢p + 40 y p = 10e - t fi A - 4 A + 40 A = 10 fi A = y = c1e -2 t cos 6 t + c 2e -2 t sin 6 t + y (0) = c1 + y=-

10 . Thus 37

10 - t e . From the initial conditions, we have 37

10 5 10 10 . = 0 and y ¢ (0) = -2c1 + 6c 2 = 0 fi c1 = - , c 2 = 37 111 37 37

10 5 -2 t 10 -2 t e cos 6 t e sin 6 t + e - t . 37 111 37

8 (b). lim y ( t) = 0. t Æ•

9 (a).

yC = c1e -2 t cos 6 t + c 2e -2 t sin 6 t . For the particular solution, we have y P = A cos 6 t + B sin 6 t , and differentiation yields y P¢ = -6 A sin 6 t + 6 B cos 6 t and y P¢¢ = -36 A cos 6 t - 36 B sin 6 t . Then we have -36 A cos 6 t - 36 B sin 6 t + 4 (-6 A sin 6 t + 6 B cos 6 t) + 40( A cos 6 t + B sin 6 t) = 10 sin 6 t . Solving for

A and B yields A = -

5 30 5 30 . Thus y = c1e -2 t cos 6 t + c 2e -2 t sin 6 t cos 6 t + sin 6 t . and B = 74 74 74 74

From the initial conditions, we have y (0) = c1 -

30 30 = 0 and y ¢ (0) = -2c1 + 6c 2 + = 0. 74 74

96 • Chapter 4 Second Order Linear Differential Equations

Solving these simultaneous equations yields c1 =

y=

30 5 , so and c 2 = 74 74

5 30 5 -2 t 30 -2 t e cos 6 t + e sin 6 t cos 6 t + sin 6 t . 74 74 74 74

9 (b). For sufficiently large t, y ( t) ª -

5 30 cos 6 t + sin 6 t , and so the limit does not exist. This equation 74 74

is called the steady state solution. 10 (a). On 0 £ t £ p2 :

yC = c1e -2 t cos 6 t + c 2e -2 t sin 6 t . For the particular solution, we have y P = 1 . Thus 4

y = c1e -2 t cos 6 t + c 2e -2 t sin 6 t + y (0) = c1 +

1 . From the initial conditions, we have 4

1 1 1 = 0 and y ¢ (0) = -2c1 + 6c 2 = 0 fi c1 = - , c 2 = - . 4 12 4

1 Êp ˆ 1 1 1 1 1 1 Êp ˆ y = - e -2 t cos 6 t - e -2 t sin 6 t + , y Á ˜ = e -p + , y ¢ Á ˜ = - e -p + e -p = 0 . Ë 2¯ 4 Ë 2¯ 4 4 2 2 12 4

On

p 2

< t < •:

y = d1e -2 t cos 6 t + d2e -2 t sin 6 t . From the initial conditions, we have 1 Êp ˆ Êp ˆ y Á ˜ = - d1e -p = (1 + e -p ) and y ¢ Á ˜ = 2 d1e -p - 6 d2e -p = 0 Ë 2¯ Ë 2¯ 4

fi d1 = -

1 1 1 1 p È ˘ e + 1), d2 = - (ep + 1) . y = - (ep + 1) Íe -2 t cos 6 t + e -2 t sin 6 t ˙ . ( 3 4 12 4 Î ˚

10 (b). lim y ( t) = 0. t Æ•

11 (a). y ¢¢ + 2dy ¢ + w 0 2 y = F cosw 0 t, y (0) = 0, y ¢ (0) = 0 . l =

(

)

(

-2d ± 4d 2 - 4w 0 2 = -d ± i w 0 2 - d 2 . Thus 2

)

yC = c1e -dt cos t w 0 2 - d 2 + c 2e -dt sin t w 0 2 - d 2 . y P = A cosw 0 t + B sin w 0 t , and differentiation

yields y P¢ = -w 0 A sin w 0 t + w 0 B cosw 0 t and y P¢¢ = -w 0 2 A cosw 0 t - w 0 2 B sin w 0 t . Then we have y P¢¢ + 2dy P¢ + w 0 2 y P = 2d [-w 0 A sin w 0 t + w 0 B cosw 0 t] = F cosw 0 t . Solving for A and B yields A = 0 and B =

(

)

F . Thus 2dw 0

(

)

y = c1e -dt cos t w 0 2 - d 2 + c 2e -dt sin t w 0 2 - d 2 +

F sin w 0 t . 2dw 0

Chapter 4 Second Order Linear Differential Equations • 97

From the initial conditions, we have y (0) = c1 = 0 and y ¢ (0) = c 2 w 0 2 - d 2 +

(

2 2 -dt È F Í sin w 0 t e sin t w 0 - d , and so y = c1 = 0 and c 2 = 2d Í w 0 2d w 0 2 - d 2 w 02 - d 2 ÍÎ

F

(

2 2 2 2 -dt È F Í w 0 - d sin w 0 t - w 0e sin t w 0 - d 11 (b). First, let us rewrite y = 2Í dw 0 w 0 2 - d 2 ÍÎ

F = 0 . Thus 2d

) ˘˙. ˙ ˙˚

) ˘˙ ∫ F N (d ) . To use ˙ ˙˚

2 D(d )

L’Hopital’s Rule to find the limit, we need dN 1 = w 02 - d 2 dd 2

(

)

-

1 2

(

(-2d ) sin w 0 t + w 0 te -dt sin t w 0 2 - d 2

(

)

-w 0e -dt cos t w 0 2 - d 2 ◊

1 w 2 -d2 2 0

(

dD 1 = w 0 w 0 2 - d 2 + dw 0 ◊ w 0 2 - d 2 dd 2

(

lim+

d Æ0

)

-

1 2

)

-

1 2

)

(-2d ) t . Thus

(-2d ) . Thus

dN Æ 0 + w 0 t sin w 0 t + 0 as d Æ 0 . dd

dD Æ w 0 2 as d Æ 0 . Therefore, dd

F N (d ) F = t sin w 0 t . 2 D(d ) 2w 0

11 (c). For sufficiently large t, y ª

F sin w 0 t . Knowing m and k means that we know w 0 = 2d w 0

k . m

F F of the steady state solution and knowing F = , we 2dw 0 m

Therefore, by measuring the amplitude can determine d .

12 (a). y ¢¢ + 2dy ¢ + w 0 2 y = F cosw1t, y (0) = 0, y ¢ (0) = 0 .

(

)

(

)

yC = c1e -dt cos t w 0 2 - d 2 + c 2e -dt sin t w 0 2 - d 2 . y P = A cosw1t + B sin w1t , and differentiation

yields y P¢ = -w1 A sin w1t + w1B cosw1t and y P¢¢ = -w12 A cosw1t - w12 B sin w1t . Then we have

(

)

y P¢¢ + 2dy P¢ + w 0 2 y P = w 0 2 - w12 [ A cosw1t + B sin w1t] + 2d [-w1 A sin w1t + w1B cosw1t] = F cosw1t .

Solving for A and B yields A =

(

)

(w (w

2 0

2 0

)

- w12 F

2 2 1

-w

(

) + (2dw ) 1

2

and B =

2dw1F

(w

2 0

- w12

)

2

) + (2dw )

2

. Thus

1

y = c1e -dt cos t w 0 2 - d 2 + c 2e -dt sin t w 0 2 - d 2 + A cosw1t + B sin w1t . From the initial conditions,

we have y (0) = c1 + A = 0 and y ¢ (0) = -dc1 + c 2 w 0 2 - d 2 + w1B = 0 .

98 • Chapter 4 Second Order Linear Differential Equations

Thus c1 = - A and c 2 = -

(w

[(w ) + (2dw )

2 2

2

- w1

0

2

2

- w12

)

2 0

]

)

- w12 cosw1t + 2dw1 sin w1t -

1

( ]

Ï sin t w 0 2 - d 2 Ô 2 2 2 2 2 2 w 0 - w1 cos t w 0 - d + d w 0 - w1 + w1 (2dw1 ) 2 2 Ì w 02 - d 2 + (2dw1 ) Ô Ó

Fe -dt 0

, and so

w 02 - d 2

F

y=

(w

dA + w1B

(

(

(

)

)

) [(

(

)

)

12 (b). Using d w 0 2 - w12 + w1 (2dw1 ) = d w 0 2 + w12 ,

( ]

Ï sin t w 0 2 - d 2 Fe -dt Ô 2 2dw sin w t lim y ( t) = Ì 2dw w 1 Æw 0 w 02 - d 2 (2dw 0 )2 0 0 (2dw 0 )2 ÔÓ 0 Ï sin t w 0 2 - d 2 ¸Ô F Ô sin w 0 t -dt = -e ˝ Ì 2d Ô w 0 w 02 - d 2 Ô ˛ Ó F

[

(

) ¸Ô˝ Ô ˛

)

12 (c). lim+ y ( t) =

d Æ0

=

F

(w

2 0

2 2

- w1

)

{(w

2 0

)

}

- w12 cosw1t -

F

(w

2 0

- w1

2 2

)

{(w

2 0

)

}

- w12 cosw 0 t

F F cosw1t - cosw 0 t} = {cosw 0t - cosw1t}. 2 2 { w 0 - w1 w1 - w 0 2

(

)

2

(

)

Ê pˆ 13 (a). mx ¢¢ = - kx - mg cosÁ ˜ , x (0) = -10, x ¢ (0) = 0 . Ë 4¯

13 (b). x ¢¢ +

150 g k k = = 32s-2 . Thus the complementary solution is x=. 150 m 2 m 32

(

(

)

)

(

)

xC = c1 cos t 32 + c 2 sin t 32 . x P = A, so

(

)

(

)

x = c1 cos t 32 + c 2 sin t 32 -

g k 1 A=, and so A = and m 2 2

1 . 2

From the initial conditions, we have x (0) = c1 c1 =

1 = -10 and x ¢ (0) = 32c 2 = 0 . Thus 2

1 1 Ê 1 ˆ - 10˜ cos t 32 . - 10, c 2 = 0 , and x = Á Ë 2 ¯ 2 2

(

)

) ¸Ô˝ Ô ˛

Chapter 4 Second Order Linear Differential Equations • 99

1 ˆ Ê Differentiation gives us x ¢ ( t) = 32 Á10 ˜ sin t 32 . We need x ¢ when x = 0 , so Ë 2¯

(

(

)

cos t 32 =

1 2 1 - 10 2

(

)

)

. Solving for sin t 32 = .9971, and so x ¢ = 52.416 ft.s .

13 (c). Letting x and y represent horizontal and vertical coordinates which have their origin at the mouth o the “cannon,” we have y ¢¢ = - g, y (0) = 0, y ¢ (0) = initial value problem, we have y ¢ = - gt + y = 0 gives us t = 0,

v0 v and x ¢¢ = 0, x (0) = 0, x ¢ (0) = 0 . For the y 2 2

v0 gt 2 v 0 t and antidifferentiation yields y = . Setting + 2 2 2

2v 0 . Substituting the second value for t into the solution of the initial value g

v0 v0 v 0 v 0 2 (52.416) 2 = = 85.857ft. t= 2 = problem for x gives us x ( t) = g g 32 2 2 14 (b). (i). (ii).

1 v2 30 2 = - gy + C1, = 0 + C1 fi v 2 = -2 gy + 900, v (2) = (900 - 128) 2 = 772 ft / sec . 2 2

k 772 v2 k =( y - 2) 2 - g( y - 2) + C2 , C2 = fi v 2 = - ( y - 2) 2 - 2 g( y - 2) + 772 m 2 2 2m

v ( 3) = 0 fi 0 = -

15.

Vs = L

1 k 708 k - 2 g + 772 fi = 772 - 64 = 708, m = 16 fi k = = 1.3828 lb / ft . m 32 16( 32) m

1 dI 1 t dV dVs 1 = LI ¢¢ + I , I (0) = 0, I ¢ (0) = 0 , and therefore I ¢¢ + I = s . + Ú I (l ) dl , so dt C 0 dt dt C 4

Vs = 5 sin 3t, so I ¢¢ +

1 Ê tˆ Ê tˆ I = 15 cos 3t . Thus IC = c1 cosÁ ˜ + c 2 sinÁ ˜ and I P = A cos 3t + B sin 3t . Ë 2¯ Ë 2¯ 4

Differentiation gives us I P¢¢ = -9 A cos 3t - 9 B sin 3t , and then we have 1ˆ 12 ˜ B sin 3t = 15 cos 3t . Thus B = 0 and A = - , and so 4¯ 7

Ê Á -9 + Ë

1ˆ Ê ˜ A cos 3t + Á -9 + Ë 4¯

IP = -

12 Ê tˆ Ê t ˆ 12 cos 3t and I = c1 cosÁ ˜ + c 2 sinÁ ˜ - cos 3t . From the initial conditions, we have Ë 2¯ Ë 2¯ 7 7

I (0) = c1 -

t 12 Ê 1 12 12 ˆ and c 2 = 0 and I ( t) = Á cos - cos 3t˜ . = 0 and I ¢ (0) = c 2 = 0 . Thus c1 = ¯ 7Ë 2 2 7 7

100 • Chapter 4 Second Order Linear Differential Equations

16.

I ¢¢ +

1 dV 1 Ê tˆ Ê tˆ I = s . Vs = 10 te - t , so I ¢¢ + I = 10 te - t . Thus IC = c1 cosÁ ˜ + c 2 sinÁ ˜ and Ë 2¯ Ë 2¯ 4 dt 4

I P = ( At + B)e - t . Differentiation gives us I P¢¢ = ( At - 2 A + B)e - t , and then we have At - 2 A + B +

24 1 ( At + B) = -10 t + 10 fi A = -8, B = - . 5 4

24 ˆ 24 ˆ Ê Ê tˆ Ê tˆ Ê Thus, I P = Á -8 t - ˜ e -t and I = c1 cosÁ ˜ + c 2 sinÁ ˜ + Á -8 t - ˜ e -t . From the initial conditions, Ë Ë 2¯ Ë 2¯ Ë 5¯ 5¯

we have I (0) = c1 I ( t) =

17.

IS =

24 1 24 24 32 and and c 2 = = 0 and I ¢ (0) = c 2 - 8 + = 0 . Thus c1 = 5 2 5 7 5

24 ˆ 24 Ê t ˆ 32 Ê t ˆ Ê cosÁ ˜ + sinÁ ˜ - Á 8 t + ˜ e -t . Ë 2¯ 5 Ë 2¯ Ë 5¯ 5

dV V 1 t dI 1 1 , and then we have CV ¢¢ + V ¢ + V = s , V (0) = 0, V ¢ (0) = 0 . + Ú V (l ) dl + C dt R L 0 dt R L

1 dI R = 1kW, L = 1H, C = mF, and so V ¢¢ + 2V ¢ + 2V = 2 s . For this problem, 2 dt Is = 1 - e -t fi 2

dI s -2 ± 4 - 8 = 2e - t . For the complementary solution, we have l = = -1 ± i , and dt 2

so VC = c1e - t cos t + c 2e - t sin t . VP = Ae - t , and substituting this into the original differential equation, we have A - 2 A + 2 A = 2 . Thus A = 2 , and so V = c1e - t cos t + c 2e - t sin t + 2e - t . From the initial conditions, we have V (0) = c1 + 2 = 0 and V ¢ (0) = -c1 + c 2 - 2 = 0 . Solving these simultaneous equations yields c1 = -2 and c 2 = 0 , and so V = -2e - t cos t + 2e - t . 18.

V ¢¢ + 2V ¢ + 2V = 2

dI dI s , V (0) = 0, V ¢ (0) = 0 . For this problem, I s = 5 sin t fi 2 s = 10 cos t . For the dt dt

complementary solution, VC = c1e - t cos t + c 2e - t sin t . VP = A cos t + B sin t, VP¢ = - A sin t + B cos t, VP¢¢= - A cos t - B sin t , and substituting this into the

original differential equation, we have (-2 A + B)sin t + ( A + 2 B)cos t = 10 cos t fi A = 2, B = 4 . Thus, V = c1e - t cos t + c 2e - t sin t + 2 cos t + 4 sin t . From the initial conditions, we have V (0) = c1 + 2 = 0 and V ¢ (0) = -c1 + c 2 + 4 = 0 . Solving these simultaneous equations yields c1 = -2 and c 2 = -6 , and so V = -2e - t cos t - 6e - t sin t + 2 cos t + 4 sin t .

Chapter 5 Higher Order Linear Differential Equations Section 5.1 1-5

Verify by substituting into differential equation.

6.

Discontinuities for the relevant functions exist at t = -3, -1, 3 .

y ¢¢¢ 7.

1 y ¢¢ + ln( t + 1) y ¢ + cos ty = 0 ; Initial condition at t = 0 . t -9

-1 < t < 3 .

2

Discontinuities for the relevant functions exist at t = -1 and np +

p . Since t0 = 0 , the largest 2

interval on which Theorem 5.1 guarantees a unique solution is -1 < t < 8.

Discontinuities for the relevant functions exist at t = ±4 and ± Initial condition at t = 3.

9.

p . 2

p . ( t 2 - 16) y ( 4 ) + 2 y ¢¢ + t 2 y = sec t ; 2

p < t < 4. 2

There are no discontinuities for the relevant functions and t0 = 0 . Thus the largest interval on which Theorem 5.1 guarantees a unique solution is -• < t < • .

10.

y ( 4 ) - 5 y ¢¢ + 4 y = 0 ; l4 - 5l2 + 4 = (l2 - 1)(l2 - 4 ) = 0

11.

l3 - l = l (l - 1)(l + 1) = 0 , so l = 0, ± 1.

12.

y ¢¢¢ - 2 y ¢¢ - y ¢ + 2 y = 0 ; l3 - 2l2 - l + 2 = l2 (l - 2) - (l - 2) = (l - 2)(l2 - 1) = 0

l = -1, 1, - 2, 2 .

l = -1, 1, 2 . 13.

l4 - 2l2 + 1 = (l2 - 1) 2 = (l - 1) 2 (l + 1) 2 = 0 , so l = ±1.

Section 5.2 1 (a).

y ¢¢¢ = 0 , and antidifferentiation yields y ¢¢ = c1, y ¢ = c1t + c 2 , and y =

c1 2 t + c2t + c3. 2

1 t t2 1 (b). W = 0 1 2 t = 2 π 0 , and thus the three functions form a fundamental set of solutions. 0 0 2

102 • Chapter 5 Higher Order Linear Differential Equations

e-t -e - t = 1(2) = 2 π 0 . e-t

2.

1 et W = 0 et 0 et

3.

1 t e-t W = 0 1 -e - t = e - t π 0 , and thus the three functions form a fundamental set of solutions. 0 0 e-t

4.

W =

1 t cos t 0 1 - sin t

sin t cos t

0 0 - cos t

- sin t - cos t

0 0

5.

6.

7.

sin t

= 1 ◊ 1 ◊ (cos2 t + sin 2 t) = 1 π 0 .

1 t t -1 W = 0 1 - t -2 = 2 t -3 π 0, t > 0 , and thus the three functions form a fundamental set of solutions. 0 0 2 t -3 ln t t -1 0 - t -2

1 W =0

t2 2 t = 1 ◊ ( 3t -1 ) = 3t -1 π 0, t > 0 . 2

y = c1 + c 2e t + c 3e - t , and differentiation yields y ¢ = c 2e t - c 3e - t and y ¢¢ = c 2e t + c 3e - t . From the initial conditions, we have y (0) = c1 + c 2 + c 3 = 3 , y ¢ (0) = c 2 - c 3 = -3 , and y ¢¢ (0) = c 2 + c 3 = 1 . Solving these simultaneous equations yields c1 = 2, c 2 = -1, and c 3 = 2 , and so the unique solution is y = 2 - e t + 2e - t .

8.

y = c1 + c 2 t + c 3e - t , y (1) = c1 + c 2 + c 3e -1 = 4 , y ¢ (1) = c 2 - c 3e -1 = 3, y ¢¢ (1) = c 3e -1 = 0 \ c 3 = 0, c 2 = 3, c1 = 1 and y ( t) = 1 + 3t .

9.

y = c1 + c 2 t + c 3 cos t + c 4 sin t , and differentiation yields y ¢ = c 2 - c 3 sin t + c 4 cos t , y ¢¢ = -c 3 cos t - c 4 sin t , and y ¢¢¢ = c 3 sin t - c 4 cos t . From the initial conditions, we have

p Ê pˆ Ê pˆ Ê pˆ Ê pˆ y Á ˜ = c1 + c 2 + c 4 = 2 + p , y ¢ Á ˜ = c 2 - c 3 = 3 , y ¢¢ Á ˜ = -c 4 = -3 , and y ¢¢¢ Á ˜ = c 3 = 1. Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ 2

Solving these simultaneous equations yields c1 = -1 - p, c 2 = 4, c 3 = 1, and c 4 = 3, and so the unique solution is y = -1 - p + 4 t + cos t + 3 sin t . 10.

1 1 3 y = c1 + c 2 t + c 3 t -1, y ¢ = c 2 - c 3 t -2 , y ¢¢ = 2c 3 t -3 , y (2) = c1 + 2c 2 + c 3 = -1, y ¢ (2) = c 2 - c 3 = , 2 4 2 1 2 1 3 1 y ¢¢ (2) = c 3 = - fi c 3 = -2 , c 2 = - = 1, c1 + 2(1) + (-2) = -1 fi c1 = -2 2 8 2 2 2

y = -2 + t - 2 t -1 .

Chapter 5 Higher Order Linear Differential Equations • 103

11.

y = c1 + c 2 ln t + c 3 t 2 , and differentiation yields y ¢ = c 2 t -1 + 2c 3 t and y ¢¢ = -c 2 t -2 + 2c 3 . From the initial conditions, we have y (1) = c1 + c 3 = 1, y ¢ (1) = c 2 + 2c 3 = 2 , and y ¢¢ (1) = -c 2 + 2c 3 = -6 . Solving these simultaneous equations yields c1 = 2, c 2 = 4, and c 3 = -1, and so the unique solution is y = 2 + 4 ln t - t .

12.

y ¢¢¢ - y ¢ = 0 ; pn -1 ( t) = p2 ( t) = 0. Abel’s theorem predicts W ( t) = W ( t0 ).

If t0 = -1, then W ( t) = W (-1) = constant. From exercise 2, W ( t) = 2. 13.

pn -1 ( t) = p2 ( t) = 1, and so Abel’s Theorem predicts W ( t) = W (0)e - t with t0 = 0 . From Exercise 3,

W ( t) = e - t = W (0)e - t since W (0) = 1. 14.

y ( 4 ) + y ¢¢ = 0 ; pn -1 ( t) = p3 ( t) = 0 . If t0 = 1, Abel’s theorem predicts W ( t) = W (1) = constant. From exercise 4, W ( t) = 1. 3 ds 3 W (1) pn -1 ( t) = p2 ( t) = , and so Abel’s Theorem predicts W ( t) = W (1)e Ú1 s = W (1)e{ -3[ ln t - ln 1]} = 3 t t t

15.

with t0 = 1. From Exercise 5, W ( t) = 2 t -3 = W (1) t -3 since W (1) = 2 . 16.

1 t 2 y ¢¢¢ + ty ¢¢ - y ¢ = 0 ; pn -1 ( t) = p2 ( t) = . With t0 = 2 , Abel’s theorem predicts t Ï t1 ¸ Ï Ê 2ˆ ¸ W (2) . W ( t) = W (2)expÌ- Ú ds˝ = W (2)exp{- ln t + ln 2} = W (2)expÌlnÁ ˜ ˝ = 2 Ë ¯ s t t Ó ˛ Ó 2 ˛

From exercise 6, W ( t) = 3t -1 \ W (2) =

3 W (2) and W ( t) = 2 . 2 t

t

17.

- ( -3 )ds pn -1 ( t) = p2 ( t) = -3. W ( t) = W (1)e Ú1 = e 3( t -1).

Ï t ¸ W ( t) = W (1)expÌ- Ú sin sds˝ = 0 since W (1) = 0. Ó 1 ˛

18.

pn -1 ( t) = p2 ( t) = sin t.

19.

- s -1 ds 1 pn -1 ( t) = p2 ( t) = . W ( t) = W (1)e Ú1 = 3t -1, t > 0 . t

20.

pn -1 ( t) = p2 ( t) = 0. \ W ( t) = W (1) = 3.

21.

u¢¢ - u = 0 , so l2 - 1 = (l + 1)(l - 1) = 0 . Then we have u = c1e - t + c 2e t = y ¢ . Antidifferentiation

t

yields y = -c1e - t + c 2e t + c 3 ∫ Ae - t + Be t + C (1) . 22 (a). u = y ¢ ; u¢¢ + u¢ = 0, l2 + l = 0 fi l = 0, - 1

u = y ¢ = c1 + c 2e - t

\ y = c1t - c 2e - t + c 3 ∫ c1t + c 2e - t + c 3 ◊ 1 22 (b). v = y ¢¢

v ¢ + v = 0 fi v = y ¢¢ = k1e - t fi y ¢ = - k1e - t + k2 fi y = k1e - t + k2 t + k3 .

104 • Chapter 5 Higher Order Linear Differential Equations

23.

v ¢¢ + v = 0 , so we have v = c1 cos t + c 2 sin t = y ¢¢ . Antidifferentiation twice yields y = -c1 cos t - c 2 sin t + c 3 t + c 4 ∫ A cos t + B sin t + Ct + D .

1 t2 t4 24 (a). W = 0 2 t 4 t 3 = 1 ◊ 24 t 3 - 8 t 3 = 16 t 3 . 0 2 12 t 2

[

]

Note that W (0) = 0 but W ( t) is not identically zero on

(-1, 1) . If y ¢¢¢ + p2 ( t) y ¢¢ + p1 ( t) y = 0 were to have 1, t 2 , t 4 solutions with p2 ( t), p1 ( t)

continuous on (-1, 1) , we would contradict Abel’s theorem. \ No. 24 (b). If y = 1, then y ¢¢¢ + p2 y ¢¢ + p1 y ¢ = 0 is satisfied for any p1, p2 .

If y = t 2 , y ¢ = 2 t, y ¢¢ = 2 fi 0 + 2 p2 + 2 tp1 = 0 fi p2 + tp1 = 0 . If y = t 4 , y ¢ = 4 t 3 , y ¢¢ = 12 t 2 , y ¢¢¢ = 24 t fi 24 t + 12 t 2 p2 + 4 t 3 p1 = 0 fi 6 + 3tp2 + t 2 p1 = 0 . Èt Therefore, Í 2 Ît

1 ˘È p1 ˘ È 0 ˘ È p1 ˘ 1 È 3t -1˘È 0 ˘ È 3 t 2 ˘ ˙. = fi = =Í 3t ˙˚ÍÎ p2 ˙˚ ÍÎ-6 ˙˚ ÍÎ p2 ˙˚ 2 t 2 ÍÎ- t 2 t ˙˚ÍÎ-6 ˙˚ Í-3 ˙ Î t˚

Both functions fail to be continuous at t = 0 . 25 (a). Differentiation yields y ¢ = c 2e t - c 3e - t and y ¢¢ = c 2e t + c 3e - t . From the initial conditions, we have y (0) = c1 + c 2 + c 3 = a , y ¢ (0) = c 2 - c 3 = b , and y ¢¢ (0) = c 2 + c 3 = 4 . Solving these simultaneous

1 1 equations in terms of a and b yields c1 = a - 4, c 2 = 2 + b, and c 3 = 2 - b . Then we have 2 2 1 ˆ 1 ˆ Ê Ê y ( t) = (a - 4 ) + Á 2 + b˜ e t + Á 2 - b˜ e - t . Since the third term goes to zero as t gets large, we Ë Ë 2 ¯ 2 ¯

must set a = 4 and b = -4 so that the first two terms also become zero (for all t). 25 (b). y will be bounded on 0 £ t < • if b = -4 ( a can be arbitrary). No choice of a and b will produce

1 1 a solution that is bounded on -• < t < • since 2 + b and 2 - b cannot simultaneously be zero. 2 2

Section 5.3 1.

Antidifferentiation yields y = c1 + c 2 t + c 3 t 2 . Since t0 = 0 , we have y1 = 1, y 2 = t, and y 3 = the initial conditions provided.

2.

y = c1 + c 2 t + c 3 t 2 , y ¢ = c 2 + 2c 3 t , y ¢¢ = 2c 3 t0 = 1: y1: c1 + c 2 + c 3 = 1, c 2 + 2c 3 = 0, 2c 3 = 0 \ c1 = 1, c 2 = c 3 = 0, y1 ( t) = 1. y 2 : c1 + c 2 + c 3 = 0, c 2 + 2c 3 = 1, 2c 3 = 0 fi c 3 = 0, c 2 = -1, c1 = -1, y 2 ( t) = t - 1,

t2 from 2

Chapter 5 Higher Order Linear Differential Equations • 105

1 1 y 3 : c1 + c 2 + c 3 = 0, c 2 + 2c 3 = 0, 2c 3 = 1 fi c 3 = , c 2 = -1, c1 = 2 2

1 y 3 ( t) = ( t - 1) 2 2

1 y1 ( t) = 1, y 2 ( t) = t - 1, y 3 ( t) = ( t - 1) 2 . 2 3.

Since t0 = 0 , we have y1 (0) = c1 + c 3 = 1, y1¢(0) = c 2 - c 3 = 0, and y1¢¢= c 3 = 0 from the initial conditions provided. Thus c1 = 1 and c 2 = c 3 = 0 , and y1 ( t) = 1. Similarly, we have y 2 (0) = c1 + c 3 = 0, y 2¢ (0) = c 2 - c 3 = 1, and y 2¢¢ = c 3 = 0 from the initial conditions provided. Thus c 2 = 1 and c1 = c 3 = 0 , and y 2 ( t) = t . Finally, we have y 3 (0) = c1 + c 3 = 0, y 3¢ (0) = c 2 - c 3 = 0, and y 3¢¢ = c 3 = 1 from the initial conditions provided. Thus c1 = -1 and c 2 = c 3 = 1, and y 3 ( t) = -1 + t + e - t .

4.

y = c1 + c 2 t + c 3e - t , y ¢ = c 2 - c 3e - t , y ¢¢ = c 3e - t t0 = 1: c1 + c 2 + c 3e -1 = 1, c 2 - c 3e -1 = 0, c 3e -1 = 0 fi c1 = 1, c 2 = c 3 = 0, y1 ( t) = 1 y 2 : c1 + c 2 + c 3e -1 = 0, c 2 - c 3e -1 = 1, c 3e -1 = 0, c 3 = 0, c 2 = 1, c1 = -1, y 2 ( t) = t - 1 y 3 : c1 + c 2 + c 3e -1 = 0, c 2 - c 3e -1 = 0, c 3e -1 = 1 fi c 3 = e, c 2 = 1, c1 = -2 , y 3 ( t) = -2 + t + e -( t -1) y1 ( t) = 1, y 2 ( t) = t - 1, y 3 ( t) = -2 + t + e -( t -1) .

5 (a).

{cosh t, 1 - sinh t, 2 + sinh t} is a solution set.

1 1 1 1 1 1 5 (b). cosh t = 0 ◊ 1 + e t + e - t , 1 - sinh t = 1 - e t + e - t , and 2 + sinh t = 2 + e t - e - t . Thus 2 2 2 2 2 2 È Í0 Í1 A=Í Í2 Í1 ÍÎ 2

1 1 2 1 2

˘ 2 ˙ 1 ˙ ˙. 2 ˙ 1 - ˙ 2 ˙˚

5 (c).

1 1 1 1 1 1 2 = 3 π 0 , so the three functions form a fundamental set. det A = 0 2 2 - 1 2 2 + 2 2 1 1 1 1 1 1 2 2 2 2 2 2 2

6 (a).

{1 - 2t,

-

t + 2, e -( t + 2 )} is a solution set.

6 (b). 1 - 2 t = 1 ◊ 1 - 2 ◊ t + 0 ◊ e - t , t + 2 = 2 ◊ 1 + 1 ◊ t + 0 ◊ e - t , e -( t + 2 ) = 0 ◊ 1 + 0 ◊ t + e -2 ◊ e - t

È1 2 0 ˘ ˙ Í A = Í-2 1 0 ˙ ÍÎ 0 0 e -2 ˙˚ 6 (c).

det A = e -2 (5) = 5e -2 π 0 \ fundamental set .

106 • Chapter 5 Higher Order Linear Differential Equations

7 (a).

y1 = 1 + t and y 2 =

t +1 -1 are solutions. However, y 3 = ( t + 1) is not a solution and so t

t +1 Ï -1 ¸ , ( t + 1) ˝ is not a solution set. Ì1 + t, t Ó ˛

8 (a).

{2t

2

- 1, 3, ln( t 3 )} is a solution set; ln( t 3 ) = 3 ln t.

8 (b). 2 t 2 - 1 = -1 ◊ 1 + 0 ◊ ln t + 2 ◊ t 2 , 3 = 3 ◊ 1 + 0 ◊ ln t + 0 ◊ t 2 , 3 ln t = 0 ◊ 1 + 3 ◊ ln t + 0 ◊ t 2

È-1 3 0 ˘ ˙ Í A = Í 0 0 3˙ ÍÎ 2 0 0 ˙˚ 8 (c).

È-1 3˘ det A = -3Í ˙ = 18 π 0 \ fundamental set. Î 2 0˚

9.

Setting c1 ◊ 1 + c 2 t + c 3 t 2 = 0 and evaluating at t = -1, 0, 1 we have c1 - c 2 + c 3 = 0, c1 = 0, and c1 + c 2 + c 3 = 0 . Thus c1 = c 2 = c 3 = 0 , and the three functions are

linearly independent on the interval. 10.

c1 ◊ 1 + c 2 ◊ (1 + t) + c 3 (1 + t + t 2 ) = 0 \ (c1 + c 2 + c 3 ) ◊ 1 + (c 2 + c 3 ) ◊ t + c 3 ◊ t 2 = 0 The argument of 9 leads to c1 + c 2 + c 3 = 0, c 2 + c 3 = 0, c 3 = 0 fi c1 = c 2 = c 3 = 0 \ linearly independent on - • < t < • .

11.

Setting c1 cos2 t + c 2 2 cos 2 t + c 3 2 sin 2 t = 0 and using the identity cos2 t - sin 2 t = cos 2 t , we have c1 cos2 t + 2c 2 (cos2 t - sin 2 t) + 2c 3 sin 2 t = (c1 + 2c 2 ) cos2 t + (-2c 2 + 2c 3 ) sin 2 t = 0 . Taking c 3 = 1, c 2 = 1, and c1 = -2 to be one nontrivial solution, we can conclude that the three functions are

linearly dependent on the interval. 12.

c1 ( t 2 + 2 t) + c 2 (at + 1) + c 3 ( t + a ) = c1t 2 + (2c1 + ac 2 + c 3 ) t + (c 2 + ac 3 ) = 0

Èa 1 ˘Èc 2 ˘ È0 ˘ = Í ˙. det = a 2 - 1 \ c1 = 0, ac 2 + c 3 = 0, c 2 + ac 3 = 0 or Í ˙ Í ˙ Î 1 a ˚Îc 3 ˚ Î0 ˚ \ linearly dependent on - • < t < • if a = ±1 and linearly independent on -• < t < • otherwise.

(nontrivial c 2 , c 3 if a = ±1 ). 13.

On 0 £ t < • , t t + 1 = t 2 + 1. Then we have c1 ( t 2 + 1) + c 2 ( t 2 - 1) + c 3 t = (c1 + c 2 ) t 2 + c 3 t + (c1 - c 2 ) = 0 . Thus c1 = c 2 = c 3 = 0 , and so the three

functions are linearly independent on the interval. 14.

On -• < t £ 0 , t t + 1 = - t 2 + 1 fi 1( t t + 1) + 1( t 2 - 1) + 0( t) = 0 , and so the three functions are linearly dependent on the interval.

Chapter 5 Higher Order Linear Differential Equations • 107

15.

Since the three functions are linearly independent on half of the interval (see 13), the functions are linearly independent on the entire interval.

Section 5.4 1 (a).

l3 - 4 l = 0

1 (b). l3 - 4 l = l (l + 2)(l - 2) = 0 . Thus l = 0, ± 2 . 1 (c).

y = c1 + c 2e -2 t + c 3e 2 t , since the roots are distinct.

2 (a).

l3 + l2 - l - 1

2 (b). l2 (l + 1) - (l + 1) = (l2 - 1)(l + 1) = 0

2 (c).

e-t W = -e - t e-t

y = c1e - t + c 2 te - t + c 3e t ;

l = -1, - 1, 1 te - t (1 - t)e - t (-2 + t)e - t

et et = et

e - t [ 3 - 2 t] - te - t [-2] + e t [2 - t - 1 + t]e -2 t = e - t [ 3 - 2 t + 2 t + 1] = 4 e - t π 0 3 (a).

l3 + l2 + 4 l + 4 = 0

3 (b). l3 + l2 + 4 l + 4 = l2 (l + 1) + 4 (l + 1) = (l2 + 4 )(l + 1) = 0 . Thus l = -1, ± i2 .

3 (c).

e-t y = c1e - t + c 2 cos 2 t + c 3 sin 2 t , since W = -e - t e-t

cos 2 t sin 2 t -2 sin 2 t 2 cos 2 t = 10e - t π 0 . -4 cos 2 t -4 sin 2 t

4 (a). 16l4 - 8l2 + 1 4 (b).

(4l

2

2

- 1) = 0 ;

1 1 1 1 l=- , - , , 2 2 2 2 t

4 (c).

y = c1e

-t2

W (0) =

+ c 2 te

1 - 12 1 4

- 18

-t2

t

+ c 3e + c 4 te 2

t

2

0 1 0 1 0 1 1 2 1 0 1 Æ 1 -1 4 1 0 -1 3 1 3 0 43 4 8 4

W =

e- 2 t - 12 e - 2 t

e- 2 t - 18 e - 2 1 4

t

(-1 + 4t )e ( 43 - 8t )e -

1 0 1 0 1 1 1 0 1 1 Æ 0 1 0 0 1 3 0 4 0 0 - 43

5 (a). 16l4 + 8l2 + 1 = 0 2 i i 5 (b). 16l4 + 8l2 + 1 = ( 4 l2 + 1) = 0 . Thus l = ± , ± . 2 2

t

te - 2 (1 - 2t )e - t 2 t

t

2

e2 1 t2 2e 2

1 4 1 8

t

e2 t e2

te

t

2

(1 + 2t )e (1 + 4t )e ( 43 + 8t )e

0 1 Ê 3ˆ = 1 ◊ 1Á ˜ π 0 . Ë 2¯ 2 0

t

2

t

t

2

2

108 • Chapter 5 Higher Order Linear Differential Equations

5 (c).

t t t t y = c1 cos + c 2 t cos + c 3 sin + c 4 t sin . To verify this, 2 2 2 2 cos( 2t ) sin( 2t ) t cos( 2t ) t sin( 2t ) t t 1 1 cos( 2t ) - sin( 2t ) cos( 2t ) sin( 2t ) + cos( 2t ) - sin( 2t ) 2 2 2 2 . W = 1 1 t t - cos( 2t ) - sin( 2t ) - cos( 2t ) - sin( 2t ) cos( 2t ) - sin( 2t ) 4 4 4 4 1 3 t 1 3 t t t t t t sin( 2 ) - cos( 2 ) + sin( 2 ) - cos( 2 ) - sin( 2 ) - cos( 2t ) 8 4 8 8 4 8

W (0) =

1

0

0 -

1 4

0

6 (a).

1

0 1 2

0 0

0

0

1

3 4

1 8

1

0

= -0

1

0 -

0

3 4

0 1 1 = - π 0. 4 2 1 8

l3 - 1

6 (b). l3 = e i 2 kp fi l k = e

i 2 kp

3

, k = 0, 1, 2 ;

l = 1, e

6 (c).

Ê 3 ˆ Ê 3 ˆ 1 t y = c1e t + c 2e - 2 cosÁ t˜ + c 3e - 2 sinÁ t˜ Ë 2 ¯ Ë 2 ¯

7 (a).

l3 - 2l2 - l + 2 = 0

i 2p

3

, e

i 4p

3

= 1, -

1 3 1 3 +i , - -i 2 2 2 2

7 (b). l3 - 2l2 - l + 2 = l2 (l - 2) - (l - 2) = (l + 1)(l - 1)(l - 2) . Thus l = 2, ± 1. 7 (c).

y = c1e - t + c 2e t + c 3e 2 t , since the roots are distinct.

8 (a).

l4 + 16

8 (b). l4 = -16 = 16e i(p + 2 kp ) fi l k = 2e ( 4 i

2e

i 34p

= - 2 + i 2 , l 3 = 2e

y = c1e 9 (a).

2t

( )

cos 2 t + c 2e -

i 54p

2t

p

+ k p2 )

, k = 0, 1, 2, 3 ;

= - 2 - i 2, l 4 = 2e

( )

sin 2 t + c 3e

2t

i 74p

p

l1 = 2e i 4 = 2 + i 2 , l 2 =

= 2-i 2 \ l = ± 2 ±i 2

( )

cos 2 t + c 4 e

2t

( )

sin 2 t .

l3 + 4 l = 0

9 (b). l3 + 4 l = l (l2 + 4 ) = 0 . Thus l = 0, ± i2 . 9 (c).

y = c1 + c 2 cos 2 t + c 3 sin 2 t

9 (d). Differentiation gives us y ¢ = -2c 2 sin 2 t + 2c 3 cos 2 t and y ¢¢ = -4 c 2 cos 2 t - 4 c 3 sin 2 t . From the initial conditions, we have y (0) = c1 + c 2 = 1, y ¢ (0) = 2c 3 = 6, and y ¢¢ (0) = -4 c 2 = 4 . Thus c1 = 2, c 2 = -1, and c 3 = 3, and so y = 2 - cos 2 t + 3 sin 2 t

10 (a). l3 + 3l2 + 3l + 1 = (l + 1) 3

Chapter 5 Higher Order Linear Differential Equations • 109

10 (b). l = -1, - 1, - 1 10 (c). y = c1e - t + c 2 te - t + c 3 t 2e - t 10 (d). y (0) = 0, y ¢ (0) = 1, y ¢¢ (0) = 0 y ¢ = -c1e - t + c 2 (1 - t)e - t + c 3 (2 t - t 2 )e - t , y ¢¢ = c1e - t + c 2 (-2 + t)e - t + c 3 ( t 2 - 4 t + 2)e - t y (0) = c1 = 0, y ¢ (0) = c 2 = 1, y ¢¢ (0) = c 2 (-2) + c 3 (2) = 0 fi c 3 = c 2 = 1

11.

y = ( t + t 2 )e - t

l2 (l2 + 9) = l4 + 9l2 = 0 . Thus the differential equation is y ( 4 ) + 9 y ¢¢ = 0 , and so a3 = 0, a2 = 9, a1 = 0, and a0 = 0 .

12.

y = c1 cos t + c 2 sin t + c 3 cos 2 t + c 4 sin 2 t ;

y ( 4 ) + 5 y ¢¢ + 4 y = 0 ; 13.

(l

2

+ 1)(l2 + 4 ) = l4 + 5l2 + 4 = 0

a3 = 0, a2 = 5, a1 = 0, a0 = 4 . 2

(l - 1) 2 (l + 1) 2 = (l2 - 1) = l4 - 2l2 + 1 = 0 . Thus the differential equation is y ( 4 ) - 2 y ¢¢ + y = 0 , an so a3 = 0, a2 = -2, a1 = 0, and a0 = 1.

14.

y = c1e - t sin t + c 2e - t cos t + c 3e t sin t + c 4 e t cos t; l = -1 ± i1, 1 + i1. (l + 1 - i1)(l + 1 + i1) = (l + 1) 2 + 1 = l2 + 2l + 2; (l - 1 - i1)(l - 1 + i1) = (l - 1) 2 + 1 = l2 - 2l + 2 2 (l2 + 2l + 2)(l2 - 2l + 2) = (l2 + 2) - 4l2 = l4 + 4l2 + 4 - 4l2 = l4 + 4 = 0 y ( 4 ) + 4 y = 0; 15.

a3 = 0, a2 = 0, a1 = 0, a0 = 4.

(l - 1) 4 = l4 - 4l3 + 6l2 - 4 l + 1 = 0 . Thus the differential equation is y ( 4 ) - 4 y ¢¢¢ + 6 y ¢¢ - 4 y ¢ + y = 0 , and so a3 = -4, a2 = 6, a1 = -4, and a0 = 1.

16 (a). n = 5 16 (b). {1, t, e t , cos t, sin t} 17 (a). n = 5 17 (b). {e t , e t cos 2 t, e t sin 2 t, e - t cos 2 t, e - t sin 2 t} 18 (a). n = 8 18 (b). {sin t, cos t, t sin t, t cos t, t 2 sin t, t 2 cos t, e t sin t, e t cos t} 19 (a). n = 7 19 (b). {sin t, cos t, t sin t, t cos t, e t , te t , t 2e t } 20 (a). n = 4 20 (b). {1, t, t 2 , e 2 t } 21.

n = 1, a = 1, y ( t) = Ce - t

110 • Chapter 5 Higher Order Linear Differential Equations

y ¢ ± 2 y = 0, y ( t) = Ce m t

22.

n = 1, a = ±2;

23.

n = 4, a = 0, y ( t) = c1 + c 2 t + c 3 t 2 + c 4 t 3

24.

n = 2, a = 4;

25.

Three values for l must be 3 and -

y ¢¢ + 4 y = 0, y = c1 cos 2 t + c 2 sin 2 t.

3 3 3 . Using these values to reach the characteristic ±i 2 2

equation gives us (l - 3)(l2 + 3l + 9) = l3 + 3l2 + 9l - 3l2 - 9l - 27 = l3 - 27 . Thus

n = 3 and a = -27 .

Section 5.5 1 (a).

l3 - l = l (l + 1)(l - 1) = 0 . Thus yC = c1e - t + c 2 + c 3e t .

1 (b).

y P = Ae 2 t , and substituting this into the original differential equation yields (8 A - 2 A)e 2 t = e 2 t . Thus A =

1 1 and so y P = e 2 t . 6 6

1 (c).

1 y = c1e - t + c 2 + c 3e t + e 2 t 6

2 (a).

y c = c1e - t + c 2 + c 3e t

2 (b).

y p = At + B cos 2 t + C sin 2 t, y p ¢ = A - 2 B sin 2 t + 2C cos 2 t, y ¢¢ = -4 B cos 2 t - 4 C sin 2 t, y p ¢¢¢ = 8 B sin 2 t - 8C cos 2 t

\ 8 B sin 2 t - 8C cos 2 t - A + 2 B sin 2 t - 2C cos 2 t = 4 + 2 cos 2 t 1 1 y p = -4 t - sin 2 t 10 B = 0, - 10C = 2, - A = 4 \ A = -4, B = 0, C = - ; 5 5 1 2 (c). y = c1e - t + c 2 + c 3e t - 4 t - sin 2 t 5 3 (a).

l3 - l = l (l + 1)(l - 1) = 0 . Thus yC = c1e - t + c 2 + c 3e t .

3 (b).

y P = t( At + B) = At 2 + Bt . Differentiation yields y P¢ = 2 At + B, y P¢¢ = 2 A, and y P¢¢¢= 0 , and substituting into the original differential equation yields 0 - 2 At - B = 4 t . Thus

A = -2 and B = 0 and so y P = -2 t 2 . 3 (c).

y = c1e - t + c 2 + c 3e t - 2 t 2

4 (a).

y c = c1e - t + c 2 + c 3e t

4 (b).

y p = Ate t , y ¢p = A( t + 1)e t , y ¢¢p = A( t + 2)e t , y ¢¢¢= A( t + 3)e t ; p y ¢¢¢y ¢p = A[ t + 3 - t - 1]e t = -4 e t fi 2 A = -4 or A = -2 \ y p = -2 te t p

Chapter 5 Higher Order Linear Differential Equations • 111

4 (c).

y = c1e - t + c 2 + c 3e t - 2 te t

5 (a).

l3 + l2 = l2 (l + 1) = 0 . Thus yC = c1 + c 2 t + c 3e - t .

5 (b).

y P = Ate - t . Differentiation yields y P¢ = A(1 - t)e - t , y P¢¢ = A( t - 2)e - t , and y P¢¢¢= A(- t + 3)e - t , and substituting into the original differential equation yields A(- t + 3 + t - 2)e - t = 6e - t . Thus

A = 6 and so y P = 6 te - t . 5 (c).

y = c1 + c 2 t + c 3e - t + 6 te - t

6 (a).

l3 - l2 = 0, l = 0, 0, 1;

6 (b).

y p = Ae -2 t , y ¢p = -2 Ae -2 t , y ¢¢p = 4 Ae -2 t , y ¢¢¢= -8 Ae -2 t \ (-8 A - 4 A)e -2 t = 4 e -2 t p

y c = c1 + c 2 t + c 3e t

1 1 A = - , y p = - e -2 t 3 3 6 (c).

1 y = c1 + c 2 t + c 3e t - e -2 t 3

7 (a).

l3 - 2l2 + l = l (l - 1) = 0 . Thus yC = c1 + c 2e t + c 3 te t .

7 (b).

y P = t( At + B) + Ct 2e t = At 2 + Bt + Ct 2e t . Differentiation yields

2

y P¢ = 2 At + B + C ( t 2 + 2 t)e t , y P¢¢ = 2 A + C ( t 2 + 4 t + 2)e t , and y P¢¢¢= C ( t 2 + 6 t + 6)e t , and

substituting into the original differential equation yields

[

]

C ( t 2 + 6 t + 6)e t - 2 2 A + C ( t 2 + 4 t + 2)e t + 2 At + B + C ( t 2 + 2 t)e t = t + 4 e t . Thus 1 1 A = , B = 2, and C = 2 and so y P = t 2 + 2 t + 2 t 2e t . 2 2 7 (c).

1 y = c1 + c 2e t + c 3 te t + t 2 + 2 t + 2 t 2e t 2

8 (a).

l3 - 3l2 + 3l - 1 = (l - 1) 3 = 0;

y c = c1e t + c 2 te t + c 3 t 2e t

8 (b). y p = At 3e t , y ¢p = A( t 3 + 3t 2 )e t , y ¢¢p = A( t 3 + 6 t 2 + 6 t)e t , y ¢¢¢= A( t 3 + 9 t 2 + 18 t + 6)e t p

[

]

\ A t 3 + 9 t 2 + 18 t + 6 - 3( t 3 + 6 t 2 + 6 t) + 3( t 3 + 3t 2 ) - t 3 e t = 12e t fi A ◊ 6 = 12;

y p = 2 t 3e t

8 (c).

y = c1e t + c 2 te t + c 3 t 2e t + 2 t 3e t

9 (a).

l3 - 1 = 0 . Thus l = 1, -

9 (b).

y P = Ate t . Differentiation yields y P¢ = A( t + 1)e t , y P¢¢ = A( t + 2)e t , and y P¢¢¢= A( t + 3)e t , and

t t Ê 3 ˆ Ê 3 ˆ 1 3 , and so yC = c1e t + c 2e 2 cosÁ ±i t˜ + c 3e 2 sinÁ t˜ . 2 2 Ë 2 ¯ Ë 2 ¯

substituting into the original differential equation yields A( t + 3 - t)e t = e t . Thus

A=

1 1 and so y P = te t . 3 3

112 • Chapter 5 Higher Order Linear Differential Equations

9 (c).

t Ê 3 ˆ Ê 3 ˆ 1 t 2 yC = c1e + c 2e cosÁ t˜ + c 3e sinÁ t˜ + te Ë 2 ¯ Ë 2 ¯ 3 t

-

t 2

10 (a). l3 + 1 = 0, l3 = -1 = e i(p + 2 kp ), l k = e

l 2 = -1, l 3 =

i ( p + 2 kp ) 3

, l1 = e

ip

3

=

1 3 +i , 2 2

Ê 3 ˆ Ê 3 ˆ t t y c = c1e - t + c 2e 2 cosÁ t˜ + c 3e 2 sinÁ t˜ Ë 2 ¯ Ë 2 ¯

1 3 -i ; 2 2

10 (b). y p = Ae t + B cos t + C sin t, y ¢p = Ae t - B sin t + C cos t, y ¢¢p = Ae t - B cos t - C sin t, y ¢¢¢= Ae t + B sin t - C cos t \ Ae t + B sin t - C cos t + Ae t + B cos t + C sin t = p

1 1 1 = e t + cos t \ 2 A = 1, B + C = 0, - C + B = 1, A = , B = , C = - , 2 2 2 1 1 1 y p = e t + cos t - sin t 2 2 2 Ê 3 ˆ Ê 3 ˆ 1 t 1 1 t t 10 (c). y = c1e - t + c 2e 2 cosÁ t˜ + c 3e 2 sinÁ t˜ + e + cos t - sin t 2 2 Ë 2 ¯ Ë 2 ¯ 2 2

11 (a). l3 - 4 l2 + 4 l = l (l - 2) = 0 . Thus yC = c1 + c 2e 2 t + c 3 te 2 t . 11 (b). y P = t( A3 t 3 + A2 t 2 + A1t + A0 ) + t 2 ( B2 t 2 + B1t + B0 )e 2 t = A3 t 4 + A2 t 3 + A1t 2 + A0 t + ( B2 t 4 + B1t 3 + B0 t 2 )e 2 t

12 (a). l3 - 3l2 + 3l - 1 = (l - 1) 3 = 0

y c = c1e t + c 2 te t + c 3 t 2e t

12 (b). y p = At 3e t + Be t cos 3t + Ce t sin 3t + D 13(a). l4 - 16 = (l2 + 4 )(l - 2)(l + 2) = 0 , l = ±2, ± i2 , and so

yC = c1e -2 t + c 2e 2 t + c 3 cos 2 t + c 4 sin 2 t . 13 (b). y P = t( A1t + A0 )e 2 t + t( B1t + B0 )e -2 t + t(C1t + C0 ) cos 2 t + t( D1t + D0 ) sin 2 t 2

14 (a). l4 + 8l2 + 16 = (l2 + 4 ) = 0 \ l = ±i2, ± i2 y c = c1 cos 2 t + c 2 sin 2 t + c 3 t cos 2 t + c 4 t sin 2 t

14 (b). y p = t 2 ( A1t + A0 ) cos 2 t + t 2 ( B1t + B0 ) sin 2 t 15 (a). l4 - 1 = (l2 + 1)(l + 1)(l - 1) = 0 . Thus yC = c1e - t + c 2e t + c 3 cos t + c 4 sin t . 15 (b). y P = t( A1t + A0 )e - t + t(C1t + C0 ) cos t + t( D1t + D0 ) sin t . 16.

y = c1 + c 2 t + c 3e 2 t + 4 sin 2 t; \ a = -2, b = 0, c = 0. -32 cos 2 t.

l2 (l - 2) = l3 - 2l2 = 0;

y ¢¢¢ - 2 y ¢¢ = g( t)

y p = 4 sin 2 t, y ¢p = 8 cos 2 t, y ¢¢p = -16 sin 2 t, y ¢¢¢= p

Chapter 5 Higher Order Linear Differential Equations • 113

Substitute: -32 cos 2 t - 2(-16 sin 2 t) = g( t) \ a = -2, b = 0, c = 0, g( t) = -32(cos 2 t - sin 2 t).

17.

(l

2

+ 4 )(l - 1) = l3 - l2 + 4 l - 4 = 0 . Therefore, g( t) = y ¢¢¢ - y ¢¢ + 4 y ¢ - 4 y . y P = t 2 , and

differentiation yields y P¢ = 2 t, y P¢¢ = 2, and y P¢¢¢= 0 . Then we have

g( t) = 0 - 2 + 4 (2 t) - 4 t 2 = -4 t 2 + 8 t - 2 and a = -1, b = 4, c = -4 . 18.

y = c1 + c 2 t + c 3 t 2 - 2 t 3 ;

y ¢¢p = -12 t, y ¢¢¢= -12; p 19.

l3 = 0, y ¢¢¢ = g( t);

y p = -2 t 3 , y ¢p = -6 t 2 ,

a = b = c = 0, g( t) = -12

y = c1 + c 2 t + c 3 t 3 + t 4 , and so 1, t, t 3 are solutions of the homogeneous equation. 0 + 0 + 0 + c ◊ 1 = 0, so c = 0 . 0 + 0 + bt ◊ 1 = 0, so b = 0 . t 3 ◊ 6 + at 2 (6 t) = 0, so a = -1.

t 3 y ¢¢¢ - t 2 y ¢¢ = g( t) and y P = t 4 , so g( t) = t 3 ◊ 24 t - t 2 ◊ 12 t 2 = 12 t 4 . 20.

y = c1t + c 2 t 2 + c 3 t 4 + 2ln t;

t, t 2 , t 4 are solutions of homogeneous equation.

0 + 0 + bt + ct = 0 \ b + c = 0, 0 + at 2 (2) + bt(2 t) + c ( t 2 ) = 0 \ 2 a + 2b + c = 0

(t )¢ = 4 t , (t )¢¢ = 12t , (t )¢¢¢ = 24 t. 4

3

4

2

t 3 (24 t) + at 2 (12 t 2 ) + bt( 4 t 3 ) + c ( t 4 ) = 0

4

24 + 12 a + 4 b + c = 0 \ c = -b fi 2 a + b = 0 and 24 + 12 a + 3b = 0

\ b = -2 a fi 24 + 12 a - 6 a = 0 fi a = -4, b = 8, c = -8

t 3 y ¢¢¢ - 4 t 2 y ¢¢ + 8 ty ¢ - 8 y = g( t) 2 2 4 (2 ln t)¢ = , (2 ln t)¢¢ = - 2 , (2 ln t)¢¢¢ = 3 t t t Ê 4ˆ Ê 2ˆ Ê 2ˆ t 3 Á 3 ˜ - 4 t 2 Á - 2 ˜ + 8 tÁ ˜ - 16 ln t = g fi g = 28 - 16 ln t Ë t¯ Ët ¯ Ë t ¯ a = -4, b = 8, c = -8, g( t) = 28 - 16 ln( t), t > 0 .

21 (a). The three solutions can be verified by substitution. È t t2 t 4 ˘È u1¢ ˘ È 0 ˘ Í ˙Í ˙ Í ˙ 21 (b). Í1 2 t 4 t 3 ˙Íu2¢ ˙ = Í 0 ˙ where y P = tu1 + t 2 u2 + t 4 u3 . det = t 24 t 3 - 8 t 3 - 1 12 t 4 - 2 t 4 = 6 t 4 . ÍÎ0 2 12 t 2 ˙˚ÍÎu3¢ ˙˚ ÍÎt -3 g ˙˚

[

0

Thus u1¢ = 0 t -3 g t

0 0

t2 2t

t4 1 1 1 4 t 3 ◊ 4 = t -7 g 2 t 5 = t -2 g , 3 6t 6 2 12 t 2

[ ]

t4

1 1 1 u2¢ = 1 4 t 3 ◊ 4 = - t -7 g 3t 4 = - t -3 g , 2 6t 6 0 t -3 g 12 t 2

[ ]

] [

]

114 • Chapter 5 Higher Order Linear Differential Equations

t2 0 1 1 1 -7 2 1 -5 and u3¢ = 1 2 t 0 ◊ 4 = t g t = t g . g = 2 t 2 . Making this substitution, 6t 6 6 0 2 t -3 g t

[ ]

4 -1 2 -3 2 -7 antidifferentiating the three u¢ equations yields u1 = - t 2 , u2 = t 2 , and u3 = - t 2 . Thus 3 3 21 4 - 12 2 - 32 2 2 - 72 4 16 12 y P = - t ( t) + t ( t ) - t ( t ) = - t . Finally, we have the general solution: 3 3 21 21 y = c1t + c 2 t 2 + c 3 t 4 22.

g = 2 t; u3¢ =

u1¢ =

16 12 t . 21

2 -1 2 t fi u1 = ln t; 3 3

1 -4 1 t fi u3 = - t -3 ; 3 9

yp =

u2¢ = - t -2 fi u2 = t -1; 2 1 t ln t + t - t 3 9

2 1 2 y = c1¢t + c 2¢ t 2 + c 3¢ t 4 + t ln t + t - t = c1t + c 2 t 2 + c 3 t 4 + t ln t 3 9 3 23.

g = t 3 + 2 t 2 + 1, and so u1¢ = u1 =

1 2 1 1 -2 3 t ( t + 2 t 2 + 1) = t + + t -2 , and antidifferentiation yields 3 3 3 3

1 1 1 t 2 2 1 -1 + t - t . u2¢ = - t -3 ( t 3 + 2 t 2 + 1) = - - t -1 - t -3 , and antidifferentiation yields 2 2 2 6 3 3

1 1 1 t 1 1 u2 = - - ln t + t -2 . u3¢ = t -5 ( t 3 + 2 t 2 + 1) = t -2 + t -3 + t -5 , and antidifferentiation yields 6 3 6 6 2 4 1 1 1 -4 u3 = - t -1 - t -2 t . Thus 6 6 24 Ê t2 2 1 ˆ 1 ˆ 1 1 -4 ˆ Ê t Ê 1 y P = tÁ + t - t -1˜ + t 2 Á - - ln t + t -2 ˜ + t 4 Á - t -1 - t -2 t ˜ Ë 2 Ë 6 4 ¯ 6 24 ¯ Ë6 3 3 ¯

1 1 1 = - t 3 + t 2 - - t 2 ln t . Finally, we have the general solution: 2 2 8 1 1 y = c1t + c 2 t 2 + c 3 t 4 - t 3 - - t 2 ln t . 2 8

Chapter 6 First Order Linear Systems Section 6.1 1.

2.

3. 4. 5.

6.

7.

Èt - 1 t2 ˘ È t -1 ˘ È2 t - 2 2 t 2 ˘ È3t 2 2 A( t) - 3tB( t) = 2 Í 3 t ˙ ˙- Í Í0 t + 2 ˙ = Í 4 t + 2˚ Î 0 Î ˚ Î 4 Î 2 2 t + 1˚ È2 t - 2 - 3t 2 2 t 2 + 3t ˘ =Í ˙ 4 2 - 2 t - 3t 2 ˚ Î

-3t ˘ ˙ 3t 2 + 6 t ˚

È 2 2t 2 + t + 2˘ A( t) B( t) - B( t) A( t) = Í ˙ -2 Î-4 ˚ Èt - 1 t 2 ˘Èt + 1˘ È ( t - 1)( t + 1) + t 2 (-1) ˘ È-1˘ A( t)c( t) = Í ˙Í ˙= Í ˙ ˙= Í 1 2 2 t + 1 2 ( t + 1 ) + ( 2 t + 1 )( 1 ) Î ˚ Î ˚ Î ˚ Î1˚ 3 2 det[ tA( t)] = - t - t

There are two natural ways to do this problem. We can form the matrix A( t) B( t) and then calculate det[ A( t) B( t)] . Alternatively, we can separately calculate det[ A( t)] and det[ B( t)] and use the fact that det[ A( t) B( t)] = det[ A( t)]det[ B( t)] . Taking the latter course, det[ A( t)] = ( t - 1)(2 t + 1) - 2 t 2 = -( t + 1) , and det[ B( t)] = t( t + 2) = t 2 + 2 t . Thus, det[ A( t) B( t)] = -( t + 1)( t 2 + 2 t) = -( t 3 + 3t 2 + 2 t) .

1 det[ A( t)] = 2 t + 1 and so the matrix A( t) is invertible for every value t except t = - . The 2 1 Èt + 1 - t ˘ 1 , tπ- . inverse of A( t) is given by A -1 ( t) = Í ˙ 2 t + 1 Î - t t + 1˚ 2 As noted in Example 2, a square matrix is invertible if and only if its determinant is nonzero. Now, det[ A( t)] = t( t - 3) - 4 = t 2 - 3t - 4 = ( t - 4 )( t + 1) and so the matrix A( t) is invertible for every value t except t = 4 and t = -1. The inverse of A( t) is given by Èt - 3 -2 ˘ 1 A -1 ( t) = , t π 4, t π -1. t ˙˚ ( t - 4 )( t + 1) ÍÎ -2

116 • Chapter 6 First Order Linear Systems

8.

det[ A( t)] = 2 sin t cos t = sin 2 t and so the matrix A( t) is invertible for every value t except np is given by 2 t = np fi t = , n = 0, ±1, ±2, ±3,.... T h e i n v e r s e o f A( t) 2 1 È 1 ˘ È cos t cos t ˘ Í 2 csc t 2 csc t ˙ 1 np -1 , tπ , n = 0, ±1, ±2, ±3,.... A ( t) = =Í ˙ Í ˙ 1 1 2 2 sin t cos t Î- sin t sin t ˚ Í- sec t sec t ˙ 2 Î 2 ˚

9.

In this case, det[ A( t)] = e t e 4 t - e 3 t e 2 t = e 5 t - e 5 t and so det[ A( t)] is zero for every value of t. Hence, the given matrix A( t) is never invertible. 3 ˘ È sin t t cos t Í t t + 1 ˙ = È1 0 3˘ . lim A( t) = limÍ 2 t ˙ ÍÎ1 1 0 ˙˚ tÆ0 tÆ0 Í e 3t ˙ sec t t 2 - 1˚ Î

10.

12.

te - t limtan t ˘ È 0 0 ˘ È te - t tan t ˘ È lim tÆ0 tÆ0 Í = lim A( t) = limÍ 2 2 sin t ˙ = Í ˙. sin t ˙ tÆ0 tÆ0 t - 2 lim[ t ] lim e 2 2 1 e ˙ Í Î ˚ Î ˚ Î tÆ0 ˚ tÆ0 Ècos t 3˘ Differentiating A( t) component wise, we have A¢ ( t) = Í ˙ and Î 2t 0˚ È- sin t 0 ˘ A¢¢ ( t) = Í . A( t), A¢ ( t) and A¢¢ ( t) are defined for - • < t < • . 0 ˙˚ Î 2

13.

È 0 Differentiating A( t) component wise, we have A¢ ( t) = Í -1/ 2 Î-0.5(1 - t)

11.

14.

15.

16.

t -1 ˘ ˙ and 3e 3 t ˚

È - t -2 ˘ 0 A¢¢ ( t) = Í ˙ . A( t) is defined for - • < t < 0 and 0 < t £ 1. -3 / 2 9e 3 t ˚ Î-0.25(1 - t) A¢ ( t) and A¢¢ ( t) are defined for - • < t < 0 and 0 < t < 1. È t 2 3˘ Èsec t ˘ P ( t) = Í ˙ and g( t) = Í ˙. Î -5 ˚ Îsin t t ˚ È y1¢ ˘ Èt -1 y1 + ( t 2 + 1) y 2 + t ˘ Èt -1 y1 + ( t 2 + 1) y 2 ˘ È t ˘ ˙+Í ˙= Í Íy¢ ˙ = Í ˙= -1 -1 Î 2 ˚ Î 4 y1 + t y 2 + 8 t ln t ˚ Î 4 y1 + t y 2 ˚ Î8 t ln t ˚ Èt -1 t 2 + 1˘ Èt -1 t 2 + 1˘È y1 ˘ È t ˘ È t ˘ . Therefore, P ( t ) = and g( t) = Í + Í Í ˙. ˙ Í ˙ -1 ˙ -1 ˙ Í t ˚ t ˚Î y 2 ˚ Î8 t ln t ˚ Î8 t ln t ˚ Î4 Î4

1 ˘ È 2t Let A¢ ( t) = Í 2 ˙ . Integrating component wise, we find Îcos t 3t ˚ È t 2 + C11 t + C12 ˘ A( t) = Í ˙. 3 Îsin t + C21 t + C22 ˚

Chapter 6 First Order Linear Systems • 117

È t2 + 2 t + 5 ˘ ÈC11 C12 ˘ È2 5 ˘ Since A(0) = Í ˙. ˙= Í ˙ , we obtain A( t) = Í 3 ÎC21 C22 ˚ Î1 -2 ˚ Îsin t + 1 t - 2 ˚

17.

18.

Èt -1 4 t ˘ Let A¢ ( t) = Í . Integrating component wise, we find 2˙ Î 5 3t ˚ 2 + C12 ˘ È2 5 ˘ Èln t + C11 2 t 2 + C12 ˘ È C11 A ( 1 ) = . Since A( t) = Í ˙ Í ˙ = Í1 -2 ˙ , we 3 5 + C 1 + C 5 t + C t + C Î 21 22 ˚ Î ˚ 21 22 ˚ Î 2 Èln t + 2 2 t + 3˘ obtain A( t) = Í ˙. t3 - 3 ˚ Î 5t - 4 È1 t ˘ Let A¢¢ ( t) = Í ˙ . Integrating component wise, we find Î0 0 ˚

È A¢ ( t) = Ít + C11 Í C Î 21

˘ È t2 t2 + C12 ˙ fi A( t) = Í + C11t + D11 2 Í 2C t + D C22 ˙˚ 21 21 Î

˘ t3 + C12 t + D12 ˙ . 6 C22 t + D22 ˙˚

È t2 5 ˘ t3 5 È 1 1˘ È-1 2 ˘ 1 1 t + + t + Í ˙. Since A(0) = Í ˙ and A(1) = Í-2 3˙ , we obtain A( t) = Í 2 2 6 6 2 1 Î ˚ Î ˚ 2 t + 1 ˙˚ -2 Î 19.

Integrating component wise, we obtain t t È t 2 s ds cos s ds 2 ds ˘ È t 2 t sin t 2t ˘ Ú Ú Ú 0 0 0 Í ˙= . B ( s ) ds = Í t t t Ú0 2 -1 Í 5 ds ˙ Î5 t ln t + 1 t 3 ˙˚ ( s + ) ds s ds 1 3 Ú0 Ú0 ˙˚ ÎÍ Ú0 È et - 1 ˘ 3t 2 t 20. Integrating component wise, we obtain Ú B( s) ds = Í sin 2pt 1 - cos 2pt ˙ . 0 ÍÎ 2p ˙˚ 2p È1 t˘ 21 (a). One example is A = Í 2 ˙. Ît 0 ˚ È0 t ˘ 22. One example is A = Í ˙. Î0 0 ˚

Section 6.2 1.

The given problem can be written as y ¢ ( t) = P ( t) y ( t) + g( t), y ( 3) = y 0 È t -1 tan t ˘ È0 ˘ È0 ˘ where P ( t) = Í . The coefficient functions t ˙, g( t) = Í ˙, y 0 = Í ˙ 0 1 ln t e Î ˚ Î ˚ Î ˚

p11 ( t) = t -1 and p21 ( t) = ln t are discontinuous at t = 0. The coefficient function p12 ( t) = tan t has discontinuities at ±p / 2, ± 3p / 2, K. The largest interval containing t0 = 3 but containing no discontinuities of any coefficient function is the interval p / 2 < t < 3p / 2 .

118 • Chapter 6 First Order Linear Systems

2.

3.

4.

5.

tan t ˘ È( t + 1) -2 ˘ È 1 È0 ˘ In standard form, the problem is y ¢ = Í 2 y+Í ˙, y (0) = Í ˙ . ˙ Ît - 2 4 ˚ Î0 ˚ Î 0 ˚ -2 tan t is discontinuous at t = ±p / 2 and (t +1) is discontinuous at t = -1. The largest interval containing t0 = 0 but containing no discontinuities of any coefficient function is the interval -1 < t < p / 2 . È(cos t) / t 2 1 / t 2 ˘ È1 / t 2 ˘ È0 ˘ In standard form, the problem is y ¢ = Í + y ˙ Í ˙, y (1) = Í ˙ . 2 4t ˚ Î2 ˚ Î Îsec t ˚ The only discontinuities of p11 ( t) and p12 ( t) are at t = 0, while g2 ( t) is discontinuous at t = ±p / 2, ± 3p / 2, K. The largest interval containing t0 = 1 but containing no discontinuities of any coefficient function is the interval 0 < t < p / 2 .

5 ˘ È 3t Ít + 2 t + 2˙ È0 ˘ y, y (1) = Í ˙ . In standard form, the problem is y ¢ = Í ˙ 2 4t Î2 ˚ Í ˙ Ît - 2 t - 2˚ The largest interval containing t0 = 1 but containing no discontinuities of any coefficient function is the interval -2 < t < 2 . Differentiating, y1¢ = 5c1e 5 t + 3c 2e 3 t and y 2¢ = 5c1e 5 t - 3c 2e 3 t . Calculating the right hand sides, 4 y1 + y 2 = 4 (c1e 5 t + c 2e 3 t ) + (c1e 5 t - c 2e 3 t ) = 5c1e 5 t + 3c 2e 3 t = y1¢ and y1 + 4 y 2 = (c1e 5 t + c 2e 3 t ) + 4 (c1e 5 t - c 2e 3 t ) = 5c1e 5 t - 3c 2e 3 t = y 2¢ .

È4 1 ˘ y¢ = Í ˙y Î1 4 ˚ È1˘ È1˘ 7 (b). y = c1e 5 t Í ˙ + c 2e 3 t Í ˙ Î1˚ Î-1˚ È 1 1˘ 8 (a). y ¢ = Í ˙y Î-1 1˚

7 (a).

È e t cos t ˘ È e t sin t ˘ 8 (b). y = c1 Í t ˙ + c2 Í t ˙ Î-e sin t ˚ Îe cos t ˚ È2˘ È1˘ È2˘ È1˘ 9. For y = c1e 2 t Í ˙ + c 2e 3 t Í ˙, we have y ¢ = 2c1e 2 t Í ˙ + 3c 2e 3 t Í ˙ . On the other hand, Î-1˚ Î-1˚ Î-1˚ Î-1˚ Ê È2˘ È 1 ˘ˆ È2˘ È1˘ Ay = AÁ c1e 2 t Í ˙ + c 2e 3 t Í ˙˜ = c1e 2 t A Í ˙ + c 2e 3 t A Í ˙ = Ë Î-1˚ Î-1˚¯ Î-1˚ Î-1˚ È1 -2 ˘È 2 ˘ È1 -2 ˘È 1 ˘ È4˘ È3˘ c1e 2 t Í + c 2e 3 t Í = c1e 2 t Í ˙ + c 2e 3 t Í ˙ . Thus, y ¢ = Ay for every choice of ˙ Í ˙ ˙ Í ˙ Î1 4 ˚Î-1˚ Î1 4 ˚Î-1˚ Î-2 ˚ Î-3˚ c1 and c 2 .

Chapter 6 First Order Linear Systems • 119

10.

11.

12.

13.

In order to solve the initial value problem, we first note that È2˘ È 2 1 ˘È c1 ˘ È 4 ˘ È 1 ˘ È 2 1 ˘È c1 ˘ y(0) = c1 Í ˙ + c 2 Í ˙ = Í . Thus, solving Í-1 -1˙Íc ˙ = Í-3˙ , we obtain ˙Í ˙ Î-1˚ Î Î-1˚ Î-1 -1˚Îc 2 ˚ ˚Î 2 ˚ Î ˚ 2t 2 1 ˘ ˘ È2e + 2e 3 t ˘ 2t È 3t È is the solution of the c1 = 1 and c 2 = 2 . Therefore, y( t) = e Í ˙ + 2e Í ˙ = Í 2 t 3t ˙ Î-1˚ Î-1˚ Î-e - 2e ˚ given initial value problem. È3 + 2 ˘ È-3 + 4 ˘ È5 ˘ È1˘ + c 2e - t Í For y ¢ = c1e 5 t Í ˙ + c 2e - t Í ˙ , Ay = c1e 5 t Í ˙ ˙ Î 4 + 1˚ Î-4 + 2 ˚ Î5 ˚ Î-2 ˚ È2e 5 t - 3e - t ˘ È1 -1˘È c1 ˘ È-1˘ = Solving Í , we obtain . Therefore, is the c = 2 and c = 3 y( t ) = Í 5t ˙Í ˙ Í ˙ 1 2 -t ˙ Î1 2 ˚Îc 2 ˚ Î 8 ˚ Î2e + 6e ˚ solution of the given initial value problem.

È y ( t) ˘ Let Y( t) = Í ˙. Calculating Y¢ ( t) , we find Î y ¢ ( t) ˚ 1 ˘ È y ( t) ˘ È 0 ˘ y ¢ ( t) È y ¢ ( t) ˘ È ˘ È0 =Í 2 =Í Y¢ ( t) = Í ˙ ˙ ˙+Í ˙ . Therefore, the scalar 2 ˙Í Î y ¢¢ ( t) ˚ Î- t y ¢ ( t) - 4 y ( t) + sin t ˚ Î-4 - t ˚Î y ¢ ( t) ˚ Îsin t ˚ 1 ˘ È0 È 0 ˘ and G ( t ) = equation can be written as Y¢ = P ( t) Y + G( t) where P ( t) = Í ˙ Ísin t ˙ . 2 Î-4 - t ˚ Î ˚ È y ( t) ˘ Let Y( t) = Í ˙. Calculating Y¢ ( t) , we find Î y ¢ ( t) ˚ 1 ˘ È y ( t) ˘ È 0 È y ¢ ( t) ˘ È 0 ˘ =Í +Í 2 Y¢ ( t) = Í ˙ ˙ Í ˙ ˙ . Therefore, the scalar equation can Î y ¢¢ ( t) ˚ Î- t sec t 3t sec t ˚Î y ¢ ( t) ˚ Î( t + 1)sec t ˚ 1 ˘ 0 È 0 È ˘ and G( t) = Í 2 be written as Y¢ = P ( t) Y + G( t) where P ( t) = Í ˙ ˙. Î- t sec t 3t sec t ˚ Î( t + 1)sec t ˚ È y ( t) ˘ ˙ Í Let Y( t) = Í y ¢ ( t) ˙ . We solve for y ¢¢¢ by multiplying the equation by e - t and find ÍÎ y ¢¢ ( t) ˙˚ ˘ y ¢ ( t) È ˙ Í Y¢ ( t) = Í y ¢¢ ( t) ˙ . Expressing Y¢ ( t) in matrix terms, we -t - t -1 -t -t ÍÎ-5e y ¢¢ ( t) - e t y ¢ ( t) - (e tan t) y ( t) + e ˙˚ 1 0 ˘ È y ( t) ˘ È 0 ˘ È 0 ˙ Í ˙ ˙Í Í have Y¢ ( t) = Í 0 0 1 ˙Í y ¢ ( t) ˙ + Í 0 ˙ . Therefore, the scalar equation can be ÍÎ-e - t tan t -e - t t -1 -5e - t ˙˚ÍÎ y ¢¢ ( t) ˙˚ ÍÎe - t ˙˚ written as Y¢ = P ( t) Y + G( t) where È0˘ È 0 1 0 ˘ Í ˙ ˙ Í P ( t) = Í 0 0 1 ˙ and G( t) = Í 0 ˙ . ÍÎe - t ˙˚ ÍÎ-e - t tan t -e - t t -1 -5e - t ˙˚

120 • Chapter 6 First Order Linear Systems

14.

15.

16.

17.

1 0 ˘ È y ( t) ˘ È 0 ˘ È0 È y ( t) ˘ ˙ Í ˙ ˙ ˙Í Í Í Let Y( t) = Í y ¢ ( t) ˙ . Y¢ ( t) = Í0 0 1 ˙Í y ¢ ( t) ˙ + Í 0 ˙ . Therefore, the scalar equation can ÍÎ y ¢¢ ( t) ˙˚ ÍÎ t - cos t 2 ˙˚ÍÎ y ¢¢ ( t) ˙˚ ÍÎe 3 t ˙˚ 1 0˘ È0˘ È0 Í ˙ ˙ Í be written as Y¢ = P ( t) Y + G( t) where P ( t) = Í0 0 1 ˙ and G( t) = Í 0 ˙ . ÍÎe 3 t ˙˚ ÍÎ t - cos t 2 ˙˚

È y ( t) ˘ È y ¢ ( t) ˘ so that Y¢ ( t) = Í Let Y( t) = Í ˙ ˙ . We are given that Î y ¢ ( t) ˚ Î y ¢¢ ( t) ˚ y ¢ ( t) È 0 1 ˘ È y ( t) ˘ È 0 ˘ È ˘ +Í =Í Y¢ ( t) = Í ˙ Í ˙ ˙ ˙. Î-3 2 ˚Î y ¢ ( t) ˚ Î2 cos 2 t ˚ Î-3 y ( t) + 2 y ¢ ( t) + 2 cos 2 t ˚ Therefore, equating components of the vector Y¢ ( t) , we see that the scalar equation is y ¢¢ = -3 y + 2 y ¢ + 2 cos 2 t, y (-1) = 1, y ¢ (-1) = 4 .

y ¢¢¢ - 4 y ¢¢ + 2 y = e 3 t , y (0) = 1, y ¢ (0) = -2, y ¢¢ (0) = 3 . È y ¢ ( t) ˘ È y1 ( t) ˘ È y ( t) ˘ Í y ( t) ˙ Í y ( t) ˙ Í y ¢ ( t) ˙ 2 ˙= Í ˙ so that Y¢ ( t) = Í ¢¢ ˙ . We are given that Let Y( t) = Í Í y ¢¢¢ ( t) ˙ Í y 3 ( t) ˙ Í y ¢¢ ( t) ˙ Í ˙ Í ˙ Í (4 ) ˙ Î y 4 ( t) ˚ Î y ¢¢¢ ( t) ˚ Î y ( t) ˚ y2 y¢ È ˘ È ˘ Í ˙ Í ˙ y3 y ¢¢ ˙ Í ˙ . Equating components of the vector Í Y¢ ( t) = = Í ˙ Í ˙ y4 y ¢¢¢ Í Í 2˙ 2˙ Î y 2 + y 3 sin( y1 ) + y 3 ˚ Î y ¢ + y ¢¢ sin( y ) + ( y ¢¢ ) ˚ Y¢ ( t) , we see that the scalar equation is y ( 4 ) = y ¢ + y ¢¢ sin( y ) + ( y ¢¢ ) 2 , y (1) = 0, y ¢ (1) = 0, y ¢¢ (1) = -1, y ¢¢¢ (1) = 2 .

18.

Making the indicated change of variables, the system of differential equations is Y2¢ = Y2 + Y3 + tY4 . Therefore, the system can be expressed in the form Y4¢ = 2 tY1 + Y2 + Y4 Y¢ = P ( t) Y + G( t) where È 0 1 0 0˘ È0 ˘ Í0 ˙ Í0 1 1 t˙ Í ˙ P ( t) = and G( t) = Í ˙ . Í 0 0 0 1˙ Í0 ˙ Í ˙ Í ˙ Î2 t 1 0 1 ˚ Î0 ˚

19.

Making the indicated change of variables, the system of differential equations is Y2¢ = t -1Y2 + 4Y1 - tY3 + (sin t)Y4 + e 2 t . Y4¢ = Y1 - 5Y4

Chapter 6 First Order Linear Systems • 121

20.

21.

Therefore, the system can be expressed in the form Y¢ = P ( t) Y + G( t) where 0 ˘ È0˘ È0 1 0 Í4 t -1 - t sin t ˙ Íe 2 t ˙ ˙ and G( t) = Í ˙ . P ( t) = Í Í0˙ Í0 0 0 1 ˙ Í ˙ Í ˙ Î1 0 0 -5 ˚ Î0˚ Making the indicated change of variables, the system of differential equations is Y2¢ = 4Y1 + 7Y2 - 8Y3 + 6Y4 + t 2 . Therefore, the system can be expressed in the form Y4¢ = 3Y1 - 6Y2 + 2Y3 + 5Y4 - sin t Y¢ = P ( t) Y + G( t) where È 0 ˘ È0 1 0 0 ˘ Í t2 ˙ Í4 7 -8 6 ˙ ˙. Í ˙ P ( t) = and G( t) = Í Í 0 ˙ Í0 0 0 1 ˙ ˙ Í Í ˙ Î- sin t ˚ Î 3 -6 2 5 ˚ Making the indicated change of variables, the system of differential equations is 15Y3 + 9Y2 + 3Y2¢ = 12Y1 - 6Y4 + 3t 2 . Y4 + 5Y1 - Y4¢ = 2Y3 - 6Y2 + t Writing this system in standard form, Y2¢ = 4Y1 - 3Y2 - 5Y3 - 2Y4 + t 2 . Y4¢ = 5Y1 + 6Y2 - 2Y3 + Y4 - t Therefore, the system can be expressed in the form Y¢ = P ( t) Y + G( t) where È0˘ È0 1 0 0 ˘ Í t2 ˙ Í4 -3 -5 -2 ˙ Í ˙ P ( t) = and G( t) = Í ˙ . Í0˙ Í0 0 0 1 ˙ Í ˙ Í ˙ Î- t ˚ Î5 6 -2 1 ˚

Section 6.3 1 (a).

In matrix terms, the system has the form y ¢ = Ay where È y1¢ ˘ È 9 -4 ˘È y1 ˘ È 9 -4 ˘ or y = = ¢ Í y ¢ ˙ Í15 -7 ˙Í y ˙ Í15 -7 ˙y. Î 2˚ Î ˚Î 2 ˚ Î ˚ 1 (b). We have È6e 3 t ˘ È 9 -4 ˘È2e 3 t ˘ È18e 3 t - 12e 3 t ˘ and y ¢ = Í 3 t ˙. Calculating Ay, we obtain Í ˙Í 3 t ˙ = Í 3t 3t ˙ Î15 -7 ˚Î3e ˚ Î30e - 21e ˚ Î9e ˚ È6e 3 t ˘ therefore, Ay = Í 3 t ˙ , showing that the function y(t) is a solution of y ¢ = Ay . Î9e ˚ È-3 -2 ˘ 2 (a). y ¢ = Í ˙y . Î4 3˚

122 • Chapter 6 First Order Linear Systems

3 (a). In matrix terms, the system has the form y ¢ = Ay where È y1¢ ˘ È 1 4 ˘È y1 ˘ È 1 4˘ Í y ¢ ˙ = Í-1 1 ˙Í y ˙ or y ¢ = Í-1 1 ˙y. Î 2˚ Î ˚Î 2 ˚ Î ˚ 3 (b). We have È2e t cos 2 t - 4 e t sin 2 t ˘ y¢ = Í t ˙. Calculating Ay, we obtain t Î-e sin 2 t - 2e cos 2 t ˚ È 1 4 ˘È2e t cos 2 t ˘ È2e t cos 2 t - 4 e t sin 2 t ˘ ˙ . Therefore, ˙= Í Í-1 1 ˙Í t t t Î ˚Î-e sin 2 t ˚ Î-2e cos 2 t - e sin 2 t ˚ the function y(t) is a solution of y ¢ = Ay . 1 ˘ È0 2 ˙y . 4 (a). y ¢ = Í 2 ÍÎ t 2 - t ˙˚ 5 (a). In matrix terms, the system has the form y ¢ = Ay where È 0 1 1˘ È y1¢ ˘ È 0 1 1 ˘È y1 ˘ ˙ Í ˙Í ˙ Í ˙ Í Í y 2¢ ˙ = Í-6 -3 1 ˙Í y 2 ˙ or y ¢ = Í-6 -3 1 ˙y. ÍÎ-8 -2 4 ˙˚ ÍÎ y 3¢ ˙˚ ÍÎ-8 -2 4 ˙˚ÍÎ y 3 ˙˚ 5 (b). We have È et ˘ Í ˙ y ¢ = Í-e t ˙. Calculating Ay, we obtain ÍÎ2e t ˙˚

˘ È et ˘ È 0 1 1 ˘È e t ˘ È -e t + 2e t Í t˙ ˙Í t ˙ Í Í t t t˙ Í-6 -3 1 ˙Í-e ˙ = Í-6e + 3e + 2e ˙ = Í-e ˙ and therefore ÍÎ-8 -2 4 ˙˚ÍÎ2e t ˙˚ ÍÎ-8e t + 2e t + 8e t ˙˚ ÍÎ2e t ˙˚ the function y(t) is a solution of y ¢ = Ay . È2 1 1 ˘ ˙ Í 6 (a). y ¢ = Í1 1 2 ˙y. ÍÎ1 2 1 ˙˚ È6e 3 t ˘ È 9 -4 ˘È2e 3 t ˘ È18e 3 t - 12e 3 t ˘ 7 (a). y1¢ = Í 3 t ˙ and also Ay1 = Í = y1¢ . ˙Í 3 t ˙ = Í 3t 3t ˙ Î15 -7 ˚Î3e ˚ Î30e - 21e ˚ Î9e ˚ Similarly for y ¢2 . 7 (b). The Wronskian W(t) is given by È2e 3 t 2e - t ˘ . Thus, W ( t) = 10e 2 t - 6e 2 t = 4 e 2 t . W ( t) = det[Y ( t)] where Y ( t) = Í 3 t -t ˙ 5e ˚ Î3e 7 (c).

È2e 3 t y( t) = Í 3 t Î3e

2e - t ˘È c1 ˘ ˙Í ˙ 5e - t ˚Îc 2 ˚

Chapter 6 First Order Linear Systems • 123

È2 2 ˘È c1 ˘ È1˘ 7 (d). Given the general solution in part (c), y(0) = Í ˙Í ˙ = Í ˙ . Solving, we find Î3 5 ˚Îc 2 ˚ Î1˚ È c1 ˘ È 3 / 4 ˘ Íc ˙ = Í-1 / 4 ˙ . Therefore, the solution of the initial value problem is Î 2˚ Î ˚ È6e 3 t - 2e - t ˘ È2e 3 t ˘ È2e - t ˘ . y( t) = ( 3 / 4 ) Í 3 t ˙ - (1 / 4 ) Í - t ˙ = (1 / 4 ) Í 3 t -t ˙ Î9e - 5e ˚ Î3e ˚ Î5e ˚ 8 (b). The Wronskian W(t) is given by È 2e 3 t - 4 e - t 4 e 3 t + 2e - t ˘ . Thus, W ( t) = 20e 2 t π 0 . W ( t) = det[Y ( t)] where Y ( t) = Í 3 t -t 3t -t ˙ 6e + 5e ˚ Î3e - 10e

8 (c).

È 2e 3 t - 4 e - t y( t) = Í 3 t -t Î3e - 10e

4 e 3 t + 2e - t ˘È c1 ˘ ˙Í ˙ 6e 3 t + 5e - t ˚Îc 2 ˚

È-2 6 ˘È c1 ˘ È0 ˘ 8 (d). Given the general solution in part (c), y(0) = Í ˙Í ˙ = Í ˙ . Solving, we find Î-7 11˚Îc 2 ˚ Î1 ˚ È c1 ˘ È-3 / 10 ˘ Íc ˙ = Í-1 / 10 ˙ . Therefore, the solution of the initial value problem is Î 2˚ Î ˚ 3t -t ˘ È4 e 3 t + 2e - t ˘ È -e + e È 2e 3 t - 4 e - t ˘ Í 3 5 = - (1 / 10) Í 3 t y( t) = (-3 / 10) Í 3 t 3t - t ˙. -t ˙ -t ˙ + e e + 6 5 3 e 10 e e e Í Î ˚ Î 2 Î ˚ 2 ˙˚ È-e - t ˘ È 3 2 ˘È e - t ˘ È 3e - t - 4 e - t ˘ 9 (a). y1¢ = Í - t ˙ and also Ay1 = Í = y1¢ . =Í ˙Í -t ˙ -t -t ˙ Î-4 -3˚Î-2e ˚ Î-4 e + 6e ˚ Î2e ˚ Similarly for y ¢2 . 9 (b). The Wronskian W(t) is given by È e-t -3e - t ˘ . Thus, W ( t) = 6e -2 t - 6e -2 t ∫ 0 W ( t) = det[Y ( t)] where Y ( t) = Í -t -t ˙ 2 e 6 e Î ˚ and therefore, the given set of solutions is not a fundamental set of solutions. 10 (b). The Wronskian W(t) is given by È ˘ -5e -2 t cos 3t -5e -2 t sin 3t W ( t) = det[Y ( t)] where Y ( t) = Í -2 t ˙ . Thus, -2 t Îe (cos 3t - 3 sin 3t) e ( 3 cos 3t + sin 3t) ˚ W ( t) = -15e -4 t π 0 . ˘È c1 ˘ È -5e -2 t cos 3t -5e -2 t sin 3t 10 (c). y( t) = Í -2 t ˙Í ˙ -2 t Îe (cos 3t - 3 sin 3t) e ( 3 cos 3t + sin 3t) ˚Îc 2 ˚ È-5 0 ˘È c1 ˘ È5 ˘ È c1 ˘ È-1˘ = Í ˙ . Solving, we find Í ˙ = Í ˙ . 10 (d). Given the general solution in part (c), y(0) = Í ˙ Í ˙ Î 1 3˚Îc 2 ˚ Î2 ˚ Îc 2 ˚ Î 1 ˚ È 5e -2 t (cos 3t - sin 3t) ˘ Therefore, the solution of the initial value problem is y( t) = Í -2 t ˙. Îe (2 cos 3t + 4 sin 3t) ˚ È et ˘ È-3 -2 ˘È e t ˘ È-3e t + 4 e t ˘ 11 (a). y1¢ = Í and also A = =Í t y = y1¢ . 1 Í 4 3 ˙Í t˙ t˙ t ˙ Î ˚Î-2e ˚ Î 4 e - 6e ˚ Î-2e ˚ Similarly for y ¢2 .

124 • Chapter 6 First Order Linear Systems

11 (b). The Wronskian W(t) is given by È et W ( t) = det[Y ( t)] where Y ( t) = Í t Î-2e È et e - t ˘È c1 ˘ 11 (c). y( t) = Í ˙ t - t ˙Í Î-2e -e ˚Îc 2 ˚

e-t ˘ ˙ . Thus, W ( t) = -1 + 2 = 1. -e - t ˚

È e e -1 ˘È c1 ˘ È 1 ˘ 11 (d). Given the general solution in part (c), y(1) = Í ˙ = Í ˙ . Solving, we find -1 ˙ Í Î-2e -e ˚Îc 2 ˚ Î-3˚ È c1 ˘ È2e -1 ˘ ˙ . Therefore, the solution of the initial value problem is Íc ˙ = Í Î 2 ˚ Î -e ˚ È e t ˘ È e - t ˘ È 2e t -1 - e1- t ˘ . y( t) = 2e -1 Í - eÍ -t ˙ = Í 1- t ˙ t˙ t -1 Î-2e ˚ Î-e ˚ Î-4 e + e ˚ 12 (b). The Wronskian W(t) is given by È1 e 3 t ˘ . Thus, W ( t) = -3e -3 t π 0 . W ( t) = det[Y ( t)] where Y ( t) = Í 3t ˙ Î1 -2e ˚ È1 e 3 t ˘È c1 ˘ 12 (c). y( t) = Í ˙ 3 t ˙Í Î1 -2e ˚Îc 2 ˚ È1 e -3 ˘È c1 ˘ È-2 ˘ 12 (d). Given the general solution in part (c), y(-1) = Í ˙ = Í ˙ . Solving, we find -3 ˙ Í Î1 -2e ˚Îc 2 ˚ Î 4 ˚ È-2e 3( t +1) ˘ È c1 ˘ e 3 È0 ˘ = . Therefore, the solution of the initial value problem is y( t) = Í 3( t +1) ˙ . Íc ˙ 3 ÍÎ6 ˙˚ Î 2˚ Î 4e ˚ 13 (a). È 2 t -2 1 - 2 t -1 + 2 t -2 ˘Èt 2 - 2 t ˘ È2 t - 2 ˘ y1¢ = Í and also y A = Í -2 ˙Í ˙ 1 ˙ 2 t -1 - 2 t -2 ˚Î 2 t ˚ Î 2 ˚ Î-2 t . È(2 - 4 t -1 ) + (2 t - 4 + 4 t -1 ) ˘ =Í ˙ = y1¢ -1 -1 Î (-2 + 4 t ) + ( 4 - 4 t ) ˚ Similarly for y ¢2 . 13 (b). The Wronskian W(t) is given by Èt 2 - 2 t t - 1˘ 2 W ( t) = det[Y ( t)] where Y ( t) = Í ˙ . Thus, W ( t) = - t . 1 ˚ Î 2t 2 Èt - 2 t t - 1˘È c1 ˘ 13 (c). y( t) = Í ˙Í ˙ 1 ˚Îc 2 ˚ Î 2t

È c1 ˘ È 1 ˘ È0 1˘È c1 ˘ È-2 ˘ = Í ˙ . Solving, we find Í ˙ = Í ˙ . 13 (d). Given the general solution in part (c), y(2) = Í ˙ Í ˙ Îc 2 ˚ Î-2 ˚ Î4 1˚Îc 2 ˚ Î 2 ˚ 2 2 Èt - 2 t ˘ Èt - 1˘ Èt - 4 t + 2 ˘ Therefore, the solution of the initial value problem is y( t) = Í ˙ - 2Í ˙. ˙= Í Î 2t ˚ Î 1 ˚ Î 2t - 2 ˚

Chapter 6 First Order Linear Systems • 125

14 (b). The Wronskian W(t) is given by Èe -2 t ˘ 0 0 Í ˙ t t W ( t) = det[Y ( t)] where Y ( t) = Í 0 2e cos 2 t 2e sin 2 t ˙ . Thus, W ( t) = 2 . ÍÎ 0 -e t sin 2 t e t cos 2 t ˙˚ Èe -2 t ˘È c1 ˘ 0 0 Í ˙Í ˙ 14 (c). y( t) = Í 0 2e t cos 2 t 2e t sin 2 t ˙Íc 2 ˙ . ÍÎ 0 -e t sin 2 t e t cos 2 t ˙˚ÍÎc 3 ˙˚

È1 0 0 ˘È c1 ˘ È 3 ˘ ˙Í ˙ Í ˙ Í 14 (d). Given the general solution on part (c), y(0) = Í0 2 0 ˙Íc 2 ˙ = Í 4 ˙ . ÍÎ0 0 1 ˙˚ÍÎc 3 ˙˚ ÍÎ-2 ˙˚ È c1 ˘ È 3 ˘ Í ˙ Í ˙ Solving, we find Íc 2 ˙ = Í 2 ˙ . Therefore, the solution of the initial value problem ÍÎc 3 ˙˚ ÍÎ-2 ˙˚ È ˘ 3e -2 t Í t ˙ is y( t) = Í 4 e (cos 2 t - sin 2 t) ˙ . ÍÎ2e t (- cos 2 t + sin 2 t) ˙˚ È 5e t ˘ È -21 -10 2 ˘È 5e t ˘ È 5e t ˘ ˙ Í ˙ Í ˙ ˙Í Í 15 (a). y1¢ = Í-11e t ˙ and also Ay1 = Í 22 11 -2 ˙Í-11e t ˙ = Í-11e t ˙ = y1¢ . ÍÎ 0 ˙˚ ÍÎ-110 -50 11 ˙˚ÍÎ 0 ˙˚ ÍÎ 0 ˙˚ Similarly for y ¢2 and y ¢3 . 15 (b). The Wronskian W(t) is given by È 5e t et e-t ˘ Í ˙ W ( t) = det[Y ( t)] where Y ( t) = Í-11e t 0 -e - t ˙ . Thus, W ( t) = -11e t . ÍÎ 0 11e t 5e - t ˙˚ È 5e t et Í 15 (c). y( t) = Í-11e t 0 ÍÎ 0 11e t

e - t ˘È c1 ˘ ˙Í ˙ -e - t ˙Íc 2 ˙ . 5e - t ˙˚ÍÎc 3 ˙˚

È 5 1 1 ˘È c1 ˘ È 3 ˘ ˙ ˙Í ˙ Í Í 15 (d). Given the general solution on part (c), y(0) = Í-11 0 -1˙Íc 2 ˙ = Í-10 ˙ . ÍÎ 0 11 5 ˙˚ÍÎc 3 ˙˚ ÍÎ-16 ˙˚ È c1 ˘ È 1 ˘ Í ˙ Í ˙ Solving, we find Íc 2 ˙ = Í-1˙ . Therefore, the solution of the initial value problem ÍÎc 3 ˙˚ ÍÎ-1˙˚ È 5e t ˘ È e t ˘ È e - t ˘ È 4 e t - e - t ˘ Í ˙ Í ˙ Í ˙ Í ˙ is y( t) = Í-11e t ˙ - Í 0 ˙ - Í-e - t ˙ = Í -11e t + e - t ˙ . ÍÎ 0 ˙˚ ÍÎ11e t ˙˚ ÍÎ5e - t ˙˚ ÍÎ-11e t - 5e - t ˙˚

126 • Chapter 6 First Order Linear Systems

È 5e - t et ˘ 16 (a). W ( t) = det Í ˙=2 -t -e t ˚ Î-7e 16 (b). The trace of A is equal to 6 - 6 = 0 . t

16 (c). For t0 = -1, W ( t0 )e

Út0 tr[ P ( s)]ds

t

= 2e Ú-1

0 ds

= 2.

È 5e 2 t 17 (a). W ( t) = det Í 2t Î-7e

17 (b). 17 (c). 18 (a). 18 (b). 18 (c).

e4 t ˘ = 2e 6 t 4t ˙ -e ˚ È9 5˘ The trace of A = Í ˙ is equal to 9 + (-3) = 6 . Î-7 -3˚ t t 6 ds Út0 tr[ P ( s)]ds Ú 0 For t0 = 0, W ( t0 )e = 2e = 2e 6 t . È-1 e t ˘ -1 t W ( t) = det Í -1 ˙ = -t e 0˚ Ît The trace of A is equal to 1 - t -1. t t t Ê Ú1 (1- s-1 )dsˆ Út0 tr[ P ( s)]ds s - ln s 1 = - eÁ e = -ee t - ln t -1 = -e t - ln t = - t -1e t . For t0 = 1, W ( t0 )e ˜ = -ee ¯ Ë

È2e t Í 19 (a). W ( t) = det Í-e t ÍÎ-e t

e4 t ˘ È 2 0 1˘ ˙ ˙ Í e 4 t ˙ = e t e - t e 4 t det Í-1 -1 1˙ = -6e 4 t ÍÎ-1 1 1˙˚ e 4 t ˙˚ È2 1 1 ˘ ˙ Í 19 (b). The trace of A = Í1 1 2 ˙ is equal to 2 + 1 + 1 = 4 . ÍÎ1 2 1 ˙˚ t t 4 ds Ú tr[ P ( s)]ds 19 (c). For t0 = 0, W ( t0 )e t0 = -6e Ú0 = -6e 4 t . È5 2e 3 t ˘ 20 (a). W ( t) = det Í = 3e 3 t π 0 . 3t ˙ 1 e Î ˚ t tr A ds [ ] 20 (b). 3e 3 t = 3e Ú0 fi tr[ A] = 3 . 0 -e - t e-t

È5 2e 3 t ˘ È0 6e 3 t ˘ È5 2e 3 t ˘ 20 (c). y = Í . fiy¢ = Í = AÍ 3t ˙ 3t ˙ 3t ˙ Î1 e ˚ Î0 3e ˚ Î1 e ˚ È0 6e 3 t ˘ 1 Èe 3 t -2e 3 t ˘ È-2 10 ˘ 20 (d). A = Í ◊ 3t Í ˙= Í ˙. 3t ˙ 5 ˚ Î-1 5 ˚ Î0 3e ˚ 3e Î -1 The results are consistent since tr[ A] = -2 + 5 = 3 t

21.

Ú tr[ P ( s)]ds If W(t) is constant, then by Abel’s Theorem, the function e t0 must also be constant. Therefore, g( t) = Ú t tr[ P ( s)]ds must be constant and hence the derivative of g(t) is identically t0 zero. However, by the fundamental theorem of calculus, g¢ ( t) = tr[ P ( t)] and hence the trace of P(t) must be zero. Since the trace is equal to 3 + a we conclude that a = -3 .

Chapter 6 First Order Linear Systems • 127

Section 6.4 1 (a). 1 (b). 1 (c). 2 (a). 2 (b). 2 (c). 3 (a). 3 (b). 3 (c). 4.

5.

6.

7.

8.

Èt t 2 ˘ 2 Let F ( t) = [f1 ( t), f2 ( t)] = Í ˙. Then, det[ F ( t)] = 2 t - t . Î1 2 ˚ No, because we do not know whether the functions f1 ( t) and f2 ( t) form a fundamental set of solutions for a linear system. Yes. At t = 1, the determinant is 2 - 1 = 1 π 0. Therefore, [f1 ( t), f2 ( t)]k = 0 fi k = 0. det[ F ( t)] = t 2e t - t sin t . No Yes. At t = 1, the determinant is e - sin1 π 0. È te t sin 2 t ˘ t 2 Let F ( t) = [f1 ( t), f2 ( t)] = Í ˙. Then, det[ F ( t)] = 2 te - ( t - 1)sin t . 2 ˚ Ît - 1 No, because we do not know whether the functions f1 ( t) and f2 ( t) form a fundamental set of solutions for a linear system. Yes. At t = 1, the determinant is 2e π 0. Èt 2 ˘ Èt t 2 ˘È k1 ˘ È0 ˘ Èt ˘ 2 k1 Í ˙ + k2 Í ˙ = Í ˙Í ˙ = Í ˙; det = t - t π 0 at t = 2 for example fi k = 0 . Therefore, the k 1 0 Î˚ Î 1 ˚ Î1 1 ˚Î 2 ˚ Î ˚ given set of functions is linearly independent. Èe t ˘ Èe - t ˘ È0 ˘ Èk e t + k2e - t ˘ È0 ˘ We need to solve the equation k1 Í ˙ + k2 Í ˙ ∫ Í ˙ or Í 1 ˙ ∫ Í ˙ . This vector k1 Î1 ˚ Î 0 ˚ Î0 ˚ Î ˚ Î0 ˚

equation requires k1e t + k2e - t ∫ 0 and k1 ∫ 0 . By the second equation, k1 = 0 and hence, using this fact in the first equation, k2e - t ∫ 0 . Multiplying this identity by the nonzero function e t , Èe t ˘ Èe - t ˘ È0 ˘ we see that k2 = 0 as well. Hence, the only way to satisfy k1 Í ˙ + k2 Í ˙ ∫ Í ˙ is to choose Î1 ˚ Î 0 ˚ Î0 ˚ k1 = k2 = 0 . This means the given set of functions is linearly independent. Èe t ˘ Èe - t ˘ Èsinh t ˘ È0 ˘ k1 Í ˙ + k2 Í ˙ + k3 Í ˙ = Í ˙; Let k1 = 1, k2 = -1, k3 = -2 . The given set of functions is Î 0 ˚ Î0 ˚ Î1 ˚ Î1˚ linearly dependent. È k1 ˘ È0 ˘ È 0 ˘ È0 ˘ È1 ˘ ˙ Í ˙ Í Í ˙ Í ˙ Í ˙ We need to solve the equation k1 Í t ˙ + k2 Í 1 ˙ ∫ Í0 ˙ or Ík1t + k2 ˙ ∫ Í0 ˙ . The first component of ÍÎ k2 t 2 ˙˚ ÍÎ0 ˙˚ ÍÎt 2 ˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚ this vector identity cannot be satisfied unless k1 = 0 and the third component cannot be È 0 ˘ È0 ˘ È1 ˘ Í ˙ Í ˙ Í ˙ satisfied unless k2 = 0 . Hence, the only way to satisfy the identity k1 Í t ˙ + k2 Í 1 ˙ ∫ Í0 ˙ is to ÍÎt 2 ˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚ choose k1 = k2 = 0 . This means the given set of functions is linearly independent. È0 ˘ È0 ˘ È0 ˘ È1 ˘ Í ˙ Í ˙ Í ˙ Í ˙ k1 Í t ˙ + k2 Í 1 ˙ + k3 Í0 ˙ = Í0 ˙; Let k1 = 0, k2 = 0, k3 = 1 . The given set of functions is linearly ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎt 2 ˙˚ ÍÎ0 ˙˚ dependent.

128 • Chapter 6 First Order Linear Systems

È k1 ˘ È0 ˘ È0 ˘ È0 ˘ È0 ˘ È1 ˘ ˙ Í ˙ Í Í ˙ Í ˙ Í ˙ Í ˙ 9. We need to solve the equation k1 Í t ˙ + k2 Í 1 ˙ + k3 Í0 ˙ ∫ Í0 ˙ or Í k1t + k2 ˙ ∫ Í0 ˙ . The first ÍÎk2 t 2 + k3 ˙˚ ÍÎ0 ˙˚ ÍÎ1 ˙˚ ÍÎ0 ˙˚ ÍÎt 2 ˙˚ ÍÎ0 ˙˚ component of this vector identity cannot be satisfied unless k1 = 0 and therefore the second component requires k2 = 0 . Given that k1 and k2 must both be zero, the third component then È0 ˘ È0 ˘ È0 ˘ È1 ˘ Í ˙ Í ˙ Í ˙ Í ˙ requires that k3 = 0 . Hence, the only way to satisfy the identity k1 Í t ˙ + k2 Í 1 ˙ + k3 Í0 ˙ ∫ Í0 ˙ is ÍÎ1 ˙˚ ÍÎ0 ˙˚ ÍÎt 2 ˙˚ ÍÎ0 ˙˚ to choose k1 = k2 = k3 = 0 . This means the given set of functions is linearly independent. È1 ˘ È0 ˘ ˘ È È 1 ˘ 0 1 Í ˙ Í ˙ Í Í 2 ˙ 2 ˙ 10. k1 Ísin t ˙ + k2 Í2 - 2 cos t ˙ + k3 Í0 ˙ = Í0 ˙; Let k1 = 1, k2 = - , k3 = -1. The given set of 2 ÍÎ1 ˙˚ ÍÎ0 ˙˚ ˙˚ ÍÎ ÍÎ 0 ˙˚ -2 functions is linearly dependent. Èe t t 2 ˘ t 11 (a). Let F ( t) = Í ˙ . Then, det[ F ( t)] = te . t 0 Î ˚ 11 (b). Since the determinant is zero at t = 0, F ( t) cannot be a fundamental matrix for a linear system defined on an interval containing t = 0. 11 (c). A fundamental matrix Y( t) satisfies the matrix differential equation Y ¢ = P ( t)Y . Given that Èe t t 2 ˘ Èe t 2 t ˘ Y( t) = Í ˙ , we know that Y ¢ ( t) = Í ˙ . Therefore, the equation Y ¢ = P ( t)Y implies Î0 t ˚ Î0 1 ˚ Èe t 2 t ˘ Èe t t 2 ˘ -1 -1 that Í ˙ = P ( t) Í ˙ . Postmultiplying by Y , we see that Y ¢Y = P ( t) . Therefore, Î0 1 ˚ Î0 t ˚ Èe t 1 / ( te t ) Í Î0

Ète t (2 t - t 2 )e t ˘ 2 t ˘È t - t 2 ˘ t and so 1 P ( t ) P ( t ) = / ( te ) = Í ˙ . Canceling the nonzero ˙Í ˙ 1 ˚Î0 e t ˚ et Î0 ˚ 2 È t (2 t - t ) ˘ term e t we have P ( t) = t -1 Í ˙. 1 ˚ Î0 Èt 2 2 t ˘ 2 12 (a). Let F ( t) = Í ˙ . Then, det[ F ( t)] = t . Î0 1 ˚ 12 (b). Since the determinant is zero at t = 0, F ( t) cannot be a fundamental matrix for a linear system defined on an interval containing t = 0. 12 (c). A fundamental matrix Y( t) satisfies the matrix differential equation Y ¢ = P ( t)Y . Given that Èt 2 2 t ˘ È2 t 2 ˘ Y( t) = Í ˙ , we know that Y ¢ ( t) = Í ˙ . Therefore, the equation Y ¢ = P ( t)Y implies Î 0 0˚ Î0 1 ˚ Èt 2 2 t ˘ È2 t 2 ˘ -1 -1 that Í = P ( t) Í ˙ . Postmultiplying by Y , we see that Y ¢Y = P ( t) . Therefore, ˙ Î 0 0˚ Î0 1 ˚ 2 t 2 ˘È1 -2 t ˘ È2 t -1 -2 ˘ 2 È 1 / (t )Í ˙ which is continuous on ˙Í 2 ˙ = P ( t) and so P ( t) = Í 0˚ Î 0 0 ˚Î0 t ˚ Î 0 (-•,0) and (0, •) .

Chapter 6 First Order Linear Systems • 129

Èe t -e - t ˘ 13 (a). We first show that Y ¢ = P ( t)Y . Now, Y ¢ ( t) = Í t ˙ whereas e-t ˚ Îe È0 1 ˘Èe t e - t ˘ Èe t -e - t ˘ P ( t)Y ( t) = Í =Í t ˙ . Thus, since Y ¢ = P ( t)Y , we know that Y( t) is a ˙Í t -t ˙ e-t ˚ Î1 0 ˚Îe -e ˚ Îe solution matrix. To show Y( t) is a fundamental matrix, we need to verify that det[Y ( t)] π 0 . Since det[Y ( t)] ∫ -2 , we know Y( t) is a fundamental matrix. ) Èa b ˘ È sinh t cosh t ˘ 1 Èe t - e - t e t + e - t ˘ 13 (b). Y( t) = Í . Thus, we need a matrix C = Í = Í t ˙ such that ˙ -t t -t ˙ e -e ˚ Îc d ˚ Îcosh t sinh t ˚ 2 Îe + e t -t e t + e - t ˘ Èe t e - t ˘Èa b ˘ 1 Èe - e . Expanding the right-hand side of this matrix ˙ Í ˙= Í 2 Îe t + e - t e t - e - t ˚ Îe t -e - t ˚ÍÎc d ˙˚ equation, we arrive at the requirement t -t e t + e - t ˘ Èae t + ce - t be t + de - t ˘ 1 Èe - e ˙ . Comparing entries, we see that Í ˙= Í 2 Îe t + e - t e t - e - t ˚ Î ae t - ce - t be t - de - t ˚

13 (c). 14 (a). 14 (b). 14 (c). 15 (a).

15 (b).

È 1 / 2 1 / 2˘ . a = 1 / 2, c = -1 / 2, b = 1 / 2, and d = 1 / 2 . Thus, C = Í -1 / 2 1 / 2 ˙˚ Î ) det[C ] = 1 / 2 and thus, Y( t) is a fundamental matrix. Since det[Y ( t)] π 0 , we know Y( t) is a fundamental matrix. ) È2e t - e - t e t + 3e - t ˘ Èe t -e - t ˘È2 1 ˘ È2 1 ˘ C = . Thus, = Y( t) = Í t Í ˙ ˙ Í1 -3˙ . Í ˙ -t e t - 3e - t ˚ Îe t e - t ˚Î1 -3˚ Î ˚ Î2e + e ) det[C ] = -7 and thus, Y( t) is a fundamental matrix. ) Èe t -2e -2 t ˘ We first show that Y = P ( t)Y . Now, Y ¢ ( t) = Í whereas -2 t ˙ Î 0 6e ˚ e -2 t ˘ Èe t -2e -2 t ˘ È1 1 ˘Èe t . Thus, since Y ¢ = P ( t)Y , we know that Y( t) is P ( t)Y ( t) = Í =Í ˙Í -2 t ˙ -2 t ˙ Î0 -2 ˚Î 0 -3e ˚ Î 0 6e ˚ a solution matrix. To show Y( t) is a fundamental matrix, we need to verify that det[Y ( t)] π 0 . Now det[Y ( t)] = -3e - t and thus is never zero for any value t. Therefore, Y( t) is a fundamental matrix. ) È 2e -2 t 0 ˘ Èa b ˘ C = and so we need a matrix Y( t) = Í ˙ Íc d ˙ such that -2 t 0˚ Î ˚ Î-6e È 2e -2 t 0 ˘ Èe t e -2 t ˘Èa b ˘ = ˙Í Í ˙ Í ˙ . Expanding the right-hand side of this matrix equation, we -2 t 0 ˚ Î 0 -3e -2 t ˚Îc d ˚ Î-6e arrive at the requirement È 2e -2 t 0 ˘ Èae t + ce -2 t be t + de -2 t ˘ ˙ . Comparing entries in the second column, we see that Í ˙= Í -2 t -3de -2 t ˚ 0 ˚ Î -3ce -2 t Î-6e d = 0 and b = 0 . Comparing entries in the first column, we see c = 2 and a = 0 . Thus, È0 0 ˘ C=Í ˙. Î2 0 ˚

130 • Chapter 6 First Order Linear Systems

) 15 (c). det[C ] = 0 and thus, Y( t) is a solution matrix but not a fundamental matrix. 16 (a). Since det[Y ( t)] = -6e 2 t π 0 , we know Y( t) is a fundamental matrix. Èe t + e - t 4 e 2 t e t + 4 e 2 t ˘ Èe t È1 0 1 ˘ 4 e 2 t ˘È1 0 1 ˘ e-t ) Í ˙ Í ˙ Í ˙ 2 t ˙Í -t 2t 2t -t 16 (b). Y( t) = Í -2e e e e ˙Í1 0 0 ˙ . Thus, C = Í1 0 0 ˙ . ˙ = Í 0 -2e ÍÎ 0 ÍÎ0 1 1 ˙˚ 3e 2 t 3e 2 t ˙˚ ÍÎ 0 0 3e 2 t ˙˚ÍÎ0 1 1 ˙˚ ) 16 (c). det[C ] = 1 and thus, Y( t) is a fundamental matrix. ) ) Èe t e - t ˘ , we need a matrix C such that 17. For Y( t) = Í t Y ( t ) = Y ( t ) C where Y (0) = I . This ˙ -t Îe -e ˚ ) requirement means that I = Y (0) = Y (0)C . È1 1 ˘ 1 È1 1 ˘ Equivalently, C is the inverse of Y(0) = Í . Thus, C = Y(0) -1 = Í . ˙ 2 Î1 -1˙˚ Î1 -1˚ ) ) Èe t e -2 t ˘ 18. For Y( t) = Í , we need a matrix C such that Y ( t) = Y ( t)C where Y (0) = I . This -2 t ˙ Î 0 -3e ˚ ) requirement means that I = Y (0) = Y (0)C . È1 1 ˘ 1 È-3 -1˘ Equivalently, C is the inverse of Y(0) = Í . Thus, C = Y(0) -1 = - Í . ˙ 3 Î 0 1 ˙˚ Î0 -3˚

Section 6.5 È 4 2 ˘È 1 ˘ È 2 ˘ Ax1 = Í ˙Í ˙ = Í ˙ = 2 x1. Thus, x1 is an eigenvector corresponding to the eigenvalue Î-1 1 ˚Î-1˚ Î-2 ˚ È 4 2 ˘È-2 ˘ È-6 ˘ l1 = 2 . Similarly, Ax 2 = Í ˙Í ˙ = Í ˙ = 3x 2 . Thus, x 2 is an eigenvector corresponding Î-1 1 ˚Î 1 ˚ Î 3 ˚ to the eigenvalue l 2 = 3. È1˘ È-2 ˘ 1 (b). Solutions are y1 ( t) = e 2 t Í ˙ and y 2 ( t) = e 3 t Í ˙ . Î-1˚ Î1˚ 1 (a).

È e 2 t -2e 3 t ˘ 5 t 1 (c). The Wronskian is W ( t) = det Í 2 t = e - 2e 5 t = -e 5 t . Since W(t) is nonzero for any 3t ˙ e e Î ˚ value t, the two solutions form a fundamental set of solutions. È 7 -3˘È3˘ È-3˘ 2 (a). Ax1 = Í ˙Í ˙ = Í ˙ = -1x1. Thus, x1 is an eigenvector corresponding to the eigenvalue Î16 -7 ˚Î8 ˚ Î-8 ˚ È 7 -3˘È1 ˘ È1 ˘ l1 = -1. Similarly, Ax 2 = Í ˙Í ˙ = Í ˙ = 1x 2 . Thus, x 2 is an eigenvector corresponding Î16 -7 ˚Î2 ˚ Î2 ˚ to the eigenvalue l 2 = 1. È3˘ È1 ˘ 2 (b). Solutions are y1 ( t) = e - t Í ˙ and y 2 ( t) = e t Í ˙ . Î8 ˚ Î2 ˚

Chapter 6 First Order Linear Systems • 131

2 (c). 3 (a).

3 (b). 4 (a).

4 (b). 5 (a).

5 (b). 5 (c).

6 (a).

6 (b).

È3e - t e t ˘ The Wronskian is W ( t) = det Í - t ˙ = -2 π 0 . Since W(t) is nonzero for any value t, the 2e t ˚ Î8e two solutions form a fundamental set of solutions. The vector x1 = 0 cannot be an eigenvector since an eigenvector must be nonzero. Considering 5 ˘È 1 ˘ È 1 ˘ È 11 the other vector, Ax 2 = Í ˙Í ˙ = Í ˙ = x 2 . Thus, Î-22 -10 ˚Î-2 ˚ Î-2 ˚ x 2 is an eigenvector corresponding to the eigenvalue l 2 = 1. È1˘ The solution is y 2 ( t) = e t Í ˙ . Î-2 ˚ È -5 2 ˘È1 ˘ È1 ˘ È -5 2 ˘È1 ˘ È -1˘ Ax1 = Í = Í ˙ = 1x1 , Ax 2 = Í ˙ ˙Í ˙ = Í ˙ . Thus, Í ˙ Î-18 7 ˚Î3˚ Î3˚ Î-18 7 ˚Î2 ˚ Î-4 ˚ x1 is an eigenvector corresponding to the eigenvalue l1 = 1, but x 2 is not an eigenvector. È1 ˘ The solution is y1 ( t) = e t Í ˙ . Î3˚ È0 1 ˘È 1 ˘ È-1˘ Ax1 = Í ˙Í ˙ = Í ˙ = - x1 . Thus, x1 is an eigenvector corresponding to the eigenvalue Î1 0 ˚Î-1˚ Î 1 ˚ È0 1 ˘È2 ˘ È2 ˘ l1 = -1. Similarly, Ax 2 = Í ˙Í ˙ = Í ˙ = x 2 . Thus, x 2 is an eigenvector corresponding to Î1 0 ˚Î2 ˚ Î2 ˚ the eigenvalue l 2 = 1. È1˘ È2 ˘ Solutions are y1 ( t) = e - t Í ˙ and y 2 ( t) = e t Í ˙ . Î-1˚ Î2 ˚ -t t Èe 2e ˘ The Wronskian is W ( t) = det Í - t ˙ = 2 + 2 = 4 . Since W(t) is nonzero for any value t, the 2e t ˚ Î-e two solutions form a fundamental set of solutions. È 2 -1˘È 1 ˘ È 4 ˘ Ax1 = Í ˙Í ˙ = Í ˙ = 4 x1 . Thus, x1 is an eigenvector corresponding to the eigenvalue Î-4 2 ˚Î-2 ˚ Î-8 ˚ È 2 -1˘È1 ˘ È0 ˘ l1 = 4 . Similarly, Ax 2 = Í ˙Í ˙ = Í ˙ = 0 x 2 . Thus, x 2 is an eigenvector corresponding Î-4 2 ˚Î2 ˚ Î0 ˚ to the eigenvalue l 2 = 0 . È1˘ È1 ˘ È1 ˘ Solutions are y1 ( t) = e 4 t Í ˙ and y 2 ( t) = e 0 Í ˙ = Í ˙ . Î-2 ˚ Î2 ˚ Î2 ˚

È e4 t 1˘ 4t 6 (c). The Wronskian is W ( t) = det Í ˙ = 4 e π 0 . Since W(t) is nonzero for any value t, the 4t e 2 2 Î ˚ two solutions form a fundamental set of solutions.

132 • Chapter 6 First Order Linear Systems

7.

8.

9.

10.

11.

È-4 3 ˘ È-6 3˘È x1 ˘ È0 ˘ the equation ( A - 2 I )x = 0 has the form Í For A = Í ˙ ˙Í ˙ = Í ˙ . Elementary row Î-4 4 ˚ Î-4 2 ˚Î x 2 ˚ Î0 ˚ operations [ -(1 / 3) R1 then R2 + 2 R1 ] can be used to row reduce the system to È2 -1˘È x1 ˘ È0 ˘ È x1 ˘ È x1 ˘ È1 ˘ x = = = = x or . Thus an eigenvector is 2 x = x Í0 0 ˙Í x ˙ Í0 ˙ Í x ˙ Í2 x ˙ 1 Í2 ˙, x1 π 0 . 1 2 Î Î 2 ˚ Î 1˚ ˚Î 2 ˚ Î ˚ Î ˚ 3˘ 3 ˘È x1 ˘ È0 ˘ È5 È6 For A = Í ˙ the equation ( A + I )x = 0 has the form Í-4 -2 ˙Í x ˙ = Í0 ˙ . Thus an Î-4 -3˚ Î ˚Î 2 ˚ Î ˚ È x1 ˘ È x1 ˘ È1˘ = x1 Í ˙, x1 π 0 . eigenvector is x = Í ˙ = Í ˙ Î x 2 ˚ Î-2 x1 ˚ Î-2 ˚ È 1 1˘ È-4 1˘È x1 ˘ È0 ˘ the equation ( A - 5 I )x = 0 has the form Í For A = Í ˙ ˙Í ˙ = Í ˙ . Elementary row Î-4 6 ˚ Î-4 1˚Î x 2 ˚ Î0 ˚ operations [ -(1 / 4 ) R1 then R2 + 4 R1 ] can be used to row reduce the system to È1 -1 / 4 ˘È x1 ˘ È0 ˘ È x1 ˘ È x1 ˘ È1 ˘ = Í ˙ or 4 x1 = x 2 . Thus an eigenvector is x = Í ˙ = Í ˙ = x1 Í ˙, x1 π 0 . Í0 ˙ Í ˙ 0 ˚Î x 2 ˚ Î0 ˚ Î Î x 2 ˚ Î4 x1 ˚ Î4 ˚ È 5 -7 3 ˘È x1 ˘ È0 ˘ È 1 -7 3˘ ˙Í ˙ Í ˙ Í ˙ Í For A = Í-1 -1 1 ˙ the equation ( A + 4 I )x = 0 has the form Í-1 3 1 ˙Í x 2 ˙ = Í0 ˙ . Thus ÍÎ 4 -4 4 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ ÍÎ 4 -4 0 ˙˚ È2˘ È x1 ˘ È2 x 2 ˘ Í ˙ ˙ Í ˙ Í an eigenvector is x = Í x 2 ˙ = Í x 2 ˙ = x 2 Í 1 ˙, x 2 π 0 . ÍÎ-1˙˚ ÍÎ x 3 ˙˚ ÍÎ- x 2 ˙˚ È3 1 1˘ ˙ Í For A = Í-1 1 -1˙ the equation ( A - 2 I )x = 0 has the form ÍÎ 2 1 2 ˙˚ È 1 1 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í-1 -1 -1˙Í x 2 ˙ = Í0 ˙ . Elementary row operations ( R2 + R1, then R3 - 2 R1 ) followed by ÍÎ 2 1 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È1 0 -1˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í R1 + R3 and (- R3 ) can be used to row reduce the system to Í0 0 0 ˙Í x 2 ˙ = Í0 ˙ , or ÍÎ0 1 2 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È1˘ È x1 ˘ È x 3 ˘ x1 = x 3 Í ˙ ˙ Í ˙ Í Thus an eigenvector is x = Í x 2 ˙ = Í-2 x 3 ˙ = x 3 Í-2 ˙, x 3 π 0 . x 2 = -2 x 3 . ÍÎ 1 ˙˚ ÍÎ x 3 ˙˚ ÍÎ x 3 ˙˚

12.

È1 3 1 ˘ ˙ Í For A = Í2 1 2 ˙ the equation ( A - 5 I )x = 0 has the form ÍÎ4 3 -2 ˙˚

È-4 3 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í 2 -4 2 ˙Í x 2 ˙ = Í0 ˙ . ÍÎ 4 3 -7 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚

Chapter 6 First Order Linear Systems • 133

È1˘ È x1 ˘ È x1 ˘ Í˙ Í ˙ Í ˙ Thus an eigenvector is x = Í x 2 ˙ = Í x1 ˙ = x1 Í1˙, x1 π 0 . ÍÎ1˙˚ ÍÎ x 3 ˙˚ ÍÎ x1 ˙˚

13.

1˘ È-2 3 ˙ Í For A = Í-8 13 5 ˙ the equation Ax = 0 has the form ÍÎ11 -17 -6 ˙˚ 1 ˘È x1 ˘ È0 ˘ È-2 3 ˙Í ˙ Í ˙ Í Í-8 13 5 ˙Í x 2 ˙ = Í0 ˙ . Since the arithmetic looks forbidding, we turned to MATLAB and ÍÎ11 -17 -6 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È1 0 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í used the RREF command. MATLAB says the system is equivalent to Í0 1 1 ˙Í x 2 ˙ = Í0 ˙ , or ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È-1˘ È x1 ˘ È- x 3 ˘ x1 = - x 3 Í ˙ ˙ Í ˙ Í Thus, an eigenvector is x = Í x 2 ˙ = Í- x 3 ˙ = x 3 Í-1˙, x 3 π 0 . x2 = - x3 . ÍÎ 1 ˙˚ ÍÎ x 3 ˙˚ ÍÎ x 3 ˙˚

14.

15.

16.

17.

18.

-5 - l 1 , or 0 4-l p(l ) = (l + 5)(l - 4 ). Thus, the eigenvalues are l1 = -5 and l 2 = 4 . 8-l 0 È8 0 ˘ For A = Í , the characteristic polynomial is p(l ) = , or ˙ 3 2-l Î3 2 ˚ p(l ) = (8 - l )(2 - l ). Thus, the eigenvalues are l1 = 8 and l 2 = 2 . 3- l -3 The characteristic polynomial is p(l ) = , or -6 6 - l p(l ) = (l )(l - 9). Thus, the eigenvalues are l1 = 0 and l 2 = 9 . 5-l 2 È5 2 ˘ For A = Í the characteristic polynomial is p l , ( ) = , or ˙ 4 3- l Î4 3˚ The characteristic polynomial is p(l ) =

p(l ) = (5 - l )( 3 - l ) - 8 = l2 - 8l + 7 = (l - 7)(l - 1). Thus, the eigenvalues are l1 = 7 and l 2 = 1. The characteristic polynomial is 5-l 0 0 p(l ) = 0 1- l 3 0 2 2-l 2 or p(l ) = (5 - l )(l - 3l - 4 ) = (5 - l )(l - 4 )(l + 1) = -l3 + 8l2 - 11l - 20 and the eigenvalues are l1 = -1, l 2 = 4, and l 3 = 5 .

134 • Chapter 6 First Order Linear Systems

19.

20.

21.

22.

1˘ È-2 3 ˙ Í For A = Í-8 13 5 ˙, the characteristic polynomial is ÍÎ11 -17 -6 ˙˚ -2 - l 3 1 p(l ) = -8 13 - l 5 . Given the arithmetic required to find the 11 -17 -6 - l characteristic polynomial, it is advisable to use a computer routine such as poly(A) from MATLAB. However, it is possible to find p(l ) by hand: -2 - l 3 1 13 - l 5 5 -8 -8 13 - l -8 13 - l 5 = (-2 - l ) -3 + -17 -6 - l 11 -6 - l 11 -17 11 -17 -6 - l or p(l ) = (-2 - l )(l2 - 7l + 7) - 3(8l - 7) + (11l - 7) = -l3 + 5l2 - 6l . Thus, p(l ) = -l (l2 - 5l + 6) = -l (l - 3)(l - 2) and hence the eigenvalues are l1 = 0, l 2 = 3, and l 3 = 2 . The characteristic polynomial is 1- l -7 3

p(l ) = -1 -1 - l 1 4 -4 -l 2 or p(l ) = -l (l - 16) = (-l )(l - 4 )(l + 4 ) = -l3 + 16l and the eigenvalues are l1 = 0, l 2 = 4, and l 3 = -4 . The eigenvalues are l1 = -2 and l 2 = 2 with corresponding eigenvectors È3˘ È1 ˘ x1 = Í ˙ and x 2 = Í ˙ . A fundamental set of solutions consists of the functions Î2 ˚ Î2 ˚ È3˘ È1 ˘ y1 ( t) = e -2 t Í ˙ and y 2 ( t) = e 2 t Í ˙ . Therefore, the general solution is Î2 ˚ Î2 ˚ È3˘ È3˘ È1 ˘ È1 ˘ y( t) = c1e -2 t Í ˙ + c 2e 2 t Í ˙ . The solution of the initial value problem is y( t) = e -2 t Í ˙ + e 2 t Í ˙ . Î2 ˚ Î2 ˚ Î2 ˚ Î2 ˚ The eigenvalues are l1 = -1 and l 2 = 3 with corresponding eigenvectors È1˘ È3˘ x1 = Í ˙ and x 2 = Í ˙ . A fundamental set of solutions consists of the functions Î-2 ˚ Î-2 ˚ È1˘ È3˘ y1 ( t) = e - t Í ˙ and y 2 ( t) = e 3 t Í ˙ . Therefore, the general solution is Î-2 ˚ Î-2 ˚ È1˘ È3˘ y( t) = c1e - t Í ˙ + c 2e 3 t Í ˙. The solution of the initial value problem is Î-2 ˚ Î-2 ˚ È 3e 3( t -1) - e -( t -1) ˘ y( t) = Í ˙. 3( t -1) + 2e -( t -1) ˚ Î-2e

Chapter 6 First Order Linear Systems • 135

23.

24.

25.

The eigenvalues are l1 = 3 and l 2 = 5 with corresponding eigenvectors È1˘ È1 ˘ x1 = Í ˙ and x 2 = Í ˙ . A fundamental set of solutions consists of the functions Î1˚ Î2 ˚ È1˘ È1˘ È1 ˘ È1 ˘ y1 ( t) = e 3 t Í ˙ and y 2 ( t) = e 5 t Í ˙ . Therefore, the general solution is y( t) = c1e 3 t Í ˙ + c 2e 5 t Í ˙ . Î1˚ Î1˚ Î2 ˚ Î2 ˚ È1˘ È1 ˘ The solution of the initial value problem is y( t) = 3e 3 t Í ˙ + 4 e 5 t Í ˙ . Î1˚ Î2 ˚ The eigenvalues are l1 = -0.11 and l 2 = -0.05 with corresponding eigenvectors È1˘ È1 ˘ x1 = Í ˙ and x 2 = Í ˙ . A fundamental set of solutions consists of the functions Î-1˚ Î2 ˚ È1˘ È1 ˘ y1 ( t) = e -0.11t Í ˙ and y 2 ( t) = e -0.05 t Í ˙ . Î-1˚ Î2 ˚ The eigenvalues are l1 = 1, l 2 = 2, and l 3 = 3 with corresponding eigenvectors È1 ˘ È1 ˘ È1˘ Í ˙ Í ˙ Í ˙ x1 = Í 1 ˙, x 2 = Í0 ˙, and x 3 = Í1 ˙ . A fundamental set of solutions consists of the functions ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ-1˙˚

È1 ˘ È1 ˘ È1˘ Í ˙ 2t Í ˙ 3t Í ˙ y1 ( t) = e Í 1 ˙, y 2 ( t) = e Í0 ˙, and y 3 ( t) = e Í1 ˙ . The solution of the initial value problem is ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ-1˙˚ È1 ˘ È1 ˘ È1˘ ˙ 2t Í ˙ tÍ 3t Í ˙ y( t) = e Í 1 ˙ + e Í0 ˙ + 2e Í1 ˙ . ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ-1˙˚ The eigenvalues are l1 = -2, l 2 = 1, and l 3 = 4 with corresponding eigenvectors È1 ˘ È2˘ È1˘ Í ˙ Í ˙ Í ˙ x1 = Í-3˙, x 2 = Í-3˙, and x 3 = Í0 ˙ . A fundamental set of solutions consists of the functions ÍÎ0 ˙˚ ÍÎ 0 ˙˚ ÍÎ 3 ˙˚ t

26.

È1 ˘ È2˘ È1˘ ˙ Í ˙ tÍ 4t Í ˙ y1 ( t) = e Í-3˙, y 2 ( t) = e Í-3˙, and y 3 ( t) = e Í0 ˙ . The solution of the initial value problem is ÍÎ0 ˙˚ ÍÎ 0 ˙˚ ÍÎ 3 ˙˚ È0 ˘ È-2 ˘ È1˘ ˙ t Í ˙ 4t Í ˙ -2 t Í y( t) = e Í-3˙ + e Í 3 ˙ + e Í0 ˙ . ÍÎ0 ˙˚ ÍÎ 0 ˙˚ ÍÎ 3 ˙˚ -2 t

136 • Chapter 6 First Order Linear Systems

27.

28.

The eigenvalues are l1 = -2, l 2 = 2, and l 3 = 4 with corresponding eigenvectors È3˘ È1 ˘ È3˘ Í ˙ Í ˙ Í ˙ x1 = Í 4 ˙, x 2 = Í0 ˙, and x 3 = Í2 ˙ . A fundamental set of solutions consists of the functions ÍÎ2 ˙˚ ÍÎ0 ˙˚ ÍÎ-8 ˙˚ È3˘ È1 ˘ È3˘ Í ˙ ˙ Í ˙ Í y1 ( t) = e -2 t Í 4 ˙, y 2 ( t) = e 2 t Í0 ˙, and y 3 ( t) = e 4 t Í2 ˙ . ÍÎ2 ˙˚ ÍÎ0 ˙˚ ÍÎ-8 ˙˚ The eigenvalues are l1 = 1, l 2 = 3, and l 3 = 5 with corresponding eigenvectors È1 ˘ È1 ˘ È0 ˘ Í ˙ Í ˙ Í ˙ x1 = Í0 ˙, x 2 = Í1 ˙, and x 3 = Í2 ˙ . A fundamental set of solutions consists of the functions ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ1 ˙˚

29.

È1 ˘ È1 ˘ È0 ˘ Í ˙ 3t Í ˙ 5t Í ˙ y1 ( t) = e Í0 ˙, y 2 ( t) = e Í1 ˙, and y 3 ( t) = e Í2 ˙ . ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ1 ˙˚ The eigenvalues are l1 = -1, l 2 = 1, and l 3 = 2 with corresponding eigenvectors È2˘ È1˘ È1˘ Í ˙ Í ˙ Í ˙ x1 = Í-4 ˙, x 2 = Í-2 ˙, and x 3 = Í-2 ˙ . A fundamental set of solutions consists of the functions ÍÎ-1˙˚ ÍÎ-3˙˚ ÍÎ-5 ˙˚

30.

È2˘ È1˘ È1˘ ˙ ˙ Í ˙ tÍ 2t Í y1 ( t) = e Í-4 ˙, y 2 ( t) = e Í-2 ˙, and y 3 ( t) = e Í-2 ˙ . ÍÎ-1˙˚ ÍÎ-3˙˚ ÍÎ-5 ˙˚ The eigenvalues are l1 = -2, l 2 = 1, and l 3 = 2 with corresponding eigenvectors È1˘ È1˘ È1˘ Í ˙ Í ˙ Í ˙ x1 = Í-5 ˙, x 2 = Í-2 ˙, and x 3 = Í-1˙ . A fundamental set of solutions consists of the functions ÍÎ 0 ˙˚ ÍÎ-3˙˚ ÍÎ-6 ˙˚

t

-t

y1 ( t) = e

-2 t

È1˘ È1˘ È1˘ ˙ ˙ Í ˙ tÍ 2t Í Í-5 ˙, y 2 ( t) = e Í-2 ˙, and y 3 ( t) = e Í-1˙ . ÍÎ 0 ˙˚ ÍÎ-3˙˚ ÍÎ-6 ˙˚

È2 x ˘È 1 ˘ È1˘ = l Í ˙ for some value l . Therefore, equating vectors, it We need to have Í ˙ Í ˙ Î1 -5 ˚Î-1˚ Î-1˚ follows that we 2 - x = l and 1 + 5 = -l . This requires l = -6 and x = 8 . y ˘È-1˘ È-1˘ Èx 32. We need to have Í ˙Í ˙ = 1Í ˙ . Therefore, it follows that Î2 x - y ˚Î 1 ˚ Î 1 ˚ - x + y = -1 and - 2 x - y = 1. This requires x = 0, y = -1. 39 (a). The eigenvalues are l1 = -3 and l 2 = -1. Corresponding eigenvectors are È1˘ È1˘ È1˘ È1˘ x1 = Í ˙ and x 2 = Í ˙ . The general solution is y( t) = c1e -3t Í ˙ + c 2e -t Í ˙ . Î-1˚ Î-1˚ Î1˚ Î1˚ 31.

Chapter 6 First Order Linear Systems • 137

39 (b). The solution of the initial value problem is y( t) = -

È1˘ Q0 -3t È 1 ˘ Q e Í ˙ + 3 0 e -t Í ˙ . 2 2 Î-1˚ Î1˚

Q0 Q -e -3t + 3e -t ) and Q2 ( t) = 0 (e -3t + 3e -t ) . ( 2 2 Note that 0 < Q1 (t ) < Q2 (t ) . Therefore, we need t such that Q0 -3t (e + 3e -t ) < .01Q0 fi (e -3t + 3e -t ) < .02 . Graphically, we find that a value t ª 5.011 will 2 suffice. Since t = (V / r)t = 50t , we obtain a value of t ª 250.55 sec or t ª 4.18 min . 40 (a). The eigenvalues are l1 = -1 and l 2 = l 3 = -4 . Corresponding eigenvectors are È1˘ È1˘ È1˘ Í ˙ Í ˙ Í˙ x1 = Í1˙, x 2 = Í-1˙, and x 3 = Í 0 ˙ . The general solution is ÍÎ-1˙˚ ÍÎ 0 ˙˚ ÍÎ1˙˚ Therefore, Q1 ( t) =

È1˘ È1˘ È1˘ ˙ ˙ Í˙ -4t Í -4t Í Q( t) = c1e Í1˙ + c 2e Í-1˙ + c 3e Í 0 ˙ . ÍÎ-1˙˚ ÍÎ 0 ˙˚ ÍÎ1˙˚ -t

È1˘ È1˘ ˙ Í˙ -4t Í 40 (b). The solution of the initial value problem is y( t) = 2Q0e Í1˙ - Q0e Í 0 ˙ . ÍÎ-1˙˚ ÍÎ1˙˚ r r È2e - v t - e -4 v t ˘ ˙ Í -r t Therefore, Q( t) = Q0 Í 2e v ˙. Í2e - rv t + e -4 rv t ˙ ˚ Î -t

Section 6.6 1.

2-l 1 È 2 1˘ p( l ) = = l2 - 4 l + 5 . For A = Í , the characteristic polynomial is ˙ -1 2 - l Î-1 2 ˚ Therefore, the eigenvalues are l1 = 2 + i and l 2 = 2 - i . We find an eigenvector x1 by solving ( A - l1I ) x = 0 or 1 È2 - (2 + i) È -i 1 ˘È x1 ˘ È0 ˘ ˘È x1 ˘ È0 ˘ = Í ˙ . This equation reduces to Í Í -1 ˙ ˙Í ˙ = Í ˙ . The Í ˙ 2 - (2 + i) ˚Î x 2 ˚ Î0 ˚ Î Î-1 -i ˚Î x 2 ˚ Î0 ˚ È-i 1 ˘È x1 ˘ È0 ˘ elementary row operation R2 + iR1 reduces the system to Í ˙Í ˙ = Í ˙ or -ix1 + x 2 = 0 . Î 0 0 ˚Î x 2 ˚ Î0 ˚ È x1 ˘ È x1 ˘ È1˘ Thus, an eigenvector is x = Í ˙ = Í ˙ = x1 Í ˙, x1 π 0 . Since the eigenvalues and eigenvectors Î x 2 ˚ Îix1 ˚ Îi ˚ occur in conjugate pairs, the eigenpairs are È1˘ È1˘ l1 = 2 + i, x1 = Í ˙ and l 2 = 2 - i, x 2 = Í ˙ . Îi ˚ Î-i ˚

138 • Chapter 6 First Order Linear Systems

2.

3.

4.

5.

-l

-9

= l2 + 9 . Therefore, the eigenvalues are 1 -l l1 = 3i and l 2 = -3i . We find an eigenvector x1 by solving ( A - l1I ) x = 0 or È-3i -9 ˘È x1 ˘ È0 ˘ È x1 ˘ È3ix 2 ˘ È3i ˘ x = = = = x . Thus, an eigenvector is 1 2 Í ˙, x 2 π 0 . Since the Í 1 -3i ˙Í x ˙ Í0 ˙ Íx ˙ Í x ˙ Î ˚Î 2 ˚ Î ˚ Î 2˚ Î 2 ˚ Î1 ˚ eigenvalues and eigenvectors occur in conjugate pairs, the eigenpairs are È3i ˘ È-3i ˘ l1 = 3i, x1 = Í ˙ and l 2 = -3i, x 2 = Í ˙ . Î1 ˚ Î1 ˚ 6 - l -13 È6 -13˘ p( l ) = = l2 - 6l + 13. For A = Í , the characteristic polynomial is ˙ 1 -l Î1 0 ˚ Therefore, the eigenvalues are l1 = 3 + 2i and l 2 = 3 - 2i . We find an eigenvector x1 by solving ( A - l1I ) x = 0 or -13 ˘È x1 ˘ È0 ˘ -13 ˘È x1 ˘ È0 ˘ È6 - ( 3 + 2i) È3 - 2i = Í ˙ . This equation reduces to Í = . The Í ˙ Í ˙ 1 -( 3 + 2i) ˚Î x 2 ˚ Î0 ˚ -3 - 2i ˙˚ÍÎ x 2 ˙˚ ÍÎ0 ˙˚ Î Î 1 elementary row operations R1 ´ R2 , then R2 - ( 3 - 2i) R1 reduces the system to È1 -( 3 + 2i) ˘È x1 ˘ È0 ˘ Í0 ˙Í x ˙ = Í0 ˙ or x1 - ( 3 + 2i) x 2 = 0 . Thus, an eigenvector is 0 Î ˚Î 2 ˚ Î ˚ È x1 ˘ È( 3 + 2i) x 2 ˘ È3 + 2i ˘ x= Í ˙= Í = x2 Í ˙ ˙, x 2 π 0 . Choosing x 2 = 1, we obtain the eigenvector x2 Î x2 ˚ Î ˚ Î 1 ˚ È3 + 2i ˘ x1 = Í ˙ . Since the eigenvalues and eigenvectors occur in conjugate pairs, the eigenpairs Î 1 ˚ are È3 + 2i ˘ È3 - 2i ˘ l1 = 3 + 2i, x1 = Í and l 2 = 3 - 2i, x 2 = Í ˙ ˙. Î 1 ˚ Î 1 ˚ 3- l 1 = (l - 2) 2 + 1. Therefore, the eigenvalues The characteristic polynomial is p(l ) = -2 1 - l are l1 = 2 + i and l 2 = 2 - i . We find an eigenvector x1 by solving ( A - l1I ) x = 0 or 1 ˘È x1 ˘ È0 ˘ È1 - i È 1 ˘ Í -2 -1 - i ˙Í x ˙ = Í0 ˙ . Thus, an eigenvector is x1 = Í-1 + i ˙ . Since the eigenvalues and Î Î ˚Î 2 ˚ Î ˚ ˚ eigenvectors occur in conjugate pairs, the eigenpairs are È 1 ˘ È 1 ˘ l1 = 2 + i, x1 = Í and l 2 = 2 - i, x 2 = Í ˙ ˙. Î-1 + i ˚ Î-1 - i ˚ Using the EIG command in MATLAB, we find eigenvalues l1 = 1, l 2 = 1 + i , and l 3 = 1 - i . For each eigenvalue l , we use the RREF command in MATLAB to solve ( A - lI ) x = 0 , È 3 - 2i ˘ È3 + 2i ˘ È-3˘ ˙ Í ˙ Í Í ˙ finding x1 = Í 0 ˙ , x 2 = Í 1 ˙ , and x 3 = Í 1 ˙ . ÍÎ-1 + i ˙˚ ÍÎ-1 - i ˙˚ ÍÎ 1 ˙˚ The characteristic polynomial is p(l ) =

Chapter 6 First Order Linear Systems • 139

6.

7.

8.

9.

10.

È3 + 2i ˘ È-5 + i ˘ ˙ ˙ Í Í Note that another possible eigenvector for l 2 is x 2 = (-1 + i) Í 1 ˙ = Í-1 + i ˙. ÍÎ-1 - i ˙˚ ÍÎ 2 ˙˚ The eigenvalues are l1 = 2, l 2 = 2 + 3i, and l 3 = 2 - 3i . The corresponding eigenvectors are È-4 + i ˘ È-4 - i ˘ È1˘ ˙ Í ˙ Í Í ˙ x1 = Í 0 ˙ , x 2 = Í 3i ˙ , and x 3 = Í -3i ˙ . ÍÎ 1 - i ˙˚ ÍÎ 1 + i ˙˚ ÍÎ-1˙˚

È 4 ˘ 4t È 4 ˘ = e (cos 2 t + i sin 2 t) Í As in Example 1, y( t) = e( 4 + 2 i )t Í ˙ ˙ is one solution of y ¢ = Ay . Î-1 + i ˚ Î-1 + i ˚ Expanding and collecting real and imaginary parts, we obtain 4 cos 2 t 4 sin 2 t ˘ È ˘ 4t È y( t) = e 4 t Í ie + ˙ Ícos 2 t - sin 2 t ˙ . Thus, a fundamental set of solutions can be Î- cos 2 t - sin 2 t ˚ Î ˚ 4 cos 2 t È È 4 sin 2 t ˘ ˘ and y 2 ( t) = e 4 t Í formed from y1 ( t) = e 4 t Í ˙ ˙. Î- cos 2 t - sin 2 t ˚ Îcos 2 t - sin 2 t ˚ È-2 + i ˘ È-2 + i ˘ y( t) = e it Í = (cos t + i sin t) Í ˙ ˙ is one solution of y ¢ = Ay . Expanding and collecting Î 5 ˚ Î 5 ˚ È-2 cos t - sin t ˘ Ècos t - 2 sin t ˘ real and imaginary parts, we obtain y( t) = Í ˙ + i Í 5 sin t ˙ . Thus, a 5 cos t Î ˚ Î ˚ 2 cos t sin t È ˘ fundamental set of solutions can be formed from y1 ( t) = Í ˙ and 5 cos t Î ˚ Ècos t - 2 sin t ˘ y 2 ( t) = Í ˙. Î 5 sin t ˚ È-1 - i ˘ È-1 - i ˘ = (cos 2 t + i sin 2 t) Í As in Example 1, y( t) = e 2 it Í ˙ ˙ is one solution of y ¢ = Ay . Î 1 ˚ Î 1 ˚ Expanding and collecting real and imaginary parts, we obtain È- cos 2 t + sin 2 t ˘ È- cos 2 t - sin 2 t ˘ y( t) = Í ˙ + iÍ ˙ . Thus, a fundamental set of solutions can be cos 2 t sin 2 t Î ˚ Î ˚ È- cos 2 t + sin 2 t ˘ È- cos 2 t - sin 2 t ˘ formed from y1 ( t) = Í and y 2 ( t) = Í ˙ ˙. cos 2 t sin 2 t Î Î ˚ ˚ È-1 + i ˘ y( t) = e t (cos t + i sin t) Í ˙ is one solution of y ¢ = Ay . Expanding and collecting real and Î i ˚ È- cos t - sin t ˘ Ècos t - sin t ˘ + ie t Í imaginary parts, we obtain y( t) = e t Í ˙ ˙ . Thus, a fundamental set Î - sin t ˚ Î cos t ˚ È- cos t - sin t ˘ Ècos t - sin t ˘ of solutions can be formed from y1 ( t) = e t Í and y 2 ( t) = e t Í ˙ ˙. Î - sin t ˚ Î cos t ˚

140 • Chapter 6 First Order Linear Systems

11.

È-5 + 3i ˘ È-5 + 3i ˘ ˙ Í ˙ 2t Í As in Example 1, y( t) = e Í 3 + 3i ˙ = e (cos 3t + i sin 3t) Í 3 + 3i ˙ is one solution of ÍÎ 2 ˙˚ ÍÎ 2 ˙˚ y ¢ = Ay . Expanding and collecting real and imaginary parts, we obtain È 3 cos 3t - 5 sin 3t ˘ È-5 cos 3t - 3 sin 3t ˘ ˙ ˙ 2t Í 2t Í y( t) = e Í 3 cos 3t - 3 sin 3t ˙ + ie Í3 cos 3t + 3 sin 3t ˙ . Thus, two linearly independent solutions ˙˚ ÍÎ ˙˚ ÍÎ 2 cos 3t 2 sin 3t ( 2 + 3 i )t

È-5 cos 3t - 3 sin 3t ˘ È 3 cos 3t - 5 sin 3t ˘ ˙ ˙ Í 2t Í are y1 ( t) = e Í 3 cos 3t - 3 sin 3t ˙ and y 2 ( t) = e Í3 cos 3t + 3 sin 3t ˙ . The third solution needed ˙˚ ˙˚ ÍÎ ÍÎ 2 cos 3t 2 sin 3t È1˘ ˙ 2t Í to complete the fundamental set is obtained from the real eigenvalue l = 2 , y 3 ( t) = e Í 0 ˙ . ÍÎ-1˙˚ Ècos 5 t ˘ È- sin 5 t ˘ Èi ˘ Í cos 5 t ˙ Í sin 5 t ˙ Í1 ˙ t tÍ tÍ ˙ ˙ . Also, Í ˙ + ie e (cos 5 t + i sin 5 t) =e Í 0 ˙ Í 0 ˙ Í0 ˙ Í ˙ Í ˙ Í ˙ Î 0 ˚ Î 0 ˚ Î0 ˚ È 0 ˘ È 0 ˘ È0 ˘ Í 0 ˙ Í ˙ Í0 ˙ ˙ + ie t Í 0 ˙ Thus, a fundamental set of solutions can be e t (cos 2 t + i sin 2 t) Í ˙ = e t Í Í- sin 2 t ˙ Ícos 2 t ˙ Íi ˙ Í ˙ Í ˙ Í ˙ Î cos 2 t ˚ Î sin 2 t ˚ Î1 ˚ È- sin 5 t ˘ Ècos 5 t ˘ È 0 ˘ È 0 ˘ Í cos 5 t ˙ Í sin 5 t ˙ Í 0 ˙ Í 0 ˙ tÍ ˙, e t Í ˙. ˙, e t Í ˙, e t Í formed from e Í 0 ˙ Í 0 ˙ Í- sin 2 t ˙ Ícos 2 t ˙ ˙ Í ˙ ˙ Í Í ˙ Í Î 0 ˚ Î 0 ˚ Î cos 2 t ˚ Î sin 2 t ˚ Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is È cos t ˘ È1 ˘ È sin t ˘ È0 ˘ È4 ˘ y( t) = c1e 2 t Í + c 2e 2 t Í . Imposing the initial condition, y(0) = c1 Í ˙ + c 2 Í ˙ = Í ˙ , we ˙ ˙ Î- sin t ˚ Î0 ˚ Îcos t ˚ Î1 ˚ Î 7 ˚ È 4 cos t + 7 sin t ˘ obtain the solution y( t) = e 2 t Í ˙. Î-4 sin t + 7 cos t ˚ Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is È3 sin 3t ˘ È0 ˘ È3 cos 3t ˘ È3˘ È6 ˘ y( t) = c1 Í + c2 Í . Imposing the initial condition, y(0) = c1 Í ˙ + c 2 Í ˙ = Í ˙ , we ˙ ˙ Î cos 3t ˚ Î1 ˚ Î sin 3t ˚ Î0 ˚ Î2 ˚ È-6 sin 3t + 6 cos 3t ˘ obtain the solution y( t) = Í ˙. Î 2 cos 3t + 2 sin 3t ˚ 2t

12.

13.

14.

Chapter 6 First Order Linear Systems • 141

15.

16.

17.

18.

22.

Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is È3 cos 2 t - 2 sin 2 t ˘ È2 cos 2 t + 3 sin 2 t ˘ y( t) = c1e 3 t Í + c 2e 3 t Í ˙ ˙ . Imposing the initial condition, cos 2 t sin 2 t Î ˚ Î ˚ 3 2 1 È ˘ È ˘ È ˘ y(0) = c1 Í ˙ + c 2 Í ˙ = Í ˙ , we find c1 = 3 and c 2 = -4 and the solution is Î1 ˚ Î0 ˚ Î3˚ È cos 2 t - 18 sin 2 t ˘ y( t) = e 3 t Í ˙. Î3 cos 2 t - 4 sin 2 t ˚ Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is cos t È ˘ È sin t ˘ y( t) = c1e 2 t Í + c 2e 2 t Í ˙ ˙ . Imposing the initial condition, Î- cos t - sin t ˚ Îcos t - sin t ˚ È1˘ È8 cos t + 14 sin t ˘ È0 ˘ È8 ˘ y(0) = c1 Í ˙ + c 2 Í ˙ = Í ˙ , we obtain the solution y( t) = e 2 t Í ˙. Î-1˚ Î6 cos t - 22 sin t ˚ Î1 ˚ Î6 ˚ Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is È2 cos t + 3 sin t ˘ È3 cos t - 2 sin t ˘ È-3˘ ˙ ˙ Í ˙ Í tÍ y( t) = c1e t Í 0 ˙ + c 2e t Í cos t sin t ˙. ˙ + c 3e Í ÍÎ - cos t - sin t ˙˚ ÍÎ - cos t + sin t ˙˚ ÍÎ 1 ˙˚

È 2 ˘ È6 ˘ È3˘ È-3˘ Í ˙ Í ˙ Í ˙ Í ˙ Imposing the initial condition, y(0) = c1 Í 0 ˙ + c 2 Í 1 ˙ + c 3 Í 0 ˙ = Í1 ˙ , we find ÍÎ-1˙˚ ÍÎ2 ˙˚ ÍÎ-1˙˚ ÍÎ 1 ˙˚ È27 - 21cos t - 38 sin t ˘ ˙ tÍ cos t - 12 sin t c1 = -9, c 2 = 1, and c 3 = -12 , and the solution y( t) = e Í ˙. ÍÎ -9 + 11cos t + 13 sin t ˙˚ Proceeding as in Exercises 7-12, we find the general solution of y ¢ = Ay is È- cos 3t - 4 sin 3t ˘ È-4 cos 3t + sin 3t ˘ È1˘ ˙ ˙ ˙ 2t Í 2t Í 2t Í -3 sin 3t 3 cos 3t y( t) = c1e Í 0 ˙ + c 2e Í ˙. ˙ + c 3e Í ÍÎ cos 3t + sin 3t ˙˚ ÍÎ cos 3t - sin 3t ˙˚ ÍÎ-1˙˚ È-1˘ È-1˘ È-4 ˘ È1˘ Í ˙ Í ˙ Í ˙ Í ˙ Imposing the initial condition, y(0) = c1 Í 0 ˙ + c 2 Í 0 ˙ + c 3 Í 3 ˙ = Í 9 ˙ , we find ÍÎ 1 ˙˚ ÍÎ 4 ˙˚ ÍÎ 1 ˙˚ ÍÎ-1˙˚ È-2 + cos 3t - 13 sin 3t ˘ ˙ Í c1 = -2, c 2 = -1, and c 3 = 3 , and the solution y( t) = e 2 t Í 9 cos 3t + 3 sin 3t ˙ . ÍÎ 2 + 2 cos 3t + 4 sin 3t ˙˚ 1 The eigenvalues of A are l = -1 ± 9 + 12m / 2 . If 9 + 12m < 0, l = - ± ib (b π 0) , 2 therefore distinct and y ( t) Æ 0 . If 0 < 9 + 12m < 1, , the eigenvalues are distinct, real and 2 negative. Therefore, -• < 9 +12m < 1 - • < m < - . 3

(

)

142 • Chapter 6 First Order Linear Systems

23.

(

(

24.

)

The eigenvalues of A are l = -5 ± 1 + 4 m / 2 . In order that both components of y( t) go to zero as t Æ • , we need each of these (real) eigenvalues to be negative. Therefore, we need -5 + 1 + 4 m / 2 < 0 or 1 + 4 m < 5 . This inequality holds if and only if 1 + 4 m < 25 or - • < m < 6 .

)

(

)

The eigenvalues of A are l = -2 ± 16 - 4 m 2 / 2 = -1 ± 4 - m 2 . Require 2

2

-• < 4 - m < 1 fi 3 < m < • . Therefore, -• < m < - 3 and 3 < m < • . The eigenvalues of A are l = -1 ± 4 + m 2 . In order that both components of y( t) go to zero as t Æ • , we need each of these (real) eigenvalues to be negative. Therefore, we need -1 + 4 + m 2 < 0 or 4 + m 2 < 1 . This inequality cannot hold for any real value of m . È 0 1˘ È cos t sin t ˘È c1 ˘ d v= Í v fi v( t) = Í 26 (a). ˙ ˙Í ˙ . dt Î-1 0 ˚ Î- sin t cos t ˚Îc 2 ˚ Èsin t - 2 cos t ˘ È d1 ˘ È c1 ˘ È1 ˘ È cos t + 2 sin t ˘ 26 (b). v(0) = Í ˙ = Í ˙ fi v( t) = Í . r( t) = Í ˙ + Í ˙. ˙ Îcos t + 2 sin t ˚ Îd2 ˚ Îc 2 ˚ Î2 ˚ Î- sin t + 2 cos t ˚ È-2 + d1 ˘ È2 ˘ Èsin t - 2 cos t + 4 ˘ Ê 3p ˆ È-2 ˘ Ê 3p ˆ È 3 ˘ r (0) = Í = Í ˙ fi r ( t) = Í . vÁ ˜ = Í ˙ and r Á ˜ = Í ˙ . ˙ ˙ Ë 2 ¯ Î-2 ˚ Î 1 + d2 ˚ Î1 ˚ Î cos t + 2 sin t ˚ Ë 2 ¯ Î 1 ˚

25.

27 (d). If the charged particle is launched with initial velocity parallel to the magnetic field, it will move with constant velocity. g g g 28 (b). The eigenpairs are - + l1, x1 and - + l 2 , x 2 and - + l 3 , x 3 . m m m g -mt The corresponding fundamental matrix is e y ( t) .

Section 6.7 È2 1 ˘ 2 1 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = (l - 2) . The eigenvalue l1 = 2 has 0 2 Î ˚ algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 2 I ) x = 0 or È0 1 ˘È x1 ˘ È0 ˘ Í0 0 ˙Í x ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 2 have the form Î ˚Î 2 ˚ Î ˚ x È 1˘ È1 ˘ x = Í ˙ = x1 Í ˙ , x1 π 0 . The geometric multiplicity of l1 = 2 is 1. Î0 ˚ Î0 ˚ 1 (b). We find a generalized eigenvector corresponding to l1 = 2 by solving the equation È1 ˘ È x1 ˘ ( A - 2 I ) x = x1 where x1 = Í ˙ . The solution is x = Í ˙ where x1 is arbitrary. Choosing x1 = 0 , Î0 ˚ Î1 ˚ È0 ˘ we obtain the generalized eigenvector x 2 = Í ˙ . Î1 ˚

Chapter 6 First Order Linear Systems • 143

1 (c).

2 (a).

2 (b).

2 (c).

3 (a).

3 (b).

Thus, we have solutions y1 ( t) = e 2 t x1 and, as in equation (6), y 2 ( t) = te 2 t x1 + e 2 t x 2 . A Èe 2 t te 2 t ˘ fundamental matrix is Y( t) = [ y1 ( t), y 2 ( t)] = Í . 2t ˙ Î0 e ˚ The general solution is Y( t)c . Imposing the initial condition, Y(0)c = y 0 . We find È1 0 ˘È c1 ˘ È 1 ˘ È1˘ Í0 1 ˙Íc ˙ = Í-1˙ or c = Í-1˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ È1 - t ˘ y( t) = e 2 t Í ˙. Î -1 ˚ The characteristic polynomial is p(l ) = ( 3 - l ) 2 . The eigenvalue l1 = 3 has algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 3I ) x = 0 . Therefore, È x1 ˘ È1 ˘ all the eigenvectors corresponding to l1 = 3 have the form x = Í ˙ = x1 Í ˙ , x1 π 0 . The Î0 ˚ Î0 ˚ geometric multiplicity of l1 = 3 is 1. È1 ˘ È1 ˘ È0 ˘ Èt˘ y1 ( t) = e 3 t Í ˙ and y 2 ( t) = te 3 t Í ˙ + e 3 t Í 1 ˙ = e 3 t Í 1 ˙ . A fundamental matrix is Î0 ˚ Î0 ˚ Î2˚ Î2˚ 3t 3t Èe ˘ te . Y( t) = [ y1 ( t), y 2 ( t)] = Í 1 3t ˙ Î 0 2e ˚ The general solution is Y( t)c . Imposing the initial condition, Y(0)c = y 0 . We find È1 0 ˘È c1 ˘ È4 ˘ È4 ˘ Í0 1 ˙Íc ˙ = Í1 ˙ or c = Í2 ˙ . Thus, the solution of the initial value problem is 2 ˚Î 2 ˚ Î Î ˚ Î ˚ È2 t + 4 ˘ y( t) = e 3 t Í ˙. Î 1 ˚ È6 0 ˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = (l - 6) . The eigenvalue l1 = 6 has 2 6 Î ˚ algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 6 I ) x = 0 or È0 0 ˘È x1 ˘ È0 ˘ Í2 0 ˙Í x ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 6 have the form Î ˚Î 2 ˚ Î ˚ È0˘ È0 ˘ x = Í ˙ = x 2 Í ˙ , x 2 π 0 . The geometric multiplicity of l1 = 6 is 1. Î x2 ˚ Î1 ˚ We find a generalized eigenvector corresponding to l1 = 6 by solving the equation È0 ˘ È0.5 ˘ ( A - 6 I ) x = x1 where x1 = Í ˙ . The solution is x = Í ˙ where x 2 is arbitrary. Choosing Î1 ˚ Î x2 ˚ È0.5 ˘ x 2 = 0 , we obtain the generalized eigenvector x 2 = Í ˙ . Thus, we have solutions y1 ( t) = e 6 t x1 Î0 ˚ 6t 6t and, as in equation (6), y 2 ( t) = te x1 + e x 2 . A fundamental matrix is È 0 0.5e 6 t ˘ Y( t) = [ y1 ( t), y 2 ( t)] = Í 6 t ˙. te 6 t ˚ Îe

144 • Chapter 6 First Order Linear Systems

3 (c). The general solution is Y( t)c . Imposing the initial condition requires Y(0)c = y 0 . We find È0 0.5 ˘È c1 ˘ È-2 ˘ È0˘ Í1 0 ˙Íc ˙ = Í 0 ˙ or c = Í-4 ˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ 2 È ˘ y( t) = e 6 t Í ˙ . Î-4 t ˚ 4 (a). The characteristic polynomial is p(l ) = ( 3 - l ) 2 . The eigenvalue l1 = 3 has algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 3I ) x = 0 . Therefore, È0˘ È0 ˘ all the eigenvectors corresponding to l1 = 3 have the form x = Í ˙ = x 2 Í ˙ , x 2 π 0 . The Î x2 ˚ Î1 ˚ geometric multiplicity of l1 = 3 is 1. È0 ˘ È0 ˘ È1 ˘ È1˘ 4 (b). y1 ( t) = e 3 t Í ˙ and y 2 ( t) = te 3 t Í ˙ + e 3 t Í ˙ = e 3 t Í ˙ . A fundamental matrix is Î1 ˚ Î1 ˚ Î0 ˚ Ît ˚ È 0 e 3t ˘ . Y( t) = [ y1 ( t), y 2 ( t)] = Í 3 t 3t ˙ e te Î ˚ 4 (c). The general solution is Y( t)c . Imposing the initial condition, Y(0)c = y 0 . We find È0 1 ˘È c1 ˘ È 2 ˘ È-3˘ Í1 0 ˙Íc ˙ = Í-3˙ or c = Í 2 ˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ È 2 ˘ y( t) = e 3 t Í ˙. Î2 t - 3˚ È5 -1˘ 5 (a). For A = Í , the characteristic polynomial is p(l ) = (l - 3) 2 . The eigenvalue l1 = 3 has ˙ Î4 1 ˚ algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 3I ) x = 0 or È2 -1˘È x1 ˘ È0 ˘ Í4 -2 ˙Í x ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 3 have the form Î ˚Î 2 ˚ Î ˚ x È 1˘ È1 ˘ x = Í ˙ = x1 Í ˙ , x1 π 0 . The geometric multiplicity of l1 = 3 is 1. Î2 x1 ˚ Î2 ˚ 5 (b). We find a generalized eigenvector corresponding to l1 = 3 by solving the equation È.5 x 2 + .5 ˘ È1 ˘ ( A - 3I ) x = x1 where x1 = Í ˙ . The solution is x = Í ˙ where x 2 is arbitrary. Choosing Î x2 ˚ Î2 ˚ È.5 ˘ x 2 = 0 , we obtain the generalized eigenvector x 2 = Í ˙ . Thus, we have solutions y1 ( t) = e 3 t x1 Î0 ˚ 3t 3t and, as in equation (6), y 2 ( t) = te x1 + e x 2 . A fundamental matrix is È e 3 t ( t + .5)e 3 t ˘ Y( t) = [ y1 ( t), y 2 ( t)] = Í 3 t ˙. 2 te 3 t ˚ Î2e

Chapter 6 First Order Linear Systems • 145

5 (c). The general solution is Y( t)c . Imposing the initial condition requires Y(0)c = y 0 . We find È1 .5 ˘È c1 ˘ È1˘ È.5 ˘ Í2 0 ˙Íc ˙ = Í1˙ or c = Í 1 ˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ t + 1 È ˘ y( t) = e 3 t Í ˙. Î2 t + 1˚ 6 (a). The characteristic polynomial is p(l ) = ( 3 - l ) 2 . The eigenvalue l1 = 3 has algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 3I ) x = 0 . Use È6˘ x1 = Í ˙ . The geometric multiplicity of l1 = 3 is 1. Î-1˚ È6˘ È6˘ È-1˘ È6 t - 1˘ 6 (b). y1 ( t) = e 3 t Í ˙ and y 2 ( t) = te 3 t Í ˙ + e 3 t Í ˙ = e 3 t Í ˙ . A fundamental matrix is Î-1˚ Î-1˚ Î0˚ Î -t ˚ È6e 3 t (6 t - 1)e 3 t ˘ Y( t) = [ y1 ( t), y 2 ( t)] = Í 3 t ˙. - te 3 t ˚ Î-e 6 (c). The general solution is Y( t)c . Imposing the initial condition, Y(0)c = y 0 . We find È 6 -1˘È c1 ˘ È0 ˘ È -2 ˘ Í-1 0 ˙Íc ˙ = Í2 ˙ or c = Í-12 ˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ È -72 t ˘ y( t) = e 3 t Í ˙. Î12 t + 2 ˚ È1 -1˘ 2 7 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = (l - 2) . The eigenvalue l1 = 2 has 1 3 Î ˚ algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 2 I ) x = 0 or È-1 -1˘È x1 ˘ È0 ˘ Í 1 1 ˙Í x ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 2 have the form Î ˚Î 2 ˚ Î ˚ È x1 ˘ È1˘ x = Í ˙ = x1 Í ˙ , x1 π 0 . The geometric multiplicity of l1 = 2 is 1. Î- x1 ˚ Î-1˚ 7 (b). We find a generalized eigenvector corresponding to l1 = 2 by solving the equation È1˘ È-1 - x 2 ˘ ( A - 2 I ) x = x1 where x1 = Í ˙ . The solution is x = Í ˙ where x 2 is arbitrary. Choosing Î-1˚ Î x2 ˚ È-1˘ x 2 = 0 , we obtain the generalized eigenvector x 2 = Í ˙ . Thus, we have solutions y1 ( t) = e 2 t x1 Î0˚ 2t 2t and, as in equation (6), y 2 ( t) = te x1 + e x 2 . A fundamental matrix is È e 2 t ( t - 1)e 2 t ˘ Y( t) = [ y1 ( t), y 2 ( t)] = Í 2 t ˙. - te 2 t ˚ Î-e 7 (c). The general solution is Y( t)c . Imposing the initial condition requires Y(0)c = y 0 . We find È 1 -1˘È c1 ˘ È 4 ˘ È1˘ Í-1 0 ˙Íc ˙ = Í-1˙ or c = Í-3˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ t 4 3 È ˘ y( t) = e 2 t Í ˙. Î 3t - 1 ˚

146 • Chapter 6 First Order Linear Systems

8 (a). The characteristic polynomial is p(l ) = (l - 5) 2 . The eigenvalue l1 = 5 has algebraic multiplicity 2. Corresponding eigenvectors are obtained by solving ( A - 5 I ) x = 0 . Use È1˘ x1 = Í ˙ . The geometric multiplicity of l1 = 5 is 1. Î-1˚ È1˘ È1˘ È1 ˘ Èt + 1˘ 8 (b). y1 ( t) = e 5 t Í ˙ and y 2 ( t) = te 5 t Í ˙ + e 5 t Í ˙ = e 5 t Í ˙ . A fundamental matrix is Î-1˚ Î-1˚ Î0 ˚ Î -t ˚ È e 5 t ( t + 1)e 5 t ˘ Y( t) = [ y1 ( t), y 2 ( t)] = Í 5 t ˙. - te 5 t ˚ Î-e 8 (c). The general solution is Y( t)c . Imposing the initial condition, Y(0)c = y 0 . We find È 1 1 ˘È c1 ˘ È 4 ˘ È4 ˘ Í-1 0 ˙Íc ˙ = Í-4 ˙ or c = Í0 ˙ . Thus, the solution of the initial value problem is Î ˚Î 2 ˚ Î ˚ Î ˚ È4˘ y( t) = e 5 t Í ˙ . Î-4 ˚ È2 1 0 ˘ ˙ Í 9 (a). For A = Í0 2 1 ˙ , the characteristic polynomial is p(l ) = (l - 2) 3 . The eigenvalue l1 = 2 has ÍÎ0 0 2 ˙˚ algebraic multiplicity 3. Corresponding eigenvectors are obtained by solving ( A - 2 I ) x = 0 or È0 1 0 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í0 0 1 ˙Í x 2 ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 2 have the form ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚

È1 ˘ È x1 ˘ Í ˙ Í ˙ x = Í 0 ˙ = x1 Í0 ˙ , x1 π 0 . The geometric multiplicity of l1 = 2 is 1. ÍÎ0 ˙˚ ÍÎ 0 ˙˚ È2 1 0 ˘ ˙ Í 9 (b). For A = Í0 2 0 ˙ , the characteristic polynomial is p(l ) = (l - 2) 3 . The eigenvalue l1 = 2 has ÍÎ0 0 2 ˙˚ algebraic multiplicity 3. Corresponding eigenvectors are obtained by solving ( A - 2 I ) x = 0 or È0 1 0 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í0 0 0 ˙Í x 2 ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 2 have the form ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚

13.

È0 ˘ È1 ˘ È x1 ˘ Í ˙ Í ˙ Í ˙ x = Í 0 ˙ = x1 Í0 ˙ + x 3 Í0 ˙ , where x is nonzero. The geometric multiplicity of l1 = 2 is 2. ÍÎ1 ˙˚ ÍÎ0 ˙˚ ÍÎ x 3 ˙˚ È5 0 0 ˘ ˙ Í For A = Í1 5 0 ˙ , the characteristic polynomial is p(l ) = (l - 5) 3 . The eigenvalue l1 = 5 has ÍÎ1 1 5 ˙˚ algebraic multiplicity 3.

Chapter 6 First Order Linear Systems • 147

14.

15.

16.

17.

È0 0 0 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Corresponding eigenvectors are obtained by solving ( A - 5 I ) x = 0 or Í1 0 0 ˙Í x 2 ˙ = Í0 ˙ . ÍÎ1 1 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ Therefore, all the eigenvectors corresponding to l1 = 5 have the form È0 ˘ È0˘ Í ˙ Í ˙ x = Í 0 ˙ = x 3 Í0 ˙ , x 3 π 0 . The geometric multiplicity of l1 = 5 is 1, so A does not have a full ÍÎ1 ˙˚ ÍÎ x 3 ˙˚ set of eigenvectors. The characteristic polynomial is p(l ) = (l - 5) 3 . The eigenvalue l1 = 5 has algebraic multiplicity 3. Corresponding eigenvectors are obtained by solving ( A - 5 I ) x = 0 . Use È0 ˘ È0 ˘ Í ˙ Í ˙ x1 = Í1 ˙, x 2 = Í0 ˙ . The geometric multiplicity of l1 = 5 is 2, so A does not have a full set of ÍÎ1 ˙˚ ÍÎ0 ˙˚ eigenvectors. È5 0 1 ˘ ˙ Í For A = Í0 5 0 ˙ , the characteristic polynomial is p(l ) = (l - 5) 3 . The eigenvalue l1 = 5 has ÍÎ0 0 5 ˙˚ algebraic multiplicity 3. Corresponding eigenvectors are obtained by solving ( A - 5 I ) x = 0 or È0 0 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í0 0 0 ˙Í x 2 ˙ = Í0 ˙ . Therefore, all the eigenvectors corresponding to l1 = 5 have the form ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È0 ˘ È1 ˘ È x1 ˘ Í ˙ Í ˙ Í ˙ x = Í x 2 ˙ = x1 Í0 ˙ + x 2 Í1 ˙ , where x is nonzero. The geometric multiplicity of l1 = 5 is 2, so A ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ 0 ˙˚ does not have a full set of eigenvectors. The characteristic polynomial is p(l ) = (l - 5) 3 . The eigenvalue l1 = 5 has algebraic multiplicity 3. Corresponding eigenvectors are obtained by solving ( A - 5 I ) x = 0 x = 0 . Use È0 ˘ È0 ˘ È1 ˘ Í ˙ Í ˙ Í ˙ x1 = Í0 ˙, x 2 = Í1 ˙, x 3 = Í0 ˙ . The geometric multiplicity of l1 = 5 is 3, so A does have a full ÍÎ1 ˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚ set of eigenvectors. È2 0 0 0 ˘ Í1 2 0 0 ˙ ˙ , the characteristic polynomial is p(l ) = (l - 2) 2 (l - 3) 2 . The For A = Í Í0 0 3 0 ˙ Í ˙ Î0 0 1 3˚ eigenvalue l1 = 2 has algebraic multiplicity 2 as does l 2 = 3.

148 • Chapter 6 First Order Linear Systems

18.

19.

20. 21. 22.

È0˘ È0 ˘ Íx ˙ Í1 ˙ 2˙ Í = x 2 Í ˙ , x 2 π 0 . Therefore, the Corresponding eigenvectors for l1 = 2 have the form x = Í0 ˙ Í0˙ Í ˙ Í ˙ Î0˚ Î0 ˚ geometric multiplicity of l1 = 2 is 1. Similarly, eigenvectors corresponding to l 2 = 3 have the È0˘ È0 ˘ Í0˙ Í0 ˙ form x = Í ˙ = x 4 Í ˙ , x 4 π 0 and so l 2 = 3 has geometric multiplicity 1. A does not have a Í0 ˙ Í0˙ Í ˙ Í ˙ Î x4 ˚ Î1 ˚ full set of eigenvectors. The characteristic polynomial is p(l ) = (2 - l ) 4 . The eigenvalue l1 = 2 has algebraic multiplicity 4. Corresponding eigenvectors are obtained by solving ( A - 2 I ) x = 0 . Use È0 ˘ È0 ˘ È1 ˘ Í1 ˙ Í0 ˙ Í0 ˙ ˙ Í ˙ Í x1 = , x = , x = Í ˙ . The geometric multiplicity of l = 2 is 3, so A does not have a full Í0 ˙ 2 Í0 ˙ 3 Í0 ˙ Í ˙ Í ˙ Í ˙ Î1 ˚ Î0 ˚ Î0 ˚ set of eigenvectors. È2 0 0 0 ˘ Í0 2 0 0 ˙ ˙ , the characteristic polynomial is p(l ) = (l - 2) 3 (l - 3). The eigenvalue For A = Í Í0 0 2 0 ˙ Í ˙ Î0 0 1 3˚ l1 = 2 has algebraic multiplicity 3 while l 2 = 3 has algebraic multiplicity 1. Corresponding È0˘ È x1 ˘ È1 ˘ È0 ˘ Í0˙ Íx ˙ Í0 ˙ Í1 ˙ 2 ˙ ˙ Í ˙ Í Í + x 3 Í ˙ . Therefore, the =x + x2 eigenvectors for l1 = 2 have the form x = Í1˙ Í x 3 ˙ 1 Í0 ˙ Í0 ˙ Í ˙ Í ˙ Í ˙ Í ˙ Î-1˚ Î- x 3 ˚ Î0 ˚ Î0 ˚ geometric multiplicity of l1 = 2 is 3. Eigenvectors corresponding to l 2 = 3 have the form È0˘ È0 ˘ Í0˙ Í0 ˙ Í ˙ = x 4 Í ˙ , x 4 π 0 and so l 2 = 3 has geometric multiplicity 1. Since every eigenvalue x= Í0˙ Í0 ˙ Í ˙ Í ˙ Î x4 ˚ Î1 ˚ of A has geometric multiplicity equal to its algebraic multiplicity, A has a full set of eigenvectors. A must have l1 = a + ib and l 2 = a - ib as two distinct eigenvalues. Therefore, A cannot have a repeated eigenvalue and cannot be defective. In order for A to be symmetric, a12 = x must be the same as a21 = 9 . Thus, x = 9 . Similarly, a23 = y must equal a32 = 4 . x = 6, y = 1.

Chapter 6 First Order Linear Systems • 149

In order for A to be symmetric, a13 = x 2 - 1 must be the same as a31 = 0 . Thus, we can have either x = 1 or x = -1. Similarly, a21 = 2 / y must equal a12 = 1. Hence, y = 2. 24. In order for A to be Hermitian, a12 = x - 3i = 9 - 3i fi x = 9 and a23 = 2 - yi = 2 + 5i fi y = -5 . 25. In order for A to be Hermitian, a11 = 2 + xi must be the same as a11 = 2 - xi . Thus, we need x = 0. Similarly, a21 = 1 + yi must equal a12 = 1 - 2i. Hence, y = 2. These choices are consistent with the remaining undetermined entries, a22 and a23 . È0 0 ˘ 26 (a). A = Í ˙ , for example. Î0 1 ˚ È1 1˘ 26 (b). A = Í ˙ , for example. Î0 1˚ È0 1 0 ˘È x1 ˘ È1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í ˙ 30. The equation ( A - 2 I ) x = v1 is Í0 0 1 ˙Í x 2 ˙ = Í0 ˙ . Choose v 2 = Í1 ˙ . ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ ÍÎ0 ˙˚ 23.

31.

È0 1 0 ˘È x1 ˘ È0 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í ˙ The equation ( A - 2 I ) x = v 2 is Í0 0 1 ˙Í x 2 ˙ = Í1 ˙ . Choose v 3 = Í0 ˙ . A fundamental set of ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ ÍÎ1 ˙˚ 2 È t2 ˘ È1 ˘ Èt˘ Í ˙ Í ˙ Í ˙ solutions can be formed from y1 ( t) = e 2 t Í0 ˙ , y 2 ( t) = e 2 t Í1 ˙ , and y 3 ( t) = e 2 t Í t ˙ . Í1 ˙ ÍÎ0 ˙˚ ÍÎ0 ˙˚ Î ˚ È4 0 0 ˘ È0 0 0 ˘È x1 ˘ È0 ˘ ˙ ˙Í ˙ Í ˙ Í Í For A = Í2 4 0 ˙ , the equation ( A - 4 I ) x = v1 is Í2 0 0 ˙Í x 2 ˙ = Í0 ˙ . The solution is ÍÎ1 3 4 ˙˚ ÍÎ1 3 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ1 ˙˚ È 0 ˘ È 0 ˘ Í ˙ Í ˙ x = Í1 / 3˙ where x 3 is arbitrary. Choosing x 3 = 0 we have v 2 = Í1 / 3˙ . ÍÎ x 3 ˙˚ ÍÎ 0 ˙˚ È0 0 0 ˘È x1 ˘ È 0 ˘ È 1/6 ˘ ˙Í ˙ Í ˙ ˙ Í Í The equation ( A - 4 I ) x = v 2 is Í2 0 0 ˙Í x 2 ˙ = Í1 / 3˙ . The solution is x = Í-1 / 18 ˙ where x 3 ÍÎ1 3 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ 0 ˙˚ ÍÎ x 3 ˙˚ È 1/6 ˘ ˙ Í is arbitrary. Choosing x 3 = 0 we have v 3 = Í-1 / 18 ˙ . By equation (12), a fundamental set of ÍÎ 0 ˙˚

È0 ˘ È 0 ˘ ˙ Í ˙ 4t 4t Í solutions can be formed from y1 ( t) = e Í0 ˙ , y 2 ( t) = e ( v 2 + tv1 ) = e Í1 / 3˙ , and ÍÎ1 ˙˚ ÍÎ t ˙˚ È 3 ˘ e4 t Í ˙ 4t 2 y 3 ( t) = e ( v 3 + tv 2 + 0.5 t v1 ) = -1 + 6 t ˙ . Í 18 ÍÎ 9 t 2 ˙˚ 4t

150 • Chapter 6 First Order Linear Systems

32.

33.

È 12 ˘ -8 -10 ˘È x1 ˘ È 1 ˘ ˙Í ˙ Í ˙ Í ˙ 6 8 ˙Í x 2 ˙ = Í-1˙ . Choose v 2 = Í0 ˙ . ÍÎ0 ˙˚ -6 -8 ˙˚ÍÎ x 3 ˙˚ ÍÎ 1 ˙˚ 1 È- 43 ˘ -8 -10 ˘È x1 ˘ È 2 ˘ ˙Í ˙ Í ˙ Í ˙ 6 8 ˙Í x 2 ˙ = Í0 ˙ . Choose v 3 = Í- 14 ˙ . A fundamental set ÍÎ 0 ˙˚ -6 -8 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ 2 È- 43 + 2t + t2 ˘ È1˘ È 12 + t ˘ ˙ Í 2 ˙ Í ˙ Í of solutions can be formed from y1 ( t) = e t Í-1˙ , y 2 ( t) = e t Í - t ˙ , and y 3 ( t) = e t Í - 14 - t2 ˙ . t2 ˙ Í ÍÎ 1 ˙˚ ÍÎ t ˙˚ 2 ˚ Î È-6 -8 22 ˘ È-8 -8 22 ˘È x1 ˘ È 1 ˘ ˙ ˙Í ˙ Í ˙ Í Í For A = Í 2 4 -4 ˙ , the equation ( A - 2 I ) x = v1 is given by Í 2 2 -4 ˙Í x 2 ˙ = Í-1˙ . A ÍÎ-2 -2 8 ˙˚ ÍÎ-2 -2 6 ˙˚ÍÎ x 3 ˙˚ ÍÎ 0 ˙˚ È2 Í The equation ( A - I ) x = v1 is Í-2 ÍÎ 2 È2 Í The equation ( A - I ) x = v 2 is Í-2 ÍÎ 2

È-1.5 ˘ ˙ Í convenient solution is v 2 = Í 0 ˙ . The equation ( A - 2 I ) x = v 2 ÍÎ-0.5 ˙˚ È-8 -8 22 ˘È x1 ˘ È-1.5 ˘ È -0.5 ˘ ˙ ˙Í ˙ Í ˙ Í Í is Í 2 2 -4 ˙Í x 2 ˙ = Í 0 ˙ . One solution is v 3 = Í 0 ˙ . A fundamental set consists of ÍÎ-2 -2 6 ˙˚ÍÎ x 3 ˙˚ ÍÎ-0.5 ˙˚ ÍÎ-0.25 ˙˚ È1˘ Èt - 1.5 ˘ ˙ Í ˙ 2t 2t Í y1 ( t) = e Í-1˙ , y 2 ( t) = e ( v 2 + tv1 ) = e Í - t ˙ , and ÍÎ 0 ˙˚ ÍÎ -0.5 ˙˚ 2t

È-2 - 6 t + 2 t 2 ˘ e Í ˙ y 3 ( t) = e 2 t ( v 3 + tv 2 + 0.5 t 2 v1 ) = -2 t 2 Í ˙. 4 ÍÎ -1 - 2 t ˙˚ È1˘ È1˘ Í ˙ Í ˙ 36 (a). Two linearly independent solutions are x1 = Í-1˙, x 2 = Í 0 ˙ . ÍÎ-1˙˚ ÍÎ 0 ˙˚ È1˘ Í˙ 36 (b). Choose x 3 = Í1˙ . ÍÎ1˙˚ È1˘ È1˘ È1˘ ˙ ˙ -4 t Í -4 t Í -t Í ˙ 36 (c). Q( t) = c1e Í-1˙ + c 2e Í 0 ˙ + c 3e Í1˙ . ÍÎ1˙˚ ÍÎ-1˙˚ ÍÎ 0 ˙˚ 2t

Chapter 6 First Order Linear Systems • 151

Section 6.8 È-2 1 ˘ 1 (a). For A = Í , the characteristic polynomial is p(l ) = l2 + 4 l + 3 = (l + 3)(l + 1) . The ˙ Î 1 -2 ˚ È1˘ È1˘ eigenvalues are l1 = -3 and l 2 = -1, with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î-1˚ Î1˚ È e -3 t e - t ˘È c1 ˘ Thus, the complementary solution of y ¢ = Ay is y C = Í -3 t ˙Í ˙. e - t ˚Îc 2 ˚ Î-e È a1 ˘ 1 (b). Inserting the suggested trial form y P = Í ˙ into the nonhomogeneous equation leads to Îa2 ˚ È0 ˘ È-2 1 ˘È a1 ˘ È1˘ È1˘ y = + . Solving this system, we obtain y ¢P = Ay P + g( t) or Í ˙ = Í P Í1˙ . ˙Í ˙ Í ˙ Î0 ˚ Î 1 -2 ˚Îa2 ˚ Î1˚ Î˚ -3 t -t c Èe e ˘È 1 ˘ È1˘ 1 (c). The general solution of the nonhomogeneous problem is y C + y P = Í -3 t ˙Í ˙ + Í ˙ . e - t ˚Îc 2 ˚ Î1˚ Î-e È 1 1˘È c1 ˘ È1˘ È3˘ 1 (d). Imposing the initial condition, Í ˙Í ˙ + Í ˙ = Í ˙ . Solving, we find c1 = 1 and c 2 = 1. Î-1 1˚Îc 2 ˚ Î1˚ Î1 ˚ È e -3 t + e - t + 1 ˘ Thus, y( t) = Í -3 t ˙ is the unique solution of the given initial value problem. -t Î-e + e + 1˚ È2 1 ˘ 2 2 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = l - 4 l + 3 = (l - 1)(l - 3) . The 1 2 Î ˚ È1˘ È1˘ eigenvalues are l1 = 1 and l 2 = 3 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î-1˚ Î1˚ 3t t Èe e ˘È c1 ˘ Thus, the complementary solution of y ¢ = Ay is y C = Í t ˙. 3 t ˙Í Î-e e ˚Îc 2 ˚ È a1 ˘ 2 (b). Inserting the suggested trial form y P = e - t Í ˙ into the nonhomogeneous equation and solving Îa2 ˚ 3 È- 8 ˘ the system, we obtain y P = e - t Í 1 ˙ . Î 8 ˚ È e t e 3 t ˘È c1 ˘ - t È- 38 ˘ 2 (c). The general solution of the nonhomogeneous problem is y C + y P = Í t ˙ + e Í 1 ˙. 3 t ˙Í Î 8 ˚ Î-e e ˚Îc 2 ˚ È 1 1˘È c1 ˘ È- 38 ˘ È0 ˘ 1 1 + Í 1 ˙ = Í ˙ . Solving, we find c1 = and c 2 = . 2 (d). Imposing the initial condition, Í ˙ Í ˙ 4 8 Î-1 1˚Îc 2 ˚ Î 8 ˚ Î0 ˚ 1 t 1 3t 3 -t È e + e - e ˘ Thus, y( t) = Í 41 t 81 3 t 8 1 - t ˙ is the unique solution of the given initial value problem. Î- 4 e + 8 e + 8 e ˚

152 • Chapter 6 First Order Linear Systems

È0 1 ˘ 2 3 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = l - 1 = (l + 1)(l - 1) . The eigenvalues 1 0 Î ˚ È1˘ È1˘ are l1 = -1 and l 2 = 1 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Thus, the Î-1˚ Î1˚ È e - t e t ˘È c1 ˘ complementary solution of y ¢ = Ay is y C = Í - t ˙Í ˙ . e t ˚Îc 2 ˚ Î-e È a1 ˘ È b1 ˘ 3 (b). Inserting the suggested trial form y P = t Í ˙ + Í ˙ into the nonhomogeneous equation leads to Îa2 ˚ Îb2 ˚ È a1 ˘ È0 1 ˘È ta1 + b1 ˘ È t ˘ È0˘ y = + . Solving this system, we obtain y ¢P = Ay P + g( t) or Í ˙ = Í P Í- t ˙ . ˙Í ˙ Í ˙ Îa2 ˚ Î1 0 ˚Îta2 + b2 ˚ Î-1˚ Î ˚ È e - t e t ˘È c1 ˘ È 0 ˘ 3 (c). The general solution of the nonhomogeneous problem is y C + y P = Í - t ˙Í ˙ + Í ˙ . e t ˚Îc 2 ˚ Î- t ˚ Î-e È 1 1˘È c1 ˘ È0 ˘ È 2 ˘ 3 (d). Imposing the initial condition, Í ˙Í ˙ + Í ˙ = Í ˙ . Solving, we find c1 = 1.5 and Î-1 1˚Îc 2 ˚ Î0 ˚ Î-1˚ È 3e - t + e t ˘ c 2 = 0.5 . Thus, y( t) = 0.5 Í - t ˙ is the unique solution of the given initial value t Î-3e + e - t ˚ problem. È 0 -1˘ 2 4 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = l - 1. The eigenvalues are 1 0 Î ˚ È1˘ È1˘ l1 = -1 and l 2 = 1 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Thus, the Î1˚ Î-1˚ Èe - t e t ˘È c1 ˘ complementary solution of y ¢ = Ay is y C = Í - t ˙Í ˙ . -e t ˚Îc 2 ˚ Îe È a1 ˘ È b1 ˘ È c1 ˘ 4 (b). Inserting the suggested trial form y P = e 2 t Í ˙ + t Í ˙ + Í ˙ into the nonhomogeneous Îa2 ˚ Îb2 ˚ Îc 2 ˚ È- 1 e 2 t - 1˘ equation and solving the system, we obtain y P = Í 2 3 2 t ˙. Î 3e + t ˚ 4 (c). The general solution of the nonhomogeneous problem is Èe - t e t ˘È c1 ˘ È- 13 e 2 t - 1˘ yC + y P = Í -t ˙Í ˙ + Í ˙. -e t ˚Îc 2 ˚ Î 23 e 2 t + t ˚ Îe È1 1 ˘È c1 ˘ È- 43 ˘ È0 ˘ 5 1 + Í 2 ˙ = Í ˙ . Solving, we find c1 = and c 2 = . 4 (d). Imposing the initial condition, Í ˙ Í ˙ 6 2 Î1 -1˚Îc 2 ˚ Î 3 ˚ Î1 ˚ 5 -t 1 t 1 2t È e + e - e - 1˘ Thus, y( t) = Í 56 - t 12 t 23 2 t ˙ is the unique solution of the given initial value problem. Î 6 e - 2 e + 3 e + t˚

Chapter 6 First Order Linear Systems • 153

È-3 -2 ˘ 2 5 (a). For A = Í ˙ , the characteristic polynomial is p(l ) = l - 1 = (l + 1)(l - 1) . The 4 3 Î ˚ È1˘ È1˘ eigenvalues are l1 = -1 and l 2 = 1 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î-1˚ Î-2 ˚ È e-t e t ˘È c1 ˘ Thus, the complementary solution of y ¢ = Ay is y C = Í - t ˙Í ˙ . -2e t ˚Îc 2 ˚ Î-e È a1 ˘ È b1 ˘ 5 (b). Inserting the suggested trial form y P = sin t Í ˙ + cos t Í ˙ into the nonhomogeneous equation Îa2 ˚ Îb2 ˚ È a1 cos t - b1 sin t ˘ È-3 -2 ˘È a1 sin t + b1 cos t ˘ Èsin t ˘ leads to y ¢P = Ay P + g( t) or Í ˙= Í ˙Í ˙+Í ˙ . Solving Îa2 cos t - b2 sin t ˚ Î 4 3 ˚Îa2 sin t + b2 cos t ˚ Î 0 ˚ È3 sin t - cos t ˘ this system, we obtain y P = 0.5 Í ˙. Î -4 sin t ˚ 5 (c). The general solution of the nonhomogeneous problem is È e-t e t ˘È c1 ˘ È3 sin t - cos t ˘ yC + y P = Í -t + 0.5 Í ˙ ˙. t ˙Í -2e ˚Îc 2 ˚ Î -4 sin t ˚ Î-e È 1 1 ˘È c1 ˘ È-.5 ˘ È0 ˘ 5 (d). Imposing the initial condition, Í ˙Í ˙ + Í ˙ = Í ˙ . Solving, we find c1 = 1 and Î-1 -2 ˚Îc 2 ˚ Î 0 ˚ Î0 ˚ È2e - t - e t + 3 sin t - cos t ˘ c 2 = -0.5 . Thus, y( t) = 0.5 Í ˙ is the unique solution of the given initial -t t Î -2e + 2e - 4 sin t ˚ value problem. È1 1˘ , the characteristic polynomial is p(l ) = l2 - 2l . The eigenvalues are 6 (a). For A = Í ˙ Î1 1˚ È1˘ È1˘ l1 = 0 and l 2 = 2 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Thus, the Î-1˚ Î1˚ È 1 e 2 t ˘È c1 ˘ complementary solution of y ¢ = Ay is y C = Í ˙. 2 t ˙Í Î-1 e ˚Îc 2 ˚ È 1 + sin 2 t ˘ È 1 + sin p ˘ È 1 ˘ it follows that y 0 = y (p / 2) = Í p / 2 7. Given y( t) = Í t ˙ ˙ = Í p / 2 ˙ . Inserting Îe + cos 2 t ˚ Îe + cosp ˚ Îe - 1˚ y( t) into the differential equation, we see that y ¢ ( t) = Ay ( t) + g( t) and thus È -2e t ˘ È 2 cos 2 t ˘ È 0 2 ˘È 1 + sin 2 t ˘ ˙. Íe t - 2 sin 2 t ˙ = Í-2 0 ˙Íe t + cos 2 t ˙ + g( t) . Solving for g(t), we obtain g( t) = Í t Î ˚ Î ˚Î ˚ Îe + 2 ˚ 8.

È t +a ˘ È1 + a ˘ È 2 ˘ it follows that y(1) = Í Given y( t) = Í 2 ˙ ˙ = Í ˙ fi a = 1, b = -2 . Inserting y( t) into Ît + b ˚ Î1 + b ˚ Î-1˚ the differential equation, we see that y ¢ ( t) = Ay ( t) + g( t) and thus È ˘ È 1 ˘ È 1 t ˘È t + 1 ˘ -t3 + t y ¢ = Í ˙ = Í 2 ˙Í 2 + g( t) . Solving for g(t), we obtain g( t) = Í 3 ˙. ˙ 2 Î2 t ˚ Ît 1˚Ît - 2 ˚ Î- t - 2 t + 2 t + 2 ˚

154 • Chapter 6 First Order Linear Systems

9.

Following the hint, we form [ y1¢, y ¢2 ] = P ( t)[ y1, y 2 ] + [g1 ( t), g 2 ( t)] which has the form È 1 e t ˘ È-2 e t ˘ È 0 et ˘ = ( ) P t ˙ . Solving for P(t), we have Í -t ˙+Í Í -t ˙ -1˚ Î 0 -1˚ 0˚ Îe Î-e -1 0 ˘È 1 e t ˘ 0˘ È -1 -e t ˘ È1 e t ˘ È 2 È 2 P ( t) = Í - t = ( / ) 1 2 ˙. Í ˙ Í -t ˙= Í Í-e - t 1 ˙ 1 ˙˚Îe - t -1˚ 1 ˚ Î0 -1˚ Î-e Î ˚ Î-e

10.

If A -1 exists, y 2 = - A -1b is the unique solution. If A -1 does not exist, the matrix equation Ay = -b will either have no solution or a nonunique solution. Therefore, either no equilibrium solution or a non-unique equilibrium solution. An equilibrium solution of y ¢ = Ay + b is a constant solution. Therefore, since y ¢ = 0 we need È1 4 ˘ È-3 -4 ˘È2 ˘ È10 ˘ È2 ˘ and b = Í ˙ we see that y e = - A -1b = - Í Ay e = -b . For A = Í ˙ ˙Í ˙ = Í ˙ . Î-1 -3˚ Î 1 1 ˚Î1 ˚ Î-3˚ Î1 ˚ È1 1 ˘È-2 ˘ È-1˘ A -1 exists and y e = Í ˙Í ˙ = Í ˙ . Î1 2 ˚Î 1 ˚ Î 0 ˚ È 1 -1˘ È2˘ and b = Í ˙ we As noted in the solution of Exercise 11, we need Ay e = -b . For A = Í ˙ Î-1 1 ˚ Î-2 ˚ È0 ˘ È1˘ see that y e = Í ˙ + a Í ˙ where a is arbitrary. Î2 ˚ Î1˚ È-1˘ È-2 ˘ È1 1 0 ˘ Í ˙ ˙ Í ˙ Í A -1 exists and Í0 -1 2 ˙y e = Í-3˙ fi y e = Í-1˙ . ÍÎ-2 ˙˚ ÍÎ-2 ˙˚ ÍÎ0 0 1 ˙˚

11.

12. 13.

14.

15.

16.

È-2 ˘ È1 0 0 ˘ Í ˙ ˙ Í As noted in the solution of Exercise 11, we need Ay e = -b . For A = Í0 1 1 ˙ and b = Í 0 ˙ ÍÎ 0 ˙˚ ÍÎ0 1 1 ˙˚ È0˘ È2 ˘ Í ˙ Í ˙ we see that y e = Í0 ˙ + a Í-1˙ where a is arbitrary. ÍÎ 1 ˙˚ ÍÎ0 ˙˚ The characteristic polynomial is p(l ) = l2 - 2l . The eigenvalues are l1 = 0 and l 2 = 2 , with È1˘ È1˘ corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î1˚ Î-1˚ 2t ) È1 e ˘ Thus, one fundamental matrix is Y( t) = Í . Set 2t ˙ 1 e Î ˚ -1 1 ) ) ) -1 È1 e 2 ˘ È 12 ˘ 2 . Y = YC . Y (1) = I = Y (1)C . \ C = Y (1) = Í = Í 1 -2 2˙ 1 -2 ˙ - 2e ˚ Î2 e Î1 -e ˚ 1 È1 e 2 t ˘È 12 ˘ È 12 (1 + e 2( t -1) ) 12 (1 - e 2( t -1) ) ˘ 2 Y( t) = Í ˙. ˙= Í 2 t ˙ Í 1 -2 - 12 e -2 ˚ ÍÎ 12 (1 - e 2( t -1) ) 12 (1 + e 2( t -1) ) ˙˚ Î1 -e ˚Î 2 e

Chapter 6 First Order Linear Systems • 155

17.

18.

19.

È 0 2˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = l + 4 = (l + 2i)(l - 2i) . The 2 0 Î ˚ eigenvalues are l1 = -2i and l 2 = 2i , with corresponding eigenvectors È-1˘ È-1˘ x1 = Í ˙ and x 2 = Í ˙ . Converting to real solutions, we have Îi˚ Î-i ˚ È-1˘ È-1˘ y( t) = e -2 it Í ˙ = (cos 2 t - i sin 2 t) Í ˙ . Therefore, a fundamental set of solutions is Îi˚ Îi˚ È- cos 2 t ˘ È sin 2 t ˘ y1 ( t ) = Í and y 2 ( t) = Í ˙ ˙. Î sin 2 t ˚ Îcos 2 t ˚ È- cos 2 t sin 2 t ˘ Thus, one fundamental matrix is Y( t) = Í ˙ . The solution of the given initial Î sin 2 t cos 2 t ˚ ) È- cos 2 t sin 2 t ˘ value problem has the form Y ( t) = Y ( t)C = Í ˙C where C is a (2 ¥ 2) matrix Î sin 2 t cos 2 t ˚ ) È1 -1˘ chosen so that Y(p / 4 ) = Í ˙ . Imposing this condition, we have Î0 1 ˚ È1 -1˘ È0 1 ˘È1 -1˘ È0 1 ˘ È0 1 ˘ Í0 1 ˙ = Y(p / 4 )C = Í1 0 ˙C . Solving for C, we obtain C = Í1 0 ˙Í0 1 ˙ = Í1 -1˙ and Î Î ˚ ˚Î Î ˚ Î ˚ ˚ ) ) È- cos 2 t sin 2 t ˘È0 1 ˘ È sin 2 t - cos 2 t - sin 2 t ˘ thus Y( t) = Í or Y( t) = Í ˙ Í ˙. ˙ Î sin 2 t cos 2 t ˚Î1 -1˚ Îcos 2 t sin 2 t - cos 2 t ˚ The characteristic polynomial is p(l ) = l2 - 2l . The eigenvalues are l1 = 0 and l 2 = 2 , with È1˘ È1˘ corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î1˚ Î-1˚ 2t ) È1 e ˘ Thus, one fundamental matrix is Y( t) = Í . Set 2t ˙ Î1 -e ˚ -1 1 ) ) -1 È1 0 ˘ ) È1 1 ˘ È1 0 ˘ È 32 2 ˘ Y = YC . Y (0) = Í Y Y Y ( 0 ) C . C ( 0 ) ( 0 ) = \ = = = . Í ˙ Í ˙ Í ˙ 1 1˙ Î2 1 ˚ Î1 -1˚ Î2 1 ˚ Î- 2 - 2 ˚ 1 È 32 - 12 e 2 t 12 - 12 e 2 t ˘ È1 e 2 t ˘È 32 2 ˘ = Y( t) = Í Í 3 1 2t 1 1 2t ˙ . 2 t ˙Í 1 1˙ 2 + 2e ˚ Î1 -e ˚Î- 2 - 2 ˚ Î 2 + 2 e È3 -4 ˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = l - 1 = (l + 1)(l - 1) . The 2 3 Î ˚ È1˘ È2 ˘ eigenvalues are l1 = -1 and l 2 = 1 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î1˚ Î1 ˚ È1˘ È2 ˘ Therefore, a fundamental set of solutions is y1 ( t) = e - t Í ˙ and y 2 ( t) = e t Í ˙ . Thus, one Î1˚ Î1 ˚ Èe - t fundamental matrix is Y( t) = Í - t Îe

2e t ˘ ˙. et ˚

156 • Chapter 6 First Order Linear Systems

) Èe - t 2e t ˘ The solution of the given initial value problem has the form Y ( t) = Y ( t)C = Í - t ˙C et ˚ Îe ) È1 0 ˘ where C is a (2 ¥ 2) matrix chosen so that Y(0) = Í ˙ . Imposing this condition, we have Î0 1 ˚ È-1 2 ˘ È1 0 ˘ È1 2 ˘ Í0 1 ˙ = Y(0)C = Í1 1 ˙C . Solving for C, we obtain C = Í 1 -1˙ and thus Î Î ˚ ˚ Î ˚ -t t -t t -t t ) ) È-e + 2e 2e - 2e ˘ Èe 2e ˘È-1 2 ˘ or Y( t) = Í - t Y( t) = Í - t ˙. ˙ t ˙Í t e ˚Î 1 -1˚ 2e - t - e t ˚ Î -e + e Îe

20.

21.

The characteristic polynomial is p(l ) = l2 - 2l + 5 . The eigenvalues are È-2i ˘ l1 = 1 + 2i and l 2 = 1 - 2i , with corresponding eigenvectors x = Í ˙ . Then Î 1 ˚ È-2i ˘ È2e t sin 2 t ˘ È-2e t cos 2 t ˘ y( t) = e t (cos 2 t + i sin 2 t) Í ˙ = Í t ˙ + iÍ t ˙. Î 1 ˚ Î e cos 2 t ˚ Î e sin 2 t ˚ ) È2e t sin 2 t -2e t cos 2 t ˘ Thus, one fundamental matrix is Y( t) = Í t ˙ . Set e t sin 2 t ˚ Î e cos 2 t p p È 12 e - 4 ) 0 ˘ Ê p ˆ È1 0 ˘ È2e 4 0 ˘ . Y = YC . Y Á ˜ = Í =Í p ˙C . \ C = Í ˙ -p ˙ Ë 4 ¯ Î0 1 ˚ ÍÎ 0 e 4 ˙˚ e 4 ˙˚ ÍÎ 0 p p È e( t - 4 ) sin 2 t -2e( t - 4 ) cos 2 t ˘ Y( t) = Í 1 ( t - p ) ˙. t -p e( 4 ) sin 2 t ˙˚ ÍÎ 2 e 4 cos 2 t È1 1˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = l - 2l = l (l - 2) . The eigenvalues are 1 1 Î ˚ È1˘ È1˘ l1 = 0 and l 2 = 2 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Therefore, a Î-1˚ Î1˚ È1˘ È1˘ fundamental set of solutions is y1 ( t) = Í ˙ and y 2 ( t) = e 2 t Í ˙ . A fundamental matrix is Î-1˚ Î1˚ -1 ˘ È 2s -e 2s ˘ È 1 e 2t ˘ È 1 -1 -2 s e and therefore, . From equation = Y ( s ) = 0 . 5 e 0 . 5 Y( t) = Í Í ˙ ˙ Í 2 s 2t e -2 s ˙˚ 1 ˚ Îe Î1 Î-1 e ˚ t

(11), the solution is y ( t) = Y ( t)Y -1 ( t0 ) y 0 + Y ( t) Ú Y -1 ( s)g( s) ds. Since y 0 = 0 and t0 = 0 , we t0

2t

t Èe 2 t - 1˘ È 1 e 2t ˘ È 1 e ˘ t Èe 2 s ˘ = y ( t) = Y ( t) Ú Y -1 ( s)g( s) ds = Í . ds . 0 5 0 25 Í ˙ Í Í ˙ 2 t ˙Ú 2t ˙ 0 Î 2t ˚ Î-1 e ˚ Î-1 e ˚ 0 Î 1 ˚ have È e 2 t - 1 + 2 te 2 t ˘ = 0.25 Í 2 t . 2t ˙ Î-(e - 1) + 2 te ˚

Chapter 6 First Order Linear Systems • 157

22.

23.

È 9 -4 ˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = l - 2l - 3 = (l - 3)(l + 1) . The 15 7 Î ˚ È2 ˘ È2 ˘ eigenvalues are l1 = -1 and l 2 = 3 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î5 ˚ Î3˚ È2e - t 2e 3 t ˘ A fundamental matrix is Y( t) = Í - t ˙ and therefore, 3e 3 t ˚ Î5e 1 s ˘ È 3e 3 s -2e 3 s ˘ È- 43 e s 1 2e . = Y -1 ( s) = - e -2 s Í Í 5 -3 s -s -s ˙ 1 -3 s ˙ 4 - 2e ˚ 2e ˚ Î 4 e Î-5e 3 2s È - 38 (e 2 t - 1) ˘ t t È- e ˘ 4 -1 Ú0 Y (s)g(s) ds = Ú0 ÍÎ 45 e -2s ˙˚ds = ÍÍ- 58 (e -2t - 1)˙˙ . Î ˚ -t t t 3 5 t È - (e - e ) - 4 (e - e 3 t ) ˘ È 43 e - t - 2e t + 45 e 3 t ˘ = Í15 - t 15 t 15 3 t ˙ . Y ( t) Ú Y -1 ( s)g( s) ds = Í 154 t -t 3t ˙ t 15 0 Î- 8 (e - e ) - 8 (e - e ) ˚ Î 8 e - 4 e + 8 e ˚ t È2 ˘ Then, y ( t) = Y ( t) y 0 + Y ( t) Ú Y -1 ( s)g( s) ds, y (0) = Y (0) y 0 + 0 = Í ˙ fi 0 Î5 ˚ È2 ˘ È2 2 ˘È c1 ˘ È c1 ˘ È1 ˘ Í5 ˙ = Í5 3˙Íc ˙ fi Íc ˙ = Í0 ˙ . Therefore, Î ˚ Î ˚Î 2 ˚ Î 2 ˚ Î ˚ È2e - t ˘ È 3 e - t - 2e t + 5 e 3 t ˘ È 11 e - t - 2e t + 5 e 3 t ˘ y( t) = Í - t ˙ + Í154 - t 15 t 415 3 t ˙ = Í 554 - t 15 t 415 3 t ˙ . Î5e ˚ Î 8 e - 4 e + 8 e ˚ Î 8 e - 4 e + 8 e ˚ È 0 1˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = l + 1 = (l - i)(l + i) . The 1 0 Î ˚ È1˘ È1˘ eigenvalues are l1 = i and l 2 = -i , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Îi ˚ Î-i ˚ È-1˘ È1˘ È-1˘ È1˘ x1 = Í ˙ and x 2 = Í ˙ . Converting to real solutions, we have y( t) = e it Í ˙ = (cos t + i sin t) Í ˙ . Îi˚ Îi ˚ Î-i ˚ Îi ˚ È cos t ˘ È sin t ˘ and y 2 ( t) = Í Therefore, a fundamental set of solutions is y1 ( t) = Í ˙ ˙ . A fundamental Î- sin t ˚ Îcos t ˚ È cos t sin t ˘ Ècos s - sin s˘ . We have Y -1 ( s) = Í matrix is Y( t) = Í ˙ ˙. Î- sin t cos t ˚ Î sin s cos s ˚ t

From equation (11), the solution is y ( t) = Y ( t)Y -1 ( t0 ) y 0 + Y ( t) Ú Y -1 ( s)g( s) ds. Since t0

È0 ˘ È2 ˘ y 0 = Í ˙, g( s) = Í ˙, and t0 = 0 , we have Î1 ˚ Î1 ˚ t È sin t ˘ È cos t sin t ˘ t È2 cos s - sin s ˘ y ( t) = Y ( t)Y -1 ( t0 ) y 0 + Y ( t) Ú Y -1 ( s)g( s) ds = Í ˙+Í ˙Ú Í ˙ ds 0 Îcos t ˚ Î- sin t cos t ˚ 0 Î2 sin s + cos s˚ È sin t ˘ È cos t sin t ˘È 2 sin t + cos t - 1 ˘ È 1 - cos t + 3 sin t ˘ =Í ˙+Í ˙Í ˙= Í ˙. Îcos t ˚ Î- sin t cos t ˚Î-2 cos t + sin t + 2 ˚ Î-2 + 3 cos t + sin t ˚

158 • Chapter 6 First Order Linear Systems

24.

È1 1˘ 2 For A = Í ˙ , the characteristic polynomial is p(l ) = (1 - l ) . The eigenvalue is l = 1, with 0 1 Î ˚ È1 ˘ corresponding eigenvector x = Í ˙ . Then Î0 ˚ Èe t ˘ È1 ˘ È0 ˘ Èt ˘ y1 = Í ˙. Let y 2 = e t ( tv1 + v 2 ). v1 = Í ˙, v 2 = Í ˙ fi y 2 = e t Í ˙ . Î0 ˚ Î1 ˚ Î1˚ Î0 ˚ Èe t A fundamental matrix is Y( t) = Í Î0

te t ˘ ˙ and therefore, et ˚

-s t Èe (1 - s) ˘ È te - t ˘ Èe s - se s ˘ Èe - s - se - s ˘ t -1 . Y ( s) = e Í ˙ . Y ( s)g( s) ds = Ú0 Í ˙= Í ˙ds = Í -s -t ˙ es ˚ Î 0 e - s ˚ Ú0 Î1 - e ˚ Î0 Î e ˚ t Èt + te t - t ˘ È te t ˘ È0 ˘ -1 Since y(0) = Í ˙ , y ( t) = Y ( t) Ú Y ( s)g( s) ds = Í t ˙ = Í t ˙. 0 Î0 ˚ Î e - 1 ˚ Îe - 1˚ -1

-2 s

25 (a). Q( t) is an equilibrium solution if

r V

È-2 1 ˘ Ècr ˘ + Q Í 1 -2 ˙ Í 0 ˙ = 0 . Solving for Q, we obtain Î ˚ Î ˚

Cv È2 ˘ . 3 ÍÎ1 ˙˚ 25 (b). The characteristic polynomial is p(l ) = l2 + ( 4 r / V )l + 3r 2 / V 2 = (l + 3r / V )(l + r / V ) . The eigenvalues are l1 = -3r / V and l 2 = - r / V , with corresponding eigenvectors È1˘ È1˘ x1 = Í ˙ and x 2 = Í ˙ . Therefore, a fundamental set of solutions is Î-1˚ Î1˚ È1˘ È1˘ y1 ( t) = e -3 rt /V Í ˙ and y 2 ( t) = e - rt /V Í ˙ . The complementary solution is Î-1˚ Î1˚ -3 rt /V - rt /V Èe ˘È c1 ˘ e QC ( t) = Í -3 rt /V ˙Í ˙ . e - rt /V ˚Îc 2 ˚ Î-e 25 (c). Finding a constant particular solution is equivalent to finding an equilibrium solution, as in part (a). È e -3 rt /V e - rt /V ˘È c1 ˘ cV È2 ˘ 25 (d). The general solution is Q( t) = QC ( t) + Q e ( t) = Í -3 rt /V ˙Í ˙ + Í ˙ . Imposing the e - rt /V ˚Îc 2 ˚ 3 Î1 ˚ Î-e È 1 1˘È c1 ˘ cV È2 ˘ È0 ˘ initial condition leads to Q(0) = 0 or Í ˙Í ˙ + Í ˙ = Í ˙ . The solution of the initial Î-1 1˚Îc 2 ˚ 3 Î1 ˚ Î0 ˚ Q e ( t) =

cV value problem is Q( t) = 6 c È2 ˘ 1 25 (e). lim Q( t) = Í ˙ . V t Æ• 3 Î1 ˚

È4 - e -3 rt /V - 3e - rt /V ˘ Í ˙. -3 rt /V - 3e - rt /V ˚ Î2 + e

Chapter 6 First Order Linear Systems • 159

1 1 26 (a). -Vs + I1¢ + I1 + 2( I1 - I 2 ) = 0, 2( I 2 - I1 ) + I 2 + I 2¢ = 0 . Therefore, 2 2 È I1 (0) ˘ È0 ˘ d È I1¢ ˘ È-6 I1 + 4 I 2 + 2Vs ˘ È-6 4 ˘È I1 ˘ È2Vs ˘ =Í =Í + Í ˙, t > 0, Í Í ˙ ˙ ˙ Í ˙ ˙= Í ˙ dt ÎI 2¢ ˚ Î 4 I1 - 6 I 2 ˚ Î 4 -6 ˚ÎI 2 ˚ Î 0 ˚ ÎI 2 (0) ˚ Î0 ˚ 26 (b). The characteristic polynomial is p(l ) = l2 + 12l + 20 = (l + 10)(l + 2) . The eigenvalues are È1˘ È1˘ l1 = -10 and l 2 = -2 , with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Therefore, a Î-1˚ Î1˚ È e -10 t e -2 t ˘ fundamental matrix is Y( t) = Í -10 t ˙. e -2 t ˚ Î-e -2 s t È2Vs ˘ -e -2 s ˘ 1 Èe10 s -e10 s ˘ 1 Èe 26 (c). I( t) = Y ( t) Ú Y -1 ( s) Í ˙ ds since I(0) = 0 . Y -1 ( s) = -12 s Í -10 s ˙. ˙= Í 0 2e Îe e -10 s ˚ 2 Î e 2 s e 2 s ˚ Î0 ˚ 10s t Èe t ˘ È101 (e10 t - 1) ˘ È2Vs ˘ -1 With Vs ( t) = 1, t > 0, Ú Y ( s) Í ˙ ds = Ú Í 2s ˙ = Í 1 2 t ˙ , Therefore, 0 0 e Î 0 ˚ Î ˚ Î 2 (e - 1) ˚ È 101 (1 - e -10 t ) + 12 (1 - e -2 t ) ˘ È- 101 e -10 t - 12 e -2 t + 35 ˘ = Í 1 -10 t 1 -2 t 2 ˙ . I( t) = Í 1 -10 t -2 t ˙ 1 Î- 10 (1 - e ) + 2 (1 - e ) ˚ Î 10 e - 2 e + 5 ˚

27 (a). In the vector system v¢ = - v + ( v ¥ k) + f , the term v ¥ k is given by Èv x¢ ˘ È-1 1 ˘Èv x ˘ È f x ˘ (v x i + v y j) ¥ k = -v x j + v y i . Therefore, the system is Í ˙ = Í ˙Í ˙ + Í ˙ . Îv y¢ ˚ Î-1 -1˚Îv y ˚ Î f y ˚

È1 ˘ 27 (b). For f = 0.5 Í ˙ , we seek a constant solution v; that is, an equilibrium solution. Thus, we need Î 3˚ È-1 1 ˘Èv x ˘ È 1 ˘ È0 ˘ to solve, if possible, Í Ív ˙ + 0.5 Í ˙ = Í ˙ . This system does indeed have a solution, ˙ Î-1 -1˚Î y ˚ Î 3 ˚ Î0 ˚ È 1+ 3 ˘ namely v e = 0.25 Í ˙ . If we choose the initial velocity equal to the “equilibrium Î-1 + 3 ˚ velocity,” v e , then the particle will move at that constant velocity. 28 (b). The characteristic polynomial is p(l ) = -l (l2 + w c2 ) . The eigenvalues are È1 ˘ È0 ˘ Í ˙ Í ˙ l1 = 0, l 2 = iw c , l 3 = -iw c , with corresponding eigenvectors x1 = Í0 ˙ and x = Í i ˙ . Therefore, ÍÎ0 ˙˚ ÍÎ1 ˙˚

È0 cosw c t sin w c t ˘ ˙ Í A fundamental matrix is Y( t) = Í0 - sin w c t cosw c t ˙ . ÍÎ1 0 0 ˙˚

160 • Chapter 6 First Order Linear Systems

È0 cosw c t sin w c t ˘È0 ˙Í Í -1 28 (c). F( t) = Y ( t)Y (0) = Í0 - sin w c t cosw c t ˙Í0 ÍÎ1 0 0 ˙˚ÍÎ1 È0 cosw c t sin w c t ˘È0 0 1 ˘ È cosw c t ˙ Í ˙Í Í Í0 - sin w c t cosw c t ˙Í1 0 0 ˙ = Í- sin w c t ÍÎ1 0 0 ˙˚ÍÎ0 1 0 ˙˚ ÍÎ 0

-1

1 0˘ ˙ 0 1˙ = 0 0 ˙˚ sin w c t 0 ˘ ˙ cosw c t 0 ˙ . 0 1 ˙˚

t

28 (d). From equation 11, v( t) = F( t) v 0 + F( t) Ú F -1 ( s)g( s) ds , using F( t) as a fundamental matrix 0

and noting that F -1 (0) = I . Therefore, v( t) = F( t) v 0 + f ( t) . t t t t ) ) t 28 (e). r ( t) = Ú v( t) dt = ÈÍÚ F( t) dt ˘˙v 0 + Ú f ( t) dt = r . \ ÈÍÚ F( t) dt ˘˙v 0 = r - Ú f ( t) dt 0 0 0 Î 0 ˚ Î 0 ˚ -1 -1 È w c sin w c t w c (1 - cosw c t) 0 ˘ t Í -1 ˙ -1 Ú0 F(t)dt = Í-w c (1 - cosw c t) w c sinw c t 0˙ . ÍÎ 0 0 t ˙˚ t wt 2t 4t 2 np Êw tˆ D = det Ú F( t) dt = 2 (1 - cosw ct ) = 2 sin 2 Á c ˜ \ c π np fi t π . 0 Ë ¯ wc wc wc 2 2

{

}

Section 6.9 1 (a). For y ¢ = P ( t) y + g( t), y ( t0 ) = y 0 , Euler’s method has the form y n +1 = y n + h[ P ( tn ) y n + g( tn )]. È1 2 ˘ È1˘ È-1˘ , g( t) = Í ˙, y 0 = Í ˙ , and t0 = 0 the iteration is For P ( t) = Í ˙ Î2 3˚ Î1˚ Î1˚ ÈÈ1 2 ˘ È1˘˘ È-1˘ y n +1 = y n + h ÍÍ y n + Í ˙˙, y 0 = Í ˙ . ˙ Î1˚˚ Î1˚ ÎÎ2 3˚ 1 (b). In general, tk = t0 + kh, k = 0, 1,K . Since t0 = 0, we have tk = kh, k = 0, 1, K. In general, h = (b - a) / N . So, for a = 0, b = 1, and h = 0.01, we obtain N = 1 / h = 100 . tn ˘ ÈÈ 1 È 1 ˘˘ È2 ˘ 2 (a). y n +1 = y n + h ÍÍ y n + Í ˙˙, y 0 = Í ˙ . ˙ Îtn ˚˚ Î1 ˚ ÎÎ2 + tn 2 ˚ 2 (b). tk = 1 + kh, k = 0, 1,K . N = .5 / h = 50 . 3 (a). For y ¢ = P ( t) y + g( t), y ( t0 ) = y 0 , Euler’s method has the form y n +1 = y n + h[ P ( tn ) y n + g( tn )]. È -t2 t ˘ È1˘ È2 ˘ For P ( t) = Í ˙, g( t) = Í ˙, y 0 = Í ˙, and t0 = 1 the iteration is Ît ˚ Î0 ˚ Î2 - t 0 ˚

ÈÈ - tn2 tn ˘ È 1 ˘˘ È2 ˘ y n +1 = y n + h ÍÍ ˙y n + Í ˙˙, y 0 = Í ˙ . ÍÎÎ2 - tn 0 ˚ Îtn ˚˙˚ Î0 ˚ 3 (b). In general, tk = t0 + kh, k = 0, 1,K . Since t0 = 1, we have tk = 1 + kh, k = 0, 1,K . In general, h = (b - a) / N . So, for a = 1, b = 4, and h = 0.01, we obtain N = 3 / h = 300 .

Chapter 6 First Order Linear Systems • 161

ÈÈ1 0 1 ˘ È0 ˘ È 0 ˘˘ ÍÍ Í ˙ Í ˙˙ ˙ 4 (a). y n +1 = y n + h ÍÍ3 2 1 ˙y n + Í 2 ˙˙, y 0 = Í0 ˙ . ÍÍÎ1 2 0 ˙˚ ÍÎ1 ˙˚ ÍÎtn ˙˚˙˚ Î 4 (b). tk = -1 + kh, k = 0, 1,K. N = 1 / h = 100 . 5 (a). For y ¢ = P ( t) y + g( t), y ( t0 ) = y 0 , Euler’s method has the form y n +1 = y n + h[ P ( tn ) y n + g( tn )]. È t -1 sin t ˘ È0 ˘ È0 ˘ For P ( t) = Í ˙, g( t) = Í 2 ˙, y 0 = Í ˙, and t0 = 1 the iteration is Ît ˚ Î0 ˚ Î1 - t 1 ˚

ÈÈ tn-1 sin tn ˘ È 0 ˘˘ È0 ˘ y n +1 = y n + h ÍÍ ˙y n + Í 2 ˙˙, y 0 = Í ˙ . 1 ˚ ÍÎÎ1 - tn Îtn ˚˙˚ Î0 ˚ 5 (b). In general, tk = t0 + kh, k = 0, 1,K . Since t0 = 1, we have tk = 1 + kh, k = 0, 1,K . In general, h = (b - a) / N . So, for a = 1, b = 6, and h = 0.01, we obtain N = 5 / h = 500 . ÈÈ1 2 ˘È-1˘ È1˘˘ È-0.98 ˘ È-1˘ 6. y1 = Í ˙ + 0.01ÍÍ ˙Í ˙ + Í ˙˙ = Í ˙ and Î1˚ ÎÎ2 3˚Î 1 ˚ Î1˚˚ Î 1.02 ˚

7.

8.

ÈÈ1 2 ˘È-0.98 ˘ È1˘˘ È-0.9594 ˘ È-0.98 ˘ y2 = Í + 0.01ÍÍ ˙ ˙Í 1.02 ˙ + Í1˙˙ = Í 1.041 ˙ 2 3 Î 1.02 ˚ Î ˚Î ˚ Î ˚˚ Î ˚ Î tn ˘ ÈÈ 1 È 1 ˘˘ È2 ˘ The iteration has the form y n +1 = y n + h ÍÍ y n + Í ˙˙, y 0 = Í ˙ where ˙ Îtn ˚˚ Î1 ˚ ÎÎ2 + tn 2 ˚ ÈÈ1 1 ˘È2 ˘ È1˘˘ È2 ˘ È2 ˘ È4 ˘ È2.04 ˘ t0 = 1 and t1 = 1.01 . Therefore, y1 = Í ˙ + 0.01ÍÍ + Í ˙˙ = Í ˙ + 0.01Í ˙ = Í ˙ Í ˙ ˙ and Î1 ˚ Î9 ˚ Î1.09 ˚ ÎÎ3 2 ˚Î1 ˚ Î1˚˚ Î1 ˚

ÈÈ 1 1.01˘È2.04 ˘ È 1 ˘˘ È2.04 ˘ È2.04 ˘ È4.1409 ˘ y2 = Í + 0.01ÍÍ +Í + 0.01Í ˙= Í ˙ ˙ Í ˙ ˙ ˙ ˙ Î1.09 ˚ Î9.3304 ˚ ÎÎ3.01 2 ˚Î1.09 ˚ Î1.01˚˚ Î1.09 ˚ È2.081409 ˘ =Í ˙. Î1.183304 ˚ ÈÈ-1 1 ˘È2 ˘ È1˘˘ È1.99 ˘ È2 ˘ y1 = Í ˙ + 0.01ÍÍ ˙Í0 ˙ + Í1˙˙ = Í0.03˙ and 1 0 Î0 ˚ Î ˚Î ˚ Î ˚˚ Î ˚ Î

ÈÈ-(1.01) 2 1.01˘È1.99 ˘ È 1 ˘˘ È1.9800030 ˘ È1.99 ˘ y2 = Í 0 01 + . ÍÍ ˙Í ˙ ˙+Í ˙˙ = Í ˙ ÍÎÎ 2 - 1.01 0 ˚Î0.03˚ Î1.01˚˙˚ Î 0.059801 ˚ Î0.03˚ 9.

ÈÈ1 0 1 ˘ È0 ˘ È 0 ˘˘ ÍÍ Í ˙ Í ˙˙ ˙ The iteration has the form y n +1 = y n + h ÍÍ3 2 1 ˙y n + Í 2 ˙˙, y 0 = Í0 ˙ where ÍÍÎ1 2 0 ˙˚ ÍÎ1 ˙˚ ÍÎtn ˙˚˙˚ Î t0 = -1 and t1 = -0.99 . Therefore, ÈÈ1 0 1 ˘È0 ˘ È 0 ˘˘ È0 ˘ È 1 ˘ È0.01˘ È0 ˘ ÍÍ ˙ Í ˙ Í ˙Í ˙ Í ˙˙ Í ˙ Í ˙ y1 = Í0 ˙ + 0.01ÍÍ3 2 1 ˙Í0 ˙ + Í 2 ˙˙ = Í0 ˙ + 0.01Í 3 ˙ = Í0.03˙ ÍÍÎ1 2 0 ˙˚ÍÎ1 ˙˚ ÍÎ-1˙˚˙ ÍÎ1 ˙˚ ÍÎ-1˙˚ ÍÎ0.99 ˙˚ ÍÎ1 ˙˚ Î ˚

162 • Chapter 6 First Order Linear Systems

and

ÈÈ1 0 1 ˘È0.01˘ È 0 ˘˘ È0.01˘ È 1 ˘ È0.01˘ ÍÍ ˙ Í ˙ ˙˙ Í ˙ Í ˙Í ˙ Í y 2 = Í0.03˙ + 0.01ÍÍ3 2 1 ˙Í0.03˙ + Í 2 ˙˙ = Í0.03˙ + 0.01Í 3.08 ˙ ÍÍÎ1 2 0 ˙˚ÍÎ0.99 ˙˚ ÍÎ-0.99 ˙˚˙ ÍÎ0.99 ˙˚ ÍÎ-0.92 ˙˚ ÍÎ0.99 ˙˚ Î ˚

10.

È 0.02 ˘ ˙ Í = Í0.0608 ˙ . ÍÎ0.9808 ˙˚ ÈÈ1 sin(1) ˘È0 ˘ È0 ˘˘ È 0 ˘ È0 ˘ and y1 = Í ˙ + 0.01ÍÍ + ˙= 1 ˙˚ÍÎ0 ˙˚ ÍÎ1 ˙˚˚ ÍÎ0.01˙˚ Î0 ˚ ÎÎ0

ÈÈ 1.101 sin(1.01) ˘È 0 ˘ È 0 ˘˘ È0.00008468318 ˘ È 0 ˘ y2 = Í 0 01 + + . Í ˙= Í ˙ ˙ 1 ˚ÍÎ0.01˙˚ ÍÎ(1.01) 2 ˙˚˙˚ ÍÎ 0.020301 ˙˚ ÍÎÎ1 - 1.01 Î0.01˚ È z1 ( t) ˘ È y ( t) ˘ 11 (a). Let z( t) = Í ˙= Í ˙ . With this, Î z 2 ( t) ˚ Î y ¢ ( t) ˚ È z1¢( t) ˘ È y ¢ ( t) ˘ È z2 ( t) ˘ È 0 1 ˘È z1 ( t) ˘ È 0 ˘ È1 ˘ 0 = = = + = z ¢ ( t) = Í , z ( ) ˙ Í ˙ Í Í ˙Í ˙ Í 3/2 ˙ Í0 ˙ . 3/2 ˙ Îz2¢ ( t) ˚ Î y ¢¢ ( t) ˚ Î- z1 ( t) + t ˚ Î-1 0 ˚Îz2 ( t) ˚ Ît ˚ Î ˚ ÈÈ 0 1 ˘ È 0 ˘˘ È1 ˘ 11 (b). Thus, the iteration has the form z n +1 = z n + h ÍÍ z n + Í 3 / 2 ˙˙, z 0 = Í ˙ where ˙ Îtn ˚˚ Î0 ˚ ÎÎ-1 0 ˚ t0 = 0 and t1 = 0.01 . ÈÈ 0 1 ˘È1 ˘ È0 ˘˘ È1 ˘ È1 ˘ È0˘ È 1 ˘ 11 (c). Therefore, z1 = Í ˙ + 0.01ÍÍ + Í ˙˙ = Í ˙ + 0.01Í ˙ = Í ˙ Í ˙ ˙ and Î0 ˚ Î-1˚ Î-0.01˚ ÎÎ-1 0 ˚Î0 ˚ Î0 ˚˚ Î0 ˚

ÈÈ 0 1 ˘È 1 ˘ È 0 ˘˘ È 1 ˘ È 1 ˘ È -.01 ˘ z2 = Í + 0.01ÍÍ +Í + 0.01Í ˙= Í ˙ ˙ Í ˙ ˙ ˙ ˙ Î-0.01˚ Î-.999 ˚ ÎÎ-1 0 ˚Î-0.01˚ Î0.001˚˚ Î-0.01˚ È .9999 ˘ =Í ˙. Î-.01999 ˚ 1 ˘È z1 ( t) ˘ È0 ˘ z 2 ( t) È z1¢( t) ˘ È ˘ È0 È1˘ =Í 2 =Í 2 + Í ˙ , z(1) = Í ˙ . 12 (a). z¢ ( t) = Í ˙ ˙ ˙ Í ˙ Îz2¢ ( t) ˚ Î- t z1 ( t) - z2 ( t) + 2 ˚ Î- t -1˚Îz2 ( t) ˚ Î2 ˚ Î1˚ 1˘ ÈÈ 0 È0 ˘˘ È1˘ 12 (b). z n +1 = z n + h ÍÍ 2 z n + Í ˙˙, z 0 = Í ˙ where t0 = 1. ˙ Î2 ˚˚ Î1˚ ÎÎ- t -1˚ ÈÈ 0 1 ˘È1˘ È0 ˘˘ È1.01˘ È1˘ 12 (c). z1 = Í ˙ + 0.01ÍÍ ˙Í ˙ + Í ˙˙ = Í ˙ and Î1˚ ÎÎ-1 -1˚Î1˚ Î2 ˚˚ Î 1 ˚ ÈÈ 0 È1.01˘ z2 = Í + 0.01ÍÍ ˙ 2 Î 1 ˚ ÎÎ-(1.01)

1 ˘È1.01˘ È0 ˘˘ È 1.02 ˘ + Í ˙˙ = Í ˙ Í ˙ -1˚Î 1 ˚ Î2 ˚˚ Î0.999696699 ˙˚

Chapter 6 First Order Linear Systems • 163

È z1 ˘ È y ˘ Í ˙ Í ˙ 13 (a). Let z = Íz2 ˙ = Í y ¢ ˙ . With this, ÍÎz3 ˙˚ ÍÎ y ¢¢ ˙˚ z2 ˘ È 0 1 0 ˘È z1 ˘ È 0 ˘ È1˘ È z1¢ ˘ È y ¢ ˘ È ˙Í ˙ Í ˙ Í Í ˙ ˙ Í ˙ Í ˙ Í z¢ = Íz2¢ ˙ = Í y ¢¢ ˙ = Í z3 ˙ = Í 0 0 1 ˙Íz2 ˙ + Í 0 ˙ , z(0) = Í-1˙ . ÍÎ 0 ˙˚ ÍÎz3¢ ˙˚ ÍÎ y ¢¢¢ ˙˚ ÍÎ-2 z2 - tz1 + t + 1˙˚ ÍÎ- t -2 0 ˙˚ÍÎz3 ˙˚ ÍÎt + 1˙˚ ÈÈ 0 È1˘ È 0 ˘˘ 1 0˘ ÍÍ Í ˙ ˙˙ Í ˙ 13 (b). Thus, the iteration has the form z n +1 = z n + h ÍÍ 0 0 1 ˙z n + Í 0 ˙˙, z 0 = Í-1˙ where ÍÍÎ- tn -2 0 ˙˚ ÍÎ 0 ˙˚ ÍÎtn + 1˙˚˙˚ Î t0 = 0 and t1 = 0.01 . ÈÈ0 1 0 ˘È 1 ˘ È0 ˘˘ È 1 ˘ È-1˘ È0.99 ˘ È1˘ ÍÍ ˙ Í ˙ Í ˙Í ˙ Í ˙˙ Í ˙ Í ˙ 13 (c). Therefore, z1 = Í-1˙ + 0.01ÍÍ0 0 1 ˙Í-1˙ + Í0 ˙˙ = Í-1˙ + 0.01Í 0 ˙ = Í -1 ˙ and ÍÍÎ0 -2 0 ˙˚ÍÎ 0 ˙˚ ÍÎ1 ˙˚˙ ÍÎ 0 ˙˚ ÍÎ 3 ˙˚ ÍÎ0.03˙˚ ÍÎ 0 ˙˚ Î ˚ ÈÈ 0 È -1 ˘ È0.99 ˘ 1 0 ˘È0.99 ˘ È 0 ˘˘ È0.99 ˘ Í ˙ Í ˙ ˙˙ Í ˙ Í ˙Í Í ˙ Í 0 1 ˙Í -1 ˙ + Í 0 ˙˙ = Í -1 ˙ + 0.01Í 0.03 ˙ z 2 = Í -1 ˙ + 0.01ÍÍ 0 ÍÍÎ-0.01 -2 0 ˙˚ÍÎ0.03˙˚ ÍÎ1.01˙˚˙ ÍÎ0.03˙˚ ÍÎ3.0001˙˚ ÍÎ0.03˙˚ Î ˚ È 0.98 ˘ ˙ Í = Í -0.9997 ˙ . ÍÎ0.060001˙˚ z 2 ( t) 1 ˘È z1 ( t) ˘ È0 ˘ È z1¢( t) ˘ È ˘ È 0 È-1˘ , z ( ) 0 = = + = 14 (a). z¢ ( t) = Í ˙ Í -t ˙ Í - t -1˙Íz ( t) ˙ Í2 ˙ Í 1 ˙. Îz2¢ ( t) ˚ Î-e z1 ( t) - z2 ( t) + 2 ˚ Î-e ˚Î 2 ˚ Î ˚ Î ˚ 1˘ 0 1 ÈÈ 0 ˘ È ˘ È ˘ 14 (b). z n +1 = z n + h ÍÍ - t z n + Í ˙˙, z 0 = Í ˙ where t0 = 0 . ˙ -1˚ Î2 ˚˚ Î1˚ ÎÎ-e ÈÈ 0 1 ˘È-1˘ È0 ˘˘ È-0.99 ˘ È-1˘ 14 (c). z1 = Í ˙ + 0.01ÍÍ ˙Í ˙ + Í ˙˙ = Í ˙ and Î1˚ ÎÎ-1 -1˚Î 1 ˚ Î2 ˚˚ Î 1.02 ˚

18.

20.

1 ˘È-0.99 ˘ È0 ˘˘ È -0.9798 ˘ ÈÈ 0 È-0.99 ˘ z2 = Í + 0.01ÍÍ -0.01 + ˙= ˙ -1˙˚ÍÎ 1.02 ˙˚ ÍÎ2 ˙˚˚ ÍÎ1.039601493˙˚ Î 1.02 ˚ ÎÎ-e È-0.00807508729 ˘ Actual error: y (1) - y 200 = Í ˙ Î 0.0759433736... ˚ È-0.0086591617...˘ Estimated error: y 200 - y100 = Í ˙ Î 0.0764878206... ˚ È 0.0027112167... ˘ Actual error: y (1) - y 200 = Í ˙ Î-0.0027112167...˚ È 0.0027202379... ˘ Estimated error: y 200 - y100 = Í ˙ Î-0.0027202379...˚

164 • Chapter 6 First Order Linear Systems

-15 5 5 dQ1 Q = -15 1 + Q2 = Q1 + Q 200 - 10 t 500 - 20 t 2 dt V1 V2 dQ2 15 35 = Q1 Q dt 200 - 10 t 500 - 20 t 2 21 (b). t=0:.01:19.9; Q1(1)=40;Q2(1)=40; h=0.01; V1=200-10*t; V2=500-20*t; N=19.9/h; for i=1:N Q1(i+1)=Q1(i)+h*(-(15/V1(i))*Q1(i)+(5/V2(i))*Q2(i)); Q2(i+1)=Q2(i)+h*((15/V1(i))*Q1(i)-(35/V2(i))*Q2(i)); end plot(t,Q1,t,Q2,':') ylabel('Amount of salt in tanks (lb)') xlabel('time (minutes)') title('Chapter 6.9 Problem 21: two-tank draining problem') 21 (c).

21 (a).

15 are not continuous at t = 20 . Therefore, Existence-Uniqueness 200 - 10t Theorem 6.1 does not apply to any interval containing t = 20 . 1 -t 22 (a). my ¢¢ + gy ¢ + ky = 0, m = 1, g = 2 te 2 , k = 4p 2 , y (0) = meters, y ¢ (0) = 0 . 5 1 ˘È y1 ( t) ˘ È y1¢( t) ˘ È 0 È0.2 ˘ y ¢ ( t) = Í , y (0) = Í ˙ . =Í - 2t ˙ Í ˙ ˙ 2 Î y 2¢ ( t) ˚ Î-4p -2 te ˚Î y 2 ( t) ˚ Î0 ˚

21 (d). The coefficients ±

Chapter 6 First Order Linear Systems • 165

22 (b). t=0:.005:10; h=.005; N=2000; y1(1)=0.2;y2(1)=0; gamma=2*t.*exp(-0.5*t); k=4*(pi^2); for i=1:N y1(i+1)=y1(i)+h*y2(i); y2(i+1)=y2(i)+h*(-k*y1(i)-gamma(i)*y2(i)); end plot(t,y1)

22 (c). The amplitude of displacement decreases significantly during the time when damping is significant. As damping dimishes, the vibration amplitude seems to settle down to a constant value. 23 (b). t=0:0.01:3; h=0.01; N=300; y1(1)=2;y2(1)=0; for i=1:N y1(i+1)=y1(i)+h*y2(i); y2(i+1)=y2(i)+h*(((pi/4)^2)*(t(i)^2)*y1(i)-0.5*y2(i)); end plot(t,y1) xlabel('time (s)'); ylabel('radial position (cm)'); title('Chapter 6.9 problem 23 radial position vs time') y1(301)

166 • Chapter 6 First Order Linear Systems

23 (c).

r( 3) = 15.2268...cm

Section 6.10 1.

2.

3.

È5 -6 ˘ 2 For A = Í ˙ the characteristic polynomial is p(l ) = l - l - 2 . Eigenvalues are 3 4 Î ˚ È1˘ È2 ˘ l1 = -1 and l 2 = 2 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . As in Example 1, Î1˚ Î1 ˚ È1 2 ˘ we can construct a diagonalizing matrix T from the eigenvectors of A, T = Í ˙ . The Î1 1 ˚ È-1 0 ˘ corresponding matrix of eigenvalues, D = Í , is such that T -1 AT = D . ˙ Î 0 2˚ 2 The characteristic polynomial is p(l ) = l - 1. Eigenvalues are l1 = -1 and l 2 = 1 . È1 2˘ È1˘ È2˘ Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . Therefore, T = Í ˙ and Î-1 -1˚ Î-1˚ Î-1˚ È-1 0 ˘ D= Í ˙. Î 0 1˚ È1 1˘ For A = Í the characteristic polynomial is p(l ) = l2 - 2l . Eigenvalues are ˙ Î1 1˚ È-1˘ È1˘ l1 = 0 and l 2 = 2 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . As in Example 1, Î1˚ Î1˚ È-1 1˘ we can construct a diagonalizing matrix T from the eigenvectors of A, T = Í ˙ . The Î 1 1˚ È0 0 ˘ -1 corresponding matrix of eigenvalues, D = Í ˙ , is such that T AT = D . 0 2 Î ˚

Chapter 6 First Order Linear Systems • 167

4.

5.

6.

7.

8.

9.

The characteristic polynomial is p(l ) = l2 - 5l . Eigenvalues are l1 = 0 and l 2 = 5 . È 3 1˘ È3˘ È1˘ Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . Therefore, T = Í ˙ and Î-2 1˚ Î-2 ˚ Î1˚ È0 0 ˘ D= Í ˙. Î0 5 ˚ È2 3˘ 2 For A = Í ˙ the characteristic polynomial is p(l ) = l - 4 l - 5 . Eigenvalues are 3 2 Î ˚ È-1˘ È1˘ l1 = -1 and l 2 = 5 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . As in Example 1, Î1˚ Î1˚ È-1 1˘ we can construct a diagonalizing matrix T from the eigenvectors of A, T = Í ˙ . The Î 1 1˚ È-1 0 ˘ -1 corresponding matrix of eigenvalues, D = Í ˙ , is such that T AT = D . 0 5 Î ˚ 2 The characteristic polynomial is p(l ) = l - 2l - 3 = (l + 1)(l - 3) . Eigenvalues are È1˘ È1˘ l1 = -1 and l 2 = 3 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . Therefore, Î-1˚ Î1˚ È 1 1˘ È-1 0 ˘ T=Í and D = Í ˙ ˙. Î-1 1˚ Î 0 3˚ È2 0 ˘ 2 For A = Í ˙ the characteristic polynomial is p(l ) = l - 3l + 2 . Eigenvalues are 1 1 Î ˚ È0 ˘ È1˘ l1 = 1 and l 2 = 2 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . As in Example 1, we Î1 ˚ Î1˚ È0 1˘ can construct a diagonalizing matrix T from the eigenvectors of A, T = Í ˙ . The Î1 1˚ È1 0 ˘ corresponding matrix of eigenvalues, D = Í , is such that T -1 AT = D . ˙ Î0 2 ˚ 2 The characteristic polynomial is p(l ) = l - l - 6 = (l + 2)(l - 3) . Eigenvalues are È1 ˘ È2 ˘ l1 = -2 and l 2 = 3 . Corresponding eigenvectors are x1 = Í ˙ and x 2 = Í ˙ . Therefore, Î0 ˚ Î1 ˚ È-2 0 ˘ È1 2 ˘ T=Í and D = Í ˙. ˙ Î 0 3˚ Î0 1 ˚ È 25 -8 30 ˘ ˙ Í For A = Í 24 -7 30 ˙ , the eigenvalues are l1 = 1 and l 2 = 2 . From the characteristic ÍÎ-12 4 -14 ˙˚ polynomial given, it follows that l1 has algebraic multiplicity 2 and l 2 has algebraic multiplicity 1.

168 • Chapter 6 First Order Linear Systems

In order to find the eigenvectors corresponding to l1 , we solve ( A - l1I ) x = 0 or È 24 -8 30 ˘È x1 ˘ È0 ˘ È12 -4 15 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ ˙Í ˙ Í ˙ Í Í Í 24 -8 30 ˙Í x 2 ˙ = Í0 ˙ . This system reduces to Í 0 0 0 ˙Í x 2 ˙ = Í0 ˙ and hence ÍÎ-12 4 -15 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ ÍÎ 0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ eigenvectors corresponding to l1 all have the form È( 4 x 2 - 15 x 3 ) / 12 ˘ È x 2 / 3˘ È-5 x 3 / 4 ˘ ˙ ˙ Í ˙ Í Í x=Í x2 ˙ = Í x 2 ˙ + Í 0 ˙ . Thus, we find two linearly independent ˙˚ ÍÎ 0 ˙˚ ÍÎ x 3 ˙˚ ÍÎ x3

È-5 ˘ È1 ˘ Í ˙ Í ˙ eigenvectors, x1 = Í3˙ and x 2 = Í 0 ˙ corresponding to l1 and therefore, l1 has geometric ÍÎ 4 ˙˚ ÍÎ0 ˙˚ multiplicity 2. Since l 2 has algebraic multiplicity 1, it also has geometric multiplicity 1. Thus, A is not defective (that is, A is diagonalizable). In order to find the eigenvectors corresponding È 23 -8 30 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í to l 2 , we solve ( A - l 2 I ) x = 0 or Í 24 -9 30 ˙Í x 2 ˙ = Í0 ˙ . Solving this system, we obtain ÍÎ-12 4 -16 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚

10.

11.

È2˘ È1 -5 2 ˘ ˙ Í ˙ Í an eigenvector corresponding to l 2 , x = Í 2 ˙ . Therefore, if T = Í3 0 2 ˙ , and ÍÎ-1˙˚ ÍÎ0 4 -1˙˚ È1 0 0 ˘ ˙ Í D = Í0 1 0 ˙ , then T -1 AT = D . ÍÎ0 0 2 ˙˚ l1 = -1 has algebraic multiplicity 1 and l 2 = 3 has algebraic multiplicity 2. The corresponding È1 ˘ È1˘ È1˘ Í ˙ Í ˙ Í ˙ eigenvectors are x = Í 2 ˙ for l1 and x1 = Í 0 ˙ and x 2 = Í2 ˙ for l 2 . Therefore, l1 has ÍÎ0 ˙˚ ÍÎ-2 ˙˚ ÍÎ-2 ˙˚ geometric multiplicity 1 and l 2 has geometric multiplicity 2. A is diagonalizable and È 1 1 1˘ È-1 0 0 ˘ ˙ ˙ Í Í T = Í 2 0 2 ˙ , and D = Í 0 3 0 ˙ . ÍÎ-2 -2 0 ˙˚ ÍÎ 0 0 3˙˚ È1 0 1 ˘ ˙ Í For A = Í2 2 -3˙ , the eigenvalues are l1 = 1 and l 2 = 2 . From the characteristic polynomial ÍÎ0 0 1 ˙˚ given, it follows that l1 has algebraic multiplicity 2 and l 2 has algebraic multiplicity 1.

Chapter 6 First Order Linear Systems • 169

l1 , we solve ( A - l1I ) x = 0 or 0 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ 1 0 ˙Í x 2 ˙ = Í0 ˙ and hence eigenvectors 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È1˘ È x1 ˘ Í ˙ ˙ Í corresponding to l1 all have the form x = Í-2 x1 ˙ = x1 Í-2 ˙ . ÍÎ 0 ˙˚ ÍÎ 0 ˙˚ Thus, there is only one linearly independent eigenvector corresponding to l1 . Therefore, l1 has geometric multiplicity 1 and consequently A is defective (not diagonalizable). In order to find the eigenvectors corresponding to È0 0 1 ˘È x1 ˘ È0 ˘ È0 ˙Í ˙ Í ˙ Í Í Í2 1 -3˙Í x 2 ˙ = Í0 ˙ . This system reduces to Í2 ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ ÍÎ0

12.

13.

14.

15.

l1 = 2 has algebraic multiplicity 2 and l 2 = 3 has algebraic multiplicity 1. The corresponding È1 ˘ È0 ˘ Í ˙ Í ˙ eigenvectors are x = Í4 ˙ for l1 and x = Í1 ˙ for l 2 . Therefore, l1 has geometric multiplicity ÍÎ1 ˙˚ ÍÎ1 ˙˚ 1 and l 2 has geometric multiplicity1 and A is not diagonalizable. È 4 -1 1 ˘ ˙ Í For A = Í10 -2 3˙ , the only eigenvalue is l1 = 1. From the characteristic polynomial given, ÍÎ 1 0 1 ˙˚ it follows that l1 has algebraic multiplicity 3. In order to find the eigenvectors corresponding È 3 -1 1 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í to l1 , we solve ( A - l1I ) x = 0 or Í10 -3 3˙Í x 2 ˙ = Í0 ˙ . This system reduces to ÍÎ 1 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È1 0 0 ˘È x1 ˘ È0 ˘ ˙Í ˙ Í ˙ Í Í0 1 -1˙Í x 2 ˙ = Í0 ˙ and hence eigenvectors corresponding to l1 all have the form ÍÎ0 0 0 ˙˚ÍÎ x 3 ˙˚ ÍÎ0 ˙˚ È0 ˘ È0˘ Í ˙ Í ˙ x = Í x 3 ˙ = x 3 Í1 ˙ . Thus, there is only one linearly independent eigenvector corresponding to ÍÎ1 ˙˚ ÍÎ x 3 ˙˚ l1 . Therefore, l1 has geometric multiplicity 1 and consequently A is defective (not diagonalizable). All four matrices are diagonalizable. Matrices (a) and (d) have distinct eigenvalues. Matrix (b) is a real, symmetric matrix. Matrix (c) is lower triangular and has distinct eigenvalues. È6 -6 ˘ For A = Í ˙ the eigenvalues are l1 = 2 and l 2 = 3 with corresponding eigenvectors Î2 -1˚ È3 2 ˘ È3˘ È2 ˘ x1 = Í ˙ and x 2 = Í ˙ . Make the substitution y = Tz = Í ˙z to obtain Tz¢ = ATz + g( t) . Î2 1 ˚ Î2 ˚ Î1 ˚

170 • Chapter 6 First Order Linear Systems

Multiplying by T -1 gives z¢ = T -1 ATz + T -1g( t) or È2 0 ˘ È-1 2 ˘È4 + 3e t ˘ È2 0 ˘ Èe t ˘ -1 z¢ = Dz + T g( t) = Í =Í ˙z + Í ˙Í ˙z + Í ˙ . Thus, the system uncouples t˙ Î0 3˚ Î 2 -3˚Î2 + 2e ˚ Î0 3˚ Î 2 ˚ È z1¢ ˘ È2 z1 + e t ˘ into Í ˙ = Í ˙ . Solving these uncoupled first order equations, we obtain Îz2¢ ˚ Î 3z2 + 2 ˚ È -e t + c1e 2 t ˘ . Finally, forming y = Tz , we obtain the general solution z=Í 3t ˙ Î-(2 / 3) + c 2e ˚ È3 2 ˘È -e t + c1e 2 t ˘ È3e 2 t 2e 3 t ˘È c1 ˘ È3e t + 4 / 3˘ y=Í = Í 2t ˙Í ˙ - Í ˙. ˙Í 3t ˙ e 3 t ˚Îc 2 ˚ Î2e t + 2 / 3 ˚ Î2 1 ˚Î-(2 / 3) + c 2e ˚ Î2e

16.

17.

È2˘ The eigenvalues are l1 = -1 and l 2 = 2 with corresponding eigenvectors x1 = Í ˙ and Î-1˚ È2 1˘ È1˘ x1 = Í ˙ . Make the substitution y = Tz = Í ˙z . Î-1 -1˚ Î-1˚ È-1 0 ˘ È 1 1 ˘È e 2 t - 2e t ˘ È-e t ˘ z¢ = Dz + T -1g( t) = Í = Í 2t ˙ . ˙z + Í ˙Í 2 t t˙ Î 0 2 ˚ Î-1 -2 ˚Î-e + e ˚ Î e ˚ È-(1 / 2)e t + c1e - t ˘ Solving these first order equations, we obtain z = Í ˙ . Finally, forming y = Tz , 2t 2t Î te + c 2e ˚ t -t -t 2t e ˘È c1 ˘ È-e t + te 2 t ˘ È 2 1 ˘È-(1 / 2)e + c1e ˘ È2e we obtain the solution y = Í ˙Í ˙ + Í ˙ = Í -t ˙. ˙Í 2t 2t -e 2 t ˚Îc 2 ˚ Î 12 e t - te 2 t ˚ Î-1 -1˚Î te + c 2e ˚ Î-e È1 1 ˘ For A = Í ˙ the eigenvalues are l1 = 0 and l 2 = 3 with corresponding eigenvectors Î2 2 ˚ È 1 1˘ È1˘ È1 ˘ x1 = Í ˙ and x1 = Í ˙ . Make the substitution y = Tz = Í ˙z to obtain Tz¢ = ATz + g( t) . Î-1 2 ˚ Î-1˚ Î2 ˚ Multiplying by T -1 gives z¢ = T -1 ATz + T -1g( t) or È0 0 ˘ È2 / 3 -1 / 3˘È t ˘ È0 0 ˘ Èt - 1˘ z¢ = Dz + T -1g( t) = Í ˙z + Í ˙Í ˙= Í ˙z + Í ˙. Î0 3˚ Î1 / 3 1 / 3 ˚Î3 - t ˚ Î0 3˚ Î 1 ˚ È z1¢ ˘ È t - 1 ˘ Thus, the system uncouples into Í ˙ = Í ˙ . Solving these uncoupled first order Îz2¢ ˚ Î3z2 + 1˚ È(1 / 2) t 2 - t + c1 ˘ equations, we obtain z = Í . Finally, forming y = Tz , we obtain the solution 3t ˙ Î -(1 / 3) + c 2e ˚ È 1 1 ˘È(1 / 2) t 2 - t + c1 ˘ È 1 e 3 t ˘È c1 ˘ È (1 / 2) t 2 - t - (1 / 3) ˘ y=Í =Í ˙. ˙Í ˙- Í 2 3t ˙ 3 t ˙Í Î-1 2 ˚Î -(1 / 3) + c 2e ˚ Î-1 2e ˚Îc 2 ˚ Î-(1 / 2) t + t - (2 / 3) ˚

Chapter 6 First Order Linear Systems • 171

18.

19.

20.

21.

È2˘ The eigenvalues are l1 = 2 and l 2 = 5 with corresponding eigenvectors x1 = Í ˙ and Î-1˚ È 2 1˘ È1˘ x1 = Í ˙ . Make the substitution y = Tz = Í ˙z . Î-1 1˚ Î1˚ È2 0 ˘ È 13 - 13 ˘È 4 t + 4 ˘ È2 t + 1˘ -1 z¢ = Dz + T g( t) = Í ˙z + Í 1 2 ˙Í ˙= Í ˙. Î0 5 ˚ Î 3 3 ˚Î-2 t + 1˚ Î 2 ˚ È- t - 1 + c1e 2 t ˘ Solving these first order equations, we obtain z = Í 2 . Finally, forming y = Tz , we 5t ˙ Î - 5 + c 2e ˚ È 2 1˘È- t - 1 + c1e 2 t ˘ È2e 2 t e 5 t ˘È c1 ˘ È-2 t - 125 ˘ obtain the solution y = Í = Í 2t ˙Í ˙ + Í ˙. ˙Í 2 5t ˙ e 5 t ˚Îc 2 ˚ Î t + 35 ˚ Î-1 1˚Î - 5 + c 2e ˚ Î-e È-9 -5 ˘ For A = Í ˙ the eigenvalues are l1 = -1 and l 2 = -4 with corresponding eigenvectors Î8 4˚ È5 1˘ È5˘ È1˘ x1 = Í ˙ and x 2 = Í ˙ . Make the substitution y = Tz = Í ˙z to obtain Tz¢¢ = ATz . Î-8 -1˚ Î-8 ˚ Î-1˚ È-1 0 ˘ Multiplying by T -1 gives z¢¢ = T -1 ATz or z¢¢ = Dz = Í ˙z . Thus, the system uncouples Î 0 -4 ˚ È z1¢¢˘ È - z1 ˘ into Í ˙ = Í ˙. Îz2¢¢˚ Î-4 z2 ˚ È c1 cos t + d1 sin t ˘ Solving these uncoupled equations, we obtain z = Í ˙ . Finally, forming Îc 2 cos 2 t + d2 sin 2 t ˚ y = Tz , we obtain the solution È 5 1 ˘È c1 cos t + d1 sin t ˘ È 5(c1 cos t + d1 sin t) + c 2 cos 2 t + d2 sin 2 t ˘ y=Í ˙Í ˙= Í ˙. Î-8 -1˚Îc 2 cos 2 t + d2 sin 2 t ˚ Î-8(c1 cos t + d1 sin t) - c 2 cos 2 t - d2 sin 2 t ˚ The eigenvalues are l1 = -9 and l 2 = -1 with corresponding eigenvectors 1˘ È 7 È 7 ˘ È1˘ x1 = Í and x 2 = Í ˙ . Make the substitution x = Tz = Í ˙z to obtain ˙ Î-15 -1˚ Î-15 ˚ Î-1˚ Èc1e -3 t + c 2e 3 t ˘ È-9 0 ˘ z¢¢ + Í . Finally, we obtain ˙z = 0 . Solving the equations, we obtain z = Í - t t ˙ Î 0 -1˚ Î k1e + k2e ˚ 1 ˘Èc1e -3 t + c 2e 3 t ˘ È 7(c1e -3 t + c 2e 3 t ) + k1e - t + k2e t ˘ È 7 the solution x = Í . =Í ˙Í - t t ˙ -3 t -t t ˙ 3t Î-15 -1˚Î k1e + k2e ˚ Î-15(c1e + c 2e ) - ( k1e + k2e ) ˚ È-2 -1˘ For A = Í ˙ the eigenvalues are l1 = -1 and l 2 = 1 with corresponding eigenvectors Î3 2˚ È1 1 ˘ È1˘ È1˘ x1 = Í ˙ and x 2 = Í ˙ . Make the substitution y = Tz = Í ˙z to obtain Tz¢¢ = ATz . Î-1 -3˚ Î-1˚ Î-3˚

172 • Chapter 6 First Order Linear Systems

22.

È-1 0 ˘ Multiplying by T -1 gives z¢¢ = T -1 ATz or z¢¢ = Dz = Í ˙z . Thus, the system uncouples Î 0 1˚ È z1¢¢˘ È- z1 ˘ into Í ˙ = Í ˙ . Îz2¢¢˚ Î z2 ˚ Èc1 cos t + d1 sin t ˘ Solving these uncoupled equations, we obtain z = Í ˙ . Finally, forming y = Tz , -t t Î c 2e + d2e ˚ È 1 1 ˘Èc1 cos t + d1 sin t ˘ È c1 cos t + d1 sin t + c 2e - t + d2e t ˘ we obtain the solution y = Í . ˙Í ˙= Í -t t -t t ˙ Î-1 -3˚Î c 2e + d2e ˚ Î-(c1 cos t + d1 sin t) - 3(c 2e + d2e ) ˚ The eigenvalues are l1 = 0 and l 2 = 5 with corresponding eigenvectors È 1 2˘ È1˘ È0 0 ˘ È2 ˘ x1 = Í ˙ and x 2 = Í ˙ . Make the substitution x = Tz = Í z to obtain z¢¢ + Í ˙ ˙z = 0 . Î-2 1 ˚ Î-2 ˚ Î0 5 ˚ Î1 ˚ c1t + c 2 È ˘ Solving the equations, we obtain z = Í ˙ . Finally, we obtain the solution Îk1 cos( 5 t) + k2 sin( 5 t) ˚

[ [

] ]

c1t + c 2 È 1 2 ˘È ˘ È (c1t + c 2 ) + 2 k1 cos( 5 t) + k2 sin( 5 t) ˘ ˙. x=Í ˙Í ˙= Í Î-2 1 ˚Îk1 cos( 5 t) + k2 sin( 5 t) ˚ ÍÎ-2(c1t + c 2 ) + k1 cos( 5 t) + k2 sin( 5 t) ˙˚ È 500 -200 ˘ 27 (a). For A = Í ˙ the eigenvalues are l1 = 100 and l 2 = 600 with corresponding Î-200 200 ˚ È1 ˘ È2˘ eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î2 ˚ Î-1˚ È1 2 ˘ 27 (b). Make the substitution y = Tz = Í ˙z to obtain Tz¢¢ + ATz = 0 . Multiplying by Î2 -1˚ È100 0 ˘ È0 ˘ z = Í ˙ . Thus, the system uncouples into T -1 gives z¢¢ + T -1 ATz = 0 or z¢¢ + Í ˙ Î 0 600 ˚ Î0 ˚ È0.2 0.4 ˘È 0.1 ˘ È0.08 ˘ È z1¢¢+ 100 z1 ˘ È0 ˘ -1 z ( 0 ) = T y ( 0 ) = = . The initial condition is Í0.4 -0.2 ˙Í0.15 ˙ = Í0.01˙ . Íz¢¢+ 600 z ˙ Í0 ˙ Î 2˚ Î 2 ˚Î Î ˚ ˚ Î ˚ È c1 cos10 t + d1 sin10 t ˘ 27 (c). Solving the uncoupled equations z¢¢ + Dz = 0 , we obtain z = Í ˙. Îc 2 cos10 6 t + d2 sin10 6 t ˚ È 0.08 cos10 t ˘ Imposing the initial condition, we find z = Í ˙ . Finally, forming y = Tz , we Î0.01cos10 6 t ˚ obtain the solution of the initial value problem: È1 2 ˘È 0.08 cos10 t ˘ È0.08 cos10 t + 0.02 cos 10 6 t ˘ ˙. y=Í ˙Í ˙= Í Î2 -1˚Î0.01cos10 6 t ˚ ÍÎ 0.16 cos10 t - 0.01cos 10 6 t ˙˚

( (

) )

Chapter 6 First Order Linear Systems • 173

Section 6.11 È5 -4 ˘ 1 (a). We proceed as in Example 2. For A = Í , the characteristic polynomial is p(l ) = l2 - l . ˙ Î5 -4 ˚ È4 ˘ È1˘ Eigenvalues are l1 = 0 and l 2 = 1 with corresponding eigenvectors x1 = Í ˙ and x 2 = Í ˙ . Î5 ˚ Î1˚ tA -1 Since A is diagonalizable, we obtain from equation (7) e = TL( t)T where Èel 1 t 0 ˘ È1 0 ˘ È4 1˘ T = [ x1, x 2 ] = Í =Í ˙ and L( t) = Í t ˙ . Thus, l 2t ˙ Î5 1˚ Î 0 e ˚ Î0 e ˚ È4 1˘È1 0 ˘È-1 1 ˘ È-4 + 5e t 4 - 4 e t ˘ . F( t) = e tA = Í ˙Í ˙= Í t ˙Í t t˙ Î5 1˚Î0 e ˚Î 5 -4 ˚ Î-5 + 5e 5 - 4 e ˚ 1 (b). The solution of y ¢ = Ay, y (-1) = y 0 is given by y ( t) = e( t +1)A y 0 . Therefore, È1 ˘ È-4 + 5e 3 4 - 4 e 3 ˘È1 ˘ È-4 + 5e 3 ˘ . y (2) = e( 2 +1)A y 0 = e 3 A Í ˙ = Í =Í 3 3 ˙Í ˙ 3˙ Î0 ˚ Î-5 + 5e 5 - 4 e ˚Î0 ˚ Î-5 + 5e ˚

2 (a). The characteristic polynomial is p(l ) = (l - 2) 2 . Eigenvalues are l1 = l 2 = 2 with È1 ˘ corresponding eigenvector x1 = Í ˙ . Therefore, Î0 ˚ 2t Èe ˘ È1 ˘ È0 ˘ Èt ˘ y1 ( t) = Í ˙. Let y 2 ( t) = e 2 t ( tx + 7), x = Í ˙, ( A - 2I)h = x fi h = Í ˙ fi y 2 ( t) = e 2 t Í ˙ . Î0 ˚ Î1 ˚ Î1˚ Î0˚ Èe 2 t Y ( t) = Í Î0

2 (b).

te 2 t ˘ ˙ = F( t) since Y (0) = I. e 2t ˚

Èe 2 y (2) = F(1) y (1) = Í Î0

e 2 ˘È1 ˘ È3e 2 ˘ ˙Í ˙ = Í ˙ . e 2 ˚Î2 ˚ Î2e 2 ˚

È6 5 ˘ 3 (a). We proceed as in Example 2. For A = Í ˙ , the characteristic polynomial is Î1 2 ˚ p(l ) = l2 - 8l + 7 . Eigenvalues are l1 = 1 and l 2 = 7 with corresponding eigenvectors È1˘ È5 ˘ x1 = Í ˙ and x 2 = Í ˙ . Since A is diagonalizable, we obtain from equation (7) e tA = TL( t)T -1 Î-1˚ Î1 ˚ Èel 1 t 0 ˘ Èe t 0 ˘ È 1 5˘ where T = [ x1, x 2 ] = Í . Thus, =Í ˙ and L( t) = Í l 2t ˙ 7t ˙ Î-1 1 ˚ Î 0 e ˚ Î0 e ˚ Èe t + 5e 7 t -5e t + 5e 7 t ˘ È 1 5 ˘Èe t 0 ˘È1 / 6 -5 / 6 ˘ F( t) = e = Í ˙. ˙Í ˙ = (1 / 6) Í t 7 t ˙Í 7t 5e t + e 7 t ˚ Î-1 1 ˚Î 0 e ˚Î1 / 6 1 / 6 ˚ Î-e + e 3 (b). The solution of y ¢ = Ay, y (0) = y 0 is given by y ( t) = e tA y 0 . Therefore, Èe -1 + 5e -7 -5e -1 + 5e -7 ˘È1˘ È-4 e -1 + 10e -7 ˘ È1˘ . y(-1) = e( -1)A Í ˙ = (1 / 6) Í -1 ( 1 / 6 ) = ˙Í ˙ Í -7 -1 -7 ˙ 5e -1 + e -7 ˚Î1˚ Î1˚ Î-e + e Î 4 e + 2e ˚ tA

174 • Chapter 6 First Order Linear Systems

4 (a). The characteristic polynomial is p(l ) = (1 - l )(2 - l )(-1 - l ) . Eigenvalues are È1 ˘ È1 ˘ È1˘ Í ˙ Í ˙ Í ˙ l1 = -1, l 2 = 1, l 3 = 2 with corresponding eigenvectors x1 = Í 1 ˙, x 2 = Í0 ˙, x 3 = Í1 ˙ . ÍÎ0 ˙˚ ÍÎ0 ˙˚ ÍÎ-3˙˚ È e-t et Í Therefore, Y ( t) = Í e - t 0 t ÍÎ-3e 0 È e-t e t e 2 t ˘È0 Í ˙Í and F( t) = Í e - t 0 e 2 t ˙Í1 ÍÎ-3e - t 0 0 ˙˚ÍÎ0

4 (b).

5 (a).

5 (b). 6.

9 (a).

9 (b). 9 (c). 10.

11. 12.

È0 0 - 13 ˘ e 2t ˘ È 1 1 1˘ ˙ ˙ Í ˙ Í e 2 t ˙, Y (0) = Í 1 0 1 ˙ fi Y -1 (0) = Í1 -1 0 ˙ 1 ˙ ÍÎ0 1 ÍÎ-3 0 0 ˙˚ 0 ˙˚ 3 ˚ 2t 2t t t -t 1 1 + e )˘ 0 - 3 ˘ Èe -e + e 3 (- e ˙ ˙ Í 2t -t 1 + e 2t )˙ -1 0 ˙ = Í 0 e 3 (- e 1 ˙ ÍÎ 0 ˙˚ 1 0 e-t 3 ˚

È1 ˘ Èe 2 ˘ Í ˙ Í ˙ y (1) = F(1) y (0) = F(1) Í1 ˙ = Íe 2 ˙ . ÍÎ0 ˙˚ ÍÎ 0 ˙˚ Èt t 2 ˘È 2 s-1 -1˘ From Theorem 6.15, F( t, s) = Y ( t)Y ( s) = Í ˙Í -2 -1 ˙ , and thus s ˚ Î1 2 t ˚Î- s È2 s-1t - s-2 t 2 - t + t 2 s-1 ˘ ; F( t, s) is not a function of t - s. F( t, s) = Í -1 -2 -1 ˙ Î 2 s - 2 s t -1 + 2 ts ˚ È6 - 9 -3 + 9 ˘È 1 ˘ È-9 ˘ From Theorem 6.15, y ( 3) = F( 3,1) y (1) = Í ˙Í ˙ = Í ˙ . Î2 - 6 -1 + 6 ˚Î-1˚ Î-9 ˚ B = T -1 p( A)T = T -1 (2 A 3 - A + 3I )T = 2T -1 A 3T - T -1 AT + 3T -1T = 2 D3 - D + 3I . È2l3 - l1 + 3 ˘ 0 Therefore, B = Í 1 ˙. 3 0 2l 2 - l 2 + 3˚ Î As we saw in equation (6), if T -1 AT = D then A n = TDn T -1. (For this present case, n n È 3 2˘ È1 0 ˘ È1 0 ˘ -1 È1 0 ˘ n T=Í ˙ and D = Í0 -1˙ .) Since A = T Í0 -1˙ T and since Í0 -1˙ = I when n is Î4 3˚ Î ˚ Î Î ˚ ˚ n even, it follows that A = I . n È1 0 ˘ -1 n n -1 -1 A = TD T = T Í ˙ T = TDT = A when n is odd. 0 1 Î ˚ As in parts (a) and (b), we see that A - n = I when n is even and A - n = A when n is odd. È1 0 ˘ -1 ȱ1 0 ˘ A = TÍ T D = . The four matrices are: Í 0 ±i ˙ ˙ Î0 -1˚ Î ˚ È1 0 ˘ È-1 0 ˘ È1 0 ˘ È-1 0 ˘ D1 = Í ˙, D2 = Í 0 i ˙, D3 = Í0 -i ˙, D4 = Í 0 -i ˙ . Î0 i ˚ Î ˚ Î ˚ Î ˚ -1 1/ 2 2 For the given matrix, A = A . Thus, if B = A , then B = A = A -1 as requested. Exercise 10 asks for four different square roots of A and any one of these will serve as B. 1 1 È1 0 ˘ ȱ1 0 ˘ A 2 + A 2 = TBT -1 fi B = T -1 A 2T + T -1 A 2 T . Since A 2 = I , B = I + D = Í ˙+Í ˙. Î0 1 ˚ Î 0 ±i ˚ -1

Chapter 6 First Order Linear Systems • 175 3

13. 14.

15.

16.

17.

È1 1˘È-2 0 ˘ È-1 1 ˘ È24 -16 ˘ Since A = TDT , it follows that A = TD T = Í ˙Í ˙ Í ˙= Í ˙. Î2 1˚Î 0 2 ˚ Î 2 -1˚ Î32 -24 ˚ 0 Ècos(pl1 ) ˘ -1 È2 1 ˘ -1 È 3 -1˘ f1 ( A) = cos(pA) = T Í T . T=Í ˙ ˙, T = Í-5 2 ˙ . cos(pl 2 ) ˚ Î 0 Î5 3˚ Î ˚ 1 Êp ˆ Êp ˆ cos(pl1 ) = cosÁ ˜ = , cos(pl 2 ) = cosÁ ˜ = 0 . Therefore, Ë 4¯ Ë 2¯ 2 1 È2 1 ˘È 2 0 ˘È 3 -1˘ È 3 2 - 2 ˘ f1 ( A) = cos(pA) = Í ˙. ˙Í ˙Í ˙ = Í15 5 Î5 3˚Î 0 0 ˚Î-5 2 ˚ Î 2 2 - 2 2 ˚ 1 Êp ˆ Êp ˆ sin(pl1 ) = sinÁ ˜ = , sin(pl 2 ) = sinÁ ˜ = 1. Therefore, Ë 4¯ Ë 2¯ 2 1 È2 1 ˘È 2 0 ˘È 3 -1˘ È 3 2 - 5 - 2 +2 ˘ f 2 ( A) = sin(pA) = Í = Í15 ˙. Í ˙ ˙ Í ˙ 5 Î5 3˚Î 0 1 ˚Î-5 2 ˚ Î 2 2 - 15 - 2 2 + 6 ˚ 0 ˘ -1 Ècos l1t T when A is a (2 ¥ 2) As we saw following Theorem 6.16, cos( tA) = T Í cos l 2 t ˙˚ Î 0 diagonalizable matrix with eigenvalues l1 and l 2 . Thus, with t = p and the given eigenvalues, 0 Ècos(p / 3) ˘ -1 È1 / 2 0 ˘ -1 cos(pA) = T Í T T = Í 0 1 / 2 ˙T cos( 7p / 3) ˙˚ 0 Î Î ˚ we have È1 0 ˘ -1 = (1 / 2)T Í ˙T = (1 / 2) I . Î0 1 ˚ -1

3

3

-1

Similarly, we find sin(pA) = ( 3 / 2) I . È1 1˘ Let T = Í ˙ . Make the substitution y = Tz . Premultiplying by Î-2 -1˚ 1 È-1˘ È-1 -1˘ È-1 0 ˘ -1 È ˘ T -1 = Í D T D gives z z , + = = = ¢¢ ˙ Í0 ˙ Í 2 ˙ Í 0 1 ˙ . The solution is Î2 1 ˚ Î ˚ Î ˚ Î ˚ È c1e - t + c 2e t + 1 ˘ z( t) = Í ˙ . Converting to the original variables, we obtain Îk1 cos t + k2 sin t + 2 ˚ È 1 1 ˘È c1e - t + c 2e t + 1 ˘ È c1e - t + c 2e t + k1 cos t + k2 sin t + 3 ˘ y ( t) = Tz( t) = Í ˙. ˙= Í ˙Í -t t Î-2 -1˚Îk1 cos t + k2 sin t + 2 ˚ Î-2c1e - 2c 2e - k1 cos t - k2 sin t - 4 ˚ È1 1˘ È1˘ Let T = Í . Making the substitution y = Tz , the system becomes ATz¢ + Tz = Í ˙ . ˙ Î-2 -1˚ Î1˚ È-1 0 ˘ È-1 -1˘ È-2 ˘ È-2 ˘ -1 T AT gives z z + = + = z z Premultiplying by T -1 = Í or ¢ ¢ Í 0 1˙ Í 3 ˙ . Thus, the ˙ Í3˙ Î Î2 1 ˚ ˚ Î ˚ Î ˚ t + 2 z z ¢ È ˘ È 1 1˘ È ˘ ce -2 = Í ˙ . The solution is z( t) = Í 1 - t system uncouples into Í ˙ . Converting to the ˙ Î z2¢ + z2 ˚ Î 3 ˚ Îc 2e + 3˚ È 1 1 ˘È c1e t - 2 ˘ È c1e t + c 2e - t + 1 ˘ original variables, we obtain y = Tz = Í ˙= Í ˙. ˙Í - t t -t Î-2 -1˚Îc 2e + 3˚ Î-2c1e - c 2e + 1˚

176 • Chapter 6 First Order Linear Systems

18.

Make the substitution y = Tz . z¢¢ + z¢ + Dz = 0.. The solution is (- 1 - 5 )t (- 1 + 5 )t È ˘ c1e 2 2 + c 2e 2 2 ˙ . Converting to the original variables, we obtain z( t) = Í - t t Ík1e 2 cos 23 t + k2e - 2 sin 23 t ˙ Î ˚ 5 5 1 1 È c e( - 2 - 2 ) t + c e( - 2 + 2 ) t + k e - 2t cos 3 t + k e - 2t sin 3 t ˘ 1 2 1 2 2 2 ˙. y ( t) = Tz( t) = Í 5 5 1 1 t t t t + Í-2c e( 2 2 ) - 2c e( 2 2 ) - k e - 2 cos 3 t - k e - 2 sin 3 t ˙ 2 1 2 2 2 ˚ Î 1

( )

( )

( ) ( )

( ) ( )

È1 1˘ Let T = Í ˙ . Making the substitution y = Tz , the system becomes Tz¢¢ + 2 ATz¢ = 0 . Î-2 -1˚ È-1 0 ˘ È-1 -1˘ È0 ˘ gives z¢¢ + 2T -1 ATz¢ = 0 or z¢¢ + 2 Í z¢ = Í ˙ . Thus, the Premultiplying by T -1 = Í ˙ ˙ Î 0 1˚ Î2 1 ˚ Î0 ˚ È c1 + c 2e 2 t ˘ È z1¢¢- 2 z1¢ ˘ È0 ˘ = system uncouples into Í . The solution is . Converting to the z( t ) = Í ˙ Í ˙ -2 t ˙ Îz2¢¢+ 2 z2¢ ˚ Î0 ˚ Îd1 + d2e ˚ original variables, we obtain È 1 1 ˘È c1 + c 2e 2 t ˘ È c1 + c 2e 2 t + d1 + d2e -2 t ˘ . y ( t) = Tz( t) = Í =Í ˙Í -2 t ˙ -2 t ˙ 2t Î-2 -1˚Îd1 + d2e ˚ Î-2(c1 + c 2e ) - ( d1 + d2e ) ˚ 20 (a). m1 x1¢¢= k1 ( x 2 - x1 ); m2 x 2¢¢ = k2 ( x 3 - x 2 ) - k1 ( x 2 - x1 ); m3 x 3¢¢ = - k2 ( x 3 - x 2 ) . Therefore, m1 x1¢¢+ k1 ( x1 - x 2 ) = 0; m2 x 2¢¢- k1 x1 + ( k1 + k2 ) x 2 - k2 x 3 = 0; m3 x 3¢¢- k2 x 2 + k2 x 3 = 0 . The result follows. È1˘ È1˘ Í˙ Í˙ 20 (b). Kv 0 = 0 , where v 0 is any nonzero multiple of Í1˙ . Therefore, 0, Í1˙ is an eigenpair. ÍÎ1˙˚ ÍÎ1˙˚ 20 (c). Let x = f ( t) v 0 . Mx ¢¢ + Kx = M ( f ¢¢ ( t) v 0 ) + Kf ( t) v 0 = 0 . Therefore, since K ( f ( t) v 0 ) = f ( t)Kv 0 = 0, Mf ¢¢ ( t) v 0 = 0 or m j f ¢¢ ( t) = 0, j = 1, 2, 3 fi f ¢¢ ( t) = 0. Therefore, 19.

f ( t) = c1t + c 2 and x ( t) = (c1t + c 2 ) v 0 . x (0) = c 2 v 0 = 0 fi c 2 = 0, x ( t) = c1v 0 = v 0 fi c1 = 1. Therefore, x ( t) = tv 0 . The system is executing motion at constant velocity v 0 . There is no relative motion; the three-mass system is translating like a rigid body. È 1 -1 0 ˘ kÍ ˙ -1 21 (a). For this case, we have A = M K = Í-1 2 -1˙ . Using MATLAB, we find the m ÍÎ 0 -1 1 ˙˚ È 1 -1 0 ˘ ˙ Í eigenvalues of B = Í-1 2 -1˙ are g 1 = 0, g 2 = 1, and g 3 = 3 with corresponding ÍÎ 0 -1 1 ˙˚ È1˘ È1˘ È1˘ Í ˙ Í ˙ Í˙ eigenvectors u1 = Í1˙, u2 = Í 0 ˙, and u3 = Í-2 ˙ . Since A = ( k / m) B , the eigenvalues of A are ÍÎ 1 ˙˚ ÍÎ-1˙˚ ÍÎ1˙˚ multiples of k/m times the eigenvalues of B while corresponding eigenvectors can be chosen to be the same as those of B.

Chapter 6 First Order Linear Systems • 177

21 (b). Making the substitution x = Tz , the system becomes Tz¢¢ + ATz = 0 . Premultiplying by È0 ˘ 0 0 ˘ È0 Í ˙ ˙ Í -1 -1 0 ˙z = Í0 ˙ . Thus, the system uncouples into gives z¢¢ + T ATz = 0 or z¢¢ + Í0 km ÍÎ0 ˙˚ ÍÎ0 0 3km -1 ˙˚

˘ ˘ È0 ˘ È c1t + c 2 z1¢¢ È ˙ ˙ Í ˙ Í Í -1 -1 Í z2¢¢+ km z2 ˙ = Í0 ˙ . The solution is z( t) = Í d1 coswt + d2 sin wt ˙ , where w = km . ÍÎe1 cos 3wt + e2 sin 3wt ˙˚ ÍÎz3¢¢+ 3km -1z3 ˙˚ ÍÎ0 ˙˚ Converting to the original variables, we obtain ˘ È1 1 1 ˘È c1t + c 2 ˙ ˙Í Í x( t) = Í1 0 -2 ˙Í d1 coswt + d2 sin wt ˙ ÍÎ1 -1 1 ˙˚ÍÎe1 cos 3wt + e2 sin 3wt ˙˚ È c1t + c 2 + d1 coswt + d2 sin wt + e1 cos 3wt + e2 sin 3wt ˘ ˙ Í =Í c1t + c 2 - 2(e1 cos 3wt + e2 sin 3wt) ˙. Íc t + c - ( d coswt + d sin wt) + e cos 3wt + e sin 3wt ˙ 2 1 2 1 2 ˚ Î1

Chapter 7 Laplace Transforms Section 7.1 1. 2.

-e - st L {1} = lim Ú 1 ◊ e dt = lim T Æ• 0 T Æ• s T

3t

L {e } = lim

3.

Ú

T

T Æ• 0

=

1 1 1 - e - sT ) = , s > 0 ( T Æ• s s

= lim 0

T

e 3 t ◊ e - st dt = lim Ú e -( s- 3) t dt = lim T Æ• 0

T Æ•

1 -( s - 3 )t T 1 = lim e (1 - e -( s- 3)T ) 0 T Æ• s - 3 s- 3

1 , s > 3. s- 3 T

T

T Æ• 0

T Æ• 0

L {te - t } = lim Ú te - t ◊ e - st dt = lim Ú te -( s+1) t dt . For integration by parts, we will use u = t, du = dt, dv = e T

lim Ú te

-( s +1) t

T Æ• 0

=

4.

T

- st

-( s +1) t

e -( s+1) t . Then we have dt, and v = s +1

ÏÔ - te -( s+1) t dt = lim Ì T Æ• ÔÓ s + 1

T

+ 0

Ï -Te -( s+1)T ¸ 1 T -( s+1) t ¸Ô 1 -( s +1)T + e e dt = lim 1 ( ) ˝ Ì ˝ s + 1 Ú0 (s + 1) 2 Ô˛ T Æ• Ó s + 1 ˛

1 , s > -1. (s + 1) 2

L {t - 5} =

T

lim Ú ( t - 5) ◊ e - st dt . For integration by parts, we will use

T Æ• 0

u = t - 5, du = dt, dv = e - st dt, and v = -

lim Ú

T

T Æ• 0

=

ÏÔ -( t - 5)e - st ( t - 5)e dt = lim Ì T Æ• s ÔÓ

T

- st

0

e - st . Then we have s

ÏÔ -(T - 5)e - sT + 5 e - st 1 T - st ¸Ô - 2 + Ú e dt˝ = lim Ì s s s 0 Ô˛ T Æ• ÔÓ

1 5 - , s > 0. s2 s

5.

L { f ( t)} does not exist because lim te t t e - st = • for all s > 0 .

6.

L { f ( t)} does not exist because lim e( t -1) e - st = • for all s > 0 .

t Æ•

2

t Æ•

T

¸Ô ˝ 0Ô ˛

Chapter 7 Laplace Transforms • 179

7.

•

1

L { t - 1 } = Ú (1 - t)e - st dt + Ú ( t - 1)e - st dt . 0 4 1 4244 3 (1)

1 4 1 4244 3 ( 2)

(1): u = 1 - t, du = - dt, dv = e - st dt, v = -

e - st . Then we have s

1

1

- st 1e e - st 1 1 1 1 dt = + 2 e - st = + 2 (e - s - 1) (1) = -(1 - t) -Ú s 0 0 s s s s s 0

(2): u = t - 1, du = dt, dv = e - st dt, v = -

e - st . Then we have s

•

e-s 1 • t - 1 - st e (2) = + Ú e - st dt = 2 s s s 1 1

1 1 2 (1) + (2) = - 2 + 2 e - s, s > 0 . s s s 8.

L {( t - 2)

2

T

} = lim Ú ( t - 2) 2 ◊ e - st dt . Using T Æ• 0

u = ( t - 2) 2 , du = 2( t - 2) dt, dv = e - st dt, and v = -

lim Ú

T

T Æ• 0

ÏÔ -( t - 2) 2 e - st ( t - 2) ◊ e dt = lim Ì T Æ• s ÔÓ

T

- st

2

+ 0

e - st , we have s

¸ 4 2 T 2 T - st Ô = + t e dt ( 2 ) lim ( t - 2)e - st dt. Using ˝ Ú Ú 0 0 Æ• T s Ô˛ s s

e - st , we have parts with u = t - 2, du = dt, dv = e dt, and v = s - st

lim Ú

T

T Æ• 0

T ÏÔÊ -( t - 2)e - st 1 ¸ 4 4 2 4 2 - st ˆ Ô ( t - 2) ◊ e dt = + lim ÌÁ + 2 e ˜ ˝ = - 2 + 3 , s > 0. s s T Æ• ÔË s s s ¯ 0Ô s s Ó ˛ - st

2

T

9.

L{ f ( t)} = lim Ú

10.

L{ f ( t)} =

T

T Æ• 1

lim Ú

T

T Æ• 1

=

e-s -e - st 1 1 ◊ e dt = lim = lim (e - s - e - sT ) = , s > 0. T Æ• s s 1 T Æ• s - st

¸Ô ÏÔ -( t - 1)e - st T 1 T Ï (T - 1)e - sT e - s - e - sT ¸ + ( t - 1) ◊ e dt = lim Ì + Ú e - st dt˝ = lim Ì˝ T Æ• s s2 s s 1 Ô˛ T Æ• Ó ÔÓ ˛ 1 - st

e-s , s > 0. s2 2

11.

-e - st 1 L{ f ( t)} = Ú 1 ◊ e dt = = (e - s - e -2 s ), s π 0; = 1, s = 0 . 1 s 1 s 2

- st

180 • Chapter 7 Laplace Transforms 2

12.

13.

L{ f ( t)} =

Ú

2

1

e -2 s 1 - s -2 s -( t - 1)e - st 1 2 ( t - 1) ◊ e dt = + Ú e - st dt = + 2 (e - e ), s π 0. s s s s 1 1 - st

Ú

t n e - st dt . u = t n , du = nt n -1dt, dv = e - st dt, and v = -

-

t n e - st n n -1 - st + Ú t e dt, s > 0 . s s

e - st . Then we have s

14 (a). Since lim+ t n = 0 and lim+ e - st = 1, lim+ t n e - st = 0 . tÆ0

tÆ0

tÆ0

14 (b). Using L’Hopital’s rule: tn nt n -1 n! = lim st st = ... = lim n st = 0. t Æ• e t Æ• se t Æ• s e

lim t n e - st = lim t Æ•

15 (a). Using 13 and 14,

ÏÔ - t n e - st L {t } = lim Ì T Æ• ÔÓ s

T

n

+ 0

T n n n T n -1 - st ¸Ô t e dt˝ = 0 + lim Ú t n -1e - st dt = L {t n -1}, s > 0 . Ú s s T Æ• 0 s 0 Ô˛

15 (b).

2 2 3 3! 4 4! 3 L {t 2 } = 4 , L {t 4 } = L {t 3 } = 5 , 3 , L {t } = s s s s s s ! 5 5 L {t 5 } = L {t 4 } = 6 , s > 0 . s s

L {t 2 } = L {t} =

15 (c). L {t m } =

m! , s > 0. sm +1

16.

ÏÔ - st Ê - s coswt - w sin wt ˆ T ¸Ô s L { f ( t)} = lim Ìe Á , s > 0. ˜ ˝= 2 2 2 T Æ• Ô Ë ¯ 0 Ô s + w2 s +w Ó ˛

17.

ÏÔ - st Ê - s sin wt - w coswt ˆ T ¸Ô w L { f ( t)} = lim Ìe Á , s > 0. ˜ ˝= 2 2 2 T Æ• Ô Ë ¯ 0 Ô s + w2 s + w Ó ˛

18.

f ( t) = cos(w ( t - 2)) = cos(wt)cos(2w ) + sin(wt)sin(2w ) . Using 16 and 17, L { f ( t)} =

19.

f ( t) = sin(w ( t - 2)) = sin wt cos 2w - coswt sin 2w . Then we have

L { f ( t)} = 20.

s cos(2w ) + w sin(2w ) , s > 0. s2 + w 2

w cos 2w - s sin 2w , s > 0. s2 + w 2

L { f ( t)} = lim

Ú

T

T Æ• 0

e

-( s - 3 )t

T ÏÔ ¸ 1 -( s - 3 )t Ê - ( s - 3)sin t - cos t ˆ Ô = , s > 3. sin t dt = lim Ìe Á ˜ ˝ 2 T Æ• Ë ¯ 0 Ô ( s - 3) 2 + 1 ( s - 3) + 1 ÔÓ ˛

Chapter 7 Laplace Transforms • 181

21.

T

L { f ( t)} = lim Ú e

-( s + 2) t

T Æ• 0

=

T ¸ Ï Ô -( s+ 2) t Ê -( s + 2) cos 4 t + 4 sin 4 t ˆ Ô cos 4 tdt = lim Ìe Á ˜ ˝ T Æ• (s + 2) 2 + 16 Ë ¯ 0Ô ÔÓ ˛

s+2 , s > -2 . (s + 2) 2 + 16 -5 t

2 6 2 6 , s > -5 . L {6 t} = 2 , s > 0 . Then L {r( t)} = + 2 , s > 0. s+5 s s+5 s

}=

22.

L {2e

23.

L {5e -7 t } = L {r( t)} =

1 2 5 , s > 2 . Then , s > -7 . L {t} = 2 , s > 0 , and L {2e 2 t } = s s- 2 s+7

5 1 2 + 2+ , s > 2. s+7 s s- 2

24 (a). The function is discontinuous on 0 £ t < • because the one-sided limits do not exist at the vertical asymptotes. 24 (b). The function is not exponentially bounded on 0 £ t < • . 25 (a). The function is continuous on 0 £ t < • . 25 (b). The function is exponentially bounded on 0 £ t < • . f ( t) £ e t , so we can take M = 1, a = 1. 26 (a). The function is continuous on 0 £ t < • . 26 (b). The function is exponentially bounded on 0 £ t < • .

If f ( t) = t 2e - t , then f ¢ ( t) = (2 t - t 2 )e - t = 0 fi t = 2 , is a maximum point, so we can take M = f (2) = 4 e -2 , a = 0 . 27 (a). The function is continuous on 0 £ t < • . e 2t + 1 2t 27 (b). The function is exponentially bounded on 0 £ t < • , since cosh2 t £ £ e on 0 £ t < • , 2 so we can take M = 1, a = 2 . 28 (a). The function is piecewise continuous on 0 £ t < • . 28 (b). Consider g( t) = te - t , g¢ ( t) = (1 - t)e - t = 0 fi t = 1. \ t = 1 is a maximum and

g( t) £ e -1 fi t £ e -1e t , 0 £ t < •. Since [[ t]] £ t, 0 £ t < • , the function is exponentially bounded on 0 £ t < • , taking M = e -1, a = 1. 29 (a). The function is piecewise continuous on 0 £ t < • . 29 (b). f ( t) £ e 2 t , and so the function is exponentially bounded on 0 £ t < • , taking M = 1, a = 2 . 30 (a). The function is continuous on 0 £ t < • .

182 • Chapter 7 Laplace Transforms

30 (b). The function is not exponentially bounded on 0 £ t < • because 2

t2 et 1 t 2 - 2t t 2 - 2t 2 e e e and f ( t) ≥ 2 t = > , t > 4. e + e 2t 2

31 (a). The function is discontinuous and not piecewise continuous on 0 £ t < • . 31 (b). The function is not exponentially bounded on 0 £ t < • . 32. 33.

lim Ú

T

T Æ• 0

Ú

T

0

1 p -1 T -1 , so the improper integral converges. 2 dt = lim tan t 0 = lim (tan T ) = T Æ• T Æ• 1+ t 2

(

)

T

1 t 1 1 ln(1 + t 2 ) = ln(1 + T 2 ) . Since lim ln(1 + T 2 ) = • , the improper integral 2 dt = T Æ• 2 2 1+ t 2 0

diverges. 34.

T

lim Ú e - t cos(-e - t ) dt = lim Ú

T Æ• 0

e-T

T Æ• 1

1

1

- cos( u) du = Ú cos u du = sin u 0 = sin(1) , so the improper 0

integral converges. 35.

Ú

•

0

T T2 2 2 2 1 T2 1 1 Ê1 ˆ te - t dt = lim Ú te - t dt = lim Ú e - u Á du˜ = lim Ú e - u du = lim 1 - e -T = , so the 0 0 0 T Æ• T Æ• T Æ• 2 Ë 2 ¯ T Æ• 2 2

(

integral converges to

1 . 2

36.

Ï 1 ¸ 2t f ( t) = 3L -1 Ì ˝ = 3e , t ≥ 0 . Ós - 2˛

37.

Ï1¸ Ï 1 ¸ -t f ( t) = -2 L -1 Ì 2 ˝ + L -1 Ì ˝ = -2 t + e , t ≥ 0 . Ós ˛ Ó s + 1˛

38.

Ï 1 ¸ -1 Ï 1 ¸ -2 t 2t f ( t) = 2 L -1 Ì ˝ = 2e + 2e = 4 cosh(2 t), t ≥ 0 . ˝ + 2L Ì Ós - 2˛ Ós + 2˛

39.

Ï 1 ¸ -1 Ï 1 ¸ t -t f ( t) = L -1 Ì ˝ = e - e = 2 sinh t, t ≥ 0 . ˝- L Ì Ó s + 1˛ Ó s - 1˛

Section 7.2 Ê 2ˆ Ës ¯

1.

L { f ( t)} = 3Á 3 ˜ +

2.

L { f ( t)} =

3.

L { f ( t)} = +

4.

L { f ( t)} = e L {e

2 1 6 2 1 + = + + , s>0 s2 s s3 s2 s

2 5 + , s>1 s-1 s 1 s

-s

3 , s>0 s +9 2

3t

}=

e-s , s> 3 s- 3

)

Chapter 7 Laplace Transforms • 183

2e - s , s>0 s3

5.

L{ f ( t)} = e - sL {t 2 } =

6.

L {sin

7.

L { f ( t)} = L {2 t} sÆ s+ 2 =

8.

3 Ï1 ¸ L {sin 3t cos 3t} = L Ì sin 6 t ˝ = 2 , s>0 Ó2 ˛ s + 36

2

1 1 Ó2 2

1 1 1 1 Ê1 s s ˆ ◊ - ◊ 2 Á - 2 2 = 2 ˜, s > 0 2 s 2 s + 4w 2 Ë s s + 4w ¯

w t} = L ÏÌ - cos(2w t ) ¸˝ = ˛

2 , s > -2 (s + 2) 2

L { f ( t)} = L {2( t - 2) h ( t - 2) + 4 h ( t - 2)} 9.

È 2 4˘ = e -2 s[ L {2 t} + L {4}] = e -2 s Í 2 + ˙, s > 0 s˚ Îs 2t

cos 3t} = L {cos 3t } sÆ s- 2 =

s- 2 , s>2 ( s - 2) 2 + 9

10.

L {e

11.

L { f ( t)} = e L {e

12.

L { f ( t)} = L t 2 + 3t + 5

13.

L -1{F ( s)} = 3 +

14.

L

15.

L

16.

L

17.

L -1{F ( s)} = sin( 3( t - 2)) h ( t - 2), t ≥ 0

18.

L

19.

L -1{F ( s)} = L -1 Á

20.

L {F ( s)} = Í2 cos( 4 ( t - 3)) +

21.

L -1{F ( s)} =

3

3( t -1)

{

h ( t - 1)} = e 3e - sL {e 3 t } =

}

sÆ s - 4

=

e 3- s , s> 3 s- 3

2 3 5 , s> 4 3 + 2 + (s - 4) (s - 4) (s - 4)

24 3 t = 3 + 4 t 3, t ≥ 0 3!

-1

{F ( s)} = 2 sin 5 t + 4 e 3 t , t ≥ 0

-1

{F ( s)} = 2e 2 t cos 3t, t ≥ 0 t 3 5 3t 3 = e t , t≥0 3! 6

-1

{F ( s)} = 5e 3 t

-1

{F ( s)} = e 9( t - 2 )h ( t - 2), t ≥ 0 Ê 4 ( s - 1) - 2 ˆ 2 ˆ tÊ 2 ˜ = e ÁË 4 cos 3t - 3 sin 3t˜¯ , t ≥ 0 Ë ( s - 1) + 9 ¯

È Î

-1

4

7 ˘ sin( 4 ( t - 3)) ˙h ( t - 3), t ≥ 0 4 ˚

48 (t - 3) 4 h(t - 3) + 2(t - 5) 4 h( t - 5) 4!

(

)

4

= 2( t - 3) h ( t - 3) + 4 ( t - 5) h ( t - 5), t ≥ 0

184 • Chapter 7 Laplace Transforms

e - s + e -3 s , s>0 s

22.

L { f ( t)} =

23.

L { f ( t)} = e -2 ps

24.

f ( t) = ( t - 1) h ( t - 1) + h ( t - 1) - ( t - 3) h ( t - 3) - 3h ( t - 3) , and so -s

1 , s>0 s +1 2

3ˆ Ê 1 1ˆ -3 s Ê 1 Á 2 + ˜ , s π 0 (= 4, s = 0) 2 + ˜ -e Ës Ës s¯ s¯

L { f ( t)} = e Á

(1 - e ) , L { f ( t)} = -3 s

25. 26.

s

L { f ( t)} =

s>0

3(e - s - e -4 s ) s

, s π 0 (= 9, s = 0)

Chapter 7 Laplace Transforms • 185

27.

f ( t) = -( t - 1) h ( t - 1) + h ( t - 1) + ( t - 3) h ( t - 3) + h ( t - 3) , and so e - s e - s e -3 s e -3 s L { f ( t)} = - 2 + + 2 + , s π 0 (= 0, s = 0) s s s s

28.

f ( t) = -( t - 1) h ( t - 1) + h ( t - 1) + ( t - 2) h ( t - 2) + ( t - 2) h ( t - 2) - ( t - 3) h ( t - 3) - h ( t - 3) , and so L { f ( t)} = -

e - s e - s 2e -2 s e -3 s e -3 s + + 2 - 2 , s π 0 (= 1, s = 0) s2 s s s s

e - s - 2e -2 s + e -3 s , s π 0 (= 0, s = 0) s

29.

L { f ( t)} =

30.

e - st 1 + e -2 s L { f ( t)} = Ú e dt = = , s π 0 (= 2, s = 0) 0 s 0 s

2

2

- st

Note: f ( t) = 1 - h ( t - 2) for all t except t = 2 .

1

1

-( s - 2) t

e -( s+ 2) t 1 - e -( s+ 2) dt = = , s π -2 (= 1, s = -2) s+2 0 s+2

31.

L { f ( t)} =

Úe

32.

L { f ( t)} =

1 + e - s - e -4 s , s>0 s

0

Note: f ( t) = 1 + [ h ( t - 1) - h ( t - 4 )] for all t except t = 4 .

186 • Chapter 7 Laplace Transforms

2

33. 34.

2

- st

L { f ( t)} = - Ú e dt + Ú 0

•

3

e - st e - st e dt = s 0 s

•

- st

3

-1 + e -2 s + e -3 s = , s>0 s

f ( t) = ( t - 2)[ h ( t - 2) - h ( t - 3)] + [ h ( t - 3) - h ( t - 4 )]

= ( t - 2) h ( t - 2) - [( t - 3) + 1]h ( t - 3) + [ h ( t - 3) - h ( t - 4 )] e -2 s - e -3 s e -4 s = ( t - 2) h ( t - 2) - ( t - 3) h ( t - 3) - h ( t - 4 ) and L { f ( t)} = , s>0 s2 s e - s + e -2 s - 2e -3 s , s π 0; = 3, s = 0 s

35.

f ( t) = h ( t - 1) + h ( t - 2) - 2 h ( t - 3) and L { f ( t)} =

36.

f ( t) = ( t - 1)[ h ( t - 1) - h ( t - 2)] + ( 3 - t)[ h ( t - 2) - h ( t - 3)]

= ( t - 1) h ( t - 1) - [( t - 2) + 1]h ( t - 2) + [-( t - 2) + 1]h ( t - 2) + ( t - 3) h ( t - 3) e - s e -2 s e -2 s e -2 s e -2 s e -3 s e - s - 2e -2 s + e -3 s and L { f ( t)} = 2 - 2 - 2 + + 2 = , s π 0 (= 1, s = 0) s s s s s s s2 37.

f ( t) = (1 - t)[ h ( t) - h ( t - 1)] + (2 - t)[ h ( t - 1) - h ( t - 2)] = 1 - t + h ( t - 1) + ( t - 2) h ( t - 2), t ≥ 0 1 1 e - s e -2 s and L { f ( t)} = - 2 + + 2 , s π 0; = 1, s = 0 s s s s

38.

F ( s) =

A1 A + 2 , and so A1 ( s + 1) + A2 ( s - 3) = 12 and A1 + A2 = 0, A1 - 3 A2 = 12 . Thus s- 3 s +1

A1 = 3, A2 = -3, and F ( s) = 39.

F ( s) =

3 3 , and so f ( t) = -3e - t + 3e 3 t , t ≥ 0 . s- 3 s +1

A1 A + 2 , and so A1 ( s + 2) + A2 s = 4 and A1 + A2 = 0, 2 A1 = 4 . Thus s s+2

A1 = 2, A2 = -2, and F ( s) =

2 2 , and so f ( t) = 2 - 2e -2 t , t ≥ 0 . s s+2

Chapter 7 Laplace Transforms • 187

40.

A ˆ Ê A F ( s) = 24 e -5 s Á 1 + 2 ˜ , and so A1 ( s + 3) + A2 ( s - 3) = 1 and A1 + A2 = 0, 3 A1 - 3 A2 = 1. Ë s - 3 s + 3¯ 1 ˆ 1 1 Ê 1 Thus A1 = , A2 = - , and F ( s) = 4 e -5 s Á ˜ , and so Ë s - 3 s + 3¯ 6 6

[

]

f ( t) = 4 e 3( t - 5 ) - e -3( t - 5 ) h ( t - 5), t ≥ 0 .

Ê 3 ˆ From (6), Table 7.1: L -1{F ( s)} = L -1 Á 2 ˜ = sinh 3t fi f ( t) = 8 sinh( 3( t - 5)) h ( t - 5), t ≥ 0 . Ë s - 9¯

Answers agree since sinh( 3( t - 5)) = 41.

e 3( t - 5 ) - e -3( t - 5 ) . 2

È A A2 ˘ - s F ( s) = Í 1 + ˙10e and so A1 ( s - 3) + A2 ( s - 2) = 1 and A1 + A2 = 0, - 3 A1 - 2 A2 = 1. Î ( s - 2) ( s - 3) ˚ 1 ˘ È -1 , and so Thus A1 = -1, A2 = 1, and F ( s) = 10e - s Í + Î s - 2 s - 3 ˙˚ f ( t) = 10(e 3( t -1) - e 2( t -1) ) h ( t - 1), t ≥ 0 .

42.

g( t) = 12[ h ( t - 1) - h ( t - 3)], and so sY ( s) - y (0) + 4Y = Y=

12 - s -3 s (e - e ) . Therefore, s

1 A A 2 12 = 1 + 2 . Thus + e - s - e -3 s ) , and so ( s( s + 4 ) s s + 4 s + 4 s( s + 4 )

A1 ( s + 4 ) + A2 s = 1 and A1 + A2 = 0, 4 A1 = 1. Solving these simultaneous equations yields A1 =

1 ˘ 2 1 1 È1 . Thus + 3(e - s - e -3 s ) Í , A2 = - , and so Y = s+4 4 4 Î s s + 4 ˙˚

[

]

y ( t) = 2e -4 t + 3[ h ( t - 1) - h ( t - 3)] - 3 e -4 ( t -1)h ( t - 1) - e -4 ( t - 3 )h ( t - 3) , t ≥ 0 .

43.

g( t) = e 3 t h ( t - 4 ) = e12e 3( t - 4 ) h ( t - 4 ) , and so sY - 1 - Y = e12

e -4 s . Therefore, s- 3

1 e12e -4 s 1 A A Y= + = 1 + 2 . Thus , and so s - 1 ( s - 1)( s - 3) (s - 1)(s - 3) s - 1 s - 3

A1 ( s - 3) + A2 ( s - 1) = 1 and A1 + A2 = 0, - 3 A1 - A2 = 1. Solving these simultaneous equations 1 1 1 1 ˘ e12 -4 s È -1 yields A1 = - , A2 = , and so Y = . Thus + + e Í s-1 2 2 2 Î s - 1 s - 3 ˙˚ 1 y ( t) = e t + e12 (-e t - 4 + e 3( t - 4 ) ) h ( t - 4 ), t ≥ 0 . 2

188 • Chapter 7 Laplace Transforms

44.

s2Y - sy (0) - y ¢ (0) - 4Y =

1 1 . Therefore, Y = , and so (s - 3)(s - 2)(s + 2) s- 3

1 A A A = 1 + 2 + 3 . Thus, (s - 3)(s + 2)(s - 2) s - 3 s + 2 s - 2

1 1 1 1 1 1 = , A2 = = , A3 = = - . Therefore, 4 ( s - 2)( s + 2) s= 3 5 ( s - 3)( s - 2) s=-2 20 ( s - 3)( s + 2) s= 2

A1 =

1 1 1 y ( t) = e 3 t + e -2 t - e 2 t , t ≥ 0 . 5 20 4 45.

s2Y - s(0) - 1 - 2( sY - 0) - 8Y = Y ( s2 - 2 s - 8) - 1 = Y=

1 . Therefore, s-1

1 1 s , which means that fi Y ( s2 - 2 s - 8) = +1= s-1 s-1 s-1

s s A A A = 1 + 2 + 3 . Thus , and so (s - 1)(s + 2)(s - 4) s - 1 s + 2 s - 4 (s - 1)(s + 2)(s - 4)

1 1 2 A1 ( s2 - 2 s - 8) + A2 ( s2 - 5 s + 4 ) + A3 ( s2 + s - 2) = s , and so A1 = - , A2 = - , A3 = . 9 9 9 -1 1 È -1 2 ˘ 1 t 1 -2 t 2 4 t , and so + + y ( t ) = e - e + e , t ≥ 0. 9 ÍÎ s - 1 s + 2 s - 4 ˙˚ 9 9 9

Finally, we have Y = 46 (a).

d d F ( s) = Ú0• (e- st ) f ( t) dt = Ú0• - te- st f ( t) dt = - Ú0• e- st ( tf ( t)) dt = - L {tf (t )} . ds ds

1 d 1 dÊ w ˆ Ï t sin w t ¸ 46 (b). L { f ( t)} = L Ì L {sin w t} = Á ˜ ˝= 2w ds 2w ds Ë s2 + w 2 ¯ Ó 2w ˛ -2 s 1 = - (-1)( s2 + w 2 ) (2 s) = 2 2 , s > 0. 2 (s + w 2 )

47.

t l t 1 1 L ÊË Ú ÈÍÚ f (s ) ds ˘˙dlˆ¯ = L ÊË Ú f (s ) ds ˆ¯ = 2 F ( s), s > max{a, 0}

48.

L

0

Î

0

˚

s

s

0

{Ú f (l )dl} = {Ú f (l )dl - Ú t

2

L

t

2

0

0

}

1 1 f (l ) dl = F ( s) - L {3} = [ F ( s) - 3], s > max{a, 0} s s

Ï1, 0 £ t £ 3 Ï1, 0 £ t < 3 49 (a). f ( t) = h ( t) h ( 3 - t) = Ì , and g( t) = h ( t) - h ( t - 3) = Ì , and so the two Ó 0, 3 < t Ó 0, 3 £ t functions are equal for all t π 3 and hence not identical. 3

49 (b). L { f ( t)} = L {g( t)} = Ú e - st dt = 0

are identical.

1 - e -3 s , s π 0; = 3, s = 0, and so the transformed functions s

Chapter 7 Laplace Transforms • 189

Section 7.3 1.

F ( s) =

A B2 B1 + 2 + s - 1 ( s - 2) s- 2

2.

F ( s) =

A3 A2 A1 B2 B1 + 3 + 2 + 2 + (s - 1) (s - 1) s - 1 (s - 2) s - 2

3.

F ( s) =

A2 A1 Bs + C + 2 + s s ( s + 1) 2 + 9

4.

F ( s) =

A Bs + C + 2 s - 2 s + 16

5.

F ( s) =

A2 A1 B2 B1 + 2 + 2 + (s - 3) s - 3 (s + 3) s + 3

6.

F ( s) =

A2 A1 B s + C2 B1s + C1 + 22 + 2 + (s + 4) s + 4 (s + 1) 2 s2 + 1

7.

F ( s) =

Bs + C Ds + E + 2 (s + 4) + 1 (s + 3) 2 + 4

8.

F ( s) =

A2 A1 B2 s + C2 B1s + C1 + + 2 + 2 (s - 2) s - 2 ((s + 4) 2 + 1) (s + 4) 2 + 1

9.

f ( t) = 2e 3 t , t ≥ 0

10.

f ( t) =

11.

5 Ê s ˆ 5Ê 3 ˆ F ( s) = 4 Á 2 ˜+ Á 2 ˜ , and so f ( t) = 4 cos 3t + sin 3t, t ≥ 0 . Ë s + 9¯ 3 Ë s + 9¯ 3

12.

F ( s) =

2s - 3 2s - 3 A B = 1 and B = = 1. Therefore, f ( t) = e t + e 2 t , t ≥ 0 . . A= + s - 2 s=1 s - 1 s= 2 s-1 s- 2

13.

F ( s) =

-2 3s + 7 3s + 7 4 A B = 1 and B = = = 2 . Thus = . A= + s + 3 s=-1 2 s + 1 s=-3 -2 s + 3 s +1

F ( s) =

1 2 and so f ( t) = e -3 t + 2e - t , t ≥ 0 . + s + 3 s +1

14.

1 2 -t t e , t≥0 2

A Bs + C As2 + A + Bs2 + Cs . Then we have A + B = 4, C = 1, A = 1, B = 3 . = F ( s) = + 2 s s +1 s( s2 + 1)

Therefore, F ( s) =

1 3s 1 and so f ( t) = 1 + 3 cos t + sin t, t ≥ 0 . + 2 + 2 s s +1 s +1

190 • Chapter 7 Laplace Transforms

15.

F ( s) =

A Bs + C . Then we have A + B = 3, C = 1, and 4 A = 8 , so A = 2, B = 1, C = 1. Thus + s s2 + 4

2 s +1 1 and so f ( t) = 2 + cos 2 t + sin 2 t, t ≥ 0 . + 2 2 s s +4 B s + C2 B1s + C1 . Then we have B1 = 0, C1 = 1, B2 = 6, C2 = 4 . Therefore, F ( s) = 22 2 + ( s + 4 ) ( s2 + 4 )

F ( s) = 16.

F ( s) =

6s + 4

(s

2

+ 4)

2

+

1 and so s +4 2

1 3 3 Ê1 ˆ 1 Êt ˆ f ( t) = 6Á sin 2 t˜ + 4 Á [sin 2 t - 2 t cos 2 t]˜ + sin 2 t = t sin 2 t + sin 2 t - t cos 2 t, t ≥ 0 . Ë 16 ¯ 2 Ë4 ¯ 2 4 2

s 1 1 (s - 1) + 1 1 2 t t t e , t ≥ 0. 3 = 3 = 2 + 3 , so f ( t) = te + 2 (s - 1) (s - 1) (s - 1) (s - 1)

17.

F ( s) =

18.

3 3 Ê 3 ˆ . sY - 3 + 2Y = 26Á 2 + 26 ˜ , and thus Y = Ë s + 9¯ s+2 (s + 2)(s2 + 9) s 1 A Bs + C 1 = . , and so A = 2 = + 2 2 s + 9 s=-2 13 (s + 2)(s + 9) s + 2 s + 9 1 - 113 ( s - 2) Bs + C - 113 ( s2 - 4 ) 1 1 - 113 ( s2 + 9) 13 . Then, = 2 = = = 2 2 2 ( s + 2)( s + 9) s + 2 ( s + 2)( s + 9) ( s + 2)( s + 9) ( s2 + 9) s +9

B=Y=

1 2 , C = , and so 13 13

3 1 s 2 1 ˆ 9 Ê 113 Ê s ˆ Ê 3 ˆ + 26 ◊ 3Á - ◊ 2 + ◊ 2 - 6Á 2 ˜= ˜ + 4Á 2 ˜ . Finally, we Ë s + 2 13 s + 9 13 s + 9 ¯ s + 2 Ë s + 9 ¯ Ë s + 9¯ s+2

have y ( t) = 9e -2 t - 6 cos 3t + 4 sin 3t . 19.

1 s Ê s ˆ . sY - 1 - 3Y = 13Á 2 + 13 ˜ , and thus Y = Ë s + 4¯ s- 3 (s - 3)(s2 + 4) s 3 s A Bs + C = . Setting s = 0 gives us , and so A = 2 = + 2 2 s + 4 s= 3 13 (s - 3)(s + 4) s - 3 s + 4

0=-

1 C 4 1 1Ê 3ˆ =- Á ˜ + + , C = . Setting s = 1 gives us -2 ◊ 5 2 Ë 13¯ 13 4 13

B+ 5

4 13 . Solving for B yields

-3 4ˆ Ê 3 s + 4 1 3 Ê s ˆ Ê 2 ˆ Á ˜ + 13Á 13 + 13 2 13 ˜ = - 3Á 2 B = - , and so Y = ˜ + 2Á 2 ˜ . Finally, we Ë ¯ Ë s + 4¯ s- 3 s + 4 ˜ s- 3 s +4 13 Ás- 3 Ë ¯ have y ( t) = 4 e 3 t - 3 cos 2 t + 2 sin 2 t .

Chapter 7 Laplace Transforms • 191

20.

4 3 4 3s2 + 4 A B C , and thus , and so = 2+ + Y = + = 2 2 2 s + 2 s ( s + 2) s ( s + 2) s s s+2 s

sY - 3 + 2Y = C=

7 4 3s2 + 4 3s2 + 4 A = 2 . Setting s = 1 gives us = 2 + B + fi B = -1 . = , = 4 2 3 3 s + 2 s= 0 s s =-2

Therefore, Y =

2 1 4 . Finally, we have y ( t) = 4 e -2 t + 2 t - 1. + 2 s s s+2

1 1 1 3t + , so Y = + te 3 t . 2 and thus y ( t) = e s - 3 ( s - 3) s- 3

21.

sY - 1 - 3Y =

22.

s2Y - s(1) - 2 + 3( sY - 1) + 2Y =

6 s+5 6 + , and thus Y = . ( s + 1)( s + 2) ( s + 1) 2 ( s + 2) s +1

s2 + 6 s + 11 A B C = + + , and so 2 ( s + 1) ( s + 2) s + 2 s + 1 ( s + 1) 2 s2 + 6 s + 11 s2 + 6 s + 11 = 6 . Setting s = 0 gives us = 3, C = A= s+2 ( s + 1) 2 s=-2 s =-1 3 2 6 11 3 + . Finally, we have = + B + 6 fi B = -2 . Therefore, Y = s + 2 s + 1 ( s + 1) 2 2 2

y ( t) = 3e -2 t - 2e - t + 6 te - t . 23.

s2Y - s(2) - 6 + 4Y =

8 2s + 6 8 . + 2 2 2 , so Y = 2 s s + 4 s (s + 4)

A( s2 + 4 ) + Bs( s2 + 4 ) + Cs3 + Ds2 A B Cs + D 8 = 2+ + 2 If 2 2 , then , and so = 2 2 2 2 s (s + 4) s s s +4 s (s + 4) s (s + 4)

8

we have B + C = 0, A + D = 0, 4 B = 0, 4 A = 8 , which means that B = C = 0, A = 2, D = -2 . Ê s ˆ Ê 2 ˆ 2 Ê 2 ˆ Then Y = 2Á 2 ˜ + 3Á 2 ˜ + -Á ˜ and y ( t) = 2 cos 2 t + 2 sin 2 t + 2 t . Ë s + 4¯ Ë s + 4 ¯ s2 Ë s2 + 4 ¯

24.

s2Y - s(1) - 1 + 4Y =

s +1 s s , so Y = 2 . Therefore, + 2 s + 4 (s + 4)2 s +4 2

1 t y ( t) = cos 2 t + sin 2 t + sin 2 t . 2 4 25.

s2Y - s(1) - 0 + 4Y = y ( t) = cos 2 t +

s 2 2 , so Y = 2 . Therefore, + 2 s + 4 (s + 4)2 s +4 2

1 2 t (sin 2t - 2t cos 2 t) = cos 2 t + sin 2 t - cos 2t . 8 4 2◊8

192 • Chapter 7 Laplace Transforms

26.

s2Y - s(0) - 0 - 2( sY - 0) + Y = so A =

1 ( s - 1) 2

= 1, C = s= 2

Therefore, Y =

1 A B C 1 + + , and thus Y = , and 2 = ( s - 2)( s - 1) s - 2 s - 1 ( s - 1) 2 s- 2

1 1 1 = -1. Setting s = 0 gives us - = - - B - 1 fi B = -1. s - 2 s=1 2 2

1 1 1 2t t t 2 . Finally, we have y ( t) = e - e - te . s - 2 s - 1 ( s - 1) 1 1 1 t2 -t -t + , so Y = . Therefore, y ( t ) = te + e . s +1 2 (s + 1) 2 (s + 1) 3

27.

s2Y - 1 + 2 sY - 0 + Y =

28.

g( t) = 6[ h ( t) - h ( t - p )] , and then we have s2Y - s(1) - 3 + 9Y =

Y=

6 (1 - e -ps ) , so s

s+3 6 A Bs + C As2 + 9 A + Bs2 + Cs 6 -ps . . Thus + e 1 = = + ( ) s2 + 9 s( s2 + 9) s( s2 + 9) s s2 + 9 s( s2 + 9)

s s ˆ 3 2 2 Ê 23 2 -ps + 2 +Á - ◊ 2 A + B = 0, C = 0, A = , B = - , and so Y = 2 ˜ (1 - e ) . Then, Ë ¯ s +9 s +9 s 3 s +9 3 3

2 2 (1 - cos 3t) - (1 - cos 3( t - p )) h( t - p ) 3 3 2 2 = cos 3t + sin 3t + (1 - cos 3t) - (1 + cos 3t) h ( t - p ) . 3 3 y ( t) = cos 3t + sin 3t +

29.

g( t) = t[1 - h ( t - 2)] = t - ( t - 2) h ( t - 2) - 2 h ( t - 2) , and then we have

s2Y - s(1) - 0 + Y =

1 1 -2 s 2 -2 s s 1 2 - e , so Y = 2 + 2 2 e -2 s . 1 - e -2 s ) - 2 ( 2 - 2 e s s s s + 1 s ( s + 1) s( s + 1)

1 1 1 A Bs + C As2 + A + Bs2 + Cs -2 and . Thus = = = + 2 s +1 s2 ( s2 + 1) s2 s2 + 1 s( s2 + 1) s s( s2 + 1) A + B = 0, C = 0, A = -2 fi B = 2 , and so

Y=

2s ˆ s 1 ˆ Ê1 Ê 2 + Á 2 - 2 ˜ (1 - e -2 s ) + Á - + 2 ˜ e -2 s . The inverse Laplace transform yields Ë s s + 1¯ s +1 Ë s s + 1¯ 2

y ( t) = cos t + t - ( t - 2) h ( t - 2) - sin t + sin( t - 2) h ( t - 2) - 2 h ( t - 2) + 2 cos( t - 2) h ( t - 2) = cos t - sin t + t + [-( t - 2) + sin( t - 2) - 2 + 2 cos( t - 2)]h ( t - 2) = cos t - sin t + t + [- t + sin( t - 2) + 2 cos( t - 2)]h ( t - 2) . 30.

s2Y - sy 0 - y 0¢ + a ( sY - y 0 ) + bY = 0 fi Y =

sy 0 + ( y 0¢ + ay 0 ) 2s - 1 = 2 , so 2 s + as + b s +s+2

a = 1, b = 2, y 0 = 2, y 0¢ + ay 0 = y 0¢ + 1(2) = -1 fi y 0¢ = -3. 31.

sy 0 + ( y 0¢ + ay 0 ) 3 = 2 , so a = 0, b = -4, y 0 = 0, y 0¢ = 3 . 2 s + as + b s -4

Chapter 7 Laplace Transforms • 193

32.

sy 0 + ( y 0¢ + ay 0 ) s = , so a = 2, b = 1, y 0 = 1, y 0¢ = -2 . 2 s + as + b ( s + 1) 2

Section 7.4 1.

È e - st 2 e - st T = 4. 3Ú e dt - 3Ú e dt = 3Í 0 2 ÍÎ - s 0 - s 2

4

- st

- st

4

˘ 3 2 3 ˙ = 1 - e -2 s + e -4 s - e -2 s = (1 - e -2 s ) . s 2˙ ˚ s

[

]

-2 s 2

3 (1 - e ) 3 Ê 1 - e -2 s ˆ = Therefore, L { f } = Á ˜. s 1 - e -4 s s Ë 1 + e -2 s ¯ 1

2.

1

- st

T = 2. 3Ú e dt + Ú 0

2

1

2

1 e - st e - st 1 e dt = 3 + = ( 3 - 3e - s + e - s - e -2 s ) = ( 3 - 2e - s - e -2 s ) . s -s 0 -s 1 s - st

3 - 2e - s - e -2 s -3 + e - s . Therefore, L { f } = = s(1 - e -2 s ) s(1 - e - s )

3.

2

4

3 2 3 2 T = 4. - 3Ú e dt + 2 Ú e dt = e - st - e - st = (e -2 s - 1) - (e -4 s - e -2 s ) 0 2 s s s s 0 2 2

4

- st

- st

1 -2e -4 s + 5e -2 s - 3 -3 + 2e -2 s -4 s -2 s . = (-2e + 5e - 3) . Therefore, L { f } = = s s(1 - e -4 s ) s(1 + e -2 s ) 1

4.

1

- st

T = 4. 2 Ú e dt + Ú 0

1

Ú

T = 2.

=-

1

0

3

1 e - st e - st 1 e dt = 2 + = (2 - 2e - s + e - s - e -3 s ) = (2 - e - s - e -3 s ) . s -s 0 -s 1 s - st

2 - e - s - e -3 s 2 + e - s + e -2 s . = s(1 - e -4 s ) s(1 + e -2 s )(1 + e - s )

Therefore, L { f } = 5.

3

1

2

te - st dt + Ú (2 - t)e - st dt = 1

2

2

1 1 2 - st + 2 ( st + 1)e - st + e - st 2 ( st + 1)e s s -s 0 1 1

2 1 1 2 1 -s - 1 + 2 (2 s + 1)e -2 s - ( s + 1)e - s + e - s - e -2 s = 2 (1 - e - s ) . Therefore, 2 ( s + 1)e s s s s

[

]

[

] [

]

2

-s 1 (1 - e ) 1 - e-s . = L{ f } = 2 s 1 - e -2 s s2 (1 + e - s ) 1

6.

T = 2.

2

Ú (t - 1)e 1

- st

1

dt = Ú ue

Therefore, L { f } =

- s( u +1)

0

-e - s -e - s du = e Ú ue du = 2 ( su + 1)e - su = 2 ( s + 1)e - s - 1 . 0 s s 0

[

e - s 1 - ( s + 1)e - s s (1 - e 2

-2 s

)

-s

].

1

- su

[

]

194 • Chapter 7 Laplace Transforms

7.

T = 2.

=

Ú

1

0

1

(1 - t 2 )e - st dt =

1 22 s t + 2 st + 2 - s2 e - st s3 0

[

]

1 2 1 2 -s - 2 + s2 = 3 (2 s + 2)e - s + s2 - 2 . Therefore, 3 ( s + 2 s + 2 - s )e s s

[

]

[(2s + 2)e + (s L{ f } = -s

s (1 - e 3

-2 s

2

[

]

].

- 2)

)

2

8.

-1 -1 1 - (2 s + 1)e -2 s - st -2 s T = 2. Ú te dt = 2 ( st + 1)e = 2 (2 s + 1)e - 1 . Therefore, L { f } = 2 . 0 s s s (1 - e -2 s ) 0

9.

p T = , L{ f } = 2

10.

T = 2p , L{ f } =

11.

e -( s+1) t 1 - e -( s+1) 1 . T = 1, Ú e e dt = = 1 - e -( s+1) ), L { f } = ( 0 s +1 0 s +1 (s + 1)(1 - e - s )

12.

e - st e -( s+1) t 1 1 T = 2, Ú (1 - e )e dt = + = (1 - e -2 s ) 1 - e -2( s+1) ) ( 0 s +1 0 s s +1 -s 0

2

[

- st

]

p - s

2 + 2e 2 p - sˆ Ê 2 (s + 4)ÁË1 - e 2 ˜¯

1 + e -ps 1 = 2 2 -2ps (s + 1)(1 - e ) (s + 1)(1 - e -ps )

1

1

- t - st

2

2

-t

2

- st

1 1 - e -2( s+1) . L{ f } = s ( s + 1)(1 - e -2 s )

13.

-as Ê e -as ˆ ˆ -1 Ê e =L Á L Á 1 + e -as + e -2as + e -3as + ...)˜ ( -as ˜ Ë s ¯ Ë s(1 - e ) ¯ -1

Ê e -as e -2as e -3as ˆ =L Á + + + ...˜ = h ( t - a ) + h ( t - 2a ) + ... , s s Ë s ¯ -1

Chapter 7 Laplace Transforms • 195

14.

1 1 e-s F ( s) = - 2 s s s(1 - e - s ) •

and so f ( t) = 1 - t + h ( t - 1) + h ( t - 2) + h ( t - 3) + ... = 1 - t + Â h ( t - n ) . n =1

15.

F ( s) =

• È ˘ 3 3e -2 s and so f ( t ) = 3 [ t h ( t 2 ) h ( t 4 ) h ( t 6 ) ...] = 3 t h(t - 2n)˙ . Â Í -2 s 2 s s(1 - e ) Î n =1 ˚

16.

F ( s) =

-2 ns • 1 1 1 e -2 s 1 e -2 s -2 s -4 s -6 s -8 s n e 1 1 = e + e e + e + ... = + ( ) ( ) 2s2 Â 2 s2 s2 1 + e -2 s 2 s2 s2 s2 n =1

and so f ( t) =

=

17 (a).

t - ( t - 2) h ( t - 2) + ( t - 4 ) h ( t - 4 ) - ( t - 6) h ( t - 6) + ... 2

t • + Â (-1) n ( t - 2 n ) h ( t - 2 n ) . 2 n =1

6 5(10 6 ) r dq dq 5(10 ) = rCi ( t) - q . With q in kg and t in days , we have q C ( t) , where + = v dt dt 50(10 6 ) 10 6 i

ÏÔ1, 0 £ t < 1 2, C ( t + 1) = C ( t) . Then dq + 1 q = 5C ( t), q(0) = 0 . Ci ( t) = Ì i i i dt 10 ÔÓ0, 1 2 £ t < 1

Ï 0.5 e - st dt ¸ ÔÚ Ô 17 (b). sQ + 0.1Q = 5Ì 0 -s ˝ = Ô 1- e Ô Ó ˛

s - ˆ Ê 5 Á1- e 2 ˜ 5 = . Then -s s - ˆ Ê s Á 1- e ˜ ¯ sÁ1 + e 2 ˜ Ë Ë ¯

1 ◊ s( s + 0.1)

1

Q=

5 1+ e

s 2

1 ˆ Ê1 = 50Á ˜ Ë s s + 0.1¯

1+ e

-

s 2

.

196 • Chapter 7 Laplace Transforms s 3s ˆ 1 ˆÊ 1 ˆ Ê1 Ê1 -s -2 s -0.1t 17 (c). Q = 50Á ˜ Á1 - e 2 + e - e 2 + e + ...˜ . Noting that L -1 Á ˜ = (1 - e ) , Ë s s + 0.1¯ Ë Ë ¯ s s + 0.1 ¯ Ê 1ˆ ˘ È Ê -0.1Á t - ˜ ˆ Ê 1ˆ -0.1t we have q( t) = 50 Í1 - e - Á1 - e Ë 2 ¯ ˜ h Á t - ˜ + (1 - e -0.1( t -1) ) h ( t - 1) - ...˙ . Thus for Ë ¯ Ë 2¯ ˙˚ ÍÎ Ê 1ˆ Ï È ˘ -0.1Á t - ˜ 3 Ô50 Í1 - e -0.1t + e Ë 2 ¯ - e -0.1( t -1) ˙, 1 £ t < 2 Ô Í ˙˚ 1 £ t < 2 , q( t) = Ì Î . 1 Ê ˆ ˘ È -0.1Á t - ˜ 3 Ô -0.1t + e Ë 2 ¯ - e -0.1( t -1) ˙, £ t < 2 Ô 50 Í-e ˙˚ 2 Ó ÍÎ T 2

2

18 (a). ms X ( s) =

f 0 Ú e - st dt 0

1 - e - st

V ( s) = sX ( s) =

v ( t) = =

T

-s f0 1 1 f Ê1- e 2 ˆ ◊ 3◊ = 0Á T - sT ˜ fi X ( s) = m s 1 + e-s 2 s Ë1- e ¯

and

f0 1 f0 1 1 - s T2 - s 32T - sT e e e ◊ 2◊ ◊ + + ... 1 T = 2 m s 1 + e-s 2 m s

(

)

˘ f0 È Ê T ˆ Ê T ˆ Ê 3T ˆ Ê 3T ˆ t - Á t - ˜ h Á t - ˜ + ( t - T ) h ( t - T ) - Á t - ˜ h Á t - ˜ + ...˙ Í Ë m Î Ë 2¯ Ë 2¯ 2¯ Ë 2¯ ˚

f0 • Ê nT ˆ Ê nT ˆ (-1) n Á t ˜ hÁ t ˜. Â Ë m n =0 2¯ Ë 2¯

2 2 ˘ f0 È 2 Ê T ˆ Ê T ˆ Ê 3T ˆ Ê 3T ˆ 2 Similarly, x ( t) = Ít - ÁË t - ˜¯ h ÁË t - ˜¯ + ( t - T ) h ( t - T ) - ÁË t - ˜¯ h ÁË t - ˜¯ + ...˙ 2m Î 2 2 2 2 ˚

2

=

f0 • Ê nT ˆ Ê nT ˆ (-1) n Á t ˜. ˜ hÁ t Â Ë 2m n = 0 2¯ Ë 2¯

18 (b). m = 1, f 0 = 1, T = 1, t =

3 5 Ê 5ˆ fi v Á ˜ = [ t - ( t - 1 2) + ( t - 1)] t = 5 = ( t - 1 2) t = 5 = m / s and 4 4 Ë 4¯ 4 4

Ê 5ˆ 1 x Á ˜ = t 2 - ( t - 1 2) 2 + ( t - 1) 2 Ë 4¯ 2

[

19.

]

t = 45

1 Ê 25 9 1 ˆ 17 = Á - + ˜= m. 2 Ë 16 16 16 ¯ 32

We know that ay i¢¢+ by i¢ + cy i = f i ( t), i = 1, 2 . Therefore,

a(c1 y1 + c 2 y 2 )¢¢ + b(c1 y1 + c 2 y 2 )¢ + c (c1 y1 + c 2 y 2 ) = c1[ ay1¢¢+ by1¢ + cy1 ] + c 2 [ ay 2¢¢ + by 2¢ + cy 2 ] = c1 f1 + c 2 f 2 .

20 (a). L {ay ¢¢ + by ¢ + cy} = L { f } . Since

y (0) = 0, y ¢ (0) = 0, ( as2 + bs + c )Y = F fi F( s) = a = 2, b = 5, c = 2 .

1 1 . Comparing: = 2 as + bs + c 2 s + 5 s + 2 2

Chapter 7 Laplace Transforms • 197

20 (b). If f ( t) = e - t , F ( s) =

1 1 . Since , Y ( s) = 2 s +1 (2s + 5s + 2)(s + 1)

2 s2 + 5 s + 2 = 2( s + 2)( s + 1 2), Y ( s) = A=

˘ 1È A 1È 1 B C ˘ = + + . Then Í ˙ Í 2 Î ( s + 2)( s + 1)( s + 12) ˚ 2 Î s + 2 s + 1 s + 12 ˙˚

1 4 1 2 1 and = , B= = -2, C = = 1 1 ( s + 1)( s + 2) s=-2 3 ( s + 2)( s + 2) s=-1 ( s + 2)( s + 1) s=- 1 2 3

1 2 -t y ( t) = e -2 t - e - t + e 2 . 3 3 21.

22.

f ( t) = t , F ( s) =

1 -t - 1) + t(e - t + 1) , 2 , y ( t) = 2(e s

Y ( s) =

2 2 1 2 2 1 1 -2( s + s) + 2 s + 2 s + 1 - + + = = 2 2 2 2 . Since Y ( s) = F ( s) F ( s) , 2 2 s + 1 s ( s + 1) s s ( s + 1) s ( s + 1)

F( s) =

1 . (s + 1) 2

From 21, F( s) =

A=

1 ( s + 1) 2

1 1 1 A B C and Y ( s) = + + 2 . If f ( t) = h ( t), F ( s) = 2 = s s( s + 1) s s + 1 ( s + 1) 2 (s + 1)

= 1, C = s= 0

1 1 B 1 = -1. Setting s = 1, = 1 + - fi B = -1. Therefore, s s=-1 4 2 4

1 1 1 -t Y ( s) = - te - t . 2 fi y ( t) = 1 - e s s + 1 ( s + 1) 23 (a). s2Y + 4Y = F , so F = 23 (b). F =

1 . (s + 4) 2

2 2 . 3 , so Y = 3 2 s s (s + 4)

24 (a). s2Y + sY + Y = F , so F =

1 . (s + s + 1) 2

1

24 (b).

Ú

2

0

e

- st

2

e - st e - st 1 1 1 f ( t) dt = Ú e dt - Ú e dt = = (1 - e - s ) + (e -2 s - e - s ) = (1 - e - s ) 2 . 0 1 s s -s 0 -s 1 s 1

- st

2

- st

(1 - e - s ) 2 (1 - e - s ) (1 - e - s ) Therefore, F = . = , so Y = s(1 + e - s )( s2 + s + 1) s(1 - e -2 s ) s(1 + e - s )

25 (a). s2Y + 4 sY + 4Y = F , so F =

1 1 . = (s + 4 s + 4) (s + 2)2 2

198 • Chapter 7 Laplace Transforms

25 (b).

Ú

1

0

1

te - st dt = -

1 1 1 - ( s + 1)e - s 1 - ( s + 1)e - s - st -s ( st + ) e = ( s + ) e 1 1 1 , so F = and Y = 2 . s2 s2 s2 (1 - e - s ) s2 (1 - e - s )( s + 2) 0

[

26 (a). s3Y - 4Y = F , so F =

26 (b). F =

1 . (s - 4)

28.

3

1 s2 + s - 1 1 s2 + s - 1 . , so Y = 2 + 2 = 2 s ( s - 1)( s3 - 4 ) s-1 s s ( s - 1)

27 (a). s3Y + 4 sY = F , so F = 27 (b). F =

]

1 . (s + 4 s) 3

s s 1 . = 2 , so Y = 2 3 s +4 (s + 4)(s + 4 s) (s + 4)2 2

y ¢¢ + by ¢ + cy = f fi s2Y - sy (0) - y ¢ (0) + b( sY - y (0)) + cY = F . Therefore, ( s2 + bs + c )Y - sy 0 - y 0¢ - by 0 = F fi Y = f ( t) = h ( t), F ( s) =

sy 0 + y 0¢ + by 0 F ( s) . If + 2 2 s + bs + c s + bs + c

1 s2 y 0 + s( y 0¢ + by 0 ) + 1 s2 + 2 s + 1 . Therefore, = fi Y ( s) = s3 + bs2 + cs s3 + 3s2 + 2 s s

b = 3, c = 2, y 0 = 1, y 0¢ + by 0 = y 0¢ + 3 = 2 fi y 0¢ = -1.

29.

Y=

1 sy 0 + y 0¢ + by 0 F ( s) . If f ( t) = e - t , F ( s) = and + 2 2 s +1 s + bs + c s + bs + c

(s + 1)(sy 0 + y 0¢ + by 0 ) + 1 s2 + s + 1 . Therefore, = (s + 1)(s2 + bs + c ) (s + 1)(s2 + 4) b = 0, c = 4, and ( s + 1)( sy 0 + y 0¢ ) + 1 = s2 y 0 + sy 0¢ + sy 0 + y 0¢ = s2 + s + 1 . Finally, y 0 = 1, y 0¢ = 0.

Section 7.5

1.

È s ˘ ˙ Í ÏÈcos t ˘¸ Í s2 + 1 ˙ 1 ÔÍ ˙Ô L ÌÍ t ˙˝ = Í 2 ˙ ÔÍ te t ˙Ô Í s ˙ ˚˛ Í 1 ˙ ÓÎ Í ( s - 1) 2 ˙ ˚ Î

2.

˘ È s +1 ˘ È s( s + 1) - 1˙ Ï Èe - t cos 2 t ˘¸ È1 ˘ Í Ï Èe - t cos 2 t ˘¸ 2 2 1 ˘ È ˙ Í Ôd Í Ôd Í ˙Ô Í ˙ Í ( s + 1) + 4 ˙ Í ˙ Í ( s + 1) + 4 ˙ ˙Ô 0 0 0 LÌ Í ˙˝ - Í0 ˙ = sÍ ˙˝ = sL Ì dt Í 0 ˙ ˙ - Í0 ˙ = Í Ô Í t + e t ˙Ô Í1 ˙ Í 1 + 1 ˙ Í1 ˙ Í 1 + s - 1 ˙ Ô dt Í t + e t ˙Ô Î ˚ ˚˛ Î ˚ ˚˛ 2 Ó Î Ó Î s - 1 ˚˙ ˙˚ ÍÎ s ÍÎ s s - 1

Chapter 7 Laplace Transforms • 199

3.

È 2 e -2 s ˘ ÏÔÈ2 t - h ( t - 2) ˘¸Ô Í s2 - s ˙ L ÌÍ ˙ ˙˝ = Í -2 s ÔÓÎ 2 h ( t - 2) ˚Ô˛ Í 2e ˙ ˙˚ ÍÎ s

4.

È 1 ˘ ÏÈ 1 ˘¸ Í s2 ˙ Ï È1 ˘ ¸ Ô t Í ˙ Ô 1 ÔÍ ˙Ô Í 1 ˙ ˙ L ÌÚ Í l ˙dl ˝ = L ÌÍ t ˙˝ = Í 3 0 Ô Íe -l ˙ Ô s ÔÍe - t ˙Ô Í s ˙ ÓÎ ˚˛ Í 1 ˙ Ó Î ˚ ˛ ÍÎ s( s + 1) ˙˚

5.

È e-s ˘ ÏÔÈsin( t - 1) h ( t - 1) ˘¸Ô Í s2 + 1 ˙ L ÌÍ ˙ ˙˝ = Í e -1 2˙ e t -1 - 2 t ÔÓÎ ˚Ô˛ Í ÍÎ s - 1 s2 ˙˚

6.

ÏÈ ˘¸ 1 ÔÍ ˙Ô s È 1 ˘ Ô ˙ÔÔ Í 1 1 1 2 2 ˙ Í -1 Ô -t . Therefore, Í2e - t sin t ˙ . = L 2e sin t ; = L ÌÍ 2 ˙˝; 2 s( s + 1) s s + 1 ÔÍ s + 2 s + 2 ˙Ô ( s + 1) + 1 ÍÎ 1 - e - t ˙˚ 1 ˙Ô ÔÍ ÔÓÍÎ s2 + s ˙˚Ô˛

7.

ÏÈ - s Ê 1 1 ˆ ˘¸ ÔÔÍe ÁË - 2 ˜¯ ˙ÔÔ È(1 - sin( t - 1)) h ( t - 1) ˘ s s +1 L -1 ÌÍ ˙˝ = Í ˙ -s 2 e ˙Ô Î 2 sin( t - 1) h ( t - 1) ˚ Í Ô ÔÓÍÎ ˙˚Ô˛ s2 + 1

8.

Èt 3 - e 2 t + 2 sin t ˘ Í ˙ -1 3 L {Y( s)} = Í 2 t + 3 sin t ˙ ÍÎ t 3 - 2e 2 t + sin t ˙˚

9.

È 5s - 4 ˘ 4 ˘ Ès - 5 È5 ˘ È5 ˘ 1 Ès + 4 -4 ˘È5 ˘ Í s2 - s ˙ sY - Í ˙ = AY , so Í = , and ˙Y = Í6 ˙ . Thus Y = s2 - s Í 5 s - 5 ˙˚ÍÎ6 ˙˚ ÍÍ 6 s - 5 ˙˙ Î Î -5 s + 4 ˚ Î6 ˚ Î ˚ Î s2 - s ˚

{[

]}

È4 + e t ˘ 5s - 4 A B 6s - 5 5 1 since 2 , we have y ( t) = Í . = + fi A = 4, B = 1 and 2 = + t˙ s - s s s-1 s - s s s -1 Î5 + e ˚

10.

È -4 ˘ 4 ˘ È 0 ˘ Ès - 5 È0 ˘ 1 Ès + 4 -4 ˘È 0 ˘ Í s2 ( s - 1) ˙ sY - 0 = AY + Í 1 ˙ = Í Y Y + fi = =Í ˙ , and ˙ Í1 ˙ s - 5 ˙˚ÍÎ 1 s ˙˚ Í s + 4 ˙ s2 - s ÍÎ 5 Î s ˚ Î -5 s + 4 ˚ Î s˚ ÍÎ s2 ( s - 1) ˙˚ since

-4 -4 4 4 4 A B C = 2+ + fi A = 4, B = 4, C = -4. then 2 = 2+ s ( s - 1) s s s-1 s ( s - 1) s s s -1 2

200 • Chapter 7 Laplace Transforms

and

-5 1 5 5 5 1 1 5 4 4 s- 5 + = 2+ + - = 2+ = 2 . Therefore, s ( s - 1) s ( s - 1) s( s - 1) s s s -1 s -1 s s s s -1 2

4 ˘ È4 4 È4 t + 4 - 4 e t ˘ Í s2 + s - s - 1 ˙ Y( s) = Í fi y ( t) = Í . t˙ 5 4 4 ˙ Î5 t + 4 - 4 e ˚ Í 2+ ˙ Îs s s - 1˚

11.

È1˘ È1˘ 4 ˘ Í s2 ˙ Í 2˙ Ès - 5 sY = AY + Í ˙ , so Í Y = Í s ˙ . Thus ˙ 1 1 Î -3 s + 2 ˚ Í ˙ Í ˙ Îs˚ Îs˚ È1˘ È -3s + 2 ˘ Ès + 2 -4 ˘Í s2 ˙ 1 1 Y= 2 = 2 . Í ˙ Í ˙ s - 5 ˚Í 1 ˙ s ( s - 1)( s - 2) ÍÎs2 - 5 s + 3˙˚ s - 3s + 2 Î 3 Îs˚ Y1 =

-3s + 2 A B C D = 2+ + + fi A = 1, B = 0, C = 1, D = -1, so y1 = t + e t - e 2 t . s ( s - 1)( s - 2) s s s-1 s- 2

Y2 =

3 s2 - 5 s + 3 A B C D 3 1 = 2+ + + fi A = , B = - , C = 1, D = - , so 2 4 s ( s - 1)( s - 2) s s s-1 s- 2 2 4

2

È ˘ t + e t - e 2t 3 1 3 2t t 3 y 2 = t - + e - e . Finally, we have y ( t) = Í 3 1 t 2t ˙ . 2 4 4 ÍÎ 2 t - 4 + e - 4 e ˙˚

12.

From 11, Y =

È3s - 2 ˘ Ès + 2 -4 ˘È3˘ 1 1 = . Í ˙ Í ˙ s - 5 ˚Î2 ˚ ( s - 1)( s - 2) ÍÎ 2 s - 1 ˙˚ ( s - 1)( s - 2) Î 3

Y1 =

A B 3s - 2 = + fi A = -1, B = 4 , so y1 = -e t + 4 e 2 t . (s - 1)(s - 2) s - 1 s - 2

Y2 =

A B 2s - 1 = + fi A = -1, B = 3 , so y 2 = -e t + 3e 2 t . Finally, we have (s - 1)(s - 2) s - 1 s - 2

È-e t + 4 e 2 t ˘ . y ( t) = Í t 2t ˙ Î-e + 3e ˚

13.

Ès - 1 4 ˘È2 ˘ Ès - 1 -4 ˘ È2 ˘ È2 ˘ 1 sY = AY + Í ˙ , so Í Y = Í ˙ . Thus Y = . 2 ˙ s - 1˚ (s - 1) + 4 ÍÎ -1 s - 1˙˚ÍÎ0˙˚ Î 1 Î0 ˚ Î0 ˚

Y1 =

2( s - 1) -2 , so y1 = 2e t cos 2 t . Y2 = , so y 2 = -e t sin 2 t . Finally, we have 2 2 (s - 1) + 4 (s - 1) + 4

È2e t cos 2 t ˘ y ( t) = Í t ˙. Î-e sin 2 t ˚

Chapter 7 Laplace Transforms • 201

14.

È 0 ˘ È 0 ˘ Ès - 1 4 ˘È 0 ˘ Ès - 1 -4 ˘ È3˘ 1 3 ˙ Í 3 ˙. Í . Thus sY - Í ˙ = AY + Í 3 ˙ , so Í Y = Y = s - 1˙˚ (s - 1) 2 + 4 ÍÎ -1 s - 1˙˚ÍÎ s - 1˙˚ ÍÎ s - 1 ˙˚ ÍÎ s - 1 ˙˚ Î 1 Î0 ˚ Y1 =

3( s - 1) + s12-1 3( s - 1) 12 . For = + 2 2 (s - 1) + 4 (s - 1) + 4 (s - 1) (s - 1) 2 + 4

[

12

[

2

( s - 1) ( s - 1) + 4

Y1 =

]

=

]

A B( s - 1) + C + , A = 3, B = -3, C = 0. Therefore, s - 1 ( s - 1) 2 + 4

3( s - 1) 3 3( s - 1) 3 + = , so y1 = 3e t . Y2 = 0 , so y 2 = 0 . Finally, we have 2 2 (s - 1) + 4 s - 1 (s - 1) + 4 s - 1

È3e t ˘ y ( t) = Í ˙ . Î0 ˚

15.

Letting t = t + 1; t = t - 1, we have

È5˘ È5˘ dy È6 -3˘ =Í y, y t = 0 = Í ˙ . Then sY = AY + Í ˙ , so ˙ dt Î8 -5 ˚ Î10 ˚ Î10 ˚

3 ˘ Ès - 6 Ès + 5 -3 ˘È 5 ˘ È5˘ 1 . Í -8 s + 5 ˙Y = Í10 ˙ . Thus Y = s2 - s - 6 Í 8 s - 6 ˙˚ÍÎ10 ˙˚ Î Î ˚ Î ˚ Y1 =

A B 5s - 5 = + fi A = 3, B = 2 , so y1 = 3e -2t + 2e 3t . (s - 3)(s + 2) s + 2 s - 3

Y2 =

A B 10 s - 20 = + fi A = 8, B = 2 , so y 2 = 8e -2t + 2e 3t .Finally, we have (s - 3)(s + 2) s + 2 s - 3

È3e -2( t -1) + 2e 3( t -1) ˘ y ( t) = Í -2( t -1) ˙. + 2e 3( t -1) ˚ Î8e

16.

2 Ès2 + 3 È1 ˘ È0 ˘ 2 ˘ Ès˘ 1 Ès - 3 -2 ˘Ès˘ s2 Y - sÍ ˙ - Í ˙ = AY , so Í . Thus = Y = Y ˙ Í ˙Í ˙ . Í1˙ s4 - 1 Î 4 s2 - 3˚ s2 + 3˚Î1˚ Î0 ˚ Î1 ˚ Î ˚ Î -4

s3 - 3s - 2 A B Cs + D fi A = 0, B = -1, C = 2, D = 1, so = + + 2 Y1 = 2 ( s - 1)( s + 1)( s + 1) s + 1 s - 1 s + 1

y1 = -e t + 2 cos t + sin t . Y2 =

s2 + 4 s + 3 s+3 A Bs + C fi A = 2, B = -2, C = -1, so = + 2 = 2 2 ( s - 1)( s + 1)( s + 1) ( s - 1)( s + 1) s - 1 s + 1

È-e t + 2 cos t + sin t ˘ y 2 = 2e t - 2 cos t - sin t . Finally, we have y ( t) = Í t ˙. Î 2e - 2 cos t - sin t ˚

202 • Chapter 7 Laplace Transforms

17.

È1˘ È1˘ È1˘ 2 2 2˙ 2˙ 1 1 È ˘ È ˘ 1 1 + s 1 1 s Í s2 ˙ Í Í È ˘ 1 Y = Í s ˙ . Thus Y = 4 Í s2 Y = Í Y + Í s ˙ , so Í ˙ ˙ Í 1 ˙. ˙ 2 2 1 1 s 1 1 1 s + 1 s Î1 -1˚ Î ˚ Î ˚ Í ˙ Í ˙ Í ˙ Îs˚ Îs˚ Îs˚ 1Ê 1 1ˆ 1Ê1 1ˆ t5 t4 t3 1 3 1 5 1 4 Y1 = 4 Á1 + 2 - ˜ , so y1 = t + t - t = + . Y2 = 4 Á 2 + s - ˜ , so Ë ¯ Ë s s s s s s¯ 3! 5! 4! 120 24 6

y2 =

t5 t4 t2 1 5 1 2 1 4 + . Finally, we have t + t - t = 5! 2! 4! 120 24 2

È t5 t4 t3 ˘ 5 4 3 + Í 24 6 ˙˙ = 1 Èt - 5 t + 20 t ˘ . y ( t) = Í120 Í5 5 4 2 4 2˙ Í t - t + t ˙ 120 Ît - 5 t + 60 t ˚ ÍÎ120 24 2 ˙˚

18.

È2 ˘ 0 1 1 Í ˙ È ˘ È ˘ s2 Y - sÍ ˙ = Í Y + Í s ˙ , so ˙ 1 Î1 ˚ Î1 -1˚ Í ˙ Îs˚

È 2 ˘ Ès2 - 1 ˘ 1 Í s ˙ Í ˙Y = Í1 ˙ . Thus 2 Î -1 s + 1˚ Í + s˙ Îs ˚

È 2 ˘ 2 1 Ès + 1 -1 ˘Í s ˙ 1Ê 2 1 ˆ 1 1 t4 t2 t4 t2 Y= 4 Í . , so Y s s = 2 + = + = + = + . y Á ˜ ˙Í ˙ 1 s4 Ë 1 s Î 1 s s ¯ s3 s5 4! 2! 24 2 s2 - 1˚Í1 + s˙ Îs ˚ È t2 t4 ˘ Í + ˙ 1 Ê2 1 t4 ˆ 1 1 . Finally, we have y ( t) = Í 2 24 Y2 = 4 Á + s - + s3 - s˜ = + 5 , so y 2 = 1 + 4 ˙. ¯ s s s Ës s 24 Í 1+ t ˙ ÍÎ 24 ˙˚

19.

0 ˘ È2˘ Ès - 6 -5 Í ˙ ˙ Í 0 ˙Y = Í-4 ˙ . Thus s+6 sY - y (0) = AY , so Í 7 ÍÎ -1˙˚ ÍÎ 0 0 s + 2 ˙˚

Ès+6 Í s2 - 1 Í -7 Y= Í 2 Ís -1 Í 0 ÍÎ

5 s -1 s- 6 s2 - 1 2

0

È 2s - 8 ˘ ˘ 0 ˙ 2 È 2 ˘ Í s -1 ˙ ˙Í ˙ Í -4 s + 10 ˙ 0 ˙Í-4 ˙ = Í 2 ˙. ˙Í ˙ Í s - 1 ˙ 1 ˙Î -1˚ Í -1 ˙ ÍÎ s + 2 ˙˚ s + 2 ˙˚

Y1 =

A B 2s - 8 = + fi A = 5, B = -3 , so y1 = 5e - t - 3e t . (s + 1)(s - 1) s + 1 s - 1

Y2 =

-4 s + 10 A B = + fi A = -7, B = 3 , so y 2 = -7e - t + 3e t . (s + 1)(s - 1) s + 1 s - 1

Chapter 7 Laplace Transforms • 203

È 5e - t - 3e t ˘ -1 Í ˙ , so y 3 = -e -2 t . Finally, we have y ( t) = Í-7e - t + 3e t ˙ . Y3 = s+2 ÍÎ -e -2 t ˙˚

20.

È 1 ˘ Í s - 1˙ Í 1 ˙ ˙ , so sY = AY + Í Í s ˙ Í- 2 ˙ ÍÎ s2 ˙˚ È 1 Ís -1 Í Y= Í 0 Í Í ÍÎ 0 Y2 =

˘È 1 ˘ ˙Í s - 1 ˙ ˙Í 1 ˙ 1 1 1 ˙Í , so y1 = te t . ˙ . Y1 = s + 1 ( s + 1)( s - 2) ˙Í s ˙ (s - 1) 2 1 ˙Í - 2 ˙ 0 ˙˚ÍÎ s2 ˙˚ s- 2 0

0

1 1 1 s( s - 2) - 2 A B C D = 2+ + + fi A = 1, B = , C = - , D = - , so 2 3 6 s ( s + 1)( s - 2) s s s +1 s- 2 2

y2 = t + Y3 =

È 1 ˘ Í s - 1˙ 0 ˘ Ès - 1 0 Í 1 ˙ ˙ Í 0 1 1 ˙ . Thus Í = s Y + ˙ Í s ˙ Í ÍÎ 0 0 s - 2 ˙˚ Í- 2 ˙ ÍÎ s2 ˙˚

1 1 - t 1 2t - e - e . 2 3 6

-2 1 1 A B C 1 1 = 2+ + fi A = 1, B = , C = - , so y 3 = t + - e 2 t . Finally, we have 2 2 s ( s - 2) s s s- 2 2 2 2

˘ È ˙ Í te t Í 1 1 - t 1 2t ˙ y ( t) = Í t + - e - e ˙ . 6 ˙ Í 2 3 1 1 ˙ Í t + - e 2t ˙˚ ÍÎ 2 2

21 (a). s2 - 9 s + 18 = ( s - 3)( s - 6) = 0 fi l = 3, 6 . -1

21 (b). sY - y (0) = AY fi Y = ( sI - A) y 0 . Then - A -1 =

1 È-2 -1˘ 1 È 2 1˘ fi A -1 = Í , and Í ˙ 18 Î 4 -7 ˚ 18 Î-4 7 ˙˚

2 -1 18 È 7 -1˘ È 7 -1˘ 1 Ê 1ˆ = det A = Á ˜ ◊ 18 = . Thus A = ( A -1 ) = Í . Ë 18 ¯ 18 Î4 2 ˙˚ ÍÎ4 2 ˙˚ 18 -1

22 (a). sY1 = AY1 + G( s), sY2 = AY2 + Y1 fi Y1 = ( sI - A) -1 G, Y2 = ( sI - A) -1 Y1 and

Y2 = ( sI - A) -2 G( s) \ W( s) = ( sI - A) -2 .

204 • Chapter 7 Laplace Transforms

Ès - 1 1 ˘ 1 Ès + 1 -1 ˘ fi ( sI - A) -1 = 2 Í 22 (b). ( sI - A) = Í and ˙ s - 1˙˚ s Î 1 Î -1 s + 1˚ W( s) = ( sI - A) -2 =

2 -2 s ˘ 1 Ès + 1 -1 ˘Ès + 1 -1 ˘ 1 Ès + 2 s = ˙. ˙ Í ˙ 4 Í 4 Í 2 s - 1˚Î 1 s - 1˚ s Î 2 s s Î 1 s - 2 s˚

2˘ È 2 2 1 ˘ È1˘ È s + 2 Ís˙ ˙ Í- s5 + s4 + s3 ˙ 1Í s G( s) = Í ˙ fi Y2 ( s) = 4 Í = Therefore, 1 s Í2 + 1 - 2 ˙˙ ÍÍ - 2 + 3 ˙˙ Í 2˙ Îs ˚ Î s˚ Î s5 s4 ˚ È 2t 4 2t 3 t 2 ˘ È t 4 t 3 t 2 ˘ Í- 4! + 3! + 2! ˙ Í- 12 + 3 + 2 ˙ Y2 ( t) = Í ˙. ˙= Í 4 3 4 3 2 t 3 t t t ˙ ˙ Í - + Í + ÍÎ 4! 3! ˙˚ ÍÎ 12 2 ˙˚

23.

Y1 =

1 È1˘ , s - 3 ÍÎ1˙˚

È 1 Ís - 3 Í Í 1 ÍÎ s - 3

È 3 Í Y2 = Í s + 2 8 Í Îs + 2

-15 ˘ (s + 2)(s - 3) ˙˙ -1 È1 0 ˘ = ( sI - A) Í ˙ , and so 5 s - 30 ˙ Î1 5 ˚ (s + 2)(s - 3) ˚˙

È -1 1 0 È ˘Í 3 A = -Í ˙Í Î1 5 ˚Í -1 Î3

24.

-15 ˘ 3 ˘ È Í ˙ s - 3 ˙ = Í ( s + 2)( s - 3) ˙ . Therefore, 3 ˙ Í 5 s - 30 ˙ ˙ s - 3 ˚ ÎÍ ( s + 2)( s - 3) ˙˚

-1

5˘ È 5 ˙ 6 2 = Í ˙ 5 ÍÍ 20 5˙ ˚ Î3

È -1 Í3 Í -1 Í Î3

5˘ 2 ˙ = - A -1 È1 0 ˘ . Thus ˙ Í1 5 ˙ Î ˚ 5˙ ˚

-5 ˘ 2 ˙ = È6 -3˘ . -25 ˙ ÍÎ8 -5 ˙˚ ˙ 6 ˚

È 1 ˘ È2 ˘ È1 ˘ 2˙ Í Y1 = Í ( s + 2) ˙, Y2 = Í s ˙, G( s) = Í s ˙ . Therefore, Í0 ˙ Í0 ˙ Í 1 ˙ Î ˚ Î ˚ Î s+2 ˚ -1

È 1 Í ( s + 2) 2 Í 1 Í Î s+2

1˘ 2˘ È s ˙ = ( sI - A) -1 Í0 1 + ˙ , and so ˙ s ˙ Í1 0 0˙ ˚ Î ˚

1 ˘ 1˘ È È 1 + 2 s + 2 ˘Í s 0 ˘ È È 2 Í 0 ( s + 2) s ˙ = - s( s + 2) Í0 s ˙ ( sI - A) = Í ˙ ˙ s ˙Í 1 s ˙Í 1 1 ˙ Í ˙Í Í1 1 0 0 Í ˙ 0 ˚ s + 2 ( s + 2) 2 ˙ Î ˚ Î ˚ Î ˚ Î s+2 1 ˘ È 1 Í- s s( s + 2) ˙ Ès + 2 -1 ˘ È-2 1 ˘ È-2 1 ˘ =Í = sI - Í = - s( s + 2) Í ; A=Í ˙ ˙ ˙ ˙. 1 + 0 2 0 2 0 2 s Î ˚ Î ˚ Î ˚ ˙ Í0 Î s ˚

Chapter 7 Laplace Transforms • 205

25 (a). -V1 ( s) + R1I1 + LsI1 + R2 ( I1 - I 2 ) = 0 , R2 ( I 2 - I1 ) +

1 I + R3 I 2 + V2 ( s) = 0 . Cs 2

- R2 ˘È I ˘ È V ˘ ÈR1 + R2 + sL 1 1 Í 1 ˙Í ˙ = Í ˙ , so + + R R R I V ˙Î 2 ˚ Î 2 ˚ Í 2 2 3 Î Cs ˚ 1 È È I1 ˘ 1 R2 + R3 + Í = Cs ÍI ˙ Î 2 ˚ ( R + R + sL)ÊÁ R + R + 1 ˆ˜ - R 2 ÍÎ R 2 1 2 3 2 Ë 2 Cs¯

˘È V ˘ ˙Í 1 ˙. R1 + R2 + sL ˙˚Î-V2 ˚ R2

È 1 ˘ 1 2˙ ˘ È Í È I1 ˘ s 1 (s + 1) -2 ÈÍ s + 1 ˘˙ 1 ˙Í ( s + 1) ˙ Í2 + s = 25 (b). Í ˙ = = s 4 -1 ˙ 2Í ˙ 2( s + 1) ÎI 2 ˚ (2 + s)ÊÁ 2 + 1ˆ˜ - 1 ÍÎ 1 2 + s˙˚Í s ( ) + 1 s + + 4 2 Î ˚ ÍÎ ( s + 1) 2 ˙˚ Ë s s¯

I1 ( s) =

È s +1 ˘ Í s ˙. Í-( s + 1) ˙ Î ˚

t2 -t 1 fi i ( t ) = e and 3 1 4 2( s + 1)

-s -( s + 1) + 1 -1 t -t t2 -t 1 I 2 ( s) = + e . 3 = 3 = 2 + 3 fi i2 ( t) = - e 2 4 2( s + 1) 2( s + 1) 2( s + 1) 2( s + 1)

Section 7.6 1.

First, let s = t - l , and differentiation yields ds = - dl . Then, t

0

t

0

t

0

t

t

0

0

f * g = Ú f ( t - l ) g(l ) dl = Ú f (s ) g( t - s )(- ds ) = Ú g( t - s ) f (s ) ds = g * f . 2 (a).

f * g = Ú h ( t - l ) h (l ) dl = Ú 1dl = t

2 (b).

1 Ê 1ˆ F = G = , so f * g = L-1 Á 2 ˜ = t Ës ¯ s

3 (a).

Ê l3 l4 ˆ Ê1 f * g = Ú ( t - l )l dl = Á t - ˜ = t 4 Á 0 Ë3 Ë 3 4¯0

3 (b).

F=

t

t

2

1 ˆ t4 ˜= 4 ¯ 12

1 2 2t 4 t 4 -1 Ï 2 ¸ , G , so f * g L = = = = ˝ Ì 5 s2 s3 Ó s ˛ 4! 12 t

t

4 (a).

f *g= Ú e

4 (b).

F=

0

( t -l ) -2l

e

Ê t e -3l ˆ et e t - e -2 t -3 t dl = Á e ˜ = (1 - e ) = 3 Ë -3 ¯ 0 3

1 1 1 1 ˆ¸ 1 t ¸ Ï -1 Ï 1 Ê 1 -2 t , G= , so f * g = L-1 Ì ˜ ˝ = (e - e ) ˝ = L Ì ÁË ¯ s-1 s+2 Ó ( s - 1)( s + 2) ˛ Ó3 s - 1 s + 2 ˛ 3

206 • Chapter 7 Laplace Transforms

5 (a).

t

t

t

f * g = Ú ( t - l ) sin ldl = - t cos l 0 - Ú l sin ldl 0

0

t

= - t(cos t - 1) - (-l cos l + sin l ) 0 = - t cos t + t + t cos t - sin t = t - sin t ¸Ô ÏÔ 1 1 1 1 ¸ -1 Ï 1 , so f * g = L-1 Ì 2 2 ˝ = L Ì 2 - 2 ˝ = t - sin t 2, G = 2 s + 1˛ s s +1 Ós ÔÓ s ( s + 1) Ô˛

5 (b).

F=

6 (a).

f * g = Ú sin( t - l ) cos ldl =

t

0

1È 1 = Ít sin t + coss 2Î 2

-t 1È 1 t Ê 1 ˆ˘ [sin t + sin( t - 2l )]dl = Ít sin t + Ú (sin s )Á - ds ˜ ˙ Ú t Ë 2 ¯˚ 2Î 2 0

t

˘ t ˙ = sin t, where s = t - 2l . 2 -t ˚

Ï s ¸Ô t s 1 -1 Ô = f g = , G , so * L sin t Ì 2 2˝= s2 + 1 2 s2 + 1 s + 1 ) ˛Ô ÓÔ (

6 (b).

F=

7 (a).

2 t Ï t t - l) t2 ( Ô , 0 £ t £1 ( t l ) d l = = Ú0 2 2 t Ô 0 f * g = Ú ( t - l )[ h (l ) - h (l - 1)]dl = Ì 2 1 0 2 t (t - l ) ( t - 1) 2 Ô 1 , 1£ t < • = ÔÚ0 ( t - l ) dl = - 2 2 2 Ó 0

7 (b).

2 1 - e-s) ( 1 e - s ¸ t 2 ( t - 1) -1 Ï 1 F = 2, G= , so f * g = L Ì 3 - 3 ˝ = h ( t - 1) 2 s s s ˛ 2 Ós

8.

ÏÈ 1 ÔÍ P * y = L-1 ÌÍ s ÔÍ 0 ÓÎ

ÏÈ 1 1 ˘¸ 1 ˘È 1 ˘¸ + 2 Ô Í ( s - 1)( s + 1) ˙ÔÔ s - 1 ˙Í s ˙Ô˝ = L-1 ÔÌÍ s ˙˝ . 1 1 ˙Í 1 ˙ ˙Ô Í Ô Ô ˙Í ˙ ÔÓÍÎ ˙˚Ô˛ s2 ˚Î s + 1 ˚˛ s2 ( s + 1)

1 1Ê 1 1 ˆ 1 A B C = Á = 2+ + fi A = 1, B = -1, C = 1. Therefore, ˜; 2 ( s - 1)( s + 1) 2 Ë s - 1 s + 1¯ s ( s + 1) s s s +1

È 1 t 1 -t ˘ t+ e - e ˙ P*y = Í 2 2 Í t - 1 + e-t ˙ Î ˚

9.

ÏÈ 1 ˘¸ 4 Ô ˙ÔÔ Í t È ˘ t3 s -1 Ô -1 Ï 1 ¸ . . Then we have t*Í L = = L ˙ Í Ì ˝ ˝ Ì ˙ 4 s Ós ˛ 6 Îcos t ˚ ˙Ô ÔÍ 2 2 ÔÓÍÎ s ( s + 1) ˙˚Ô˛ s ¸ A Bs + C 1 Ï1 fi A = 1, B = -1, C = 0 . L -1 Ì - 2 ˝ = 1 - cos t , = + 2 s +1 s( s + 1) s Ó s s + 1˛ 2

Chapter 7 Laplace Transforms • 207 3 ˘ È t ˘ ÈÍ t so t * Í ˙ = Í 6 ˙˙ . cos t Î ˚ Î1 - cos t ˚

10.

1 1 e -2 s 2e -2 s g( t) = th ( t) - ( t - 2) h ( t - 2) - 2 h ( t - 2), F ( s) = , G( s) = 2 - 2 , s s s s 1 e -2 s 2e -2 s t 2 ( t - 2) 2 so FG = 3 - 3 - 2 . Then f * g = h ( t - 2) - 2( t - 2) h ( t - 2) . 2 2 s s s

11.

F=G=

e -2 s - 2e -3 s + e -4 s e - s - e -2 s . Then , so FG = s2 s

f * g = ( t - 2) h ( t - 2) - 2( t - 3) h ( t - 3) + ( t - 4 ) h ( t - 4 ) . 12.

F ( s) =

e - s - 3e -2 s + 2e -3 s e - s - 2e -2 s 1 - e-s . Then , G( s) = , so FG = s2 s s

f * g = ( t - 1) h ( t - 1) - 3( t - 2) h ( t - 2) + 2( t - 3) h ( t - 3) .

Ê 1ˆ Ës ¯

3

13.

L {t * t * t} = Á 2 ˜ =

14.

L{h ( t) * e

-t

* e -2 t } =

h ( t) * e - t * e -2 t =

1 t5 t5 , so . t * t * t = = s6 5! 120 1 A B C 1 1 = + + . Thus A = , B = -1, C = and s( s + 1)( s + 2) s s + 1 s + 2 2 2

1 - t 1 -2 t -e + e . 2 2

208 • Chapter 7 Laplace Transforms

15.

L{t * e - t * e t } =

1 A B C D 1 1 = 2+ + + . Thus A = -1, B = 0, C = , D = s ( s + 1)( s - 1) s s s-1 s +1 2 2 2

1 1 ¸ Ï Ô Ô -1 1 and t * e - t * e t = L -1 Ì 2 + 2 - 2 ˝ = - t + (e t - e - t ) . 2 s - 1 s + 1Ô Ôs ˛ Ó 16.

L{h ( t) * h ( t) * ...* h ( t)} =

Therefore, 17.

Ï

Ï 1 ¸ t n -1 - t 1 -1 e = Ct 4 eat . Thus , L = Ì n n˝ n ( ) ! 1 (s + 1) Ó ( s + 1) ˛

¸

n times

n = 5, C = 18.

1 t n -1 = Ct 8 fi n = 9, C = . ( n - 1)! 8!

-t -t t L Ìe1 * e -2 *44 ...* e3 ˝= 44

Ó

1 t n -1 -1 Ï 1 ¸ and L = . Ì ˝ sn Ó s n ˛ ( n - 1)!

˛

1 , a = -1. 4!

t

Ú sin(t - l ) y (l )dl = t = sin t * y . Therefore, 2( s + 1) 2 2 2 1 = Y fiY = = + fi y ( t) = 2 + t . 2

0

2

2

s

19.

3

2

s +1

s

3

s

s

3

t

t 2e - t = Ú cos( t - l ) y (l ) dl = cos t * y . Therefore, 0

2( s2 + 1) A B C D s 2 Y fiY = + + . Thus 3 = 2 3 = 2 + s s + 1 ( s + 1) s( s + 1) (s + 1) s + 1 (s + 1) 3 A = 2, B = -2, C = 0, D = -4 , and so we have Y =

2 2 4 . Finally, s s + 1 ( s + 1) 3

y ( t) = 2 - 2e - t - 2 t 2e - t . 20.

t

y ( t) - Ú e t -l y (l ) dl = t fi y - e t * y = t, Y 0

1 1 Y = 2 . Therefore, s-1 s

s-1 A B C 1 s- 2 1 1 1 = + 2+ Y fiY = 2 . Thus A = - , B = , C = , and so we have 2 = s ( s - 2) s s s- 2 s s-1 4 2 4

Y= 21.

t

1 - 14 12 1 t 1 4 . Finally, y ( t) = - + + e 2 t . + 2+ 4 2 4 s s s- 2

Ú y (t - l ) y (l )dl = 6t 0

3

= y * y fi Y 2 = (6)

3! 36 6 4 = 4 fi Y = ± 2 and y ( t) = ±6 t . s s s

Chapter 7 Laplace Transforms • 209

22.

1 2 2 2 2 s2 . Y = fi Y = s2 s3 ( s + 1) 3 s ( s + 1) 3

s2 = ( s2 + 2 s + 1) - (2 s + 1) = ( s + 1) 2 - (2 s + 2 - 1) = ( s + 1) 2 - 2( s + 1) + 1. Therefore, Ê -t Ê t2 -t ˆ t2 ˆ -t 2 Ê 1 2 1 ˆ -t , so y ( t) = 2 - 2Á e - 2 te + e ˜ = 2 - 2Á1 - 2 t + ˜ e . Y = - 2Á 2 + 2 ¯ 2¯ ( s + 1) 3 ˜¯ s Ë Ë Ë s + 1 ( s + 1) 23.

sY - 0 + Y=

24.

1 1 1Ê s + 2 ˆ A B C . Thus A = 2, B = -2, C = -1, and + + Y = fiY = Á 2˜ = s+2 s s Ë ( s + 1) ¯ s s + 1 ( s + 1) 2

2 2 1 -t - te - t . 2 , so y ( t) = 2 - 2e s s + 1 ( s + 1)

È1 ˘ 1 È1 ˘ s È1 ˘ 1 Ê 1 1 ˆ È1 ˘ Ê 1ˆ sY - Í ˙ = Y fi Á s - ˜ Y = Í ˙ fi Y = 2 = Á + ˜ , so Í ˙ Ë s¯ s - 1 Î2 ˚ 2 Ë s - 1 s + 1¯ ÍÎ2 ˙˚ Î2 ˚ s Î2 ˚ y ( t) =

È1 ˘ È1 ˘ 1 t e + e - t ) Í ˙ = cosh t Í ˙ . ( 2 Î2 ˚ Î2 ˚ 1 1 1 1 t4 t4 . ◊ fi Y = + fi y ( t ) = 1 + = 1 + s2 s2 s s5 4! 24

25.

sY - 1 =

26.

sY - (-1) - Y =

1 1 1 1 -1 + 2 fi ( s - 1)Y = -1 + 2 fiY = . 2 ◊ s s -1 s ( s - 1) s - 1 s ( s - 1) 2

1 A B C D fi A = 2, B = 1, C = -2, D = 1 . Thus, + 2+ + 2 = s ( s - 1) s s s - 1 ( s - 1) 2 2

Y =-

1 2 1 2 1 2 1 3 1 t t + + 2+ + 2+ 2 = 2 , so y ( t) = 2 + t - 3e + te . s-1 s s s - 1 ( s - 1) s s s - 1 ( s - 1)

Section 7.7 1 (a).

3

Ú (1 + e )d (t - 2)dt = 1 + e -t

-2

0

1

1 (b).

Ú (1 + e )d (t - 2)dt = 0 since t=2 lies outside the integration interval.

1 (c).

Ècos2 t ˘ È1 ˘ d ( ) = t dt Í Í0 ˙ ˙ t Ú-1 Î te ˚ Î ˚

1 (d).

Èe -4 - 2 ˘ Èd ( t + 2) ˘ Í 2 ˙ ˙ Í 2t Ú-3 (e + t)Í d (t - 1) ˙dt = Í e + 1 ˙ ÍÎ 0 ˙˚ ÍÎd ( t - 3) ˙˚

2.

From Equation 7b, f * d = f ( t) .

-t

-2

2

2

210 • Chapter 7 Laplace Transforms

3.

1

Ê

1ˆ

È Ê1

ˆ˘

3

È Ê1

ˆ˘

Ú sin [p(t - t )]d ÁË t - 2˜¯ dt = sin ÍÎpÁË 2 - t ˜¯ ˙˚ = 4 fi sinÍÎpÁË 2 - t ˜¯ ˙˚ = 2

2

0

0

0

0

3 2

1 1 1 Ê1 ˆ p One possible t0 : p Á - t0 ˜ = fi - t0 = fi t0 = . Ë2 ¯ 3 2 3 6 5

t nd ( t - 2) dt = 2 n = 8 fi n = 3.

4.

Ú

5.

f ( t) = 1 - h (1 - t) = h ( t - 1) for all t except t=1.

6.

t Ï 0, t < 1 g( t) = Ú h (l - 1) dl = Ì . Therefore, g( t) = ( t - 1) h ( t - 1) . 0 Ót - 1, t ≥ 1

7.

k = h (2 - t) - h (1 - t) = h ( t - 1) - h ( t - 2) for all t except t=1, 2.

8.

t Ï 0, g( t) = Ú eatd ( t - t0 ) dt = Ì at 0 0 Óe ,

9 (a).

(e y )¢ = e

1

-t

-t

0 £ t £ t0 . Therefore, t0 = 2, e 2a = e -2 fi a = -1. t0 < t < •

fi e - t y = -e - t + C fi y = -1 + Ce t . From the initial condition, we have

y (0) = 0 = -1 + C . Thus C=1 and y = -1 + e t .

9 (b).

sF - F = 1 fi F =

1 . Therefore, f ( t) = e t . s-1

t

t

t

0

0

0

9 (c). f * g = Ú e( t -l ) h (l ) dl = Ú e t -l dl = e t (-e -l ) = -1 + e t 10 (a). (e - t y )¢ = 1 fi e - t y = t + C fi y = te t + Ce t . From the initial condition, we have y (0) = 0 = C . Thus y = te t . 10 (b). From 9b, f ( t) = e t . t

t

0

0

10 (c). f * g = Ú e( t -l )el dl = e t Ú dl = te t . 11 (a). (e - t y )¢ = te - t fi e - t y = - te - t - e - t + C fi y = -( t + 1) + Ce t . From the initial condition, we have y (0) = 0 = -1 + C . Thus C=1 and y = e t - ( t + 1) .

11 (b). s F - F = 1 fi F =

1 . Therefore, f ( t) = e t . s-1

t

t

0

0

11 (c). f * g = Ú e( t -l )ldl = e t (-le -l - e -l ) = e t - t - 1 12.

sY + Y =

2 2 2 e-s A B , + e-s fi Y = + = + fi A = 2, B = -2 . Therefore, s s( s + 1) s + 1 s( s + 1) s s + 1

y ( t) = 2 - 2e - t + e -( t -1)h ( t - 1) .

Chapter 7 Laplace Transforms • 211

e-s e -2 s . Therefore, y ( t) = e -( t -1) h ( t - 1) - e -( t - 2) h ( t - 2) . s +1 s +1

13.

sY + Y = e - s - e -2 s fi Y =

14.

s2Y = e - s - e -3 s fi Y =

15.

(s

16.

s2Y - s - 2( sY - 1) = e - s fi Y =

2

e - s e -3 s - 2 . Therefore, y ( t) = ( t - 1) h ( t - 1) - ( t - 3) h ( t - 3) . s2 s

+ 4 p 2 )Y = 2 pe -2 s fi Y =

2p e -2 s . Therefore, y ( t) = sin(2 p( t - 2)) h ( t - 2) . s + 4p2 2

s- 2 e-s 1 e-s . + = + s( s - 2) s( s - 2) s s( s - 2)

1 1 ˆ 1 A B 1 1 Ê 1 = + fi A = - , B = . Therefore, Y = + e - s Á - + ˜ and Ë 2 s s( s - 2) ¯ s s( s - 2) s s - 2 2 2 1 1 y ( t) = 1 - h ( t - 1) + e 2( t -1)h ( t - 1) . 2 2

212 • Chapter 7 Laplace Transforms

17. 18.

(s

2

+ 2 s + 2)Y = e - s fi Y = e - s

2

s Y - 1 + 2 sY + Y = e

-2 s

1 . Therefore, y ( t) = e -( t -1) sin( t - 1) h ( t - 1) . 2 (s + 1) + 1

fi ( s + 2 s + 1)Y = 1 + e 2

-2 s

1 e -2 s fiY = + . Therefore, (s + 1) 2 (s + 1) 2

y ( t) = te - t + ( t - 2)e -( t - 2 )h ( t - 2) .

19.

Ès - 1 -1 ˘ È1 ˘ -1 È1 ˘ sY = AY + e - s Í ˙ fi Y = e - s ( sI - A) Í ˙ . sI - A = Í ˙ , so Î -1 s - 1˚ Î0 ˚ Î0 ˚

(sI - A)

-1

1 Ès - 1 1 ˘ e -2 s Ès - 1 1 ˘È1 ˘ e -2 s Ès - 1˘ = 2 = . Then Y = . s - 1˙˚ÍÎ0 ˙˚ s( s - 2) ÍÎ 1 ˙˚ s - 1˙˚ s( s - 2) ÍÎ 1 s - 2 s ÍÎ 1

1 1 1 1 1 1 1 1 s-1 s = 2 + 2 , so , = + = 2+ 2 fi s( s - 2) s s( s - 2) s( s - 2) s s- 2 s( s - 2) s s - 2

y1 ( t) =

1 1 1 + e 2( t -1) ) h ( t - 1) and y 2 ( t) = (-1 + e 2( t -1) ) h ( t - 1) . Finally, we have ( 2 2

È1 2( t -1) )h(t - 1) ˘˙ Í 2 (1 + e y ( t) = Í ˙. 1 2( t -1) Í (-1 + e )h(t - 1)˙˚ Î2

Chapter 7 Laplace Transforms • 213

20.

Ê È0 ˘ È0 ˘ Ès - 2 -1 ˘ È1 ˘ È1 ˘ˆ -1 sY = AY + Í1 ˙ - e - s Í ˙ fi Y = ( sI - A) Á Í1 ˙ - e - s Í ˙˜ . sI - A = Í ˙ , so ÁÍ ˙ ˜ s 0 1 0 ÍÎ s ˙˚ Î ˚ Î0 ˚ Î ˚ ËÎs˚ ¯

(sI - A) -1 =

Ès - 1 1 ˘ 1 . Then s - 2 ˙˚ ( s - 1)( s - 2) ÍÎ 0

e-s ˘ 1 È1 - s ˘ È e s ( 1 ) Ís ˙ Í s( s - 1)( s - 2) ( s - 2) ˙ 1 Y= ˙. ˙= Í s- 2 1 ( s - 1)( s - 2) ÍÍ ˙ ˙ Í ˙˚ s Î ˚ ÎÍ s( s - 1) A B C 1 1 1 1 1 1 = + + fi A = , B = -1, C = ; =- + , so s( s - 1)( s - 2) s s - 1 s - 2 2 s( s - 1) s s-1 2

y1 ( t) =

1 t 1 2 t 2( t -1) - e + e - e h ( t - 1) and y 2 ( t) = -1 + e t . 2 2

Chapter 8 Nonlinear Systems Section 8.1 1 (a). F o r

1 (b).

1 (c). 2 (a).

2 (b).

2 (c). 3 (a).

y ¢¢ + ty = sin y ¢, y (0) = 0, y ¢ (0) = 1,

È y1 ( t) ˘ È y ( t) ˘ y( t) = Í ˙= Í ˙. Î y 2 ( t) ˚ Î y ¢ ( t) ˚ È y1 (0) ˘ È y (0) ˘ È0 ˘ y (0) = Í ˙= Í ˙ = Í ˙. Î y 2 (0) ˚ Î y ¢ (0) ˚ Î1 ˚

let

Thus,

y¢ y2 È y1¢ ˘ È y ¢ ˘ È ˘ È ˘ =Í y¢ = Í ˙ = Í ˙ = Í ˙ ˙, Î y 2¢ ˚ Î y ¢¢ ˚ Î- ty + sin y ¢ ˚ Î- ty1 + sin y 2 ˚ y2 È f1 ( t, y1, y 2 ) ˘ È ˘ =Í From part (a), f ( t, y ) = Í ˙ ˙ . Therefore, the requested partial Î f 2 ( t, y1, y 2 ) ˚ Î- ty1 + sin y 2 ˚ ∂f ∂f ∂f ∂f derivatives are 1 = 0, 1 = 1, 2 = - t, 2 = cos y 2 . ∂y 2 ∂y1 ∂y 2 ∂y1 There are no points in 3-dimensional space where the hypotheses of Theorem 8.1 fail to be satisfied. È y1 ( t) ˘ È y ( t) ˘ For y ¢¢ + ( y ¢ ) 3 + y1/ 3 = tan( t / 2), y (1) = 1, y ¢ (1) = -2 , let y( t) = Í ˙= Í ˙ . Thus, Î y 2 ( t) ˚ Î y ¢ ( t) ˚ y2 È y1¢ ˘ È ˘ È1˘ y¢ = Í ˙ = Í ˙. 1/ 3 3 ˙ , y (1) = Í Î y 2¢ ˚ Îtan( t / 2) - y1 - y 2 ˚ Î-2 ˚ È f1 ( t, y1, y 2 ) ˘ f ( t, y ) = Í For the requested partial derivatives are ˙, Î f 2 ( t, y1, y 2 ) ˚ ∂f ∂f ∂f 1 ∂f1 = 0, 1 = 1, 2 = - y1-2 / 3 , 2 = -3 y 22 . ∂y 2 ∂y1 ∂y 2 3 ∂y1 The hypotheses of Theorem 8.1 are not satisfied at t = ±(2 n + 1)p / 2 and y1 = 0 . For y ¢¢ + t -1 (1 + y + 2 y ¢ ) -1 = t -1e - t , y (2) = 2, y ¢ (2) = 1, let È y1 ( t) ˘ È y ( t) ˘ y( t) = Í ˙= Í ˙ . Thus, Î y 2 ( t) ˚ Î y ¢ ( t) ˚ y¢ y2 È y1¢ ˘ È y ¢ ˘ È ˘ È ˘ y ¢ = Í ˙ = Í ˙ = Í -1 -1 -1 - t ˙ = Í -1 -1 -1 - t ˙ , Î y 2¢ ˚ Î y ¢¢ ˚ Î- t (1 + y + 2 y ¢ ) + t e ˚ Î- t (1 + y1 + 2 y 2 ) + t e ˚

È y1 (2) ˘ È y (2) ˘ È2 ˘ y (2) = Í ˙= Í ˙. ˙= Í Î y 2 (2) ˚ Î y ¢ (2) ˚ Î1 ˚ y2 È f1 ( t, y1, y 2 ) ˘ È ˘ = Í -1 3 (b). From part (a), f ( t, y ) = Í ˙ -1 -1 - t ˙ . Î f 2 ( t, y1, y 2 ) ˚ Î- t (1 + y1 + 2 y 2 ) + t e ˚ Therefore, the requested partial derivatives are ∂f ∂f ∂f ∂f1 = 0, 1 = 1, 2 = t -1 (1 + y1 + 2 y 2 ) -2 , 2 = 2 t -1 (1 + y1 + 2 y 2 ) -2 . ∂y 2 ∂y1 ∂y 2 ∂y1

Chapter 8 Nonlinear Systems • 215

3 (c). The hypotheses of Theorem 8.1 are satisfied everywhere except on the planes t = 0 and 1 + y1 + 2 y 2 = 0 . 4 (a). For y ¢¢¢ + cos( ty ¢ ) = t( y ¢¢ ) 2 , y (0) = 1, y ¢ (0) = 1, y ¢¢ (0) = -2 , let y2 ˘ È1˘ È y1¢ ˘ È È y1 ( t) ˘ È y ( t) ˘ ˙ Í , y (0) = Í 1 ˙ . y( t) = Í y 2 ( t) ˙ = Í y ¢ ( t) ˙ . Thus, y ¢ = Í y 2¢ ˙ = y3 ˙ Í ˙ Í ˙ Í ˙ Í Í ˙ 2 Î-2 ˚ Î y 3¢ ˚ ÎÍ- cos( ty 2 ) + y 3 ˚˙ Î y 3 ( t) ˚ Î y ¢¢ ( t) ˚ È f1 ( t, y1, y 2 , y 3 ) ˘ 4 (b). For f ( t, y ) = Í f 2 ( t, y1, y 2 , y 3 ) ˙ , the requested partial derivatives are ˙ Í Î f 3 ( t, y1, y 2 , y 3 ) ˚ ∂f ∂f ∂f ∂f ∂f ∂f1 = 0, 1 = 1, 1 = 0, 2 = 0, 2 = 0, 2 = 1, ∂y 3 ∂y 2 ∂y1 ∂y 3 ∂y 2 ∂y1 . ∂f 3 ∂f 3 ∂f 3 = 2 ty 3 = t sin( ty 2 ), = 0, ∂y 3 ∂y 2 ∂y1 4 (c). The hypotheses of Theorem 8.1 are satisfied in all of ty1 y 2 y 3 - space . 5 (a). For y ¢¢¢ + 2 t1/ 3 ( y - 2) -1 ( y ¢¢ + 2) -1 = 0, y (0) = 0, y ¢ (0) = 2, y ¢¢ (0) = 2 , let È y1 ( t) ˘ È y ( t) ˘ ˙ ˙ Í Í y( t) = Í y 2 ( t) ˙ = Í y ¢ ( t) ˙ . Thus, ÍÎ y 3 ( t) ˙˚ ÍÎ y ¢¢ ( t) ˙˚

˘ ˘ È y¢ y2 È y1¢ ˘ È y ¢ ˘ È ˙ ˙ Í Í ˙ Í ˙ Í y ¢ = Í y 2¢ ˙ = Í y ¢¢ ˙ = Í y ¢¢ y3 ˙, ˙= Í 1/ 3 1/ 3 -1 -1 -1 -1 ÍÎ y 3¢ ˙˚ ÍÎ y ¢¢¢ ˙˚ ÍÎ-2 t ( y - 2) ( y ¢¢ + 2) ˙˚ ÍÎ-2 t ( y1 - 2) ( y 3 + 2) ˙˚ È y1 (0) ˘ È y (0) ˘ È0 ˘ ˙ Í ˙ ˙ Í Í y (0) = Í y 2 (0) ˙ = Í y ¢ (0) ˙ = Í2 ˙ . ÍÎ y 3 (0) ˙˚ ÍÎ y ¢¢ (0) ˙˚ ÍÎ2 ˙˚ ˘ È f1 ( t, y1, y 2 , y 3 ) ˘ È y2 ˙ ˙ Í Í 5 (b). From part (a), f ( t, y ) = Í f 2 ( t, y1, y 2 , y 3 ) ˙ = Í y3 ˙. 1/ 3 -1 -1 ÍÎ f 3 ( t, y1, y 2 , y 3 ) ˙˚ ÍÎ-2 t ( y1 - 2) ( y 2 + 2) ˙˚ Therefore, the requested partial derivatives are ∂f ∂f ∂f1 = 0, 1 = 1, 1 = 0 ∂y 3 ∂y 2 ∂y1 ∂f ∂f ∂f 2 = 0, 2 = 0, 2 = 1 . ∂y 3 ∂y 2 ∂y1 ∂f ∂f ∂f 3 = 2 t1/ 3 ( y1 - 2) -2 ( y 3 + 2) -1, 3 = 0, 3 = 2 t1/ 3 ( y1 - 2) -1 ( y 3 + 2) -2 ∂y 3 ∂y 2 ∂y1 5 (c). The hypotheses of Theorem 8.1 are satisfied everywhere except on the “hyperplanes” y1 = 2 and y 3 = -2 . 6. Since y 2¢ = t cos2 ( y 2 ) - 3 y1 + t 4 , it follows that the scalar problem is y ¢¢ = t cos2 ( y ¢ ) - 3 y + t 4 , y (2) = 1, y ¢ (2) = -1 . 7. Since y 2¢ = y 2 tan y1 + e y 2 , it follows that the scalar problem is y ¢¢ = y ¢ tan y + e y ¢ , y (0) = 0, y ¢ (0) = 1.

216 • Chapter 8 Nonlinear Systems

8. 9. 11. 14 (a).

14 (b). 14 (c).

14 (d).

15 (a).

15 (b).

15 (c). 16 (a).

Since y 3¢ = y1 y 2 + y 32 , it follows that the scalar problem is y ¢¢¢ = yy ¢ + ( y ¢¢ ) 2 , y (-1) = -1, y ¢ (-1) = 2, y ¢¢ (-1) = -4 . Since y 3¢ = ( y 2 y 3 + t 2 )1/ 2 , it follows that the scalar problem is y ¢¢¢ = ( y ¢y ¢¢ + t 2 )1/ 2 , y (1) = 1, y ¢ (1) = 1 / 2, y ¢¢ (1) = 3 . Laplace transforms cannot be productively used because the equation is nonlinear. Let a = p / (2d ). Then tan ax = ax + (1 / 3) a 3 x 3 + (2 / 15) a 5 x 5 + L . Retaining the first term of the Maclaurin series in equation (7), we have mx ¢¢ + (2 kd / p )tan(px / 2d ) ª mx ¢¢ + (2 kd / p )(px / 2d ) = mx ¢¢ + kx . As in part (a), retaining the first two terms of the Maclaurin series in equation (7) results in equation (8). y2 È y1¢ ˘ È ˘ Equation (7) becomes y ¢ = Í ˙ = Í ˙. Î y 2¢ ˚ Î-(2 kd / mp )tan(py1 / 2d ) ˚ y2 È y1¢ ˘ È ˘ Equation (8) becomes y ¢ = Í ˙ = Í 2 2 3 ˙. Î y 2¢ ˚ Î-( k / m)( y1 + (p / 12d ) y1 ) ˚ The system version of equation (7) satisfies the hypotheses of Theorem 8.1 everywhere except along y1 = ±(2 n + 1)p / 2 . The system version of equation (8) satisfies the hypotheses of Theorem 8.1 everywhere in ty1 y 2 - space dc de Adding equations 3 and 4, we obtain + = 0 . Thus, using the linearity of differentiation, dt dt d (c + e) = 0 and hence, c ( t) + e( t) ∫ e0 is a constant function. dt Substituting e( t) = e0 - c ( t) in equations 1 and 3, we find da dc = - k1e0 a( t) + k1c ( t) a( t) + k1¢c ( t) and = k1e0 a( t) - k1c ( t) a( t) - ( k1¢ + k2 )c ( t) . dt dt The hypotheses of Theorem 8.1 are satisfied for all points in ( t, a, c ) - space . At the instant shown in the figure,

Vsub = (2 / 3)pR 3 + Ú

y( t ) 0

pr 2 dy =(2 / 3)pR 3 + Ú

y( t ) 0

p ( R 2 - y 2 ) dy

= (2 / 3)pR 3 + p [ R 2 y ( t) - (1 / 3)( y ( t)) 3 ] . 16 (b). Equation (10) is physically relevant as long as - R £ y ( t) £ R .

Section 8.2 1.

For

x ¢ = x (-1 + y ) y ¢ = y (1 - x ) , we see that x ¢ = 0 if (a) x = 0 or (b) y = 1. In Case (a), we have y ¢ = 0 only if y = 0 , yielding the equilibrium point ( x, y ) = (0, 0) . In Case (b), we have y ¢ = 0 only if x = 1, yielding the equilibrium point ( x, y ) = (11 , ).

Chapter 8 Nonlinear Systems • 217

2.

For

x ¢ = y ( x + 3)

3.

4.

5.

6.

7.

8.

y ¢ = ( x - 1)( y - 2) , we see that x ¢ = 0 if (a) x = -3 or (b) y = 0 . In Case (a), we have y ¢ = 0 only if y = 2 , yielding the equilibrium point ( x, y ) = (-3, 2) . In Case (b), we have y ¢ = 0 only if x = 1, yielding the equilibrium point ( x, y ) = (1, 0) . For x ¢ = ( x - 2)( y + 1) y¢ = x 2 - 4 x + 3 , we see that x ¢ = 0 if (a) x = 2 or (b) y = -1. In Case (a), we cannot have y ¢ = 0 . In Case (b), we have y ¢ = 0 only if x = 1 or x = 3 , yielding the equilibrium points ( x, y ) = (1, - 1) and ( x, y ) = ( 3, - 1) . For x ¢ = ( x - 1)( y + 1)

y ¢ = ( x - 2) y , we see that x ¢ = 0 if (a) x = 1 or (b) y = -1. In Case (a), we have y ¢ = 0 only if y = 0 , yielding the equilibrium point ( x, y ) = (1, 0) . In Case (b), we have y ¢ = 0 only if x = 2 , yielding the equilibrium point ( x, y ) = (2, - 1) . For x¢ = x( x - 2 y) y ¢ = y (3x - y ) , we see that x ¢ = 0 if (a) x = 0 or (b) x = 2 y . In Case (a), we have y ¢ = 0 only if y = 0 , yielding the equilibrium point ( x, y ) = (0, 0) . In Case (b), we have y ¢ = 0 only if y = 0 , yielding the same equilibrium point as in Case (a), ( x, y ) = (0, 0) . For x¢ = y( y - x) y¢ = x( x + 2 y) , we see that x ¢ = 0 if (a) y = 0 or (b) y = x . In Case (a), we have y ¢ = 0 only if x = 0 , yielding the equilibrium point ( x, y ) = (0, 0) . In Case (b), we have y ¢ = 0 only if x = 0 , yielding the same equilibrium point ( x, y ) = (0, 0) . For x¢ = x 2 + y 2 - 8 y¢ = x 2 - y 2 , we see that y ¢ = 0 if x 2 = y 2 . Using this requirement in the first equation, we see that x ¢ = 0 requires 2 x 2 - 8 = 0 or x = ±2 . Since y = ± x , we find 4 equilibrium points, (2, 2), (2, - 2), (-2, - 2), and (-2, 2). For x¢ = x 2 + 2 y 2 - 3 y¢ = 2 x 2 + y 2 - 3 , we see that x ¢ = 0 if x 2 = 3 - 2 y 2 . In this event, we have y ¢ = 0 only if 2( 3 - 2 y 2 ) + y 2 - 3 = 0 . Solving for y we obtain y = ±1 . Then, since x 2 = 3 - 2 y 2 , we see that x = ±1 for each choice of y. The equilibrium points are ( x, y ) = (11 , ), (-1, 1), (1, - 1), (-1, - 1) .

218 • Chapter 8 Nonlinear Systems

9.

For x¢ = y - 1

10.

y¢ = x( y + x) z¢ = y (2 - z) , we see that x ¢ = 0 requires y = 1. Using this requirement in the second equation, we see that y ¢ = 0 requires x (1 + x ) = 0 . Thus, we need in Case (a) x = 0 or in Case (b), x = -1. Finally, z¢ = 0 requires z = 2 since y is nonzero. We obtain 2 equilibrium points, ( x, y, z) = (0,1, 2) and ( x, y, z) = (-11 , , 2) . For x¢ = z2 - 1

y ¢ = z(1 - 2 x + y )

11.

12.

13.

14.

z¢ = -(1 - x - y ) 2 , we see that x ¢ = 0 requires z = ±1. Using this requirement in the second equation, we see that y ¢ = 0 requires 1 - 2 x + y = 0 while z¢ = 0 requires 1 - x - y = 0 . Satisfying y ¢ = 0 and z¢ = 0 therefore requires x = 2 / 3 and y = 1 / 3. Combining this requirement with z = ±1, we obtain 2 equilibrium points, ( x, y, z) = (2 / 3,1 / 3,1) and ( x, y, z) = (2 / 3,1 / 3, - 1) . Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2

y 2¢ = - y1 - y13 Since y 2¢ = - y1 (1 + y12 ) , we cannot have y 2¢ = 0 unless y1 = 0 . Similarly, from the first equation, y1¢ = 0 requires y 2 = 0 . Thus, the only equilibrium point is ( y1, y 2 ) = ( y, y ¢ ) = (0, 0) . Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2 y 2¢ = 1 - e y1 y 2 - sin 2 (py1 ) Thus, the equilibrium points are ( y1, y 2 ) = ( y, y ¢ ) = ( n + 0.5, 0), n = 0, ±1, ±2,K . Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2 y 2¢ = 1 - y12 - 2(1 + y14 ) -1 y 2 From the first equation, y1¢ = 0 requires y 2 = 0 . Thus, in the second equation, y 2¢ = 0 requires 1 - y12 = 0 or y1 = ±1. There are two equilibrium points ( y1, y 2 ) = ( y, y ¢ ) = (1, 0) and ( y1, y 2 ) = ( y, y ¢ ) = (-1, 0) . Making the substitution y1 = y, y 2 = y ¢, and y 3 = y ¢¢ the scalar equation can be expressed as the system y1¢ = y 2 y 2¢ = y 3 y 3¢ = 1 + y 3 - 2sin y1 Thus, the equilibrium points are ( y1, y 2 , y 3 ) = ((p / 6) + 2 np , 0, 0) and

( y1, y 2 , y 3 ) = ((5p / 6) + 2 np , 0, 0), n = 0, ±1, ±2,K .

Chapter 8 Nonlinear Systems • 219

15.

Making the substitution y1 = y, y 2 = y ¢ and y 3 = y ¢¢ , the scalar equation can be expressed as the system y1¢ = y 2

y 2¢ = y 3

16. 17.

18.

19.

20.

21.

y 3¢ = y 22 + ( y12 - 4 )(2 + y 22 ) -1 . From the first equation, y1¢ = 0 requires y 2 = 0 while (by the second equation) y 2¢ = 0 requires y 3 = 0 . Having these requirements, the third equation tells us that y 3¢ = 0 only if y1 = ±2 . Hence, There are two equilibrium points ( y1, y 2 , y 3 ) = ( y, y ¢, y ¢¢ ) = (2, 0, 0) and ( y1, y 2 , y 3 ) = ( y, y ¢, y ¢¢ ) = (-2, 0, 0) . Since (0, 0) is an equilibrium point, we know b = 0 and d = 0 . Similarly, since (2,1) is an equilibrium point, we know 2a + 2 = 0 and g - 6 = 0. Thus, a = -1 and g = 6 . Since (11 , ) is an equilibrium point, we know a + b + 2 = 0 and g + d - 1 = 0 . Similarly, since (2, 0) is an equilibrium point, we know 2a + 2 = 0 and 2g - 1 = 0. Thus, a = -1 and g = 1 / 2 . Using the equations derived from the equilibrium point (11 , ) , we have -1 + b + 2 = 0 and (1 / 2) + d - 1 = 0 . Therefore, b = -1 and d = 1 / 2 . The slope of a phase plane trajectory is given by y ¢ / x ¢ = g( x, y ) / f ( x, y ) , see equation (9). As given, g(2,1) / f (2,1) = 1 and g(1, - 1) / f (1, - 1) = 0 . Therefore, g(1, - 1) = 0 and so b = 2 . Knowing b = 2 and g(2,1) / f (2,1) = 1, we obtain ( 3 + b ) / (2 + a ) = 1 or 5 / (2 + a ) = 1 . Thus, we obtain a = 3. The slope of a phase plane trajectory is given by y ¢ / x ¢ = g( x, y ) / f ( x, y ) , see equation (9). As given, g(11 , ) / f (11 , ) = 0 and g(1, - 1) / f (1, - 1) = 4 . Therefore, g(11 , ) = 0 and so and 2 + g = 0 or g = -2 . Knowing g = -2 g(1, - 1) / f (1, - 1) = 4 , we obtain (2 - g ) / (a - b + 1) = 4 or 1 / (a - b + 1) = 1. Finally, since there is a vertical tangent at (0, - 1) we know f (0, - 1) = 0 , and thus -b + 1 = 0 . Using b = 1 along with the prior equation 1 / (a - b + 1) = 1, we obtain a = 1. The slope of a phase plane trajectory is given by y ¢ / x ¢ = g( x, y ) / f ( x, y ) , see equation (9). As given, g(1, 2) / f (1, 2) = 1 / 6 and thus 1 / 6 = g(1, 2) / f (1, 2) = (-1 + 0.5) / (5 - 2 n ) . Solving for n, we obtain n = 3. Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2

y 2¢ = y 2 - 2 y12 + a . Since ( y1, y 2 ) = (2, 0) is an equilibrium point, it follows that 2 y12 = 8 = a . 22 (a). v = 4 i - 3 j 22 (b). v = 15i + j 22 (a). v = - j È-9 1 ˘ 24. For A = Í ˙ , the eigenvalues are l1 = -10 and l 2 = -8 with corresponding eigenvectors Î 1 -9 ˚ È1˘ È1˘ u1 = Í ˙ and u2 = Í ˙ . The general solution is Î-1˚ Î1˚ È1˘ È1˘ y( t) = c1e -10 t Í ˙ + c 2e -8 t Í ˙ and hence all solution points are attracted to the origin. Thus, the Î-1˚ Î1˚ direction field corresponding to the given matrix is C.

220 • Chapter 8 Nonlinear Systems

È-1 -3˘ For A = Í ˙ , the eigenvalues are l1 = -4 and l 2 = 2 with corresponding eigenvectors Î-3 -1˚ È1˘ È1˘ u1 = Í ˙ and u2 = Í ˙ . The general solution is Î1˚ Î-1˚ È1˘ È1˘ y( t) = c1e -4 t Í ˙ + c 2e 2 t Í ˙ and hence solution points that begin on the line y = x are attracted Î1˚ Î-1˚ to the origin whereas those that begin on the line y = - x are repelled away from the origin. Thus, the direction field corresponding to the given matrix is B. È-4 6 ˘ 26. For A = Í ˙ , the eigenvalues are l1 = -10 and l 2 = 2 with corresponding eigenvectors Î 6 -4 ˚ È1˘ È1˘ u1 = Í ˙ and u2 = Í ˙ . The general solution is Î-1˚ Î1˚ È1˘ È1˘ y( t) = c1e -10 t Í ˙ + c 2e 2 t Í ˙ and hence solution points that begin on the line y = x are repelled Î-1˚ Î1˚ away from the origin whereas those that begin on the line y = - x are attracted to the origin. Thus, the direction field corresponding to the given matrix is D. È4 2 ˘ 27. For A = Í ˙ , the eigenvalues are l1 = 6 and l 2 = 2 with corresponding eigenvectors Î2 4 ˚ È1˘ È1˘ u1 = Í ˙ and u2 = Í ˙ . The general solution is Î1˚ Î-1˚ È1˘ È1˘ y( t) = c1e 6 t Í ˙ + c 2e 2 t Í ˙ and hence solution points that begin on the line y = x are repelled Î1˚ Î-1˚ away from the origin as are those that begin on the line y = - x . Thus, the direction field corresponding to the given matrix is A. 28. The phase plane point (a , 0) is an equilibrium point when a is a root of f ( y) = 0 . 29 (a). Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2 25.

y 2¢ = - y1 - y13 . The nullclines are the lines y1 = 0 and y 2 = 0 . The only equilibrium point is the point (0, 0) . 30 (a). Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2 y 2¢ = - y1 (1 - y12 ) . The nullclines are the lines y1 = 0, y1 = ±1, and y 2 = 0 . The equilibrium points are (0, 0), (-1, 0), (1, 0) .

Chapter 8 Nonlinear Systems • 221

31 (a). Making the substitution y1 = y and y 2 = y ¢ the scalar equation can be expressed as the system y1¢ = y 2

32 (a). 33 (a). 34 (a).

35 (a).

36 (a).

y 2¢ = 1 - 2sin 2 y1 . The nullclines are the lines y1 = ±(p / 4 ) + np , n = 0, ± 1, ± 2,Kand the line y 2 = 0 The equilibrium points are ( ±(p / 4 ) + np , 0), n = 0, ± 1, ± 2,K. The nullclines are the lines y = 3 x - 2 and y = x . These lines intersect at the point (11 ,) yielding the only equilibrium point. The nullclines are the lines y = 2 - x and y = x . These lines intersect at the point (11 , ) yielding the only equilibrium point. The nullclines are the lines y = 2 x - 2 and y = 4 - x where f = 0 and the line y = (1 / 2) x where g = 0 . The lines f = 0 and g = 0 intersect at the points ( 4 / 3, 2 / 3) and (8 / 3, 4 / 3) yielding the only equilibrium points. The nullclines are the lines y = 2 x - 6 and y = x, where f = 0 and the line y = - x, where g = 0 . The lines f = 0 and g = 0 intersect at the points (0, 0) and (2, - 2) yielding the only equilibrium points. The nullclines are the curves y = 1 - x 2 and y = -1 + x 2 . These curves intersect at the equilibrium points (-1, 0) and (1, 0) .

Section 8.3 1 (a). Given x ¢¢ + 4 x = 0 , multiply by x ¢ to obtain x ¢x ¢¢ + 4 x ¢x = 0 . Integrating, we obtain 2 0.5( x ¢ ) + 2 x 2 = C . x¢ = y 1 (b). The equation x ¢¢ + 4 x = 0 can be expressed as With this notation, the conserved y ¢ = -4 x . quantity found in part (a) is 0.5 y 2 + 2 x 2 = C . The graph passes through the point ( x, y ) = (11 ,) when C = 2.5 . 1 (c). At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i - 4 j . The velocity vector is tangent to the graph and indicates that the graph is traversed in the clockwise direction as t increases. 2 (a). Given x ¢¢ - ( x + 1) = 0 , multiply by x ¢ to obtain x ¢x ¢¢ - x ¢ ( x + 1) = 0 . Integrating, we obtain ( x ¢ ) 2 - ( x + 1) 2 = C . x¢ = y 2 (b). The equation x ¢¢ - ( x + 1) = 0 can be expressed as With this notation, the conserved y ¢ = x + 1. quantity found in part (a) is y 2 - ( x + 1) 2 = C . The graph passes through the point ( x, y ) = (11 ,) when C = -3 . 2 (c). At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i + 2 j . The velocity vector indicates that the solution point moves upward and to the right along the right branch of the hyperbola as t increases. 3 (a). Given x ¢¢ + x 3 = 0 , multiply by x ¢ to obtain x ¢x ¢¢ + x ¢x 3 = 0 . Integrating, we obtain 2 0.5( x ¢ ) + 0.25 x 4 = C .

222 • Chapter 8 Nonlinear Systems

3 (b). The equation x ¢¢ + x 3 = 0 can be expressed as

3 (c). 4 (a). 4 (b).

4 (c). 5 (a).

5 (b).

5 (c). 6 (a). 6 (b).

6 (c). 7.

8.

x¢ = y

With this notation, the conserved y¢ = - x 3 . quantity found in part (a) is 0.5 y 2 + 0.25 x 4 = C . The graph passes through the point ( x, y ) = (11 , ) when C = 0.75 . At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i - j . The velocity vector is tangent to the graph and indicates that the graph is traversed in the clockwise direction as t increases. Given x ¢¢ - ( x 3 + p sin px ) = 0 , multiply by x ¢ to obtain x ¢x ¢¢ - x ¢ ( x 3 + p sin px ) = 0 . Integrating, we obtain 2( x ¢ ) 2 - ( x 4 - 4 cospx ) = C . x¢ = y With this notation, The equation x ¢¢ - ( x 3 + p sin px ) = 0 can be expressed as y ¢ = x 3 + p sin px . the conserved quantity found in part (a) is 2 y 2 - ( x 4 - 4 cospx ) = C . The graph passes through the point ( x, y ) = (11 , ) when C = -3 . At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i + j . The velocity vector indicates that the solution point moves upward and to the right along the right branch of the graph as t increases. Given x ¢¢ + x 2 = 0 , multiply by x ¢ to obtain x ¢x ¢¢ + x ¢x 2 = 0 . Integrating, we obtain 2 0.5( x ¢ ) + (1 / 3) x 3 = C . x¢ = y The equation x ¢¢ + x 2 = 0 can be expressed as With this notation, the conserved y¢ = - x 2 . quantity found in part (a) is 0.5 y 2 + (1 / 3) x 3 = C . The graph passes through the point ( x, y ) = (11 , ) when C = 5 / 6 . At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i - j . The velocity vector is tangent to the graph and indicates that the solution point moves “down the graph” as t increases. Given x ¢¢ + x / (1 + x 2 ) = 0 , multiply by x ¢ to obtain x ¢x ¢¢ + x ¢x / (1 + x 2 ) = 0 . Integrating, we obtain ( x ¢ ) 2 + ln(1 + x 2 ) = C . x¢ = y The equation x ¢¢ + x / (1 + x 2 ) = 0 can be expressed as With this notation, the y ¢ = - x / (1 + x 2 ). conserved quantity found in part (a) is y 2 + ln(1 + x 2 ) = C . The graph passes through the point ( x, y ) = (11 , ) when C = 1 + ln 2 . At (11 , ) , the velocity vector is v = x ¢i + y ¢j = i - 0.5 j . The velocity vector indicates that the solution point moves clockwise along the curve as t increases. Rewriting the conservation law in terms of x and x ¢ , we have ( x ¢ ) 2 + x 2 cos x = C . Differentiating with respect to t, we obtain 2 x ¢x ¢¢ + 2 x ¢x cos x - x 2 x ¢ sin x = 0 or x ¢ (2 x ¢¢ + 2 x cos x - x 2 sin x ) = 0 . Therefore, the differential equation is x ¢¢ + x cos x - 0.5 x 2 sin x = 0 . 2 Rewriting the conservation law in terms of x and x ¢ , we have ( x ¢ ) 2 - e - x = C . Differentiating 2 with respect to t, we obtain 2 x ¢x ¢¢ - (e - x )(-2 xx ¢ ) = 0 . Therefore, the differential equation is 2 x ¢¢ + xe - x = 0 .

9 (a). The equation x ¢¢ + x + x 3 = 0 can be expressed as

x¢ = y

The nullclines are the lines y¢ = - x - x 3 . defined by y = 0 and - x (1 + x 2 ) = 0 ; the lines y = 0 and x = 0 . Thus, the only equilibrium point is the point ( x, y ) = (0, 0) .

Chapter 8 Nonlinear Systems • 223

9 (b). The velocity vector has the form v( x, y ) = yi - ( x + x 3 ) j. Thus, we obtain v(11 , ) = i - 2j, v(1, - 1) = -i - 2 j , v(-11 , ) = i + 2 j , and v(-1, - 1) = -i + 2 j . 9 (c). Multiplying by x ¢ , the equation becomes x ¢x ¢¢ + x ¢ ( x + x 3 ) = 0 . Integrating, we obtain 0.5( x ¢ ) 2 + 0.5 x 2 + 0.25 x 4 = C or 2 y 2 + 2 x 2 + x 4 = C1 . The graph of the conserved quantity passes through the point (11 , ) when C1 = 5 . The graph passes through the other three points and is consistent with the sketch in part (b). 10. Since x ¢¢ + ax = 0 it follows that 0.5( x ¢ ) 2 + 0.5ax 2 = C1 and hence ax 2 + y 2 = C . 10 (a). Figure A is a circle of radius 2 and thus a = 1 and x 2 + y 2 = 4 . Figure B is a hyperbola with asymptotes y = ± x . Since (0, 2) is on the graph, we see that a = -1 and y 2 - x 2 = 4 . Figure C shows horizontal lines, y = ±2 . Thus, a = 0 . 10 (b). The solution point in Figure A travels clockwise around the circle. Solution points in Figure B move to the right on the upper branch and to the left on the lower branch. Solutions points in Figure C move to the right on the upper line and to the left on the lower line. 11. In analogy with Exercise 9, multiply the equation y ¢¢¢ + f ( y ¢ ) = 0 by y ¢¢ , obtaining y ¢¢y ¢¢¢ + y ¢¢f ( y ¢ ) = 0 . Integrating, we find 0.5 y ¢¢ + F ( y ¢ ) = C where F ( u) is an antiderivative of f ( u) . Thus, the differential equation has a conservation law given by 0.5( y ¢¢ ) 2 + F ( y ¢ ) = C . dE 12. (a) From the definition of E(t), it follows that = mx ¢x ¢¢ + kxx ¢ = ( mx ¢¢ + kx ) x ¢ . From the dt differential equation, mx ¢¢ + gx ¢ + kx = 0 and hence mx ¢¢ + kx = -gx ¢ . Therefore, dE = (-gx ¢ ) x ¢ £ 0 . dt (b) Energy is not conserved. On t-intervals where x ¢ ( t) π 0 , E(t) is a decreasing function of t and energy is being lost. 13 (a). For the system x¢ = 2 x

y ¢ = -2 y we have f ( x, y ) = 2 x and g( x, y ) = -2 y . Thus, f x = 2 and gy = -2 . Since f x = - gy , the system is Hamiltonian. 13 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = 2 y . Integrating with respect to x, we obtain H ( x, y ) = 2 xy + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = 2 x + p¢ ( y ) = f ( x, y ) = 2 x . Therefore, p¢ ( y ) = 0 and hence p( y ) = C is a constant function. Dropping the constant, we obtain a Hamiltonian function, H ( x, y ) = 2 xy . 13 (c). From part (b), the phase-plane trajectories are defined by 2xy = C . If a phase-plane trajectory passes through the point (11 , ) , then C = 2 and the trajectory is given by xy = 1. 14 (a). For the system x ¢ = 2 xy y¢ = - y 2 we have f ( x, y ) = 2 xy and g( x, y ) = - y 2 . Thus, f x = 2 y and gy = -2 y . Since f x = - gy , the system is Hamiltonian. 14 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = y 2 . Integrating with respect to x, we obtain H ( x, y ) = xy 2 + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = 2 xy + p¢ ( y ) = f ( x, y ) = 2 xy .

224 • Chapter 8 Nonlinear Systems

Therefore, p¢ ( y ) = 0 and hence p( y ) = C is a constant function. Dropping the constant, we obtain a Hamiltonian function, H ( x, y ) = xy 2 . 14 (c). From part (b), the phase-plane trajectories are defined by xy 2 = C . If a phase-plane trajectory passes through the point (11 , ) , then C = 1 and the trajectory is given by xy 2 = 1. 15 (a). For the system x¢ = x - x 2 + 1

y ¢ = - y + 2 xy + 4 x we have f ( x, y ) = x - x 2 +1 and g( x, y ) = - y + 2 xy + 4 x . Thus, f x = 1 - 2 x and gy = -1 + 2 x . Since f x = - gy , the system is Hamiltonian. 15 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = y - 2 xy - 4 x . Integrating with respect to x, we obtain H ( x, y ) = xy - x 2 y - 2 x 2 + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = x - x 2 + p¢ ( y ) = f ( x, y ) = x - x 2 + 1. Therefore, p¢ ( y ) = 1 and hence p( y ) = y + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = xy - x 2 y - 2 x 2 + y . 15 (c). From part (b), the phase-plane trajectories are defined by xy - x 2 y - 2 x 2 + y = C . If a phaseplane trajectory passes through the point (11 , ) , then C = -1 and the trajectory is given by 2 2 xy - x y - 2 x + y + 1 = 0 . 16 (a). For the system x ¢ = -8 y

y¢ = 2 x we have f ( x, y ) = -8 and g( x, y ) = 2 x . Thus, f x = 0 and gy = 0 . Since f x = - gy , the system is Hamiltonian. 16 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H y ( x, y ) = f ( x, y ) = -8 y . Integrating with respect to y, we obtain H ( x, y ) = -4 y 2 + q( x ) . Differentiating with respect to x in order to determine q( x ) , we find H x ( x, y ) = q¢ ( x ) = -2 x . Therefore, q( x ) = - x 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = - x 2 - 4 y 2 . 16 (c). From part (b), the phase-plane trajectories are defined by - x 2 - 4 y 2 = C . If a phase-plane trajectory passes through the point (11 , ) , then C = -5 and the trajectory is given by 2 2 x + 4 y = 5. 17 (a). For the system x ¢ = 2 y cos x y ¢ = y 2 sin x we have f ( x, y ) = 2 y cos x and g( x, y ) = y 2 sin x . Thus, f x = -2 y sin x and gy = 2 y sin x . Since f x = - gy , the system is Hamiltonian. 17 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = - y 2 sin x . Integrating with respect to x, we obtain H ( x, y ) = y 2 cos x + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = 2 y cos x + p¢ ( y ) = f ( x, y ) = 2 y cos x . Therefore, p¢ ( y ) = 0 and hence p( y ) = C is a constant function. Dropping the constant, we obtain a Hamiltonian function, H ( x, y ) = y 2 cos x .

Chapter 8 Nonlinear Systems • 225

17 (c). From part (b), the phase-plane trajectories are defined by y 2 cos x = C . If a phase-plane trajectory passes through the point (11 , ) , then C = cos1 and the trajectory is given by 2 y cos x = cos1. 18 (a). For the system x¢ = 2 y - x + 3

y¢ = y + 4 x 3 - 2 x we have f x = -1 and gy = 1. Since f x = - gy , the system is Hamiltonian. 18 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H y ( x, y ) = f ( x, y ) = 2 y - x + 3. Integrating with respect to y, we obtain H ( x, y ) = y 2 - xy - 3 y + q( x ) . Differentiating with respect to x in order to determine q( x ) , we find H x ( x, y ) = - y + q¢ ( x ) = - y - 4 x 3 + 2 x . Therefore, q( x ) = - x 4 + x 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = y 2 - xy + 3 y - x 4 + x 2 . 18 (c). If a phase-plane trajectory H ( x, y ) = C passes through the point (11 , ) , then the trajectory is 2 4 2 given by y - xy + 3 y - x + x = 8 . 19 (a). For the system x ¢ = -2 y y ¢ = 3x 2 we have f ( x, y ) = -2 y and g( x, y ) = 3 x 2 . Thus, f x = 0 and gy = 0 . Since f x = - gy , the system is Hamiltonian. 19 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = -3 x 2 . Integrating with respect to x, we obtain H ( x, y ) = - x 3 + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = p¢ ( y ) = f ( x, y ) = -2 y . Therefore, p¢ ( y ) = -2 y and hence

p( y ) = - y 2 + C is a constant function. Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = - x 3 - y 2 . 19 (c). From part (b), the phase-plane trajectories are defined by - x 3 - y 2 = C . If a phase-plane trajectory passes through the point (11 , ) , then C = -2 and the trajectory is given by 3 2 x + y = 2. 20 (a). For the system x ¢ = xe xy y ¢ = -2 x - ye xy we have f x = e xy + xye xy and gy = -e xy - xye xy . Since f x = - gy , the system is Hamiltonian. 20 (b). Let H ( x, y ) denote the Hamiltonian function. Thus, H y ( x, y ) = f ( x, y ) = xe xy . Integrating with respect to y, we obtain H ( x, y ) = e xy + q( x ) . Differentiating with respect to x in order to determine q( x ) , we find H x ( x, y ) = ye xy + q¢ ( x ) = 2 x + ye xy . Therefore, q( x ) = x 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = e xy + x 2 . 20 (c). If a phase-plane trajectory H ( x, y ) = C passes through the point (11 , ) , then the trajectory is xy 2 given by e + x = 1 + e .

226 • Chapter 8 Nonlinear Systems

21.

22.

23.

24.

Consider the system x ¢ = x 3 + 3 sin(2 x + 3 y ) y ¢ = -3 x 2 y - 2 sin(2 x + 3 y ) . Calculating the partial derivatives, we have f x = 3 x 2 + 6 cos(2 x + 3 y ) and gy = -3 x 2 - 6 cos(2 x + 3 y ) . Since f x = - gy , the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = 3 x 2 y + 2 sin(2 x + 3 y ) . Integrating with respect to x, we obtain H ( x, y ) = x 3 y - cos(2 x + 3 y ) + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = x 3 + 3 sin(2 x + 3 y ) + p¢ ( y ) = f ( x, y ) = x 3 + 3 sin(2 x + 3 y ) . Therefore, p¢ ( y ) = 0 and hence p( y ) = C is a constant function. We obtain a Hamiltonian function, H ( x, y ) = x 3 y - cos(2 x + 3 y ) . Consider the system x ¢ = e xy + y 3 y ¢ = -e xy - x 3 . Calculating the partial derivatives, we have f x = ye xy and gy = - xe xy . Since f x π - gy , the system is not Hamiltonian. Consider the system x ¢ = - sin(2 xy ) - x

y ¢ = sin(2 xy ) + y . Calculating the partial derivatives, we have f x = -2 y cos(2 xy ) - 1 and gy = 2 x cos(2 xy ) + 1. Since f x π - gy , the system is not Hamiltonian. Consider the system x ¢ = -3 x 2 + xe y y ¢ = 6 xy + 3 x - e y . Calculating the partial derivatives, we have f x = -6 x + e y and gy = 6 x - e y . Since f x = - gy , the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = -6 xy - 3 x + e y . Integrating with respect to x, we obtain H ( x, y ) = -3 x 2 y - ( 3 / 2) x 2 + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = -3 x 2 + p¢ ( y ) = f ( x, y ) = -3 x 2 + xe y . Therefore, p¢ ( y ) = xe y and hence

25.

p( y ) = xe y + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = -3 x 2 y - ( 3 / 2) x 2 + xe y . Consider the system x¢ = y y¢ = x - x 2 . Calculating the partial derivatives, we have f x = 0 and gy = 0 . Since f x = - gy , the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x, y ) = x 2 - x . Integrating with respect to x, we obtain H ( x, y ) = (1 / 6)(2 x 3 - 3 x 2 ) + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = p¢ ( y ) = f ( x, y ) = y . Therefore, p¢ ( y ) = y and hence p( y ) = 0.5 y 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = (1 / 6)(2 x 3 - 3 x 2 + 3 y 2 ) .

Chapter 8 Nonlinear Systems • 227

26.

27.

28.

29.

30.

Consider the system x¢ = x + 2 y

y¢ = x 3 - 2 x + y . Calculating the partial derivatives, we have f x = 1 and gy = 1. Since f x π - gy , the system is not Hamiltonian. Consider the system x¢ = f ( y)

y ¢ = g( x ) . Calculating the partial derivatives, we have ∂ x [ f ( y )] = 0 and ∂ y [ g( x )] = 0 . Since ∂ x [ f ( y )] = -∂ y [ g( x )], the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x ) . Integrating with respect to x, we obtain H ( x, y ) = -G( x ) + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = p¢ ( y ) = f ( y ) . Therefore, p( y ) = F ( y ) + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = F ( y ) - G( x ) . Consider the system x¢ = f ( y) + 2 y y ¢ = g( x ) + 6 x . Calculating the partial derivatives, we have ∂ x [ f ( y ) + 2 y ] = 0 and ∂ y [ g( x ) + 6 x ] = 0 . Since ∂ x [ f ( y ) + 2 y ] = -∂ y [ g( x ) + 6 x ], the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x ) - 6 x . Integrating with respect to x, we obtain H ( x, y ) = -G( x ) - 3 x 2 + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = p¢ ( y ) = f ( y ) + 2 y . Therefore, p( y ) = F ( y ) + y 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = -G( x ) - 3 x 2 + F ( y ) + y 2 . Consider the system x ¢ = 3 f ( y ) - 2 xy y ¢ = g( x ) + y 2 + 1 . Calculating the partial derivatives, we have ∂ x [ 3 f ( y ) - 2 xy ] = -2 y and ∂ y [ g( x ) + y 2 + 1] = 2 y . Since ∂ x [ 3 f ( y ) - 2 xy ] = -∂ y [ g( x ) + y 2 + 1], the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - g( x ) - y 2 - 1. Integrating with respect to x, we obtain H ( x, y ) = -G( x ) - y 2 x - x + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = -2 yx + p¢ ( y ) = 3 f ( y ) - 2 xy . Therefore, p( y ) = 3F ( y ) + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = 3F ( y ) - G( x ) - y 2 x - x . Consider the system x¢ = f ( x - y) + 2 y

y¢ = f ( x - y) . Calculating the partial derivatives, we have ∂ x [ f ( x - y ) + 2 y ] = f ¢ ( x - y ) and ∂ y [ f ( x - y )] = - f ¢ ( x - y ) . Since ∂ x [ f ( x - y ) + 2 y ] = -∂ y [ f ( x - y )] , the system is Hamiltonian. Let H ( x, y ) denote the Hamiltonian function. Thus, H x ( x, y ) = - f ( x - y ) . Integrating with respect to x, we obtain H ( x, y ) = - F ( x - y ) + p( y ) . Differentiating with respect to y in order to determine p( y ) , we find H y ( x, y ) = f ( x - y ) + p¢ ( y ) = f ( x - y ) + 2 y .

228 • Chapter 8 Nonlinear Systems

31.

Therefore, p( y ) = y 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = - F ( x - y ) + y 2 . Consider the composition K ( x ( t), y ( t)). Differentiating with respect to t, we obtain d ∂K dx ∂K dy K ( x ( t), y ( t)) = + = -(mg) f + (mf ) g = 0 . Therefore, K ( x ( t), y ( t)) is a conserved dt ∂x dt ∂y dt quantity.

Section 8.4 1 (a). All points lying within the ellipse E having semi-major axis e and semi-minor axis e / 2 lie within the circle of radius e . Likewise, all points lying within the circle of radius e / 2 lie within the ellipse E . Therefore, given e > 0 , choose d = e / 2 . 1 (b). The origin is not an asymptotically stable equilibrium point since the solution points remain on an ellipse and do not approach the origin as t Æ • . 2. The origin is an unstable equilibrium point. Any solution point starting near the origin will follow a branch of the hyperbola and will eventually exit any circle centered at the origin. 3 (a). Making the substitution y = x ¢ , the scalar equation x ¢¢ + gx ¢ + x = 0 can be expressed as the system x¢ = y

y ¢ = - x - gy . The origin is the only equilibrium point for this system. 3 (b). We analyze stability by appealing to Theorem 8.3. The system in part (a) has the form y ¢ = Ay È0 1 ˘ where A = Í . The characteristic polynomial for A is p(l ) = l2 + gl + 1 and thus the ˙ Î-1 -g ˚

(

)

(

)

eigenvalues of A are l1 = 0.5 -g - g 2 - 4 and l 2 = 0.5 -g + g 2 - 4 . When g 2 - 4 ≥ 0 , we

4.

5.

6.

see that l1 £ l 2 . Thus, if 2 £ g , then l1 £ l 2 < 0 which shows the origin is asymptotically stable. On the other hand, if g £ -2 , then 0 < l1 £ l 2 which shows the origin is an unstable equilibrium point. For -2 < g < 2 , the eigenvalues are complex with nonzero imaginary parts. For -2 < g < 0 , the real parts of l1 and l 2 are positive, which shows the origin is an unstable equilibrium point. Likewise, for 0 < g < 2 , the origin is an asymptotically stable equilibrium point. When g = 0 , the origin is a stable (but not asymptotically stable) equilibrium point. È-3 -2 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = -1 and l 2 = 1 . Î4 3˚ Thus, by Theorem 8.3, the origin is an unstable equilibrium point. È5 -14 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = -1 and l 2 = -2 . Î3 -8 ˚ Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. È0 -2 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = 2i and l 2 = -2i . Î2 0 ˚ Thus, by Theorem 8.3, the origin is a stable equilibrium point but not an asymptotically stable equilibrium point.

Chapter 8 Nonlinear Systems • 229

7.

8.

9.

10.

11.

12.

13.

14.

15. 16. 17. 18. 19. 20. 21. 22.

È 1 4˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues Î-1 1 ˚ l1 = 1 + 2i and l 2 = 1 - 2i . Thus, by Theorem 8.3, the origin is an unstable equilibrium point. È-7 -3˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = -4 and l 2 = -2 . Î5 1˚ Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. È9 5˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = 2 and l 2 = 4 . Î-7 -3˚ Thus, by Theorem 8.3, the origin is an unstable equilibrium point. È-3 -5 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues Î 2 -1˚ l1 = -2 + 3i and l 2 = -2 - 3i . Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. È 9 -4 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = 3 and l 2 = -1 . Î15 -7 ˚ Thus, by Theorem 8.3, the origin is an unstable equilibrium point. È-13 -8 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = -3 and l 2 = -1. Î 15 9 ˚ Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. È3 -2 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = i and l 2 = -i . Thus, Î5 -3˚ by Theorem 8.3, the origin is a stable (but not asymptotically stable) equilibrium point. È1 -5 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues Î1 -3˚ l1 = -1 + i and l 2 = -1 - i . Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. È-3 3 ˘ For the system y ¢ = Í ˙y , the coefficient matrix has eigenvalues l1 = -6 and l 2 = -2 . Î 1 -5 ˚ Thus, by Theorem 8.3, the origin is an asymptotically stable equilibrium point. Eigenvalues are l1 = -2 and l 2 = 3 . Since one of the eigenvalues is real and positive, the origin is an unstable equilibrium point. Eigenvalues are l1 = 2 and l 2 = 3 . Since the eigenvalues are real and positive, the origin is an unstable equilibrium point. Eigenvalues are l1 = -4 and l 2 = -2 . Since the eigenvalues are real and negative, the origin is an asymptotically stable equilibrium point. Eigenvalues are l1 = 1 - 2i and l 2 = 1 + 2i . Since the eigenvalues are complex with positive real parts, the origin is an unstable equilibrium point. Eigenvalues are l1 = -2i and l 2 = 2i . Since the eigenvalues are purely imaginary, the origin is a stable equilibrium point but it is not an asymptotically stable equilibrium point. Eigenvalues are l1 = -2 - 2i and l 2 = -2 + 2i . Since the eigenvalues are complex with negative real parts, the origin is an asymptotically stable equilibrium point. Eigenvalues are l1 = -2 and l 2 = 3 . Since one of the eigenvalues is real and positive, the origin is an unstable equilibrium point.

230 • Chapter 8 Nonlinear Systems

Eigenvalues are l1 = -2 and l 2 = -3 . Since the eigenvalues are real and negative, the origin is an asymptotically stable equilibrium point. 24 (a). Solving 0 = Ay e + g 0 , it follows that y e = - A -1g 0 is the unique equilibrium point. 24 (b). Let z( t) = y ( t) - y e . Then, z¢ = y ¢ = Ay + g 0 = Ay - Ay e = Az . Theorem 8.3 can be applied to the new system z¢ = Az . È-2 1 ˘ È-4 ˘ y + Í ˙ , the unique equilibrium point is 25. For the system y ¢ = Í ˙ Î 1 -2 ˚ Î2˚ È-4 ˘ È-2 -1˘È-4 ˘ È-2 ˘ y e = - A -1 Í ˙ = -(1 / 3) Í ˙Í ˙ = Í ˙ . With the change of variable z( t) = y ( t) - y e the Î2˚ Î-1 -2 ˚Î 2 ˚ Î 0 ˚ È-2 1 ˘ È-2 1 ˘ È-2 1 ˘ È-4 ˘ È-4 ˘ ( z + y e ) + Í ˙ or z¢ = Í z+Í ye + Í ˙. system becomes ( z + y e )¢ = Í ˙ ˙ ˙ Î 1 -2 ˚ Î 1 -2 ˚ Î 1 -2 ˚ Î2˚ Î2˚ È-2 1 ˘ This last system reduces to the homogeneous system z¢ = Í ˙z . The coefficient matrix Î 1 -2 ˚ has eigenvalues l1 = -3 and l 2 = -1. By Theorem 8.3, the origin is an asymptotically stable equilibrium point of z¢ = Az and therefore, y e is an asymptotically stable equilibrium point of È-2 1 ˘ È-4 ˘ y + Í ˙. the nonhomogeneous system y ¢ = Í ˙ Î 1 -2 ˚ Î2˚ È 0 1˘ È2 ˘ È 1 ˘ È2 ˘ y + Í ˙ , the unique equilibrium point is y e = - A -1 Í ˙ = Í ˙ . With 26. For the system y ¢ = Í ˙ Î-1 0 ˚ Î1 ˚ Î-2 ˚ Î1 ˚ the change of variable z( t) = y ( t) - y e the system reduces to the homogeneous system È 0 1˘ z¢ = Í ˙z . The coefficient matrix has eigenvalues l1 = i and l 2 = -i . By Theorem 8.3, the Î-1 0 ˚ origin is a stable but not an asymptotically stable equilibrium point of z¢ = Az . Therefore, y e is a stable but not an asymptotically stable equilibrium point of the nonhomogeneous system. È3 2˘ È-2 ˘ y + Í ˙ , the unique equilibrium point is 27. For the system y ¢ = Í ˙ Î-4 -3˚ Î2˚ È-2 ˘ È-3 -2 ˘È-2 ˘ È 2 ˘ y e = - A -1 Í ˙ = Í ˙Í ˙ = Í ˙ . With the change of variable z( t) = y ( t) - y e the system Î 2 ˚ Î 4 3 ˚Î 2 ˚ Î-2 ˚ È3 2˘ È3 2˘ È3 2˘ È-2 ˘ È-2 ˘ ( z y ) = + + + + z z y or becomes ( z + y e )¢ = Í ¢ e Í-4 -3˙ Í-4 -3˙ e Í 2 ˙ . This ˙ Í2˙ Î-4 -3˚ Î Î ˚ ˚ Î ˚ Î ˚ È3 2˘ last system reduces to the homogeneous system z¢ = Í ˙z . The coefficient matrix has Î-4 -3˚ eigenvalues l1 = -1 and l 2 = 1 . By Theorem 8.3, the origin is an unstable equilibrium point of z¢ = Az and therefore, y e is an unstable equilibrium point of the nonhomogeneous system È3 2˘ È-2 ˘ y¢ = Í y + Í ˙. ˙ Î-4 -3˚ Î2˚ È -1 1 ˘ È1 ˘ È-3 / 5 ˘ È1 ˘ y + Í ˙ , the unique equilibrium point is y e = - A -1 Í ˙ = Í 28. For the system y ¢ = Í ˙ ˙. Î-10 5 ˚ Î2 ˚ Î-8 / 5 ˚ Î2 ˚ With the change of variable z( t) = y ( t) - y e the system reduces to the homogeneous system 23.

Chapter 8 Nonlinear Systems • 231

È -1 1 ˘ z¢ = Í ˙z . The coefficient matrix has eigenvalues l1 = 2 + i and l 2 = 2 - i . By Theorem Î-10 5 ˚ 8.3, the origin is an unstable equilibrium point of z¢ = Az . Therefore, y e is an unstable equilibrium point of the nonhomogeneous system. È2 1 1 ˘ ˙ Í 29. For the system y ¢ = Í1 1 2 ˙y , the coefficient matrix has eigenvalues ÍÎ1 2 1 ˙˚ l1 = -1, l 2 = 2, and l 3 = 3 . Thus, by the discussion following Theorem 8.3, the origin is an unstable equilibrium point. È2 ˘ È2 ˘ È4 ˘ È1 -1 0 ˘ Í ˙ ˙ Í ˙ Í ˙ Í 30. For the system y ¢ = Í0 -1 2 ˙y + Í0 ˙ , the unique equilibrium point is y e = - A -1 Í0 ˙ = Í6 ˙ . ÍÎ3˙˚ ÍÎ3˙˚ ÍÎ 3 ˙˚ ÍÎ0 0 -1˙˚ With the change of variable z( t) = y ( t) - y e the system reduces to the homogeneous system È1 -1 -2 ˘ ˙ Í z¢ = Í0 -1 -2 ˙z . The coefficient matrix has eigenvalues l1 = 1, l1 = -1, and l 3 = -1. By ÍÎ0 0 -1˙˚ Theorem 8.3, the origin is an unstable equilibrium point of z¢ = Az . Therefore, y e is an unstable equilibrium point of the nonhomogeneous system. È-3 -5 0 0 ˘ Í 2 -1 0 0 ˙ ˙y , the coefficient matrix has eigenvalues 31. For the system y ¢ = Í Í 0 0 0 2˙ ˙ Í Î 0 0 -2 0 ˚ l1 = -2 + 3i, l 2 = -2 - 3i, l 3 = 2i, and l 4 = -2i . Thus, by the discussion following Theorem 8.3, the origin is a stable (but not asymptotically stable) equilibrium point. È-1˘ È0 1 0 0 ˘ Í2˙ Í1 0 0 0 ˙ ˙ Í y + Í ˙ , unique equilibrium point is given by 32. For the system y ¢ = Í1˙ Í0 0 -1 0 ˙ Í ˙ Í ˙ Î0˚ Î0 0 0 -1˚ È-1˘ È-2 ˘ Í2˙ Í1 ˙ -1 Í ˙ = Í ˙ . With the change of variables z( t) = y ( t) - y , the system reduces to the ye = - A e Í1˙ Í1 ˙ Í ˙ Í ˙ Î0˚ Î0 ˚ È0 1 0 0 ˘ Í1 0 0 0 ˙ ˙z . The coefficient matrix has eigenvalues homogeneous system z¢ = Í Í0 0 -1 0 ˙ Í ˙ Î0 0 0 -1˚ l1 = -1, l 2 = -1, l 3 = -1, and l 4 = 1. Thus, by the discussion following Theorem 8.3, the origin is an unstable equilibrium point. 34 (a). Since the coefficient matrix A is real and symmetric, it has real eigenvalues and a full set of eigenvectors.

232 • Chapter 8 Nonlinear Systems

34 (b). From the discussion following Theorem 8.3, the equilibrium point y e = 0 is isolated if and only if det[ A] π 0 . Now, det[ A] = 1 - a 2 and therefore, y e = 0 is an isolated equilibrium point if and only if a π ±1. 34 (c). When a = 1 the equilibrium points lie on the line y = x . When a = -1 the equilibrium points lie on the line y = - x . 34 (d). No, since the eigenvalues of A are real and not purely imaginary; see Theorem 8.3. 34 (e). The eigenvalues of A are l1 = -1 + a , and l 2 = -1 - a . By part (b), if y e = 0 is an isolated equilibrium point, then a π ±1. Clearly, both eigenvalues are negative when -1 < a < 1 whereas one of the eigenvalues is positive when a > 1. È 1 a12 ˘È1 ˘ È1 ˘ 2 = 35. Since Í ˙Í ˙ Í2 ˙ , it follows that 1 + 2 a12 = 2 and a21 + 2 a22 = 4 . From the first Îa21 a22 ˚Î2 ˚ Î ˚ equation, we have a12 = 1 / 2 . Since y = 0 is not an isolated equilibrium point, it follows that det[ A] = 0 . Thus, a22 - a12 a21 = 0 or a22 - (1 / 2) a21 = 0 . This last equation, together with the È1 1 / 2 ˘ prior equation a21 + 2 a22 = 4 tells us that a21 = 2 and a22 = 1. Thus, A = Í ˙. Î2 1 ˚

Section 8.5 1 (a). For the system x ¢ = x 2 + y 2 - 32

y¢ = y - x , È4 ˘ È-4 ˘ the equilibrium points are y e = Í ˙ and y e = Í ˙ . Î4 ˚ Î-4 ˚ È2 x 2 y ˘ 1 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix A = Í ˙. Î -1 1 ˚ È 8 8˘ Thus, the linearized systems are (i) z¢ = Í ˙z Î-1 1 ˚ È-8 -8 ˘ and (ii) z¢ = Í ˙z . Î-1 1 ˚ 1 (c). In case (i), the eigenvalues are l1 = 2.438K and l 2 = 6.561K and thus the nonlinear system is unstable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are l1 = -8.815K and l 2 = 1.815K and thus the nonlinear system is unstable at the corresponding equilibrium point y e . 2 (a). For the system x¢ = x 2 + 9 y 2 - 9 y¢ = x , È0 ˘ È0˘ the equilibrium points are y e = Í ˙ and y e = Í ˙ . Î1 ˚ Î-1˚

Chapter 8 Nonlinear Systems • 233

È2 x 18 y ˘ 2 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix A = Í . 0 ˙˚ Î1 È0 -18 ˘ È0 18 ˘ z and (ii) z¢ = Í Thus, the linearized systems are (i) z¢ = Í ˙z ˙ Î1 0 ˚ Î1 0 ˚ 2 (c). In case (i), the eigenvalues are l1 = 4.242K and l 2 = -4.242K and thus the nonlinear system is unstable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are ±3 2 i and thus nothing can be inferred about the stability of the nonlinear system.

3 (a). For the system x¢ = 1 - x 2 y¢ = x 2 + y 2 - 2 ,

È1˘ È-1˘ È-1˘ È1˘ the equilibrium points are y e = Í ˙, y e = Í ˙, y e = Í ˙, and y e = Í ˙ . Î1˚ Î1˚ Î-1˚ Î-1˚ È-2 x 0 ˘ 3 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix A = Í ˙. Î 2x 2y˚ È-2 0 ˘ Thus, the linearized systems are (i) z¢ = Í ˙z , Î 2 2˚ È 2 0˘ È2 0˘ È-2 0 ˘ z , (iii) z¢ = Í z , and (iv) z¢ = Í (ii) z¢ = Í ˙ ˙ ˙z . Î-2 2 ˚ Î-2 -2 ˚ Î 2 -2 ˚ 3 (c). In cases (i) – (iii), l = 2 is an eigenvalue and thus the nonlinear system is unstable at each of the corresponding equilibrium points y e . For case (iv), the eigenvalues are l1 = -2 and l 2 = -2 and thus the nonlinear system is asymptotically stable at the corresponding equilibrium point y e . 4 (a). For the system x¢ = x - y - 1 y¢ = x 2 - y 2 + 1 , È0˘ the equilibrium point is y e = Í ˙ . Î-1˚ -1 ˘ È1 4 (b). At the equilibrium point, the linearized system z¢ = Az has coefficient matrix A = Í ˙. Î2 x -2 y ˚ È1 -1˘ Thus, the linearized system is z¢ = Í ˙z . Î0 2 ˚ 4 (c). The eigenvalues are l1 = 1 and l 2 = 2 and thus the nonlinear system is unstable at the equilibrium point y e .

234 • Chapter 8 Nonlinear Systems

5 (a). For the system x ¢ = ( x - 2)( y - 3)

y ¢ = ( x + 2 y )( y - 1) , È2˘ È2 ˘ È-6 ˘ the equilibrium points are y e = Í ˙, y e = Í ˙, and y e = Í ˙ . Î-1˚ Î1 ˚ Î3˚ 5 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix x-2 ˘ Èy - 3 È-4 0 ˘ A=Í . Thus, the linearized systems are (i) z¢ = Í ˙ ˙z , Î y - 1 x + 4 y - 2˚ Î-2 -4 ˚ È-2 0 ˘ È0 -8 ˘ z , and (iii) z¢ = Í (ii) z¢ = Í ˙ ˙z . Î 0 4˚ Î2 4 ˚ 5 (c). In case (i), the eigenvalues are l1 = -4 and l 2 = -4 and thus the nonlinear system is asymptotically stable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are l1 = -2 and l 2 = 4 and thus the nonlinear system is unstable at the corresponding equilibrium point y e . In case (iii), the eigenvalues are l1 = 2 + 2 3 i and l 2 = 2 - 2 3 i . Thus the nonlinear system is unstable at the corresponding equilibrium point y e . 6 (a). For the system x ¢ = ( x - y )( y + 1)

y ¢ = ( x + 2)( y - 4 ) , È-2 ˘ È4 ˘ È-2 ˘ the equilibrium points are y e = Í ˙, y e = Í ˙, and y e = Í ˙ . Î-2 ˚ Î4 ˚ Î-1˚ 6 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix È y + 1 x - 2 y - 1˘ È-1 1 ˘ A=Í = z . Thus, the linearized systems are (i) ¢ Í-6 0 ˙z , x + 2 ˙˚ Îy - 4 Î ˚ È5 -5 ˘ È 0 -1˘ (ii) z¢ = Í ˙z , and (iii) z¢ = Í-5 0 ˙z . Î0 6 ˚ Î ˚ 6 (c). In case (i), the eigenvalues are -0.5 ± 0.5i 23 and thus the nonlinear system is asymptotically stable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are l1 = 5 and l 2 = 6 and thus the nonlinear system is unstable at the corresponding equilibrium point y e . In case (iii), the eigenvalues are ± 5 . Thus the nonlinear system is unstable at the corresponding equilibrium point y e . 7 (a). For the system x ¢ = ( x - 2 y )( y + 4 )

y¢ = 2 x - y , È0 ˘ È-2 ˘ the equilibrium points are y e = Í ˙ and y e = Í ˙ . Î0 ˚ Î-4 ˚ 7 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix È y + 4 x - 4 y - 8˘ È4 -8 ˘ A=Í . Thus, the linearized systems are (i) z¢ = Í ˙ ˙z , -1 Î 2 Î2 -1˚ ˚ È0 6 ˘ and (ii) z¢ = Í ˙z . Î2 -1˚

Chapter 8 Nonlinear Systems • 235

7 (c). In case (i), the eigenvalues are l1 = 0.5( 3 + 39 i) and l 2 = 0.5( 3 - 39 i) and thus the nonlinear system is unstable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are l1 = -4 and l 2 = 3 and thus the nonlinear system is unstable at the corresponding equilibrium point y e . 8 (a). For the system x ¢ = xy - 1

y ¢ = ( x + 4 y )( x - 1) , È1˘ the equilibrium point is y e = Í ˙ . Î1˚ 8 (b). At the equilibrium point, the linearized system z¢ = Az has coefficient matrix y x ˘ È È1 1 ˘ A=Í . Thus, the linearized system is z¢ = Í ˙ ˙z . Î2 x + 4 y - 1 4 ( x - 1) ˚ Î5 0 ˚

(

)

8 (c). The eigenvalues are 0.5 1 ± 21 and thus the nonlinear system is unstable at the equilibrium point y e . 9 (a). For the system x¢ = y 2 - x y¢ = x 2 - y ,

È0 ˘ È1˘ the equilibrium points are y e = Í ˙ and y e = Í ˙ . Î0 ˚ Î1˚ È -1 2 y ˘ 9 (b). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix A = Í ˙. Î2 x -1˚ È-1 0 ˘ Thus, the linearized systems are (i) z¢ = Í ˙z , Î 0 -1˚ È-1 2 ˘ and (ii) z¢ = Í ˙z . Î 2 -1˚ 9 (c). In case (i), the eigenvalues are l1 = -1 and l 2 = -1 and thus the nonlinear system is asymptotically stable at the corresponding equilibrium point y e . For case (ii), the eigenvalues are l1 = -3 and l 2 = 1 and thus the nonlinear system is unstable at the corresponding equilibrium point y e . 10. At an equilibrium point, the linearized system z¢ = Az has coefficient matrix -(1 / 4 ) x È(1 / 2)[1 - x - (1 / 2) y ] ˘ A=Í . Thus, the linearized systems are: (i) at -(1 / 12) y (1 / 4 )[1 - (1 / 3) x - ( 4 / 3) y ]˙˚ Î 0 ˘ È 1/8 È0 ˘ È 0 ˘ È1 / 2 0 ˘ Í0 ˙ , z¢ = Í 0 1 / 4 ˙z , (ii) at Í3 / 2 ˙ , z¢ = Í-1 / 8 -1 / 4 ˙z , Î Î ˚ Î Î ˚ ˚ ˚ È-1 / 2 -1 / 2 ˘ È2 ˘ z . Thus, in all three of these cases, the system is (iii) at Í ˙ , z¢ = Í 1 / 12 ˙˚ Î 0 Î0 ˚ unstable at the corresponding equilibrium point.

236 • Chapter 8 Nonlinear Systems

11 (c). By Taylor’s theorem, f ( z) = f (0) + f ¢ (0) z + f ¢¢ (g ) z 2 / 2 where g is between z and 0. For f ( z) = sin z , we have sin z1 - z1 = (- sin g ) z12 / 2 where g is between z1 and 0. Now, g( z) / z = z1 - sin z1 / z12 + z22 £ z1 - sin z1 / z1 . So, by the remarks above,

12 (a).

12 (b). 12 (c).

12 (d). 13 (a).

g( z) / z £ z12 / 2 / z1 = z1 / 2 . Hence, since z1 / 2 goes to 0 as z goes to 0, the system is almost linear at both equilibrium points. È 9 -4 ˘ For the given system z¢ = Az + g( z) , the coefficient matrix A is A = Í ˙ , while Î15 -7 ˚ Èz22 ˘ g( z) = Í ˙ . Î0˚ g( z) = z22 , or using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) = r 2 sin 2 q . From part (b), g( z) / z = r 2 sin 2 q / r = r sin 2 q . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. The eigenvalues of A are l1 = -1 and l 2 = 3 . Thus, by Theorem 8.4, z = 0 is an unstable equilibrium point. For the system z¢ = Az + g( z) , z1¢ = 5 z1 - 14 z2 + z1z2

z2¢ = 3z1 - 8 z2 + z12 + z22 , È5 -14 ˘ È z1z2 ˘ , while g( z) = Í 2 the coefficient matrix A is given by A = Í ˙ 2˙. Î3 -8 ˚ Îz1 + z2 ˚ 13 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) =

(z1z2 )2 + (z12 + z22 )

2

=

(r

2

2

2

cosq sin q ) + ( r 2 ) or g( z) = r 4 (cos2 q sin 2 q + 1) .

(Also note that z = r .) 13 (c). From part (b), g( z) / z = r 4 (cos2 q sin 2 q + 1) / r £ r 2 2 / r = r 2 . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. 13 (d). The eigenvalues of A are l1 = -2 and l 2 = -1. Thus, by Theorem 8.4, z = 0 is an asymptotically stable equilibrium point. È-3 1 ˘ 14 (a). For the given system z¢ = Az + g( z) , the coefficient matrix A is A = Í ˙ , while Î 2 -2 ˚ È z12 + z22 ˘ . g( z) = Í 2 2 1/ 3 ˙ Î( z1 + z2 ) ˚ 14 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) = r 2 / 3 1 + r 8 / 3 .

14 (c). From part (b), g( z) / z = r 2 / 3 1 + r 8 / 3 / r = 1 + r 8 / 3 / r1/ 3 . Thus, g( z) / z does not exist as z Æ 0 . The system is not almost linear at z = 0.

Chapter 8 Nonlinear Systems • 237

15 (a). For the system z¢ = Az + g( z) , z1¢ = - z1 + 3z2 + z2 cos z12 + z22 z2¢ = - z1 - 5 z2 + z1 cos z12 + z22 ,

Èz cos z 2 + z 2 ˘ È-1 3 ˘ 1 2 , while g( z) = Í 2 ˙. the coefficient matrix A is given by A = Í ˙ 2 2 Î-1 -5 ˚ ÍÎ z1 cos z1 + z2 ˙˚ 15 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) =

(z

2 1

+ z22 ) cos2 z12 + z22 = r 2 cos2 r or g( z) = r cos r . (Also note that z = r .)

15 (c). From part (b), g( z) / z = r cos r / r = cos r . Thus, g( z) / z Æ 1 as z Æ 0 . Therefore, the system is not an almost linear system. È-2 2 ˘ 16 (a). For the given system z¢ = Az + g( z) , the coefficient matrix A is A = Í ˙ , while Î 1 -3˚ Èz1z2 cos z2 ˘ g( z) = Í ˙. Î z1z2 sin z2 ˚ 16 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) = r 2 cosq sin q . 16 (c). From part (b), g( z) / z = r 2 sin q cosq / r £ r . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. 16 (d). The eigenvalues of A are l1 = -4 and l 2 = -1. Thus, by Theorem 8.4, z = 0 is an asymptotically stable equilibrium point. 17 (a). For the system z¢ = Az + g( z) , z1¢ = 2 z2 + z22

z2¢ = -2 z1 + z1z2 , È z22 ˘ È 0 2˘ , the coefficient matrix A is given by A = Í while g ( z ) = Í ˙. ˙ Î-2 0 ˚ Îz1z2 ˚ 17 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) =

(z1z2 )2 + z24 = (r 2 cosq sinq )

2

+ r 4 sin 4 q or

g( z) = r 4 sin 2 q (cos2 q + sin 2 q ) = r 2 sin q . (Also note that z = r .) 17 (c). From part (b), g( z) / z = r 2 sin q / r = r sin q . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. (d) The eigenvalues of A are l1 = -2i and l 2 = 2i . No conclusion can be drawn from Theorem 8.4 relative to the stability of z¢ = Az + g( z) . È-3 -5 ˘ 18 (a). For the given system z¢ = Az + g( z) , the coefficient matrix A is A = Í ˙ , while Î 2 -1˚ È z e - z12 + z22 ˘ g( z) = Í 1 - z 2 + z 2 ˙ . ÍÎz2e 1 2 ˙˚ 18 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) = re - r .

238 • Chapter 8 Nonlinear Systems

18 (c). From part (b), g( z) / z = e - r . Thus, g( z) / z Æ 1 as z Æ 0 ; the system is not almost linear at z = 0. 19 (a). For the system z¢ = Az + g( z) , z1¢ = 9 z1 + 5 z2 + z1z2

z2¢ = -7 z1 - 3z2 + z12 , È9 5˘ Èz1z2 ˘ , while g( z) = Í 2 ˙ . the coefficient matrix A is given by A = Í ˙ Î-7 -3˚ Î z1 ˚ 19 (b). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) =

19 (c).

20 (a).

20 (b). 20 (c).

20 (d). 21 (a).

(z1z2 )2 + z14 = (r 2 cosq sinq )

2

+ r 4 cos4 q or

g( z) = r 4 cos2 q (cos2 q + sin 2 q ) = r 2 cosq . (Also note that z = r .) From part (b), g( z) / z = r 2 cosq / r = r cosq . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. (d) The eigenvalues of A are l1 = 2 and l 2 = 4 . Thus, by Theorem 8.4, z = 0 is an unstable equilibrium point of the system. È2 2˘ For the given system z¢ = Az + g( z) , the coefficient matrix A is A = Í ˙ , while Î-5 -2 ˚ È0˘ g( z) = Í 2 ˙ . Îz1 ˚ Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain g( z) = r 2 cos2 q . From part (b), g( z) / z = r cos2 q . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. The eigenvalues of A are l1 = i 6 and l 2 = -i 6 . Thus, no conclusions can be drawn by using Theorem 8.4. The system x ¢ = - x + xy + y

y ¢ = x - xy - 2 y È-1 1 ˘ can be expressed as z¢ = Az + g( z) where the coefficient matrix A is given by A = Í ˙, Î 1 -2 ˚ È z1z2 ˘ È z1 ˘ È x ˘ z = Í ˙ = Í ˙ , and g( z) = Í ˙ . Since A is invertible, the solutions of Î- z1z2 ˚ Î z2 ˚ Î y ˚ Az + g( z) = 0 are vectors z e such that 0 = - A -1g( z e ) and therefore, we need g( z e ) = 0 . Clearly, the only solution of g( z) = 0 is z e = 0 . 21 (b). The linearized system is z¢ = Az and we find that A has eigenvalues l1 = -2.618K and l 2 = -0.382K we see that z = 0 is an asymptotically stable equilibrium point of z¢ = Az .

Chapter 8 Nonlinear Systems • 239

21 (c). Using polar coordinates with z1 = r cosq and z2 = r sin q , we obtain 2

g( z) = 2( z1z2 ) = 2 r 4 cos2 q sin 2 q = 2 r 2 cosq sin q . (Also note that z = r .) Therefore, g( z) / z = 2 r 2 cosq sin q / r = 2 r cosq . Thus, g( z) / z Æ 0 as z Æ 0 . In addition to the limit requirement, the system satisfies the other necessary conditions to be an almost linear system. 21 (d). By Theorem 8.4, z = 0 is an asymptotically stable equilibrium point of the original system. 22 (a). The system has the form x¢ = y

y ¢ = 1 - (1 + x ) 3 / 2 . 22 (c). At an equilibrium point, the linearized system z¢ = Az has coefficient matrix 0 1˘ 1˘ È È 0 A=Í . Thus, at z = 0, A = Í ˙ ˙ . The eigenvalues of A are 1/ 2 0˚ Î-( 3 / 2)(1 + x ) Î-3 / 2 0 ˚ l1 = i 3 / 2 and l 2 = -i 3 / 2 and hence the linearized system is stable but not asymptotically stable at z = 0. 22 (d). Theorem 8.4 does not provide any information about the stability of the nonlinear system since the eigenvalues of the linearized system z¢ = Az are purely imaginary. 23 (a). Multiplying by x ¢ we obtain x ¢x ¢¢ = x ¢[1 - (1 + x ) 3 / 2 ]. Integrating, we obtain 0.5( x ¢ ) 2 = x - 0.4 (1 + x ) 5 / 2 . Therefore, with y = x ¢ we have y 2 = 2 x - 0.8(1 + x ) 5 / 2 + C . 24 (a). At the equilibrium point (0, 0), the linearized system z¢ = Az has coefficient matrix È 1 -1˘ A=Í ˙ . Since A is not invertible, Theorem 8.4 does not apply. Î-1 1 ˚ È- z12 / 3 ˘ Èz1 ˘ È x ˘ 24 (b). Let z = Í ˙ = Í ˙ . For the given system z¢ = Az + g( z) , g( z) = Í 1/ 3 ˙ . Using polar Î z2 ˚ Î y ˚ Î 2 z2 ˚

27.

28.

coordinates, g( z) / z = r -2 / 3 cos4 / 3 q + 4 r -4 / 3 sin 2 / 3 q . Thus, the limit of g( z) / z does not exist as z Æ 0 ; The system is not almost linear at (0, 0). In this case, a11 = 0, a12 = 1, a21 = -1, a22 = 0, g1 = ar 3 cosq, and g2 = ar 3 sin q . Thus, h ( r) = ar 2 and we obtain the system r¢ = ar 3

q ¢ = -1 . Solving, r( t) = (C1 - 2at) -1/ 2 and q ( t) = - t + C2 . Hence, x = (C1 - 2at) -1/ 2 cos(- t + C2 ) and y = (C1 - 2at) -1/ 2 sin(- t + C2 ) . So, a11 = 1, a12 = 0, a21 = 0, a22 = 1, g1 = r 2 cosq, and g2 = r 2 sin q . Thus, h ( r) = r and we obtain the initial value problem r¢ = r + r 2 , r(0) = 1 q ¢ = 0 , q (0) = 3 . The solution is r = (2 / 3)e t /[1 - (2 / 3)e t ] , q = p / 3 . However, the denominator in the expression for r, 1 - (2 / 3)e t , vanishes at 3 / 2 = e t . Solving for t, we have t = ln1.5 = 0.405K. Thus, the solution does not exist at t = 1.

240 • Chapter 8 Nonlinear Systems

29.

So, a11 = 0, a12 = 1, a21 = -1, a22 = 0, g1 = - r cosq ln r 2 , and g2 = - r sin q ln r 2 . Thus, h ( r) = - ln r 2 and we obtain the initial value problem r¢ = -2 r ln r , r(0) = 1

q ¢ = 1 , q (0) = p / 4 . The general solution is r = C1 exp(e -2 t ) , q = t + C2 . Imposing the initial conditions we arrive at r = exp(e -2 t - 1) , q = t + p / 4 . Hence, at t = 1, we find x = exp(e -2 - 1)cos(1 + p / 4 ) ª -0.0896K and y = exp(e -2 - 1)sin(1 + p / 4 ) ª 0.411K

Section 8.6 1 (a). Since the eigenvalues are real and have opposite signs, y = 0 is an unstable saddle point. Èe 2 t e - t ˘ È2e 2 t -e - t ˘ 1 (d). We have Y( t) = [el 1 t x1, el 2 t x 2 ] = Í 2 t and Y ( t ) = ¢ ˙ Í 2t ˙. -e - t ˚ e-t ˚ Îe Î2e È2e 2 t -e - t ˘È0.5e -2 t 0.5e -2 t ˘ È0.5 1.5 ˘ Therefore, A = Y ¢ ( t)Y -1 ( t) = Í 2 t ˙= Í ˙Í ˙. e - t ˚Î 0.5e t -0.5e t ˚ Î1.5 0.5 ˚ Î2e 2 (a). Since the eigenvalues are real and positive, y = 0 is an unstable node. È e t 2e 2 t ˘ È et 4e 2t ˘ 2 (d). We have Y( t) = [el 1 t x1, el 2 t x 2 ] = Í t and . = Y ( t ) ¢ Í t 2t ˙ 2t ˙ Î2e -e ˚ Î2e -2e ˚ È 9 / 5 -2 / 5 ˘ Therefore, A = Y ¢ ( t)Y -1 ( t) = Í ˙. Î-2 / 5 6 / 5 ˚ 3 (a). Since both eigenvalues are real and positive, y = 0 is an unstable improper node. È2e 2 t 0 ˘ È4 e 2 t 0 ˘ l1t l 2t 3 (d). We have Y( t) = [e x1, e x 2 ] = Í and Y ¢ ( t) = Í ˙. t˙ 2e t ˚ Î 0 2e ˚ Î 0 È4 e 2 t 0 ˘È0.5e -2 t 0 ˘ È2 0 ˘ Therefore, A = Y ¢ ( t)Y -1 ( t) = Í ˙= Í ˙. t ˙Í 2e ˚Î 0 0.5e - t ˚ Î0 1 ˚ Î 0 4 (a). Since the eigenvalues are real and negative, y = 0 is an asymptotically stable node. È-2e -2 t -e - t ˘ Èe -2 t e - t ˘ and 4 (d). We have Y( t) = [el 1 t x1, el 2 t x 2 ] = Í Y ( t ) = ¢ Í ˙. -t ˙ -e - t ˚ Î 0 Î 0 e ˚ È-2 1 ˘ Therefore, A = Y ¢ ( t)Y -1 ( t) = Í ˙. Î 0 -1˚ 5 (a). Since the eigenvalues are real and have opposite signs, y = 0 is an unstable saddle point. Èe t -2e - t ˘ Èe t 2e - t ˘ l1t l 2t 5 (d). We have Y( t) = [e x1, e x 2 ] = Í and Y ¢ ( t) = Í . -t ˙ -t ˙ Î 0 -e ˚ Î0 e ˚ Èe t -2e - t ˘Èe - t -2e - t ˘ È1 -4 ˘ Therefore, A = Y ¢ ( t)Y -1 ( t) = Í ˙= Í ˙. - t ˙Í e t ˚ Î0 -1˚ Î 0 -e ˚Î 0

È1 -6 ˘ 6 (a). For A = Í ˙ , the eigenvalues are l1 = -1 and l 2 = -2 . Î1 -4 ˚ 6 (b). Since the eigenvalues are real and negative, y = 0 is an asymptotically stable improper node.

Chapter 8 Nonlinear Systems • 241

È6 -10 ˘ 7 (a). For A = Í ˙ , the eigenvalues are l1 = 1 and l 2 = 2 . Î2 -3 ˚ 7 (b). Since the eigenvalues are real and positive, y = 0 is an unstable improper node. È-6 14 ˘ 8 (a). For A = Í ˙ , the eigenvalues are l1 = 1 and l 2 = -2 . Î-2 5 ˚ 8 (b). Since the eigenvalues have opposite sign, y = 0 is an unstable saddle point. È1 2˘ 9 (a). For A = Í ˙ , the eigenvalues are l1 = 3i and l 2 = -3i . Î-5 -1˚ 9 (b). Since the eigenvalues are complex with zero real part, y = 0 is a stable, but not asymptotically stable, center. È-1 1 ˘ 10 (a). For A = Í ˙ , the eigenvalues are l1 = -1 + i and l 2 = -1 - i . Î-1 -1˚ 10 (b). Since the eigenvalues are complex with negative real part, y = 0 is an asymptotically stable spiral point. È1 -6 ˘ 11 (a). For A = Í ˙ , the eigenvalues are l1 = -3 and l 2 = -2 . Î2 -6 ˚ 11 (b). Since the eigenvalues are real and negative, y = 0 is an asymptotically stable improper node. È2 -3˘ 12 (a). For A = Í ˙ , the eigenvalues are l1 = 2 + 3i and l 2 = 2 - 3i . Î3 2 ˚ 12 (b). Since the eigenvalues are complex with positive real part, y = 0 is an unstable spiral point. È-2 -4 ˘ 13 (a). For A = Í ˙ , the eigenvalues are l1 = 4 i and l 2 = -4 i . Î5 2˚ 13 (b). Since the eigenvalues are complex with zero real part, y = 0 is a stable, but not asymptotically stable, center. È7 -24 ˘ 14 (a). For A = Í ˙ , the eigenvalues are l1 = 1 and l 2 = -1 . Î2 -7 ˚ 14 (b). Since the eigenvalues are real with opposite sigen, y = 0 is an unstable saddle point. È-1 8 ˘ 15 (a). For A = Í ˙ , the eigenvalues are l1 = 1 and l 2 = 3 . Î-1 5 ˚ 15 (b). Since the eigenvalues are real and positive, y = 0 is an unstable improper node. È-2 1 ˘ 16 (a). For A = Í ˙ , the eigenvalues are l1 = -2 + i and l 2 = -2 - i . Î-1 -2 ˚ 16 (b). Since the eigenvalues are complex with negative real part, y = 0 is an asymptotically stable spiral point. È2 4˘ 17 (a). For A = Í ˙ , the eigenvalues are l1 = -2 and l 2 = -2 . Î-4 -6 ˚ 17 (b). Since the eigenvalues are real and negative and A is not a multiple of the identity, y = 0 is an asymptotically stable improper node. È3 0 ˘ 18 (a). For A = Í ˙ , the eigenvalues are l1 = 3 and l 2 = 3 . Î0 3˚

242 • Chapter 8 Nonlinear Systems

18 (b). Since the eigenvalues are real and positive and A is a multiple of the identity, y = 0 is an unstable proper node. È 1 2˘ 19 (a). For A = Í ˙ , the eigenvalues are l1 = 1 + 4 i and l 2 = 1 - 4 i . Î-8 1 ˚ 19 (b). Since the eigenvalues are complex with positive real part, y = 0 is an unstable spiral point. È-1 -2 ˘ 20 (a). For A = Í ˙ , the eigenvalues are l1 = 1 and l 2 = 1. Î2 3˚ 20 (b). Since the eigenvalues are real and positive and A is not a multiple of the identity, y = 0 is an unstable improper node. È-2 1 ˘ 21 (a). For A1 = Í ˙ , the eigenvalues are l1 = -3 and l 2 = -1. Since the eigenvalues are real Î 1 -2 ˚ and negative, y = 0 is an asymptotically stable equilibrium point. Therefore, A1 corresponds to Direction Field 2. È1 2˘ 21 (b). For A2 = Í ˙ , the eigenvalues are l1 = - 3i and l 2 = 3i . Since the eigenvalues are Î-2 -1˚ complex with zero real part, y = 0 is a stable, but not asymptotically stable, center. Therefore, A2 corresponds to Direction Field 4. È2 1 ˘ 21 (c). For A3 = Í ˙ , the eigenvalues are l1 = - 3 and l 2 = 3 . Since the eigenvalues are real Î-1 -2 ˚ and have opposite sign, y = 0 is an unstable saddle point. Therefore, A3 corresponds to Direction Field 1. È 1 2˘ 21 (d). For A4 = Í ˙ , the eigenvalues are l1 = 1 - 2i and l 2 = 1 + 2i . Since the eigenvalues are Î-2 1 ˚ complex with positive real part, y = 0 is an unstable spiral point. Therefore, A4 corresponds to Direction Field 3. 22. For a center, eigenvalues are purely imaginary. Therefore, a = -2 . È-4 a ˘ 2 23. Consider A = Í ˙ . The characteristic polynomial is p(l ) = l + 2l + (2a - 8) . Thus, the 2 2 Î ˚

24. 25.

eigenvalues are l = -1 ± 9 - 2a . In order to have an asymptotically stable spiral point at y = 0 , we need complex eigenvalues with negative real parts. Thus, we need 9 - 2a < 0 or 9 /2 < a. Note that l1 = -2 and l 2 = -2 no matter the value of a . Thus, y = 0 is always an asymptotically stable equilibrium point; it will be a proper node if a = 0. È 4 -2 ˘ Consider A = Í . The characteristic polynomial is p(l ) = l2 + (2a - 16) . Thus, the ˙ Îa -4 ˚ eigenvalues are l = ± 16 - 2a . In order to have a saddle point at y = 0 , we need real eigenvalues with opposite signs. Thus, we need 16 - 2a > 0 or a < 8 .

Chapter 8 Nonlinear Systems • 243

26.

27.

È 1 4˘ È3˘ y + Í ˙ . The system has a unique Consider the nonhomogeneous system y ¢ = Í ˙ Î-1 1 ˚ Î2 ˚ È1˘ equilibrium point given by y e = Í ˙ . Making the substitution z = y - y e , we obtain Î-1˚ È 1 -4 ˘ z¢ = Í ˙z . The eigenvalues of the coefficient matrix are l1 = 1 + 2i and l 2 = 1 - 2i . Î-1 1 ˚ Therefore, z = 0 is an unstable spiral point and consequently, y = y e is an unstable spiral point of the original system. È6 5˘ È4˘ + y Consider the nonhomogeneous system y ¢ = Í ˙ Í-6 ˙ . The system has a unique Î-7 -6 ˚ Î ˚ -1

28.

29.

È 6 5 ˘ È 4 ˘ È-6 -5 ˘È 4 ˘ È 6 ˘ equilibrium point given by y e = - Í ˙ Í ˙= Í ˙Í ˙ = Í ˙ . Making the Î-7 -6 ˚ Î-6 ˚ Î 7 6 ˚Î-6 ˚ Î-8 ˚ È6 5˘ substitution z = y - y e , we obtain z¢ = Í ˙z . The eigenvalues of the coefficient matrix Î-7 -6 ˚ are l1 = -1 and l 2 = 1 . Therefore, z = 0 is an unstable saddle point and consequently, y = y e is an unstable saddle point of the original system. È5 -14 ˘ È2 ˘ y + Í ˙ . The system has a unique Consider the nonhomogeneous system y ¢ = Í ˙ Î3 -8 ˚ Î1 ˚ È1 ˘ equilibrium point given by y e = Í ˙ . Making the substitution z = y - y e , we obtain Î0.5 ˚ È5 -14 ˘ z¢ = Í ˙z . The eigenvalues of the coefficient matrix are l1 = -2 and l 2 = -1. Therefore, Î3 -8 ˚ z = 0 is an asymptotically stable improper node and consequently, y = y e is an asymptotically stable improper node of the original system. È-1 0 ˘ È2˘ y + Í ˙ . The system has a unique Consider the nonhomogeneous system y ¢ = Í ˙ Î 0 2˚ Î-4 ˚ -1

30 (a). 30 (b). 30 (c). 32 (a). 32 (b).

0 ˘È 2 ˘ È2 ˘ È-1 0 ˘ È 2 ˘ È1 =Í equilibrium point given by y e = - Í ˙ Í ˙ ˙Í ˙ = Í ˙ . Making the Î 0 2 ˚ Î-4 ˚ Î0 -0.5 ˚Î-4 ˚ Î2 ˚ È-1 0 ˘ substitution z = y - y e , we obtain z¢ = Í ˙z . The eigenvalues of the coefficient matrix are Î 0 2˚ l1 = -1 and l 2 = 2 . Therefore, z = 0 is an unstable saddle point and consequently, y = y e is an unstable saddle point of the original system. The characteristic equation is l2 - ( a11 + a22 )l + a11a22 - a12 a21 = 0 . The origin is a center if the roots are purely imaginary. That is, if a11 + a22 = 0 and a11a22 - a12 a21 < 0 . Note that f ( x, y ) = a11 x + a12 y and g( x, y ) = a21 x + a22 y . Thus, f x = a11 and gy = a22 . By part (a), f x = - gy and hence the system is Hamiltonian. The converse is not true since the system can be Hamiltonian even though a11a22 - a12 a21 = 0 . È-2 1 ˘ The eigenvalues of the coefficient matrix A = Í ˙ are l1 = 3 and l 2 = -3 . Î 5 2˚ Since the eigenvalues are real with opposite sign, y = 0 is an (unstable) saddle point.

244 • Chapter 8 Nonlinear Systems

32 (c). Since the system is Hamiltonian, we know that H y ( x, y ) = -2 x + y . Therefore, H ( x, y ) = -2 xy + 0.5 y 2 + q( x ) . We determine q( x ) by differentiating H ( x, y ) with respect to x, finding H x ( x, y ) = -2 y + q¢ ( x ) = -5 x - 2 y . Thus, q¢ ( x ) = -5 x and so q( x ) = -2.5 x 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = -2.5 x 2 - 2 xy + 0.5 y 2 . The conservation law for the system is H ( x, y ) = C . È1 3˘ 33 (a). The eigenvalues of the coefficient matrix A = Í ˙ are l1 = -2 2i and l 2 = -2 2i . Î-3 -1˚ 33 (b). Since the eigenvalues are complex with zero real part, y = 0 is a stable, but not asymptotically stable, center. 33 (c). Since the system is Hamiltonian, we know that H y ( x, y ) = x + 3 y . Therefore,

H ( x, y ) = xy + 1.5 y 2 + q( x ) . We determine q( x ) by differentiating H ( x, y ) with respect to x, finding -3x - y = - H x ( x, y ) = - y - q¢ ( x ) . Thus, q¢ ( x ) = 3 x and so q( x ) = 1.5 x 2 + C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = xy + 1.5( x 2 + y 2 ) . The conservation law for the system is H ( x, y ) = C . È2 1 ˘ 34 (a). The eigenvalues of the coefficient matrix A = Í ˙ are l1 = 2 and l 2 = -2 . Î0 -2 ˚ 34 (b). Since the eigenvalues are real with opposite sign, y = 0 is an (unstable) saddle point. 34 (c). Since the system is Hamiltonian, we know that H y ( x, y ) = 2 x + y . Therefore, H ( x, y ) = 2 xy + 0.5 y 2 + q( x ) . We determine q( x ) by differentiating H ( x, y ) with respect to x, finding H x ( x, y ) = 2 y + q¢ ( x ) = 2 y . Thus, q¢ ( x ) = 0 and so q( x ) = C . Dropping the additive constant, we obtain a Hamiltonian function, H ( x, y ) = 2 xy + 0.5 y 2 . The conservation law for the system is H ( x, y ) = C .

Section 8.7 1 (a). Consider the system x ¢ = x - x 2 - xy y ¢ = y - 3 y 2 - 0.5 xy . If y = 0 , then all direction field filaments on the positive x-axis point towards x = 1. Thus, x approaches an equilibrium value of x e = 1 as t increases. Similarly, if x = 0 , then y approaches an equilibrium value of y e = 1 / 3 as t increases. In each case, the presence of the xy term causes the derivative to decrease. Therefore, the presence of the other species is harmful in each case. 1 (b). Rewriting the system as x ¢ = x (1 - x - y )

y ¢ = y (1 - 3 y - 0.5 x ) , we see that x ¢ = 0 if (i) x = 0 or (ii) 1 - x - y = 0 . In case (i), y ¢ = 0 if y = 0 or y = 1 / 3. Thus, two equilibrium points are ( x, y ) = (0, 0) and ( x, y ) = (0,1 / 3) . In case (ii), y ¢ = 0 if y = 0 (and hence, x = 1) or if 1 - 3 y - 0.5 x = 0 (and hence x + y = 1 and 0.5 x + 3 y = 1 ). Thus, case (ii) leads us to two more equilibrium points ( x, y ) = (1, 0) and ( x, y ) = (0.8, 0.2) .

Chapter 8 Nonlinear Systems • 245

È1 0 ˘ 1 (c). At the equilibrium point z = 0 , the linearized system takes the form z¢ = Í ˙z . The Î0 1 ˚ eigenvalues of the coefficient matrix are l1 = 1 and l 2 = 1. Since, z = 0 is an unstable proper node of the linearized system, the original system is also unstable at y = 0 . 2 (a). Consider the system x¢ = - x - x 2 y ¢ = - y + xy . If y = 0 , then x approaches an equilibrium value of x e = 0 as t increases. If x = 0 , then y approaches an equilibrium value of y e = 0 as t increases. The presence of y is a matter of indifference to x. The presence of x is beneficial to y. 2 (b). The only equilibrium point in the first quadrant is ( x, y ) = (0, 0) . È-1 0 ˘ 2 (c). At the equilibrium point z = 0 , the linearized system takes the form z¢ = Í ˙z . The Î 0 -1˚ eigenvalues of the coefficient matrix are l1 = -1 and l 2 = -1. Since, z = 0 is an asymptotically stable proper node of the linearized system, the original system is also asymptotically stable at y = 0. 3 (a). Consider the system x ¢ = x - x 2 - xy y ¢ = - y - y 2 + xy . If y = 0 , then all direction field filaments on the positive x-axis point towards x = 1. Thus, x approaches an equilibrium value of x e = 1 as t increases. Similarly, if x = 0 , then y approaches an equilibrium value of y e = 0 as t increases. The presence of the xy term in the first equation causes the derivative to decrease. Therefore, the presence of y is harmful to x. On the other hand, the presence of the xy term in the second equation causes the derivative to increase. Therefore, the presence of x is beneficial to y.

3 (b). Rewriting the system as x ¢ = x (1 - x - y ) y ¢ = - y (1 + y - x ) , we see that x ¢ = 0 if (i) x = 0 or (ii) 1 - x - y = 0 . In case (i), y ¢ = 0 if y = 0 or y = -1. The latter possibility has been excluded and thus case (i) leads to a single equilibrium point, ( x, y ) = (0, 0) . In case (ii), y ¢ = 0 if y = 0 (and hence, x = 1) or if 1 + y - x = 0 (and hence x + y = 1 and x - y = 1). This second set of equations also has solution x = 1 and y = 0. Thus, case (ii) leads us to one more equilibrium point ( x, y ) = (1, 0) . È1 0 ˘ 3 (c). At the equilibrium point z = 0 , the linearized system takes the form z¢ = Í ˙z . The Î0 -1˚ eigenvalues of the coefficient matrix are l1 = -1 and l 2 = 1 . Since, z = 0 is an unstable saddle point of the linearized system, the original system is also unstable at y = 0 .

246 • Chapter 8 Nonlinear Systems

4 (a). Consider the system x ¢ = x - x 2 + xy y ¢ = y - y 2 + xy . If y = 0 , then x approaches an equilibrium value of x e = 1 as t increases. If x = 0 , then y approaches an equilibrium value of y e = 1 as t increases. In both cases, the presence of one species is beneficial to the other species. 4 (b). The only equilibrium points in the first quadrant are ( x, y ) = (0, 0) , ( x, y ) = (0,1) , and ( x, y ) = (1, 0) . È1 0 ˘ 4 (c). At the equilibrium point z = 0 , the linearized system takes the form z¢ = Í ˙z . The Î0 1 ˚ eigenvalues of the coefficient matrix are l1 = 1 and l 2 = 1. Since, z = 0 is an unstable proper node of the linearized system, the original system is also unstable at y = 0 . 5 (a). When y = 0, the assumed model reduces to x ¢ = r1 (1 + a1 x ) x . In this case, we see from the x ¢ ( t) = 0.5 or x ¢ = 0.5 x . Thus, figure, that ln x ( t) = 0.5 t + ln x (0) . Differentiating, we obtain x ( t) a1 = 0 and r1 = 0.5 . Similarly, when x = 0, the model reduces to y ¢ = r2 (1 + a 2 y ) y . In this case, y ¢ ( t) = -1 or we see from the figure, that ln y ( t) = - t + ln y (0) . Differentiating, we obtain y ( t) y ¢ = - y . Thus, a 2 = 0 and r2 = -1. So far, we have deduced that the assumptions of the population model imply it has the form x ¢ = 0.5(1 + b1 y ) x

y ¢ = -(1 + b 2 x ) y . Knowing the equilibrium point ( x e , y e ) = (2, 3) , allows us to determine the last remaining model parameters, b1 and b 2 . In particular, we know from the first equation that 0.5(1 + 3b1 )2 = 0 while the second equation gives -(1 + 2b 2 ) 3 = 0 . Consequently, b1 = -1 / 3 and b 2 = -1 / 2. 5 (b). From part (a), the model is given by x ¢ = (1 / 2) x - (1 / 6) xy y ¢ = - y + (1 / 2) xy . The presence of y causes x ¢ to decrease and hence y is harmful to x. The presence of x causes y ¢ to increase and hence x is beneficial to y. 6 (a). Consider the system x ¢ = r(1 - ax - by ) x + mx y ¢ = r(1 - ay - bx ) y . The equilibrium points are ( x, y ) = (0, 0) , ( x, y ) = (0,a -1 ) , ( x, y ) = (a -1 (1 + mr -1 ), 0) , and ( x, y ) = d -1 (a (1 + mr -1 ) - b,a - b (1 + mr -1 )) where d = a 2 - b 2 . 6 (b). If m is chosen large enough so that b (1 + mr -1 ) > a then we see from part (a) that the “coexisting species” equilibrium point is moved into the fourth quadrant and is therefore physically irrelevant.

Chapter 8 Nonlinear Systems • 247

Èr + m 0 ˘ z . The point z = 0 is an unstable 6 (c). At z = 0 , the linearized system has the form z¢ = Í r ˙˚ Î 0 È 0 ˘ improper node. At the equilibrium point z = Í ˙ , the linearized system is Î1 / a ˚ Èr(1 + mr -1 - ba -1 ) 0 ˘ -1 -1 z¢ = Í ˙z . The eigenvalues are l1 = - r and l 2 = r(1 + mr - ba ) . Since -1 r r ba Î ˚ the eigenvalues have opposite sign, the equilibrium point is an unstable saddle point. The equilibrium point ( x, y ) = (a -1 (1 + mr -1 ), 0) is an asymptotically stable improper node since the eigenvalues of the linearized system are negative and different: l1 = - r(1 + mr -1 ) and l 2 = r[1 - bm (ar) -1 - ba -1 ]. 6 (d). For the nonlinear system, (0, 0) and (0,a -1 ) are unstable equilibrium points. The equilirium point ( x, y ) = (a -1 (1 + mr -1 ), 0) is stable. 6 (e). It appears that the y species will be driven to extinction with the x species approaching the limiting value a -1 (1 + mr -1 ) . 7 (a). Consider the system x ¢ = r(1 - ax - by ) x

y ¢ = r(1 - ay - bx ) y - my . We see that x ¢ = 0 if (i) x = 0 or (ii) 1 - ax - by = 0 . In case (i), y ¢ = 0 if y = 0 or y = ( r - m ) / (ar) . Thus case (i) leads to two equilibrium points, ( x, y ) = (0, 0) and ( x, y ) = (0,( r - m ) / (ar)) . In case (ii), y ¢ = 0 if y = 0 or if 1 - (m / r) - ay - bx = 0 . Thus case (ii) leads to two equilibrium points, ( x, y ) = (1 / a , 0) and ( x, y ) = (d -1[a - b (1 - mr -1 )],d -1[-b + a (1 - mr -1 )]) where d = a 2 - b 2 . 7 (b). If m > r, then 1 - mr -1 < 0 . In this case, we see from part (a) that the only physically relevant equilibrium points are ( x, y ) = (0, 0) and ( x, y ) = (1 / a , 0) . 0 ˘ Èr 7 (c). At z = 0 , the linearized system has the form z¢ = Í ˙z . Since we are assuming m > r, Î0 r - m ˚ È1 / a ˘ the point z = 0 is an unstable saddle point. At the equilibrium point z = Í ˙ , the linearized Î 0 ˚ È- r - rba -1 ˘ system is z¢ = Í z . The eigenvalues are l1 = - r and l 2 = r - m - rba -1. -1 ˙ Î 0 r - m - rba ˚ Since both eigenvalues are negative, the equilibrium point is an asymptotically stable improper node. 7 (d). For the nonlinear system, (0, 0) is unstable and (a -1, 0) is stable. 7 (e). If m > r, it appears that the y species will be driven to extinction with the x species approaching the limiting value a -1 . 8. The strategy of nurturing the desirable species leads to an equilibrium x-population of a -1 (1 + mr -1 ) . This is greater than the equilibrium x-population of a -1 that results from harvesting the undesirable species.

248 • Chapter 8 Nonlinear Systems

9.

Consider the population model x ¢ = ± a1 x ± b1 x 2 ± c1 xy ± d1 xz y ¢ = ± a2 y ± b2 y 2 ± c 2 xy ± d2 yz z¢ = ± a3 z ± c 3 xz ± d3 yz . Since x and y are mutually competitive, we need to choose a negative sign for c1 and c 2 (the presence of x reduces the growth rate y ¢ and similarly the presence of y reduces the growth rate x ¢ ). The same argument applies to the signs of d1 and d2 since the predator is harmful to x and to y. The presence of the prey is beneficial to the predator z and thus we need to choose a positive sign for c 3 and d3 . So far, we have deduced x ¢ = ± a1 x ± b1 x 2 - c1 xy - d1 xz y ¢ = ± a2 y ± b2 y 2 - c 2 xy - d2 yz z¢ = ± a3 z + c 3 xz + d3 yz . We also know that, in the absence of the other two species, x and y each evolve towards a nonzero equilibrium value. Thus, from the first equation, we know the term ± a1 x ± b1 x 2 = x ( ± a1 ± b1 x ) has a positive zero, as does the corresponding term in the second equation, ± a2 y ± b2 y 2 = y ( ± a2 ± b2 y ) . From this fact, we infer that a1 and b1 have opposite signs, as do a2 and b2 . The general solution of an equation of the form u¢ = au + bu 2 is u = Ae - at + Bt 2 + Ct + D . If a is negative, then u( t) Æ • as t Æ • . Hence, there cannot be a nonzero equilibrium solution when a is negative. Applying this observation to the equations x ¢ = ± a1 x ± b1 x 2 and y ¢ = ± a2 y ± b2 y 2 , we deduce that a1 and a2 are positive and b1 and b2 are negative. Likewise, in order that z decrease to zero in the absence of x and y, we need to have a3 negative. In summary, we arrive at the following model which will support the observations: x ¢ = a1 x - b1 x 2 - c1 xy - d1 xz

y ¢ = a2 y - b2 y 2 - c 2 xy - d2 yz z¢ = - a3 z + c 3 xz + d3 yz . 10 (a). Consider the system s¢ = -asi + gr i¢ = asi - bi r¢ = bi - gr . Summing these three equations, we obtain s¢ ( t) + i¢ ( t) + r¢ ( t) = 0 . Hence, s( t) + i( t) + r( t) is constant, say s( t) + i( t) + r( t) = N where N denotes the size of the population. 10 (b). If those who recover are permanently immunized, then s¢ = -asi i¢ = asi - bi r¢ = bi . As in part (a), we can sum these equations and again conclude that s( t) + i( t) + r( t) = N .

Chapter 8 Nonlinear Systems • 249

10 (c). If some infected members perish, then s¢ = -asi i¢ = asi - bi r¢ = bi - gr . In this case, s¢ ( t) + i¢ ( t) + r¢ ( t) = -gr( t) . Thus, the population is not constant but rather is decreasing. 11 (a). Consider the system s¢ = -asi + gr i¢ = asi - bi r¢ = bi - gr . Using the fact, from Exercise 10, that s + i + r = N , we obtain a reduced system, s¢ = -asi + g ( N - i - s)

i¢ = asi - bi . 11 (b). For the given values, a = b = g = 1 and N = 9 , the reduced system has the form s¢ = - si + (9 - i - s) i¢ = si - i . Rewriting this system slightly, s¢ = - si + 9 - i - s i¢ = i( s - 1) . We see that i¢ = 0 if (i) i = 0 or (ii) s = 1. In case (i), s¢ = 0 if s = 9 . Thus case (i) leads to the equilibrium point ( s, i) = (9, 0) . In case (ii), s¢ = 0 if i = 4 . Thus case (ii) leads to the equilibrium point ( s, i) = (1, 4 ) . È-1 -10 ˘ È9 ˘ z . The eigenvalues are 11 (c). At z = Í ˙ , the linearized system has the form z¢ = Í 8 ˙˚ Î0 Î0 ˚ È1 ˘ l1 = -1 and l 2 = 8 . This equilibrium point is an unstable saddle point. At z = Í ˙ , the Î4 ˚ È-5 -2 ˘ linearized system has the form z¢ = Í ˙z . The eigenvalues are Î4 0˚

l1 = (-5 - i 7 ) / 2 and (-5 + i 7 ) / 2 . This equilibrium point is an asymptotically stable spiral point. 11 (d). (9, 0) is an unstable equilibrium point while (1, 4 ) is stable.

Chapter 9 Numerical Methods Section 9.1 Unless indicated by K , all results are rounded to the places shown. 1 (a). Integrating y ¢ = 2 t - 1, we find y = t 2 - t + C . Imposing the initial condition y(1) = 0 , we obtain y = t 2 - t . 1 (b). Since f ( t, y ) = 2 t - 1, it follows that f ( t + h, y + hf ( t, y )) = 2( t + h ) - 1. Therefore, Heun’s method takes the form y n +1 = y n + ( h / 2)[(2 tn - 1) + (2 tn +1 - 1)] . 1 (c). As in part (b), we find the modified Euler’s method takes the form y n +1 = y n + h (2( tn + h / 2) - 1) . 1 (d).

1.0000 1.1000 1.2000 1.3000

0 0.1100 0.2400 0.3900

1 (e).

1.0000 0 1.1000 0.1100 1.2000 0.2400 1.3000 0.3900

1 (f).

1.0000 1.1000 1.2000 1.3000

0 0.1100 0.2400 0.3900

2 (a). Integrating y ¢ = - y and imposing the initial condition, we obtain y = e - t . 2 (b). Heun’s method takes the form y n +1 = (1 - h + 0.5 h 2 ) y n . 2 (c). The modified Euler’s method takes the form y n +1 = (1 - h + 0.5 h 2 ) y n . 2 (d).

0.0000 0.1000 0.2000 0.3000

1.0000 0.9050 0.8190 0.7412

Chapter 9 Numerical Methods • 251

2 (e).

0.0000 1.0000 0.1000 0.9050 0.2000 0.8190 0.3000 0.7412

2 (f).

0.0000 1.0000 0.1000 0.9048 0.2000 0.8187 0.3000 0.7408 2

3 (a). Solving the separable equation y ¢ = - ty , we find y = Ce - t / 2 . Imposing the initial condition 2 y(0) = 1, we obtain y = e - t / 2 . 3 (b). Since f ( t, y ) = - ty , it follows that f ( t + h, y + hf ( t, y )) = -( t + h )[ y + h (- ty )] . Therefore, Heun’s method takes the form y n +1 = y n + ( h / 2)[- tn y n - tn +1 ( y n - htn y n )] . 3 (c). As in part (b), we find the modified Euler’s method takes the form y n +1 = y n - h ( tn + 0.5 h )( y n - 0.5 htn y n ) . 3 (d).

3 (e).

3 (f).

0 0.1000 0.2000 0.3000 0 0.1000 0.2000 0.3000 0 0.1000 0.2000 0.3000

1.0000 0.9950 0.9802 0.9560 1.0000 0.9950 0.9801 0.9559 1.0000 0.9950 0.9802 0.9560

4 (a). Integrating y ¢ = - y + t and imposing the initial condition, we obtain y = t - 1 + e - t . 4 (b). Heun’s method takes the form y n +1 = y n + 0.5 h[- y n + tn - ( y n + h (- y n + tn )) + tn +1 ] . 4 (c). The modified Euler’s method takes the form y n +1 = y n + h[-( y n + 0.5 h (- y n + tn )) + tn + 0.5 h ]. 4 (d).

0.0000 0.1000 0.2000 0.3000

0.0000 0.0050 0.0190 0.0412

4 (e).

0.0000 0.0000 0.1000 0.0050 0.2000 0.0190 0.3000 0.0412

252 • Chapter 9 Numerical Methods

4 (f).

0.0000 0.1000 0.2000 0.3000

0.0000 0.0048 0.0187 0.0408

5 (a). Solving the separable equation y 2 y ¢ + t = 0 , we find y 3 = -1.5 t 2 + C . Imposing the initial condition y(0) = 1, we obtain y 3 = -1.5 t 2 + 1 or y = (1 - 1.5 t 2 )1/ 3 . 5 (b). Since f ( t, y ) = - ty -2 , it follows that f ( t + h, y + hf ( t, y )) = -( t + h )[ y + h (- ty )]-2 . Therefore, Heun’s method takes the form y n +1 = y n + ( h / 2)[- tn y n-2 - tn +1 ( y n - htn y n-2 ) -2 ] . 5 (c). As in part (b), we find the modified Euler’s method takes the form y n +1 = y n - h ( tn + 0.5 h )( y n - 0.5 htn y n-2 ) -2 . 5 (d). 0 1.0000 0.1000 0.9950 0.2000 0.9796 0.3000 0.9529 5 (e). 0 1.0000 0.1000 0.9950 0.2000 0.9797 0.3000 0.9531 5 (f). 0 1.0000 0.1000 0.9950 0.2000 0.9796 0.3000 0.9528

Chapter 9 Numerical Methods • 253

6 (a). Solving the separable equation y ¢ = 1 + y 2 and imposing the initial condition we obtain y = tan( t - p / 4 ) . 6 (b). T E H IE 0.0000 -1.0000 -1.0000 -1.0000 0.0500 -0.9000 -0.9047 -0.9049 0.1000 -0.8095 -0.8177 -0.8179 0.1500 -0.7267 -0.7375 -0.7378 0.2000 -0.6503 -0.6630 -0.6634 0.2500 -0.5792 -0.5933 -0.5937 0.3000 -0.5124 -0.5276 -0.5280 0.3500 -0.4493 -0.4653 -0.4657 0.4000 -0.3892 -0.4058 -0.4062 0.4500 -0.3316 -0.3486 -0.3491 0.5000 -0.2761 -0.2934 -0.2940 0.5500 -0.2223 -0.2399 -0.2404 0.6000 -0.1698 -0.1876 -0.1881 0.6500 -0.1184 -0.1362 -0.1368 0.7000 -0.0677 -0.0856 -0.0862 0.7500 -0.0175 -0.0354 -0.0360 0.8000 0.0326 0.0147 0.0140 0.8500 0.0826 0.0648 0.0641 0.9000 0.1329 0.1152 0.1145 0.9500 0.1838 0.1662 0.1654 1.0000 0.2355 0.2181 0.2173 The errors at t = 1 are, respectively, 0.0176, 1.6495e-004, and 6.8239e-004.

254 • Chapter 9 Numerical Methods

7 (a). Solving the separable equation yy ¢ + t = 0 , we find y 2 = - t 2 + C . Imposing the initial condition y(0) = 3, we obtain y 2 = - t 2 + 9 or y = (9 - t 2 )1/ 2 . 7 (b). T E H IE 0 3.0000 3.0000 3.0000 0.0500 3.0000 2.9996 2.9996 0.1000 2.9992 2.9983 2.9983 0.1500 2.9975 2.9962 2.9962 0.2000 2.9950 2.9933 2.9933 0.2500 2.9917 2.9896 2.9896 0.3000 2.9875 2.9850 2.9850 0.3500 2.9825 2.9795 2.9795 0.4000 2.9766 2.9732 2.9732 0.4500 2.9699 2.9661 2.9661 0.5000 2.9623 2.9580 2.9580 0.5500 2.9539 2.9492 2.9492 0.6000 2.9445 2.9394 2.9394 0.6500 2.9344 2.9287 2.9287 0.7000 2.9233 2.9172 2.9172 0.7500 2.9113 2.9047 2.9047 0.8000 2.8984 2.8914 2.8914 0.8500 2.8846 2.8771 2.8771 0.9000 2.8699 2.8618 2.8618 0.9500 2.8542 2.8456 2.8456 1.0000 2.8376 2.8284 2.8284 The errors at t = 1 are, respectively, -9.1466e-003, -6.9021e-007, and -1.3752e-005.

Chapter 9 Numerical Methods • 255

8 (a). Solving the equation y ¢ + 2 y = 4 and imposing the initial condition we obtain y = 2 + e -2 t . 8 (b). T E H IE 0 3.0000 3.0000 3.0000 0.0500 2.9000 2.9050 2.9050 0.1000 2.8100 2.8190 2.8190 0.1500 2.7290 2.7412 2.7412 0.2000 2.6561 2.6708 2.6708 0.2500 2.5905 2.6071 2.6071 0.3000 2.5314 2.5494 2.5494 0.3500 2.4783 2.4972 2.4972 0.4000 2.4305 2.4500 2.4500 0.4500 2.3874 2.4072 2.4072 0.5000 2.3487 2.3685 2.3685 0.5500 2.3138 2.3335 2.3335 0.6000 2.2824 2.3018 2.3018 0.6500 2.2542 2.2732 2.2732 0.7000 2.2288 2.2472 2.2472 0.7500 2.2059 2.2237 2.2237 0.8000 2.1853 2.2025 2.2025 0.8500 2.1668 2.1832 2.1832 0.9000 2.1501 2.1658 2.1658 0.9500 2.1351 2.1501 2.1501 1.0000 2.1216 2.1358 2.1358 The errors at t = 1 are, respectively, 1.3758e-002, 4.8717e-004, and 4.8717e-004.

256 • Chapter 9 Numerical Methods 2

9 (a). Solving the separable equation y ¢ + 2 ty = 0 , we find y = Ce - t . Imposing the 2 initial condition y(0) = 2 , we obtain y = 2e - t . 9 (b). T 0 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500 1.0000

E 2.0000 2.0000 1.9900 1.9701 1.9405 1.9017 1.8542 1.7986 1.7356 1.6662 1.5912 1.5117 1.4285 1.3428 1.2555 1.1676 1.0801 0.9937 0.9092 0.8274 0.7488

H 2.0000 1.9950 1.9801 1.9555 1.9216 1.8788 1.8278 1.7694 1.7043 1.6334 1.5576 1.4780 1.3955 1.3110 1.2255 1.1398 1.0549 0.9715 0.8902 0.8116 0.7364

IE 2.0000 1.9950 1.9801 1.9554 1.9215 1.8787 1.8277 1.7692 1.7040 1.6330 1.5572 1.4775 1.3949 1.3103 1.2247 1.1390 1.0541 0.9706 0.8893 0.8107 0.7354

The errors at t = 1 are, respectively, -1.3009e-002, -6.0218e-004, and 3.3293e-004. 10. 11.

12. 13.

14. 16.

The iteration is Euler’s method, with t0 = 2, T = 1, and f ( t, y ) = y + t 2 y 3 . Since tn = 1 + nh, h = 0.05, n = 0, 1, K, 99 , it follows that t0 = 1 and N - 1 = 99 . Thus, N = 100, and T = tN = 1 + Nh = 1 + 100 h = 1 + (100)(0.05) = 5 . From the form of the iteration, it must be Heun’s method. Therefore, f ( t, y ) = ty 2 + 1. The iteration is the modified Euler’s method, with t0 = 0, T = 2, and f ( t, y ) = t sin 2 y . Since tn = 2 + nh, h = 0.01, n = 0, 1, K, 99 , it follows that t0 = 2 and N - 1 = 99 . Thus, N = 100, and T = tN = 2 + Nh = 2 + 100 h = 2 + (100)(0.01) = 1. From the form of the iteration, it must be Euler’s method. Therefore, f ( t, y ) = y / ( t 2 + y 2 ) . The iteration is the modified Euler’s method, with t0 = -1, T = 10, and f ( t, y ) = sin( t + y ) . Q( t) , Q(0) = 0 where V ( t) = 90 + 5 t . (a) The initial value problem is Q¢ ( t) = 6(2 - cospt) V ( t) (c) The tank contains 100 gallons when t = 2 minutes. As estimated by Heun’s method, Q(2) = 23.7538K pounds.

Chapter 9 Numerical Methods • 257

17 (a). From Exercise 16, part (a), the problem to be solved is Q¢ = 12 - 6 cospt - Q / (90 + 5 t), Q(0) = 0, 0 £ t £ 2 . 17 (b). Using the modified Euler’s method with h = 0.05, we obtain

18.

t Q(t) 0 0 0.0500 0.3008 0.1000 0.6089 0.1500 0.9313 0.2000 1.2749 0.2500 1.6460 0.3000 2.0501 0.3500 2.4922 0.4000 2.9759 0.4500 3.5041 0.5000 4.0785 0.5500 4.6997 0.6000 5.3670 0.6500 6.0787 0.7000 6.8320 0.7500 7.6230 0.8000 8.4469 0.8500 9.2979 0.9000 10.1700 0.9500 11.0561 1.0000 11.9491 1.0500 12.8416 1.1000 13.7264 1.1500 14.5961 1.2000 15.4441 1.2500 16.2640 1.3000 17.0502 1.3500 17.7979 1.4000 18.5033 1.4500 19.1636 1.5000 19.7772 1.5500 20.3434 1.6000 20.8628 1.6500 21.3372 1.7000 21.7695 1.7500 22.1635 1.8000 22.5241 1.8500 22.8569 1.9000 23.1681 1.9500 23.4647 2.0000 23.7538 The Heun’s method estimate is P(2) = 1.5005 million individuals.

258 • Chapter 9 Numerical Methods

19. Using the modified Euler’s method, we estimate P(2) = 1.5003 million individuals. 20 (a). The results are listed below. The columns headed H1, H2, and H3 are the results obtained using step sizes h = 0.05, h = 0.025, and h = 0.0125 respectively. t H1 H2 H3 True 0.0000 1.0000 1.0000 1.0000 1.0000 0.0500 1.0526 1.0526 1.0526 1.0526 0.1000 1.1109 1.1111 1.1111 1.1111 0.1500 1.1762 1.1764 1.1765 1.1765 0.2000 1.2495 1.2499 1.2500 1.2500 0.2500 1.3326 1.3332 1.3333 1.3333 0.3000 1.4275 1.4283 1.4285 1.4286 0.3500 1.5370 1.5381 1.5384 1.5385 0.4000 1.6645 1.6661 1.6665 1.6667 0.4500 1.8151 1.8174 1.8180 1.8182 0.5000 1.9954 1.9988 1.9997 2.0000 0.5500 2.2153 2.2204 2.2218 2.2222 0.6000 2.4894 2.4972 2.4993 2.5000 0.6500 2.8402 2.8527 2.8560 2.8571 0.7000 3.3049 3.3257 3.3314 3.3333 0.7500 3.9488 3.9860 3.9964 4.0000 0.8000 4.8975 4.9714 4.9925 5.0000 0.8500 6.4264 6.5969 6.6480 6.6667 0.9000 9.2615 9.7669 9.9353 10.0000 0.9500 15.9962 18.4267 19.5053 20.0000 The error ratios are denoted, respectively, by R1 and R2 where R1 = ( H1 - True) / ( H 2 - True) and R2 = ( H 2 - True) / ( H 3 - True) t 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500

R1 3.8916 3.8879 3.8838 3.8791 3.8737 3.8674 3.8601 3.8513 3.8408 3.8279 3.8117 3.7909 3.7634 3.7255 3.6707 3.5860 3.4428 3.1683 2.5450

R2 3.9471 3.9454 3.9436 3.9414 3.9390 3.9362 3.9329 3.9291 3.9244 3.9187 3.9117 3.9026 3.8907 3.8743 3.8503 3.8126 3.7459 3.6039 3.1800

Chapter 9 Numerical Methods • 259

) ) 20 (b). An error monitor is y ( t* ) - y 2 n = ( y 2 n - y n ) / 3 . 20 (c). The column headed est gives the estimated error using the error monitor from part (b). The column headed true gives the actual error. The error monitor used step sizes of h = 0.025 and h = 0.0125. t 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500

est 4.4179e-006 1.0382e-005 1.8466e-005 2.9497e-005 4.4686e-005 6.5850e-005 9.5774e-005 1.3886e-004 2.0228e-004 2.9820e-004 4.4818e-004 6.9264e-004 1.1126e-003 1.8854e-003 3.4446e-003 7.0273e-003 1.7050e-002 5.6137e-002 3.5951e-001

true 4.4972e-006 1.0574e-005 1.8820e-005 3.0084e-005 4.5614e-005 6.7281e-005 9.7965e-005 1.4222e-004 2.0751e-004 3.0650e-004 4.6178e-004 7.1587e-004 1.1547e-003 1.9679e-003 3.6255e-003 7.4956e-003 1.8628e-002 6.4677e-002 4.9473e-001

Section 9.2 Unless indicated by K , all results are rounded to the places shown. 1 (a). From the given equation y ¢ = - y + 2 , we know y ¢ ( t) = - y ( t) + 2 , y ¢¢ ( t) = - y ¢ ( t) , y ¢¢¢ ( t) = - y ¢¢ ( t) , and y ( 4 ) ( t) = - y ¢¢¢ ( t) . We also know that y(0) = 1. Therefore, y ¢ (0) = - y (0) + 2 = -1 + 2 = 1, y ¢¢ (0) = - y ¢ (0) = -1, y ¢¢¢ (0) = - y ¢¢ (0) = 1 , and y ( 4 ) (0) = - y ¢¢¢ (0) = -1. Thus, P4 ( t) = 1 + t - (1 / 2) t 2 + (1 / 6) t 3 - (1 / 24 ) t 4 . 2 (a). From the given equation y ¢ = 2 ty , we know, y ¢ (0) = 0 , y ¢¢ (0) = 2 , y ¢¢¢ (0) = 0 , and y ( 4 ) (0) = 12 . Thus, P4 ( t) = 1 + t 2 + (1 / 2) t 4 . 3 (a). From the given equation y ¢ = ty 2 , we know y ¢ ( t) = ty 2 ( t) , y ¢¢ ( t) = y 2 ( t) + 2 ty ( t) y ¢ ( t) , y ¢¢¢ ( t) = 4 y ( t) y ¢ ( t) + 2 ty ¢ ( t) y ¢ ( t) + 2 ty ( t) y ¢¢ ( t) , and y ( 4 ) ( t) = 6 y ¢ ( t) y ¢ ( t) + 6 y ( t) y ¢¢ ( t) + 6 ty ¢ ( t) y ¢¢ ( t) + 2 ty ( t) y ¢¢¢ ( t) . We also know that y(0) = 1. Therefore, y ¢ (0) = 0 , y ¢¢ (0) = 1, y ¢¢¢ (0) = 0 , and y ( 4 ) (0) = 6 . Thus, P4 ( t) = 1 + (1 / 2) t 2 + (1 / 4 ) t 4 .

260 • Chapter 9 Numerical Methods

4 (a). From the given equation y ¢ = t 2 + y , we know, y ¢ (0) = 1, y ¢¢ (0) = 1, y ¢¢¢ (0) = 3 , and y ( 4 ) (0) = 3 . Thus, P4 ( t) = 1 + t + (1 / 2) t 2 + (1 / 2) t 3 + (1 / 8) t 4 . 5 (a). From the given equation y ¢ = y1/ 2 , we know y ¢ ( t) = y1/ 2 ( t) , y ¢¢ ( t) = (1 / 2) y -1/ 2 ( t) y ¢ ( t) , y ¢¢¢ ( t) = -(1 / 4 ) y -3 / 2 ( t) y ¢ ( t) + (1 / 2) y -1/ 2 ( t) y ¢¢ ( t) , and y ( 4 ) ( t) = ( 3 / 8) y -5 / 2 ( t) y ¢ ( t) y ¢ ( t) - ( 3 / 4 ) y -3 / 2 ( t) y ¢¢ ( t) y ¢ ( t) + (1 / 2) y -1/ 2 ( t) y ¢¢¢ ( t) . We also know that y(0) = 1. Therefore, y ¢ (0) = 1, y ¢¢ (0) = 1 / 2 , y ¢¢¢ (0) = -(1 / 4 ) + (1 / 2)(1 / 2) = 0 , and y ( 4 ) (0) = ( 3 / 8) - ( 3 / 4 )(1 / 2) = 0 . Thus, P4 ( t) = 1 + t + (1 / 4 ) t 2 . 6 (a). From the given equation y ¢ = ty -1 , we know, y ¢ (0) = 0 , y ¢¢ (0) = 1, y ¢¢¢ (0) = 0 , and y ( 4 ) (0) = -3 . Thus, P4 ( t) = 1 + (1 / 2) t 2 - (1 / 8) t 4 . 7 (a). From the given equation y ¢ = y + sin t , we know y ¢ ( t) = y ( t) + sin t , y ¢¢ ( t) = y ¢ ( t) + cos t , y ¢¢¢ ( t) = y ¢¢ ( t) - sin t , and y ( 4 ) ( t) = y ¢¢¢ ( t) - cos t . We also know that y(0) = 1. Therefore, y ¢ (0) = 1, y ¢¢ (0) = 1 + 1 = 2 , y ¢¢¢ (0) = 2 - 0 = 2 , and y ( 4 ) (0) = 2 - 1 = 1. Thus, P4 ( t) = 1 + t + t 2 + (1 / 3) t 3 + (1 / 24 ) t 4 . 8 (a). From the given equation y ¢ = y 3 / 4 , we know, y ¢ (0) = 1, y ¢¢ (0) = 3 / 4 , y ¢¢¢ (0) = 3 / 8 , and y ( 4 ) (0) = 3 / 32 . Thus, P4 ( t) = 1 + t + ( 3 / 8) t 2 + (1 / 16) t 3 + (1 / 256) t 4 . 9 (a). From the given equation y ¢ = 1 + y 2 , we know y ¢ ( t) = 1 + y 2 ( t) , y ¢¢ ( t) = 2 y ( t) y ¢ ( t) , y ¢¢¢ ( t) = 2 y ¢ ( t) y ¢ ( t) + 2 y ( t) y ¢¢ ( t) , and y ( 4 ) ( t) = 6 y ¢¢ ( t) y ¢ ( t) + 2 y ( t) y ¢¢ ( t) . We also know that y(0) = 1. Therefore, y ¢ (0) = 2 , y ¢¢ (0) = (2)(1)(2) = 4 , y ¢¢¢ (0) = (2)(2)(2) + (2)(1)( 4 ) = 16 , and y ( 4 ) (0) = (6)( 4 )(2) + (2)(1)(16) = 80 . Thus, P4 ( t) = 1 + 2 t + 2 t 2 + (8 / 3) t 3 + (10 / 3) t 4 . 10 (a). From the given equation y ¢ = -4 t 3 y , we know, y ¢ (0) = 0 , y ¢¢ (0) = 0 , y ¢¢¢ (0) = 0 , and y ( 4 ) (0) = -24 . Thus, P4 ( t) = 1 - t 4 . 11 (a). From the given equation y ¢¢ = 3 y ¢ - 2 y , we know y ¢¢ ( t) = 3 y ¢ ( t) - 2 y ( t) , y ¢¢¢ ( t) = 3 y ¢¢ ( t) - 2 y ¢ ( t) , y ( 4 ) ( t) = 3 y ¢¢¢ ( t) - 2 y ¢¢ ( t) , and y ( 5 ) ( t) = 3 y ( 4 ) ( t) - 2 y ¢¢¢ ( t) . We also know that y (0) = 1 and y ¢ (0) = 0 . Therefore, y ¢¢ (0) = ( 3)(0) - (2)(1) = -2 , y ¢¢¢ (0) = ( 3)(-2) - (2)(0) = -6 , y ( 4 ) (0) = ( 3)(-6) - (2)(-2) = -14 , and y ( 5 ) (0) = ( 3)(-14 ) - (2)(-6) = -30 . Thus, P5 ( t) = 1 - t 2 - t 3 - ( 7 / 12) t 4 - (1 / 4 ) t 5 . 12 (a). From the given equation y ¢¢ - y ¢ = 0 , we know, y ¢¢ (1) = 2 , y ¢¢¢ (1) = 2 , y ( 4 ) (1) = 2 , and y ( 5 ) (1) = 2 . Thus, P5 ( t) = 1 + 2( t - 1) + ( t - 1) 2 + (1 / 3)( t - 1) 3 + (1 / 12)( t - 1) 4 + (1 / 60)( t - 1) 5 . 13 (a). From the given equation y ¢¢¢ = y ¢ , we know y ¢¢¢ ( t) = y ¢ ( t) , y ( 4 ) ( t) = y ¢¢ ( t) , and y ( 5 ) ( t) = y ¢¢¢ ( t) . We also know that y (0) = 1, y ¢ (0) = 2 and y ¢¢ (0) = 0 . Therefore, y ¢¢¢ (0) = 2 , y ( 4 ) (0) = 0 , y ( 5 ) (0) = 2 . Thus, P5 ( t) = 1 + 2 t + (1 / 3) t 3 + (1 / 60) t 5 .

Chapter 9 Numerical Methods • 261

14 (a). From the given equation y ¢¢ + y + y 3 = 0 , we know, y ¢¢ (0) = -2 , y ¢¢¢ (0) = 0 , y ( 4 ) (0) = 8 , and y ( 5 ) (0) = 0 . Thus, P5 ( t) = 1 - t 2 + (1 / 3) t 4 . 15. The function q( h ) = sin 2 h has a Maclaurin expansion given by sin 2 h = 2 h - (1 / 6)(2 h ) 3 + L. Therefore, q( h ) = O( h ) . 16. q( h ) = O( h ) . 17. The function q( h ) = 1 - cos h has a Maclaurin expansion given by 1 - cos h = 1 - (1 - (1 / 2) h 2 + (1 / 24 ) h 4 + L = (1 / 2) h 2 + L . Therefore, q( h ) = O( h 2 ) . 18. The function q( h ) = e h - (1 + h ) has a Maclaurin expansion given by e h - (1 + h ) = [1 + h + (1 / 2) h 2 + L] - (1 + h ) = (1 / 2) h 2 + L . Therefore, q( h ) = O( h 2 ) . 20 (a). t ts-1 ts-2 ts-3 0.0000 1.0000 1.0000 1.0000 0.0500 1.0000 1.0006 1.0006 0.1000 1.0013 1.0025 1.0025 0.1500 1.0037 1.0056 1.0056 0.2000 1.0075 1.0100 1.0100 0.2500 1.0125 1.0156 1.0156 0.3000 1.0187 1.0224 1.0224 0.3500 1.0261 1.0304 1.0304 0.4000 1.0348 1.0396 1.0396 0.4500 1.0446 1.0500 1.0500 0.5000 1.0556 1.0616 1.0616 0.5500 1.0677 1.0743 1.0742 0.6000 1.0810 1.0881 1.0881 0.6500 1.0955 1.1030 1.1030 0.7000 1.1110 1.1190 1.1190 0.7500 1.1276 1.1360 1.1360 0.8000 1.1452 1.1541 1.1541 0.8500 1.1638 1.1732 1.1731 0.9000 1.1835 1.1932 1.1932 0.9500 1.2041 1.2142 1.2142 1.0000 1.2256 1.2361 1.2361 20 (b). At t = 1, the errors are (respectively): -1.0441e-002, 5.5071e-005, and -1.0500e-006.

262 • Chapter 9 Numerical Methods

21 (a). t ts-1 ts-2 ts-3 0.0000 -1.0000 -1.0000 -1.0000 0.0500 -1.0000 -0.9975 -0.9975 0.1000 -0.9950 -0.9901 -0.9901 0.1500 -0.9851 -0.9779 -0.9780 0.2000 -0.9705 -0.9614 -0.9615 0.2500 -0.9517 -0.9409 -0.9412 0.3000 -0.9291 -0.9171 -0.9174 0.3500 -0.9032 -0.8905 -0.8908 0.4000 -0.8746 -0.8616 -0.8620 0.4500 -0.8440 -0.8311 -0.8316 0.5000 -0.8120 -0.7994 -0.8000 0.5500 -0.7790 -0.7672 -0.7677 0.6000 -0.7456 -0.7347 -0.7353 0.6500 -0.7123 -0.7024 -0.7030 0.7000 -0.6793 -0.6705 -0.6711 0.7500 -0.6470 -0.6394 -0.6400 0.8000 -0.6156 -0.6092 -0.6098 0.8500 -0.5853 -0.5800 -0.5806 0.9000 -0.5562 -0.5520 -0.5525 0.9500 -0.5283 -0.5252 -0.5256 1.0000 -0.5018 -0.4996 -0.5000 21 (b). At t = 1, the errors are (respectively): -1.8055e-003, 4.0475e-004, and -6.8372e-006 22 (a). t ts-1 ts-2 ts-3 0.0000 1.0000 1.0000 1.0000 0.0500 1.0250 1.0247 1.0247 0.1000 1.0494 1.0488 1.0488 0.1500 1.0732 1.0724 1.0724 0.2000 1.0965 1.0954 1.0954 0.2500 1.1193 1.1180 1.1180 0.3000 1.1416 1.1401 1.1402 0.3500 1.1635 1.1619 1.1619 0.4000 1.1850 1.1832 1.1832 0.4500 1.2061 1.2041 1.2042 0.5000 1.2269 1.2247 1.2247 0.5500 1.2472 1.2449 1.2450 0.6000 1.2673 1.2649 1.2649 0.6500 1.2870 1.2845 1.2845 0.7000 1.3064 1.3038 1.3038 0.7500 1.3256 1.3228 1.3229 0.8000 1.3444 1.3416 1.3416 0.8500 1.3630 1.3601 1.3601 0.9000 1.3814 1.3783 1.3784 0.9500 1.3995 1.3964 1.3964 1.0000 1.4173 1.4142 1.4142

Chapter 9 Numerical Methods • 263

22 (b). At t = 1, the errors are (respectively): 3.1075e-003, -5.7087e-005, and 1.3615e-006. 23 (a). t ts-1 ts-2 ts-3 0.0000 0.0000 0.0000 0.0000 0.0500 0.0500 0.0488 0.0488 0.1000 0.0977 0.0955 0.0956 0.1500 0.1436 0.1405 0.1407 0.2000 0.1880 0.1841 0.1844 0.2500 0.2311 0.2266 0.2269 0.3000 0.2733 0.2682 0.2686 0.3500 0.3146 0.3091 0.3095 0.4000 0.3553 0.3494 0.3498 0.4500 0.3955 0.3892 0.3897 0.5000 0.4354 0.4288 0.4293 0.5500 0.4751 0.4682 0.4687 0.6000 0.5146 0.5075 0.5080 0.6500 0.5541 0.5468 0.5473 0.7000 0.5937 0.5862 0.5868 0.7500 0.6335 0.6258 0.6264 0.8000 0.6736 0.6657 0.6664 0.8500 0.7139 0.7060 0.7067 0.9000 0.7547 0.7467 0.7475 0.9500 0.7960 0.7879 0.7888 1.0000 0.8379 0.8298 0.8307 23 (b). At t = 1, the errors are (respectively): 7.2979e-003, -8.2708e-004, and 3.0263e-005. 24. We find E1 = -1.0499 ¥ 10 -6 and E 2 = -1.2939 ¥ 10 -7 . The error ratio is 0.12323 while 1/8 is equal to 0.125. Thus, the error ratio is close to 1/8. 25. We find E1 = -6.8372 ¥ 10 -6 and E 2 = -8.4649 ¥ 10 -7 . The error ratio is 0.12381 while 1/8 is equal to 0.125. Thus, the error ratio is close to 1/8. 26. We find E1 = -1.3615 ¥ 10 -6 and E 2 = -1.6598 ¥ 10 -7 . The error ratio is 0.12191 while 1/8 is equal to 0.125. Thus, the error ratio is close to 1/8. 27. We find E1 = 3.0263 ¥ 10 -5 and E 2 = 3.6501 ¥ 10 -6 . The error ratio is 0.12061 while 1/8 is equal to 0.125. Thus, the error ratio is fairly close to 1/8.

Section 9.3 Unless indicated by K , all results are rounded to the places shown. 1 (a). For the given initial value problem y ¢ = - y + 2 , y (0) = 1, we have K1 = f ( t0 , y 0 ) = f (0,1) = 1 K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = f (0.05,1 + 0.05(1)) = 0.95 K 3 = f ( t0 + h, y 0 - hK1 + 2 hK 2 ) = 0.91 y1 = y 0 + h (K1 + 4 K 2 + K 3 ) / 6 = 1 + (0.1)(1 + 3.8 + 0.91) / 6 = 1.095166K

264 • Chapter 9 Numerical Methods

1 (b). As in (a), we find K1 = 1, K 2 = 0.95, K 3 = 0.9525, K 4 = 0.90475 and thus y1 = 1.0951625 . 1 (c). A kth order Runge-Kutta method will give the exact solution if the solution is a polynomial of degree k. In this case, the Runge-Kutta method will not give the exact solution. 2 (a). For the given initial value problem y ¢ = 2 ty , y (0) = 1 , we have y1 = 1.01006667 2 (b). y1 = 1.01005017 . 2 (c). Neither Runge-Kutta method will give the exact solution. 3 (a). For the given initial value problem y ¢ = ty 2 , y (0) = 1 , we have K1 = f ( t0 , y 0 ) = f (0,1) = 0 K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = f (0.05,1 + 0.05(0)) = 0.05 K 3 = f ( t0 + h, y 0 - hK1 + 2 hK 2 ) = 0.1020K y1 = y 0 + h (K1 + 4 K 2 + K 3 ) / 6 = 1 + (0.1)(0 + 0.05 + 0.1020) / 6 = 1.0050335K 3 (b). As in (a), we find K1 = 0, K 2 = 0.05, K 3 = 0.0503K, K 4 = 0.1010K and thus y1 = 1.0050251K . 3 (c). A kth order Runge-Kutta method will give the exact solution if the solution is a polynomial of degree k. In this case, the Runge-Kutta method will not give the exact solution. 4 (a). For the given initial value problem y ¢ = t 2 + y , y (0) = 1, we have y1 = 1.10550833 4 (b). y1 = 1.10551271. 4 (c). Neither Runge-Kutta method will give the exact solution. 5 (a). For the given initial value problem y ¢ = y , y (0) = 1, we have K1 = f ( t0 , y 0 ) = f (0,1) = 1 K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = f (0.05,1 + 0.05(1)) = 1.0246K K 3 = f ( t0 + h, y 0 - hK1 + 2 hK 2 ) = 1.0511K y1 = y 0 + h (K1 + 4 K 2 + K 3 ) / 6 = 1.1024990K 5 (b). As in (a), we find K1 = 1, K 2 = 1.0246K, K 3 = 1.0252K, K 4 = 1.0500K and thus y1 = 1.1024999K . 5 (c). A kth order Runge-Kutta method will give the exact solution if the solution is a polynomial of degree k. In this case, since the solution is a quadratic polynomial, both Runge-Kutta methods will give the exact solution. 6 (a). For the given initial value problem y ¢ = t / y , y (0) = 1, we have y1 = 1.00498350 6 (b). y1 = 1.00498757 . 6 (c). Neither Runge-Kutta method will give the exact solution. 7 (a). For the given initial value problem y ¢ = y + sin t , y (0) = 1, we have K1 = f ( t0 , y 0 ) = f (0,1) = 1 K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = f (0.05,1 + 0.05(1)) = 1.0999K K 3 = f ( t0 + h, y 0 - hK1 + 2 hK 2 ) = 1.2198K y1 = y 0 + h (K1 + 4 K 2 + K 3 ) / 6 = 1.110329K 7 (b). As in (a), we find K1 = 1, K 2 = 1.0999K, K 3 = 1.1049K, K 4 = 1.2103K and thus y1 = 1.110337K . 7 (c). A kth order Runge-Kutta method will give the exact solution if the solution is a polynomial of degree k. In this case, a Runge-Kutta method will not give the exact solution. 8 (a). For the given initial value problem y ¢ = y 3 / 4 , y (0) = 1, we have y1 = 1.10381059 8 (b). y1 = 1.10381285 . 8 (c). The 4th order Runge-Kutta method will give the exact solution, but not the 3rd order.

Chapter 9 Numerical Methods • 265

9 (a). For the given initial value problem y ¢ = 1 + y 2 , y (0) = 1, we have K1 = f ( t0 , y 0 ) = f (0,1) = 2 K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = f (0.05,1 + 0.05(2)) = 2.21 K 3 = f ( t0 + h, y 0 - hK1 + 2 hK 2 ) = 2.5425K y1 = y 0 + h (K1 + 4 K 2 + K 3 ) / 6 = 1.2230427K 9 (b). As in (a), we find K1 = 2, K 2 = 2.21, K 3 = 2.2332K, K 4 = 2.4965K and thus y1 = 1.2230489K . 9 (c). A kth order Runge-Kutta method will give the exact solution if the solution is a polynomial of degree k. In this case, a Runge-Kutta method will not give the exact solution. 10 (a). For the given initial value problem y ¢ = -4 t 3 y , y (0) = 1, we have y1 = 0.99990001 10 (b). y1 = 0.99990000 . 10 (c). Neither Runge-Kutta method will give the exact solution. 11. Rewriting the given initial value problem, y ¢¢ + ty ¢ + y , y (0) = 1, y ¢ (0) = -1, as a first order system, we have

y1¢ = y 2 , y1 (0) = 1 È1˘ È y2 ˘ or y ¢ = f ( t, y ), y (0) = Í ˙, f ( t, y ) = Í ˙. y 2¢ = - ty 2 - y1, y 2 (0) = -1 , Î-1˚ Î- ty 2 - y1 ˚ Therefore,

È-1˘ K1 = f ( t0 , y 0 ) = Í ˙ Î-1˚

12. 13.

È -1.05 ˘ K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = Í ˙ Î-0.8975 ˚ È-1.0448K˘ K 3 = f ( t0 + h / 2, y 0 + ( h / 2)K 2 ) = Í ˙ Î-0.8952K˚ È-1.0895K˘ K 4 = f ( t0 + h, y 0 + hK 3 ) = Í ˙ Î-0.7865K˚ È 0.895345K ˘ y1 = y 0 + h(K1 + 2K 2 + 2K 3 + K 4 ) / 6 = Í ˙. Î-1.089534 K˚ È1.194834 K˘ y1 = Í ˙ Î1.895042K˚ È0 t˘ È1˘ È2 ˘ y + Í ˙ , y (0) = Í ˙ , we have For the given initial value problem, y ¢ = Í t ˙ Îe 0 ˚ Ît ˚ Î1 ˚ È 1 + ty 2 ˘ f ( t, y ) = Í ˙ . Therefore, t Ît + e y1 ˚ È1 ˘ K1 = f ( t0 , y 0 ) = Í ˙ Î2 ˚ È 1.055 ˘ K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = Í ˙ Î2.2051K˚

266 • Chapter 9 Numerical Methods

14. 15.

È1.0555K ˘ K 3 = f ( t0 + h / 2, y 0 + ( h / 2)K 2 ) = Í ˙ Î2.2079K˚ È1.1220K ˘ K 4 = f ( t0 + h, y 0 + hK 3 ) = Í ˙ Î2.4269K˚ È2.105718K˘ y1 = y 0 + h(K1 + 2K 2 + 2K 3 + K 4 ) / 6 = Í ˙. Î1.220886K ˚ È-0.900625K˘ y1 = Í ˙ Î 0.809968K ˚ Rewriting the given initial value problem, y ¢¢¢ = ty , y (0) = 1, y ¢ (0) = 0, y ¢¢ (0) = -1, as a first order system, we have y1¢ = y 2 , y1 (0) = 1 È y2 ˘ È1 ˘ ˙ Í or y ¢ = f ( t, y ), y (0) = 0 , f ( t, y ) = Í y 3 ˙ . y 2¢ = y 3 , y 2 (0) = 0 Í ˙ Í ˙ y 3¢ = ty1, y 3 (0) = -1 , Îty1 ˚ Î-1˚

Therefore, È0 ˘ K1 = f ( t0 , y 0 ) = Í-1˙ Í ˙ Î0˚

16. 19 (a). 19 (b). 20 (a). 20 (b).

È-0.05 ˘ K 2 = f ( t0 + h / 2, y 0 + ( h / 2)K1 ) = Í-1.0 ˙ ˙ Í Î 0.05 ˚ È -0.05 ˘ K 3 = f ( t0 + h / 2, y 0 + ( h / 2)K 2 ) = Í -0.9975 ˙ ˙ Í Î0.0498K˚ È-0.0997K˘ K 4 = f ( t0 + h, y 0 + hK 3 ) = Í-0.9950K˙ ˙ Í Î 0.0995 ˚ È0.995004 K ˘ y1 = y 0 + h(K1 + 2K 2 + 2K 3 + K 4 ) / 6 = Í-0.099833K˙ . ˙ Í Î-0.995012K˚ È1.199637K˘ y1 = Í1.988834 K˙ Í0.114991K ˙ ˚ Î For the given initial value problem y ¢ = t / (1 + y ) , y (0) = 1 and for the step size h = 0.05 , we obtain y 20 = 1.2360679786K as our estimate of y(1) . The actual value of the solution is y(1) = 1.2360679749K . For the given initial value problem y ¢ = 2 ty 2 , y (0) = -1 and for the step size h = 0.05 , we obtain y 20 = -0.5000000409K as our estimate of y(1) . The actual value of the solution is y(1) = -0.5 .

Chapter 9 Numerical Methods • 267

21 (a). For the given initial value problem y ¢ = 1 / (2 y ) , y (0) = 1 and for the step size h = 0.05 , we obtain y 20 = 1.4142135632K as our estimate of y(1) . 21 (b). The actual value of the solution is y 20 = 1.4142135623K. 22 (a). For the given initial value problem y ¢ = (1 + y 2 ) / (1 + t) , y (0) = 0 and for the step size h = 0.05 , we obtain y 20 = 0.83064092K as our estimate of y(1) . 22 (b). The actual value of the solution is y(1) = 0.83064087K . 23 (a). Rewriting the given initial value problem, y ¢¢ + 2 y ¢ + 2 y = -2 , y (0) = 0, y ¢ (0) = 1, as a first order system, we have y1¢ = y 2 , y1 (0) = 0 È0 ˘ È y2 ˘ or y ¢ = f ( t, y ), y (0) = Í ˙, f ( t, y ) = Í ˙. y 2¢ = -2 y 2 - 2 y1 - 2, y 2 (0) = 1 , Î1 ˚ Î-2 y 2 - 2 y1 - 2 ˚ È-0.810202K˘ 23 (b). Using the step size h = 0.1, we obtain y 20 = Í ˙ as our estimate to the solution Î-0.425496K˚ È-0.810199K˘ value y(2) = Í ˙. Î-0.425499K˚ È0.829662K˘ 24 (b). Using the step size h = 0.1, we obtain y10 = Í ˙ as our estimate to the solution value Î0.383398K˚ È0.829660K˘ y(1) = Í ˙. Î0.383400K˚ 25 (a). Rewriting the given initial value problem, t 2 y ¢¢ - ty ¢ + y = t 2 , y (1) = 2, y ¢ (1) = 2 , as a first order system, we have y1¢ = y 2 , y1 (1) = 2 or y 2¢ = ( ty 2 - y1 + t 2 ) / t 2 , y 2 (1) = 2 , È y2 ˘ È2 ˘ y ¢ = f ( t, y ), y (0) = Í ˙, f ( t, y ) = Í . 2 2˙ Î2 ˚ Î( ty 2 - y1 + t ) / t ˚

È4.6137054 K˘ 25 (b). Using the step size h = 0.1, we obtain y10 = Í ˙ as our estimate to the solution Î 3.3068527K˚ È4.6137056K˘ value y10 = Í ˙. Î3.3068528K˚

Chapter 10 Series Solutions of Linear Differential Equations Section 10.1 tn Consider the power series  n . Applying the ratio test at an arbitrary value of t, t π 0, we n =0 2 n n +1 2 t t t = . The limiting ratio is less than 1 if obtain lim n +1 n = lim n Æ• 2 n Æ• 2 2 t . Therefore, the radius of convergence is R = 2 . t <2 •

1.

2.

lim

n Æ•

t n +1n 2 t = t . Therefore, the radius of convergence is R = 1. n 2 = lim n Æ• t ( n + 1) (1 + 1n )2 •

3.

Consider the power series

 (t - 2)

n

. Applying the ratio test at an arbitrary value of t, t π 2,

n =0

4.

5.

6.

7.

( t - 2) n +1 = lim t - 2 = t - 2 . The limiting ratio is less than 1 if t - 2 < 1. we obtain lim n n Æ• ( t - 2) n Æ• Therefore, the radius of convergence is R = 1. 2 ( 3t - 1) n +1 = 3t - 1 < 1 fi -1 < 3t - 1 < 1 fi 0 < t < . Therefore, the radius of lim n n Æ• ( 3t - 1) 3 1 convergence is R = . 3 • ( t - 1) n Consider the power series  . Applying the ratio test at an arbitrary value of t, t π 1, we n! n =0 n!( t - 1) n +1 t -1 = 0 . The limiting ratio is less than 1 for all t, t π 1. obtain lim n = lim n Æ• ( n + 1)!( t - 1) n Æ• n + 1 Therefore, the radius of convergence is R = • . ( n + 1)!( t - 1) n +1 = lim ( n + 1)( t - 1) = •, t π 1 . Therefore, the radius of convergence is lim n Æ• n Æ• n!( t - 1) n R = 0. • (-1) n t n Consider the power series  . Applying the ratio test at an arbitrary value of t, t π 0, n n =1 nt n +1 nt = t . The limiting ratio is less than 1 if t < 1. Therefore, we obtain lim n = lim n Æ• ( n + 1) t n Æ• n + 1 the radius of convergence is R = 1.

Chapter 10 Series Solutions of Linear Differential Equations • 269

8.

(-1) n +1 ( t - 3) n +1 4 n t- 3 = < 1 fi -4 < t - 3 < 4 fi -1 < t < 7 . Therefore, the radius of n n n +1 n Æ• (-1) ( t - 3) 4 4 convergence is R = 4 . lim

•

9.

Consider the power series

 (ln n)(t + 2)

n

. Applying the ratio test at an arbitrary value of t,

n =1

10.

11.

t π -2, we obtain ln( n + 1) (ln( n + 1))( t + 2) (ln( n + 1))( t + 2) n +1 = lim = t + 2 lim = t + 2 . (The last limit lim n n Æ• n Æ• n Æ• ln n ln n (ln n )( t + 2) can be found using L’Hôpital’s Rule.) The limiting ratio is less than 1 if t + 2 < 1. Therefore, the radius of convergence is R = 1. ( n + 1) 3 ( t - 1) n +1 = t - 1 < 1 fi -1 < t - 1 < 1 fi 0 < t < 2 . Therefore, the radius of lim n Æ• n 3 ( t - 1) n convergence is R = 1. • n (t - 4)n Consider the power series  . Applying the ratio test at an arbitrary value of t, 2n n =1 2 n n + 1 ( t - 4 ) n +1 n + 1 (t - 4) t- 4 . The limiting ratio is = lim = 1 n + n n Æ• n Æ• 2 2 n (t - 4) 2 n less than 1 if t - 4 < 2 . Therefore, the radius of convergence is R = 2 .

t π 4, we obtain lim

12. 13.

14.

( t - 2) n +1 arctan(n ) pˆ Ê = t - 2 < 1 fi -1 < t - 2 < 1 fi 1 < t < 3 Á recall lim arctan(n ) = ˜ . n n Æ• ( t - 2) arctan( n + 1) n Æ• Ë 2¯ Therefore, the radius of convergence is R = 1. Applying the ratio test, we see the power series for f(t) and g(t) both have radius of convergence R = 1. Therefore, each series converges in the interval -1 < t < 1. (a) f ( t) = 1 + t + t 2 + t 3 + t 4 + t 5 + L g( t) = 0 + t + 4 t 2 + 9 t 3 + 16 t 4 + 25 t 5 + L (b) f ( t) + g( t) = 1 + 2 t + 5 t 2 + 10 t 3 + 17 t 4 + 26 t 5 + L (c) f ( t) - g( t) = 1 - 3t 2 - 8 t 3 - 15 t 4 - 24 t 5 - L (d) f ¢ ( t) = 1 + 2 t + 3t 2 + 4 t 3 + 5 t 4 + 6 t 5 + L (e) f ¢¢ ( t) = 2 + 6 t + 12 t 2 + 20 t 3 + 30 t 4 + 42 t 5 + L Applying the ratio test, we see the power series for f(t) and g(t) both have radius of convergence R = 1. Therefore, each series converges in the interval -1 < t < 1. (a) f ( t) = t + 2 t 2 + 3t 3 + 4 t 4 + 5 t 5 + 6 t 6 + L g( t) = - t + 2 t 2 - 3t 3 + 4 t 4 - 5 t 5 + 6 t 6 - L (b) f ( t) + g( t) = 4 t 2 + 8 t 4 + 12 t 6 + 16 t 8 + 20 t10 + L (c) f ( t) - g( t) = 2 t + 6 t 3 + 10 t 5 + 14 t 7 + 18 t 9 + 22 t11 + L (d) f ¢ ( t) = 1 + 4 t + 9 t 2 + 16 t 3 + 25 t 4 + 36 t 5 + L (e) f ¢¢ ( t) = 4 + 18 t + 48 t 2 + 100 t 3 + 180 t 4 + 294 t 5 + L lim

270 • Chapter 10 Series Solutions of Linear Differential Equations

15.

16.

Applying the ratio test, we see the power series for f(t) has radius of convergence R = 1 / 2 while the series for g(t) has radius of convergence R = 1. Therefore, each series converges in the interval t - 1 < 1 / 2 , or 1 / 2 < t < 3 / 2 . (a) f ( t) = 1 - 2( t - 1) + 4 ( t - 1) 2 - 8( t - 1) 3 + 16( t - 1) 4 - 32( t - 1) 5 + L g( t) = 1 + ( t - 1) + ( t - 1) 2 + ( t - 1) 3 + ( t - 1) 4 + ( t - 1) 5 + L (b) f ( t) + g( t) = 2 - ( t - 1) + 5( t - 1) 2 - 7( t - 1) 3 + 17( t - 1) 4 - 31( t - 1) 5 + L (c) f ( t) - g( t) = -3( t - 1) + 3( t - 1) 2 - 9( t - 1) 3 + 15( t - 1) 4 - 33( t - 1) 5 + L (d) f ¢ ( t) = -2 + 8( t - 1) - 24 ( t - 1) 2 + 64 ( t - 1) 3 - 160( t - 1) 4 + 384 ( t - 1) 5 L (e) f ¢¢ ( t) = 8 - 48( t - 1) + 192( t - 1) 2 - 640( t - 1) 3 + 1920( t - 1) 4 - 5376( t - 1) 5 L Applying the ratio test, we see the power series for f ( t) is 1/2 and g( t) is 1. Therefore, 1 R= . 2 (a) f ( t) = 1 + 2( t + 1) + 4 ( t + 1) 2 + 8( t + 1) 3 + 16( t + 1) 4 + 32( t + 1) 5 + L g( t) = ( t + 1) + 2( t + 1) 2 + 3( t + 1) 3 + 4 ( t + 1) 4 + 5( t + 1) 5 + 6( t + 1) 6 + L (b) f ( t) + g( t) = 1 + 3( t + 1) + 6( t + 1) 2 + 11( t + 1) 3 + 20( t + 1) 4 + 37( t + 1) 5 + L (c) f ( t) - g( t) = 1 + ( t + 1) + 2( t + 1) 2 + 5( t + 1) 3 + 12( t + 1) 4 + 27( t + 1) 5 + L (d) f ¢ ( t) = 2 + 8( t + 1) + 24 ( t + 1) 2 + 64 ( t + 1) 3 + 160( t + 1) 4 + 384 ( t + 1) 5 + L (e) f ¢¢ ( t) = 8 + 48( t + 1) + 192( t + 1) 2 + 640( t + 1) 3 + 1920( t + 1) 4 + 5376( t + 1) 5 + L •

17.

Consider the power series

Â2

n n +2

t

. Make the change of index k = n + 2 . With this change,

n =0

the lower limit of n = 0 transforms to k = 2 while the upper limit remains at • . Thus, the •

power series can be rewritten as

Â2

k -2 k

t . Finally, changing to the original summation index,

k =2 •

n, we obtain

Â2

n -2 n

t .

n =2 •

18.

 (k - 2)(k - 1)t

Make the change of index k = n + 3. The power series can be rewritten as

k

.

k =3 •

Finally, changing to the original summation index, n, we obtain

 (n - 2)(n - 1)t

n

.

n =3

•

19.

Consider the power series

Âa t

n +2

n

. Make the change of index k = n + 2 . With this change,

n =0

the lower limit of n = 0 transforms to k = 2 while the upper limit remains at • . Thus, the •

power series can be rewritten as

Âa

k -2

t k . Finally, changing to the original summation index,

k =2 •

n, we obtain

Âa

n -2

tn .

n =2 •

20.

Make the change of index k = n - 1. The power series can be rewritten as

 (k + 1)a

k +1

k =0 •

Finally, changing to the original summation index, n, we obtain

 (n + 1)a

tn .

n +1

n =0

tk .

Chapter 10 Series Solutions of Linear Differential Equations • 271 •

21.

Consider the power series

 n(n - 1)a t

n -2

n

. Make the change of index k = n - 2 . With this

n =2

change, the lower limit of n = 2 transforms to k = 0 while the upper limit remains at • . •

Thus, the power series can be rewritten as

 (k + 2)(k + 1)a

k +2

t k . Finally, changing to the

k =0 •

original summation index, n, we obtain

 (n + 2)(n + 1)a

n +2

tn .

n =0 •

22.

Make the change of index k = n + 3. The power series can be rewritten as

 (-1)

k -3

ak - 3 t k .

k =3 •

Finally, changing to the original summation index, n, we obtain

 (-1)

n -3

an - 3 t n .

n =3 •

23.

Consider the power series

 (-1)

n +1

( n + 1) an t n + 2 . Make the change of index k = n + 2 . With

n =0

this change, the lower limit of n = 0 transforms to k = 2 while the upper limit remains at • . •

Thus, the power series can be rewritten as

 (-1)

k -1

( k - 1) ak - 2 t k . Finally, changing to the

k =2 •

original summation index, n, we obtain

 (-1)

n -1

( n - 1) an - 2 t n .

n =2 • (-1) n t 2 n +1 (-1) n +1 t 2 n + 3 . Therefore, f ( t) = Â . Let f ( t) = t 2 ( t - sin t) . t - sin t = - Â n =1 (2 n + 1)! n =1 (2 n + 1)! (-1) n + 2 (2 n + 1)!( t) 2 n + 5 = 0 . Thus, the radius of convergence is R = • . lim n +1 n Æ• (-1) (2 n + 3)!( t) 2 n + 3 •

24.

•

25.

Let f ( t) = 1 - cos 3t . From the Maclaurin series for cos u we have cos u = Â (-1) n n =0

u2n . (2 n )!

9 t 2 81t 4 729 t 6 Therefore, cos 3t = 1 + + L. Hence, 2! 4! 6! 2n • 9 t 2 81t 4 729 t 6 n +1 ( 3t) + - L = Â (-1) f ( t) = . We calculate the radius of convergence by (2 n )! 2! 4! 6! n =1 using the ratio test. For an arbitrary value of t, t π 0, we have 9t 2 (2 n )!( 3t) 2 n + 2 = 0 . Thus, the radius of convergence is R = • . lim 2 n = lim n Æ• (2 n + 2)!( 3t) n Æ• (2 n + 2)(2 n + 1) 26.

27.

• • 1 (-2 t) n +1 = 2 t < 1. = Â (-2 t) n = Â (-2) n t n . lim n n Æ• (-2 t) 1 - (-2 t) n = 0 n =0 1 Thus, the radius of convergence is R = . 2 • 1 = Â u n . Therefore, Let f ( t) = 1 / (1 - t 2 ) . From the Maclaurin series for 1 / (1 - u) we have 1 - u n =0 • 1 = 1 + t 2 + t 4 + t 6 + L. Hence, f ( t) = Â t 2 n . 1 - t2 n =0

Let f ( t) =

1 1 = . 1 + 2 t 1 - (-2 t)

272 • Chapter 10 Series Solutions of Linear Differential Equations

We calculate the radius of convergence by using the ratio test. For an arbitrary value of t, t π 0, t 2n + 2 we have lim 2 n = lim t 2 = t 2 . Thus, the radius of convergence is R = 1. n Æ• n Æ• t 2 n • t t3 t4 t5 t 28 (a). e t =  = 1 + t + + + + + .... 2! 3! 4! 5! n = 0 n! n • t2 t3 t4 t5 (- t) -t e = = 1 - t + - + - + ... 2! 3! 4! 5! n = 0 n! 2 ˆ Ê ˆ¸ t t3 t4 t5 t2 t3 t4 t5 t3 t5 1 ÏÊ 28 (b). sinh( t) = ÌÁ1 + t + + + + + ...˜ - Á1 - t + - + - + ...˜ ˝ = t + + + .... 2 ÓË 2! 3! 4! 5! 2! 3! 4! 5! 3! 5! ¯ Ë ¯˛ ˆ Ê ˆ¸ t2 t3 t4 t5 t2 t3 t4 t5 t2 t4 1 ÏÊ cosh( t) = ÌÁ1 + t + + + + + ...˜ + Á1 - t + - + - + ...˜ ˝ = 1 + + + ... 2 ÓË 2! 3! 4! 5! 2! 3! 4! 5! 2! 4! ¯ Ë ¯˛

29 (a). Consider the differential equation y ¢¢ - w 2 y = 0 and assume there is solution of the form •

•

•

y ( t) = Â an t n . Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =0

n =1

n =2 •

•

 n(n - 1)an t n - 2 -w 2  an t n = 0 .

Inserting these series into the differential equation, we have

n =2

n =0

Making the change of index k = n - 2 in the series for y ¢¢ ( t) , we obtain •

•

•

 (n + 2)(n + 1)an + 2t n -w 2  an t n = 0 , or

 [(n + 2)(n + 1)a

n =0

n =0

n =0

n +2

coefficients to zero, we find the recurrence relation an + 2 =

-w 2 an ]t n = 0 . Equating the

w 2 an , n = 0, 1,K ( n + 2)( n + 1)

29 (b). The recurrence relation in part (a) leads us to a2 = w 2 a0 / 2 , a4 = w 2 a2 / 12 = w 4 a0 / 24, a6 = w 2 a4 / 30 = w 6 a0 / 720 , K a3 = w 2 a1 / 6 , a5 = w 2 a3 / 20 = w 4 a1 / 120, a7 = w 2 a5 / 42 = w 6 a1 / 5040 , K (wt) 2 (wt) 4 (wt) 6 (wt) 3 (wt) 5 (wt) 7 a1 Thus, y ( t) = a0 [1 + + + + L]+ [wt + + + + L] . w 6 120 5040 2 24 720 By Exercise 28, y1 ( t) = cosh wt and y 2 ( t) = sinh wt . 30 (a). y ( t) = Ú

t •

0

•

 nl n =1

n -1

•

•

dl + C = Â t n + C, y (0) = C = 1 fi y ( t) = 1 + Â t n = Â t n . n =1

n =1

n =0

30 (b). R = 1. 30 (c). y ( t) =

1 . 1- t

( t - 1) n , y (1) = 1. Integrating the series termwise, 31 (a). Consider the function given by y ¢ ( t) = Â n! n =0 • ( t - 1) n +1 we obtain y ( t) = C + Â . Imposing the condition y(1) = 1, it follows that C = 1. n = 0 ( n + 1)! • • ( t - 1) n ( t - 1) n =Â Adjusting the index of summation, we can write y ( t) = 1 + Â . n! n! n =1 n =0 •

Chapter 10 Series Solutions of Linear Differential Equations • 273

31 (b). Applying the ratio test, lim

n Æ•

n!( t - 1) n +1 t -1 = 0 . Therefore, the radius of n = lim n Æ• n + 1 ( n + 1)!( t - 1)

convergence is R = • . 31 (c). From the power series (7a), we see that y ( t) = e t -1 . n n n • • • t • Ï t n +1 n -1 t n t ¸ n l n dl = -1 +  (-1) = -1 +  (-1) = -Ì1 +  (-1) ˝ 32 (a). y ¢ ( t) = -1 + Ú Â (-1) 0 n! n!˛ n! ( n + 1)! n =1 n =0 n =0 Ó n =1 n n +1 n +1 n • • • • t t t t y ¢ = -  (-1) n . Then, y ( t) = -  (-1) n + 1 = 1 +  (-1) n +1 =  (-1) n . n! n! ( n + 1)! ( n + 1)! n = 0 n =0 n =0 n =0 32 (b). R = • . 32 (c). y ( t) = e - t . n • n ( t - 1) , y (1) = 0 . Integrating the series 33 (a). Consider the function given by y ¢ ( t) =  (-1) n! n =2 • ( t - 1) n +1 . Imposing the condition y(1) = 0 , it follows termwise, we obtain y ( t) = C +  (-1) n ( n + 1)! n =2 that C = 0 . Adjusting the index of summation, we can write n n • • n +1 ( t - 1) n ( t - 1) = -  (-1) y ( t) =  (-1) . n! n! n =3 n =3 (-1) n +1 n!( t - 1) n +1 t -1 = 0 . Therefore, the radius of 33 (b). Applying the ratio test, lim n n = lim n Æ• (-1) ( n + 1)!( t - 1) n Æ• n + 1 convergence is R = • . • ( t - 1) n = e -( t -1) . Thus, 33 (c). From the power series (7a), we see that  (-1) n ! n n =0 n 1 ( ) ( t - 1) ( t - 1) 2 • t 1+ +  (-1) n = e -( t -1) . Or, using the results of part (a), 1! 2! n! n =3 2 ( t - 1) ( t - 1) 1+ - e -( t -1) = y ( t) . 1! 2! • t • t 2 n +1 . 34 (a). y ( t) = Ú Â (-1) n s2 n ds =  (-1) n 0 2n + 1 n =0 n =0 (-1) n +1 t 2 n + 3 (2 n + 1) = t 2 < 1 fi R = 1. 34 (b). lim n 2 n +1 n Æ• (-1) t (2 n + 3) -1 34 (c). y ( t) = tan ( t) . • t tn 35 (a). Consider the function y(t) where Ú y ( s) ds =  . Differentiating both sides, we obtain 0 n =1 n •

•

y ( t) = Â t n -1 . Adjusting the index of summation, we can write y ( t) = Â t n . n =0

n =1

35 (b). Applying the ratio test, lim

n Æ•

t

n +1

tn

= t . Therefore, the radius of convergence is R = 1. •

35 (c). From the power series (7d), we see that y ( t) = Â t n = n =0

1 . 1- t

274 • Chapter 10 Series Solutions of Linear Differential Equations •

36.

Assume there is solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0 •

•

•

•

y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 = Â ( n + 2)( n + 1) an + 2 t n , ty ¢ = Â nan t n . n =1

n =2

n =0

n =0

•

Therefore,

 [(n + 2)(n + 1)a

n +2

- ( n + 1)an ]t n = 0 . Equating the coefficients to zero, we find

n =0

a ( n + 1) an = n The recurrence leads us to ( n + 2)( n + 1) n + 2 a0 a1 a2 a0 a a = , a5 = 3 = 1 a2 = , a3 = , a4 = 2 3 4 8 5 15 ¸ ¸ Ï t3 t5 Ï t2 t4 Therefore, y ( t) = a0 Ì1 + + + ...˝ + a1 Ìt + + + ...˝, y (0) = a0 = 1, y ¢ (0) = a1 = -1. ˛ ˛ Ó 3 15 Ó 2 8 2 4 3 5 ¸ ¸ Ï t Ï t t t Finally, y ( t) = Ì1 + + + ...˝ - Ìt + + + ...˝ . ˛ ˛ Ó 3 15 Ó 2 8 Consider the initial value problem y ¢¢ + ty ¢ - 2 y = 0 , y (0) = 0, y ¢ (0) = 1 and assume there is the recurrence relation an + 2 =

37.

•

solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0 •

•

y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . Inserting these series into the differential n =1

n =2 •

•

•

 n(n - 1)an t n - 2 + t  nan t n -1 - 2 an t n = 0 . Making the change of index

equation, we have

n =2

n =1

n =0 •

k = n - 2 in the series for y ¢¢ ( t) , we obtain

•

•

 (n + 2)(n + 1)an + 2t n +  nan t n - 2 an t n = 0 , or n =0

n =1

n =0

•

 [(n + 2)(n + 1)a

n +2

+ ( n - 2)an ]t n = 0 . Equating the coefficients to zero, we find the

n =0

-( n - 2) an , n = 0, 1,K The recurrence leads us to ( n + 2)( n + 1) a2 = 2 a0 / 2 = a0 , a4 = 0 a2 / 12 = 0, a6 = -2 a4 / 30 = 0 , K a3 = a1 / 6 , a5 = - a3 / 20 = - a1 / 120, a7 = -3a5 / 42 = a1 / 1680 , K Imposing the initial conditions, we have a0 = 0 and a1 = 1. Thus, t3 t5 t7 y ( t) = t + + + L. 6 120 1680

recurrence relation an + 2 =

•

38.

Assume there is solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0 •

y ¢ ( t) = Â nan t

• n -1

and y ¢¢ ( t) = Â n ( n - 1) an t

n =1

n =2

• n -2

•

= Â ( n + 2)( n + 1) an + 2 t , ty = Â an t n

n =0

n =0

• n +1

= Â an -1t n n =1

•

Therefore, 2 a2 + Â [( n + 2)( n + 1) an + 2 +an -1 ]t n = 0 . Equating the coefficients to zero, we find n =1

the recurrence relation an + 2 =

- an -1 , n = 1, 2,... ( n + 2)( n + 1)

Chapter 10 Series Solutions of Linear Differential Equations • 275

39.

The recurrence leads us to -a -a -a a3 = 0 , a4 = 1 , a5 = 2 = 0 3◊ 2 4◊3 5◊ 4 3 ¸ ¸ Ï t4 Ï t Therefore, y ( t) = a0 Ì1 - + ...˝ + a1 Ìt - + ...˝, a0 = 1, a1 = 2 . ˛ ˛ Ó 12 Ó 6 3 4 ¸ ¸ Ï t Ï t Finally, y ( t) = Ì1 - + ...˝ + 2Ìt - + ...˝. ˛ ˛ Ó 12 Ó 6 Consider the initial value problem y ¢¢ + (1 + t) y ¢ + y = 0 , y (0) = -1, y ¢ (0) = 1 and assume there •

is solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0 •

•

y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . Inserting these series into the differential n =1

n =2 •

•

n =2 •

•

 n(n - 1)a t n

n -2

•

 n(n - 1)an t n - 2 +(1 + t) nan t n -1 +  an t n = 0 or

equation, we have

n =1

n =0

•

+ Â nan t n -1 + Â (1 + n ) an t n = 0 . Making the change of index k = n - 2 in the

n =2

n =1

n =0

series for y ¢¢ ( t) and k = n - 1 in the series for y ¢ ( t) , we obtain •

•

 (n + 2)(n + 1)a

•

t + Â ( n + 1) an +1t + Â (1 + n ) an t n = 0 , or n

n +2

n =0 •

n

n =0

 [(n + 2)(n + 1)a

n +2

n =0

+ ( n + 1) an +1 + ( n + 1)an ]t n = 0 . Equating the coefficients to zero, we find

n =0

-( n + 1) an +1 - ( n + 1) an - an +1 - an = . The recurrence leads us to n+2 ( n + 2)( n + 1) a2 = -( a0 + a1 ) / 2 , a3 = -( a2 + a1 ) / 3, a4 = -( a3 + a2 ) / 4 = 0 , a5 = -( a4 + a3 ) / 5 . Imposing the initial conditions, we have a0 = -1 and a1 = 1 . Thus, a2 = 0 , a3 = -1 / 3, a4 = 1 / 12 , a5 = 1 / 20 and so we find 1 1 1 y ( t) = -1 + t - t 3 + t 4 + t 5 + L . 3 12 20 the recurrence relation an + 2 =

•

40.

Assume there is solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0

•

•

y ¢ ( t) = Â ( n + 1) an +1t n and y ¢¢ ( t) = Â ( n + 2)( n + 1) an + 2 t n . Inserting these series into the n =0

n =0 •

differential equation, we have

 {(n + 2)(n + 1)a

n +2

- 5( n + 1) an +1 + 6 an }t n = 0 . Equating the

n =0

coefficients to zero, we find the recurrence relation an + 2 =

5( n + 1) an +1 - 6 an , n = 0,1, 2,.... The ( n + 2)( n + 1)

recurrence leads us to 5(2) a2 - 6 a1 10(2) - 6(2) 4 5 a - 6 a0 5(2) - 6(1) = = . = = 2, a3 = a2 = 1 3◊ 2 6 3 2 2 5( 3) a3 - 6 a2 15( 4 / 3) - 6(2) 2 5( 4 ) a4 - 6 a3 20(2 / 3) - 6( 4 / 3) 4 = = = = , a5 = a4 = 4◊3 12 3 5◊ 4 20 15

276 • Chapter 10 Series Solutions of Linear Differential Equations

Therrefore, y ( t) = 1 + 2 t + 2 t 2 + 41.

4 3 2 4 4 5 t + t + t + L. 3 3 15

Consider the initial value problem y ¢¢ - 2 y ¢ + y = 0 , y (0) = 0, y ¢ (0) = 2 and assume there is •

solution of the form y ( t) = Â an t n . Differentiating, we obtain n =0 •

y ¢ ( t) = Â nan t

•

and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . Inserting these series into the differential

n -1

n =1

n =2 •

•

 n(n - 1)a t

equation, we have

n

n =2

n -2

•

- 2 Â nan t n -1 + Â an t n = 0 . Making the change of index n =1

n =0

k = n - 2 in the series for y ¢¢ ( t) and k = n - 1 in the series for y ¢ ( t) , we obtain •

•

•

 (n + 2)(n + 1)a

n n n n + 2 t - 2 Â ( n + 1) an +1t + Â an t = 0 , or

n =0 •

n =0

 [(n + 2)(n + 1)a

n +2

n =0

- 2( n + 1) an +1 +an ]t n = 0 . Equating the coefficients to zero, we find the

n =0

2( n + 1) an +1 - an . The recurrence leads us to ( n + 2)( n + 1) a2 = (2 a1 - a0 ) / 2 , a3 = ( 4 a2 - a1 ) / 6 , a4 = (6 a3 - a2 ) / 12 , a5 = (8 a4 - a3 ) / 20 . Imposing the initial conditions, we have a0 = 0 and a1 = 2 . Thus, 1 1 a2 = 2 , a3 = 1, a4 = 1 / 3, a5 = 1 / 12 and so we find y ( t) = 2 t + 2 t 2 + t 3 + t 4 + t 5 + L . 3 12

recurrence relation an + 2 =

Section 10.2 1.

2. 3.

4.

Consider the differential equation y ¢¢ + (sec t) y ¢ + t( t 2 - 4 ) -1 y = 0 . The coefficient function p( t) = sec t is not analytic at odd integer multiples of p / 2 . Thus, in the interval -10 < t < 10 , p 3p 5p . Similarly, the coefficient function q( t) = t( t 2 - 4 ) -1 is p( t) is not analytic at ± , ± ,± 2 2 2 not analytic at t = ±2 . These 8 points are the only singular points in -10 < t < 10 . 2 The function p( t) = t 3 is not analytic at t = 0 . The function q( t) = sin t is analytic everywhere. Therefore, t = 0 is the only singular point in -10 < t < 10 . Consider the differential equation (1 - t 2 ) y ¢¢ + ty ¢ + (csc t) y = 0 . Putting the differential equation into the form of equation (1), we see that the coefficient function p( t) = t(1 - t 2 ) -1 is not analytic at t = ±1 . Similarly, the coefficient function q( t) = (csc t)(1 - t 2 ) -1 is not analytic at integer multiples of p or at t = ±1 . Thus, in the interval -10 < t < 10 , the singular points are given by t = 0, ± 1, ± p , ± 2p , ± 3p . 5p p 3p et The function p( t) = is not analytic at t = 0, ± , ± p , ± , ± 2p , ± , ± 3p . The 2 2 2 sin 2 t t function q( t) = is also not analytic at t = ±5 . Therefore, (25 - t 2 )sin 2 t 5p p 3p , ± 2p , ± , ± 3p , ± 5 are the singular points in -10 < t < 10 . t = 0, ± , ± p , ± 2 2 2

Chapter 10 Series Solutions of Linear Differential Equations • 277

5.

6.

7.

8. 9.

10.

11.

12.

Consider the differential equation (1 + ln t ) y ¢¢ + y ¢ + (1 + t 2 ) y = 0 . Putting the differential equation into the form of equation (1), we see that the coefficient function p( t) = (1 + ln t ) -1 is not analytic at t = 0 or at t = ±e -1. Similarly, the coefficient function q( t) = (1 + t 2 )(1 + ln t ) -1 is not analytic t = 0 or at t = ±e -1. These three points are the only singular points in the interval -10 < t < 10 . t is not analytic at t = 0 . The function q( t) = tan t is not analytic at The function p( t) = 1+ t p p 3p 5p 3p 5p are the singular points in t= ± , ± , ± ,... . Therefore, t = 0, ± , ± , ± 2 2 2 2 2 2 -10 < t < 10 . Consider the differential equation y ¢¢ + (1 + 2 t) -1 y ¢ + t(1 - t 2 ) -1 y = 0 . Since the coefficient functions are rational functions, each is analytic with a radius of convergence R equal to the distance from t0 = 0 to its nearest singularity; see Figure 10.2. The only singularity of p( t) = (1 + 2 t) -1 is t = -1 / 2 while the only singularities of q( t) = t(1 - t 2 ) -1 are t = ±1 . Thus, the radius of convergence of the series for p(t) is R = 1 / 2 while the series for q(t) has radius of convergence R = 1. The given initial value problem is guaranteed to have a unique solution that is analytic in the interval -1 / 2 < t < 1 / 2 . 2 p( t) = 4 (1 - 9 t 2 ) -1 and q( t) = t(1 - 9 t 2 ) -1 are not analytic at t = ±1 / 3 . Thus, for t0 = 1, R = . 3 Consider the differential equation y ¢¢ + ( 4 - 3t) -1 y ¢ + 3t(5 + 30 t) -1 y = 0 . Since the coefficient functions are rational functions, each is analytic with a radius of convergence R equal to the distance from t0 = -1 to its nearest singularity; see Figure 10.2. The only singularity of p( t) = ( 4 - 3t) -1 is t = 4 / 3 while the only singularity of q( t) = 3t(5 + 30 t) -1 is t = -1 / 6 . Thus, the radius of convergence of the series for p(t) is R = - 1 - ( 4 / 3) = 7 / 3 while the series for q(t) has radius of convergence R = - 1 - (-1 / 6) = 5 / 6 . The given initial value problem is guaranteed to have a unique solution that is analytic in the interval -5 / 6 < t + 1 < 5 / 6 . i p( t) = (1 + 4 t 2 ) -1 is not analytic at t = ± and q( t) = t( 4 + t) -1 is not analytic at t = -4 . Thus, 2 1 for t0 = 0 , R = . 2 Consider the differential equation y ¢¢ + (1 + 3( t - 2)) -1 y ¢ + (sin t) y = 0 . The coefficient function p( t) = ( 3t - 5) -1 is a rational function and is analytic with a radius of convergence R equal to the distance from t0 = 2 to its nearest singularity; see Figure 10.2. The only singularity of p( t) = ( 3t - 5) -1 is t = 5 / 3 . The other coefficient function, q( t) = sin t , is analytic everywhere with an infinite radius of convergence. The radius of convergence of the series for p(t) is R = 2 - (5 / 3) = 1 / 3. Therefore, the given initial value problem is guaranteed to have a unique solution that is analytic in the interval -1 / 3 < t - 2 < 1 / 3 . p( t) = ( t + 3)(1 + t 2 ) -1 is not analytic at t = ±i and q( t) = t 2 is analytic everywhere. Thus, for t0 = 1, R = 2 . •

13 (a). Consider the differential equation y ¢¢ + ty ¢ + y = 0 . Let the solution be given by y ( t) = Â an t n . n =0 •

•

Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =1

n =2

278 • Chapter 10 Series Solutions of Linear Differential Equations

Inserting these series into the differential equation, we have •

•

 n(n - 1)a t

n -2

n

•

+ t  nan t

n =2

n -1

n =1

•

+ Â an t = 0 or n

n =0 •

Adjusting the indices, we obtain

•

 n(n - 1)a t n

n =2

n -2

•

+ Â nan t + Â an t n = 0 . n

n =1 •

 (n + 2)(n + 1)a

n +2

n =0 •

t n + Â nan t n + Â an t n = 0 or

n =0

n =1

n =0

•

2 a2 + a0 + Â [( n + 2)( n + 1) an + 2 +( n + 1) an ]t n = 0 . Consequently, the recurrence relation is n =1

given by a2 = - a0 / 2 and an + 2 = - an / ( n + 2), n = 1, 2, K. 13 (b). The recurrence leads us to a2 = - a0 / 2 , a4 = - a2 / 4 = a0 / 8 ,K a3 = - a1 / 3 , a5 = - a3 / 5 = a1 / 15 ,K Thus, the general solution is t2 t4 t3 t5 y ( t) = a0 [1 - + - L] + a1[ t - + - L] = y1 ( t) + y 2 ( t) . 2 8 3 15 13 (c). Since the coefficient functions are analytic for -• < t < • , the series converges for -• < t < • . 13 (d). The coefficient function p( t) = t is odd and the coefficient function q( t) = 1 is even. Therefore, Theorem 10.2 guarantees that the given equation has even solutions and odd solutions. •

14 (a).

 [(n + 2)(n + 1)a

n +2

+ 2 nan + 3an ]t n = 0 . Consequently, the recurrence relation is given by

n =0

-(2 n + 3) an , n = 0,1, 2,K. ( n + 2)( n + 1) 14 (b). The recurrence leads us to a2 = -3a0 / 2 , a3 = -5 a1 / 6, a4 = -7 a2 / 12 = 7 a0 / 8 , a5 = -9 a3 / 20 = 3a1 / 8 K a3 = - a1 / 3 , a5 = - a3 / 5 = a1 / 15 ,K Thus, the general solution is 3t 2 7 t 4 5 t 3 3t 5 y ( t) = a0 [1 + - L] + a1[ t + - L]. 2 8 6 8 14 (c). Since the coefficient functions are analytic for -• < t < • , R = • . 14 (d). p( t) = 2 t is odd and q( t) = 3 is even. Therefore, Theorem 10.2 guarantees that the given equation has even solutions and odd solutions. 15 (a). Consider the differential equation (1 + t 2 ) y ¢¢ + ty ¢ + 2 y = 0 . Let the solution be given by an + 2 =

•

•

•

y ( t) = Â an t n . Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =0

n =1

n =2

Inserting these series into the differential equation, we have •

•

(1 + t ) Â n ( n - 1) an t 2

n -2

+ t  nan t

n =2

+2 Â an t n = 0 or

n =1

•

•

 n(n - 1)a t n

n =2 •

• n -1

n -2

n =0 •

•

+ Â n ( n - 1) an t n + Â nan t n +2 Â an t n = 0 . Adjusting the indices, we obtain n =2

n =1 •

n =0 •

•

 (n + 2)(n + 1)an + 2t n +  n(n - 1)an t n +  nan t n +2 an t n = 0 . Consequently, the recurrence n =0

n =2

n =1

n =0

relation is given by a2 = - a0 , a3 = - a1 / 2, and an + 2 = -( n 2 + 2) an /[( n + 2)( n + 1)], n = 2, 3,K.

Chapter 10 Series Solutions of Linear Differential Equations • 279

15 (b). The recurrence leads us to a2 = - a0 , a4 = - a2 / 2 = a0 / 2 ,K a3 = - a1 / 2 , a5 = -11a3 / 20 = 11a1 / 40 ,K Thus, the general solution is t4 t 3 11t 5 y ( t) = a0 [1 - t 2 + - L] + a1[ t - + - L] = y1 ( t) + y 2 ( t) . 2 2 40 15 (c). The coefficient functions p( t) = t(1 + t 2 ) -1 and q( t) = 2(1 + t 2 ) -1 fail to be analytic at t = ±i . Therefore, the radius of convergence for each coefficient function is R = 1. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -1 < t < 1. 15 (d). The coefficient function p( t) = t(1 + t 2 ) -1 is odd and the coefficient function q( t) = 2(1 + t 2 ) -1 is even. Therefore, Theorem 10.2 guarantees that the given equation has even solutions and odd solutions. •

16 (a).

 [(n + 2)(n + 1)a

n +2

- 5( n + 1) an +1 + 6 an ]t n = 0 . Consequently, the recurrence relation is given

n =0

5( n + 1) an +1 - 6 an , n = 0,1, 2,K. ( n + 2)( n + 1) 16 (b). The recurrence leads us to a2 = (5 a1 - 6 a0 ) / 2 = 5 a1 / 2 - 3a0 , a3 = (5(2) a2 - 6 a1 ) / ( 3 ◊ 2) = 19 a1 / 6 - 5 a0 Thus, the general solution is 5 t 2 19 t 3 2 3 y ( t) = a0 [1 - 3t - 5 t - L] + a1[ t + + + L] . 2 6 16 (c). Since the coefficient functions are analytic for -• < t < • , R = • . 16 (d). p( t) = -5 and q( t) = 6 are both even. Therefore, Theorem 10.2 does not apply. 17 (a). Consider the differential equation y ¢¢ - 4 y ¢ + 4 y = 0 . Let the solution be given by by an + 2 =

•

•

•

y ( t) = Â an t n . Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =0

n =1

n =2

Inserting these series into the differential equation, we have •

•

 n(n - 1)a t n

n -2

•

- 4 Â nan t

n =2 •

n -1

n =1

+ 4 Â an t n = 0 . Adjusting the indices, we obtain n =0

•

•

 (n + 2)(n + 1)a

n n n n + 2 t - 4 Â ( n + 1) an +1t + 4 Â an t = 0 . Consequently, the recurrence relation

n =0

n =0

n =0

is given by an + 2 = [ 4 ( n + 1) an +1 - 4 an ] /[( n + 2)( n + 1)], n = 0, 1,K . 17 (b). The recurrence leads us to a2 = 2 a1 - 2 a0 , a3 = (8 a2 - 4 a1 ) / 6 = (16 a1 - 16 a0 - 4 a1 ) / 6 = 2 a1 - (8 / 3) a0 ,K Thus, the general solution is 8t 3 y ( t) = a0 [1 - 2 t 2 + L] + a1[ t + 2 t 2 + 2 t 3 L] = y1 ( t) + y 2 ( t) . 3 17 (c). The coefficient functions are constant and hence analytic everywhere. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -• < t < • . 17 (d). The coefficient function p( t) = -4 is even and hence Theorem 10.2 does not apply.

280 • Chapter 10 Series Solutions of Linear Differential Equations •

18 (a).

 [(n + 2)(n + 1)a + (n + 1)na -[( n + 1) na + a ] . by a = n +2

n +1

+ an ]t n = 0 . Consequently, the recurrence relation is given

n =0

n +1

n

n +2

( n + 2)( n + 1) 18 (b). The recurrence leads us to -[(2)(1) a2 + a1 ] a0 a1 -[( 3)(2) a3 + a2 ] -a a a = - , a4 = =- 0 + 1 a2 = 0 , a3 = 2 3◊ 2 6 6 4◊3 8 12 Thus, the general solution is t2 t3 t3 t4 y ( t) = a0 [1 - - - L] + a1[ t - + + L] . 2 6 6 12 1 18 (c). q( t) = is not analytic at t = -1, R = 1. 1+ t 1 is neither even nor odd. Therefore, Theorem 10.2 does not apply. 18 (d). q( t) = 1+ t 19 (a). Consider the differential equation ( 3 + t) y ¢¢ + 3ty ¢ + y = 0 . Let the solution be given by •

•

•

y ( t) = Â an t . Differentiating, we obtain y ¢ ( t) = Â nan t n

n =0

n -1

n =1

and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =2

Inserting these series into the differential equation, we have •

•

•

( 3 + t)  n ( n - 1) an t n - 2 + 3t  nan t n -1 +  an t n = 0 or n =2

n =1

•

•

n =0 •

•

3Â n ( n - 1) an t n - 2 + Â n ( n - 1) an t n -1 + 3Â nan t n + Â an t n = 0 . Adjusting the indices, we obtain n =2 •

n =2

n =1 •

n =0 •

•

3Â ( n + 2)( n + 1) an + 2 t + Â ( n + 1) nan +1t + 3Â nan t + Â an t n = 0 . Consequently, the n

n =0

n

n

n =1

n =1

n =0

recurrence relation is given by a2 = - a0 / 6 and an + 2 = -[ n ( n + 1) an +1 + ( 3n + 1) an ] /[ 3( n + 2)( n + 1)], n = 1, 2,K. 19 (b). The recurrence leads us to a2 = - a0 / 6 , a3 = -(2 a2 + 4 a1 ) / 18 = -(-2 a0 / 6 + 4 a1 ) / 18 = ( a0 - 12 a1 ) / 54 ,K Thus, the general solution is t2 t3 2t 3 y ( t) = a0 [1 - + + L] + a1[ t + L] = y1 ( t) + y 2 ( t) . 6 54 9 19 (c). The coefficient functions p( t) = 3t( 3 + t) -1 and q( t) = ( 3 + t) -1 fail to be analytic at t = -3 . Therefore, the radius of convergence for each coefficient function is R = 3. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -3 < t < 3 . 19 (d). The coefficient function p( t) = 3t( 3 + t) -1 is neither even nor odd. Therefore, Theorem 10.2 does not apply. •

20 (a).

 [2(n + 2)(n + 1)a

n +2

+ n ( n - 1) an + 4 an ]t n = 0 . Consequently, the recurrence relation is given

n =0

by an + 2 =

-[ n ( n - 1) + 4 ]an . 2( n + 2)( n + 1)

Chapter 10 Series Solutions of Linear Differential Equations • 281

20 (b). The recurrence leads us to a a a a2 = - a0 , a3 = - 1 , a4 = 0 , a5 = 1 3 4 12 Thus, the general solution is t4 t3 t5 y ( t) = a0 [1 - t 2 + - L] + a1[ t - + + L] . 4 3 12 20 (c). R = 2 . 4 20 (d). p( t) = 0 can be considered odd and q( t) = 2 is even. Therefore, Theorem 10.2 guarantees t +2 that the given equation has even solutions and odd solutions. •

21 (a). Consider the differential equation y ¢¢ + t 2 y = 0 . Let the solution be given by y ( t) = Â an t n . n =0 •

•

Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . Inserting these n =1

n =2 •

•

series into the differential equation, we have  n ( n - 1) an t n - 2 + t 2  an t n = 0 or n =2 •

n =0

•

 n(n - 1)a t n

n -2

+ Â an t n + 2 = 0 . Adjusting the indices, we obtain

n =2 •

n =0 •

 (n + 2)(n + 1)an + 2t n +  an - 2t n = 0 . Consequently, the recurrence relation is given by n =0

n =2

a2 = 0, a3 = 0, and an + 2 = - an - 2 /[( n + 2)( n + 1)], n = 2, 3, K . 21 (b). The recurrence leads us to a2 = 0 , a3 = 0 , a4 = - a0 / 12, a5 = - a1 / 20,K Thus, the general solution is t4 t5 y ( t) = a0 [1 - + L] + a1[ t + L] = y1 ( t) + y 2 ( t) . 12 20 21 (c). The coefficient functions are polynomials and hence analytic everywhere. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -• < t < • . 21 (d). The coefficient function p( t) = 0 can be considered an odd function while q( t) = t 2 is clearly an even function. Therefore, Theorem 10.2 guarantees that the given equation has even solutions and odd solutions. •

22 (a).

 [(n + 2)(n + 1)a

n +2

+ nan + an ]( t - 1) n = 0 . Consequently, the recurrence relation is given by

n =0

- an -( n + 1) an , n = 0,1, 2,.... = ( n + 2)( n + 1) n + 2 22 (b). The recurrence leads us to a a a a a a a2 = - 0 , a3 = - 1 , a4 = - 2 = 0 , a5 = - 3 = 1 2 3 4 8 5 15 Thus, the general solution is ( t - 1) 2 ( t - 1) 4 ( t - 1) 3 ( t - 1) 5 y ( t) = a0 [1 + + L] + a1[( t - 1) + + L] . 3 15 2 8 22 (c). The coefficient functions are analytic everywhere. Consequently, R = • . an + 2 =

282 • Chapter 10 Series Solutions of Linear Differential Equations

23 (a). Consider the differential equation y ¢¢ + y = 0 . Let the solution be given by •

y ( z) = Â an z n where z = t - 1. Differentiating, we obtain n =0 •

•

y ¢ ( z) = Â nan z n -1 and y ¢¢ ( z) = Â n ( n - 1) an z n - 2 . Inserting these series into the differential n =1

n =2 •

equation, we have

•

 n(n - 1)an z n - 2 +  an z n = 0 . Adjusting the indices, we obtain n =2

•

n =0 •

 (n + 2)(n + 1)a

n +2

z n + Â an z n = 0 . Consequently, the recurrence relation is given by

n =0

n =0

an + 2 = - an /[( n + 2)( n + 1)], n = 0, 1,K. 23 (b). The recurrence leads us to a2 = - a0 / 2, a4 = - a2 / 12 = a0 / 24,K a3 = - a1 / 6, a5 = - a3 / 20 = a1 / 120,K Thus, the general solution is ( t - 1) 2 ( t - 1) 4 ( t - 1) 3 ( t - 1) 5 y ( t) = a0 [1 + + L] + a1[( t - 1) + + L] . 6 120 2 24 23 (c). The coefficient functions are constants and hence analytic everywhere. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -• < t - 1 < • . •

24 (a).

 [(n + 1)na

n +1

- ( n + 2)( n + 1) an + 2 + ( n + 1) an +1 + an ]( t - 1) n = 0 . Consequently, the recurrence

n =0

relation is given by an + 2 =

( n + 1) 2 an +1 + an , n = 0,1, 2,.... ( n + 2)( n + 1)

24 (b). The recurrence leads us to a + a0 a1 a0 4a + a a a = + , a3 = 2 1 = 1 + 0 a2 = 1 2 2 2 3◊ 2 2 3 Thus, the general solution is ( t - 1) 2 ( t - 1) 3 ( t - 1) 2 ( t - 1) 3 y ( t) = a0 [1 + + + L] + a1[( t - 1) + + L] . 2 2 2 3 1 are not analytic at t = 2 . Consequently, R = 1. 24 (c). p( t) = q( t) = t-2 25 (a). Consider the differential equation y ¢¢ + y ¢ + ( t - 2) y = 0 or y ¢¢ + y ¢ + [( t - 1) - 1] y = 0 . Let the •

solution be given by y ( z) = Â an z n where z = t - 1. Differentiating, we obtain n =0 •

y ¢ ( z) = Â nan z

• n -1

and y ¢¢ ( z) = Â n ( n - 1) an z n - 2 . Inserting these series into the differential

n =1

n =2 •

equation, we have

n

n =2 •

indices, we obtain

•

 n(n - 1)a z

n -2

•

•

+ Â nan z n -1 + Â an z n +1 - Â an z n = 0 . Adjusting the n =1

n =0 •

n =0 •

•

 (n + 2)(n + 1)an + 2z n +  (n + 1)an +1z n +  an -1z n -  an z n = 0 . n =0

n =0

n =1

n =0

Consequently, the recurrence relation is given by a2 = ( a0 - a1 ) / 2 and an + 2 = -[( n + 1) an +1 - an + an -1 ] /[( n + 2)( n + 1)], n = 1, 2, K.

Chapter 10 Series Solutions of Linear Differential Equations • 283

25 (b). The recurrence leads us to a3 = -(2 a2 - a1 + a0 ) / 6 = -( a0 - a1 ) / 3,K Thus, the general solution is ( t - 1) 2 ( t - 1) 3 ( t - 1) 2 ( t - 1) 3 y ( t) = a0 [1 + + L] + a1[( t - 1) + + L] . 2 3 2 3 25 (c). The coefficient functions are polynomials and hence analytic everywhere. Consequently, Theorem 10.1 guarantees that the power series solution converges in the interval -• < t - 1 < • . ( n 2 - m 2 ) an 26. , n = 0,1, 2,... an + 2 = ( n + 2)( n + 1) 16 For m = 5, a3 = -4 a1, a5 = a1, a7 = a9 = ... = 0, T5 ( t) = a1[ t - 4 t 3 + 165 t 5 ]. 5 Set T5 (1) = a1[1 - 4 + 165 ] = 1 fi a1 = 5 . Therefore, T5 ( t) = 16 t 5 - 20 t 3 + 5 t For m = 6, a2 = -18 a0 , a4 = 48 a0 , a6 = -32 a0 , T6 ( t) = a0 [1 - 18 t 2 + 48 t 4 - 32 t 6 ]; a0 = -1. Therefore, T6 ( t) = 32 t 6 - 48 t 4 + 18 t 2 - 1 27 (c).

27 (d). TN ( t) £ 1 for - 1 < t < 1 . For t ≥ 1, lim TN ( t) = • . t Ʊ•

•

28 (a).

 [(n + 2)(n + 1)a

n +2

- n ( n - 1) an - 2 nan + m (m + 1) an ]t n = 0 . Therefore the recurrence relation

n =0

[ n ( n + 1) - m (m + 1)]an , n = 0,1, 2,... ( n + 2)( n + 1) 28 (b). When m = N , aN + 2 = aN + 4 = aN + 6 = ... = 0. Therefore, if m = 2M , a polynomial solution of the form a0 + a2 t 2 + ... + a2 M t 2 M exists, while if m = 2 M + 1, a polynomial solution of the form a1t + a3 t 3 + ... + a2 M +1t 2 M +1 exists. is an + 2 =

284 • Chapter 10 Series Solutions of Linear Differential Equations

28 (c). If m = 0 and y = 1, (1 - t 2 )(0) - 2 t(0) + 0(1) = 0 . If m = 1 and y = t, (1 - t 2 )(0) - 2 t(1) + 1(2)( t) = 0 . 3 1 [ n ( n + 1) - 6]an fi P2 ( t) = t 2 - . 28 (d). If m = 2, an + 2 = 2 2 ( n + 2)( n + 1) 5 3 3 [ n ( n + 1) - 12]an fi P3 ( t) = t - t . If m = 3, an + 2 = 2 2 ( n + 2)( n + 1) 35 4 15 2 3 [ n ( n + 1) - 20]an fi P4 ( t) = t - t + . If m = 4, an + 2 = 8 4 8 ( n + 2)( n + 1) 63 5 35 3 15 [ n ( n + 1) - 30]an fi P5 ( t) = t t + t. If m = 5, an + 2 = 8 4 8 ( n + 2)( n + 1) 29 (a). Consider the differential equation y ¢¢ - 2 ty ¢ + 2my = 0 . Let the solution be given by •

•

•

y ( t) = Â an t n . Differentiating, we obtain y ¢ ( t) = Â nan t n -1 and y ¢¢ ( t) = Â n ( n - 1) an t n - 2 . n =0

n =1

n =2

Inserting these series into the differential equation, we have •

•

 n(n - 1)a t n

n -2

•

- 2  nan t + 2m  an t n = 0 . Adjusting the indices, we obtain n

n =2 •

n =1

n =0 •

•

 (n + 2)(n + 1)a

n n n n + 2 t - 2  nan t + 2m  an t = 0 . Consequently, the recurrence relation is

n =0

n =1

n =0

given by a2 = -ma0 and an + 2 = (2 n - 2m ) an /[( n + 2)( n + 1)], n = 1, 2,K . 29 (d). For m = 2 , the even indexed coefficients an vanish when n > 2 . From the recurrence relation, H 2 ( t) = a0 - 2 a0 t 2 = - a0 (2 t 2 - 1) . Choosing a0 = -2 leads us to H 2 ( t) = 4 t 2 - 2 . For m = 3 , the odd indexed coefficients an vanish when n > 3. From the recurrence relation, H 3 ( t) = a1t - (2 / 3) a1t 3 = - a1[(2 / 3) t 3 - t) . Choosing a1 = -12 leads us to H 3 ( t) = 8 t 3 - 12 t . Similarly, H 4 ( t) = 16 t 4 - 48 t 2 + 12 and H 5 ( t) = 32 t 5 - 160 t 3 + 120 t . •

•

30 (a). Try y ( t) =  an t n fi  [( n + 1) nan +1 + ( n + 1) an +1 - an ]t n = 0 . n =0

n =0

• tn an t n +1 ( n + 1) 2 fi an +1 = fi y ( t) = a0 Â 2 = t and the series 2 . By the ratio test, lim n n Æ• t ( n + 2) ( n + 1) 2 n = 0 ( n + 1) converges in -1 < t < 1. •

•

30 (b). Try y ( t) =  an t fi  [ n ( n - 1) + 1]an t n = 0 fi [ n ( n - 1) + 1]an = 0 . n

n =0

n =0

1 ± 1- 4 . Since there are no positive integer roots, the 2 factor [ n ( n - 1) + 1] is nonzero for all n = 0,1, 2,... Therefore, an = 0, n = 0,1, 2,... and y ( t) = 0, The trivial solution results. The coefficient function p( t) = sin t is odd and analytic everywhere. The coefficient function q( t) = t 2 is even and analytic everywhere. Thus, Theorem 10.2(b) applies. The differential equation has a general solution of the form (15). No. p( t) = cos t is even; q( t) = t is odd. The coefficient function p( t) = 0 can be regarded as a function that is odd and analytic everywhere. The coefficient function q( t) = t 2 is even and analytic everywhere. Thus, Theorem 10.2(b) applies. The differential equation has a general solution of the form (15).

The polynomial x 2 - x + 1 has roots

33.

34. 35.

Chapter 10 Series Solutions of Linear Differential Equations • 285

No. p( t) = 1 and q( t) = t 2 are both even. The coefficient function q( t) = t is odd. Thus, Theorem 10.2(b) does not apply. No. p( t) = e t is neither even nor odd and q( t) = 1 is even. Consider the differential equation y ¢¢ + ay ¢ + by = 0 . The coefficient function p( t) = a can be regarded as an odd function if a = 0 , but is even if a is nonzero. The coefficient function q( t) = b is even. Both coefficient functions are analytic everywhere. Thus, Theorem 10.2(b) applies if a = 0 and b is arbitrary. 1 . The denominator of q( t) vanishes at t = ±i fi R = 1. 40 (a). p( t) = 0, q( t) = 1 + t2

36. 37. 38. 39.

•

•

40 (b). y ( t) =  an t n fi  [( n + 2)( n + 1) an + 2 + n ( n - 1) an + an ]t n = 0 n =0

n =0

fi r( n ) = ( n + 2)( n + 1), s( n ) = n ( n - 1) + 1 . Then lim

n Æ•

an + 2 n ( n - 1) + 1 = 1. Therefore, = lim n Æ• ( n + 2)( n + 1) an

the series diverges for t 2 > 1 fi t > 1 by the Ratio Test. 40 (c). No contradiction. The unique solution of the initial value problem exists for -• < t < • , but its Maclaurin series has a radius of convergence R = 1.

Section 10.3 1 (a). l2 + (-2a + 1 - 1)l + a 2 = l2 - 2al + a 2 = 0 1 (b). Using the technique in Section 4.5, the general solution is y = c1ta + c 2 ta ln t, t > 0 . tg cos(d ln t) tg sin(d ln t) 2. W = g -1 = dt 2g -1 π 0 t [g cos(d ln t) - d sin(d ln t)] tg -1[g sin(d ln t) + d cos(d ln t)] in 0 < t < • since d π 0 . 3. When put in standard form, the differential equation is y ¢¢ - 4 t -1 y ¢ + 6 t -2 y = 0 . Thus, t0 = 0 is the only singular point. The characteristic equation is l2 - 5l + 6 = 0 which has roots l1 = 2 and l 2 = 3 . Hence, the general solution is y = c1t 2 + c 2 t 3 , t π 0 . 4. t0 = 0 . The characteristic equation is l2 - l - 6 = 0 which has roots l1 = -2 and l 2 = 3 . Hence, the general solution is y = c1t -2 + c 2 t 3 , t π 0 . 5. When put in standard form, the differential equation is y ¢¢ - 3t -1 y ¢ + 4 t -2 y = 0 . Thus, t0 = 0 is the only singular point. The characteristic equation is l2 - 4 l + 4 = 0 which has roots l1 = 2 and l 2 = 2 . Hence, the general solution is y = c1t 2 + c 2 t 2 ln t , t π 0 . 6. t0 = 0 . The characteristic equation is l2 - 2l + 5 = 0 which has roots l1 = 1 + 2i and l 2 = 1 - 2i . Hence, the general solution is y = c1t cos(2 ln t ) + c 2 t sin(2 ln t ), t π 0 . 7. When put in standard form, the differential equation is y ¢¢ - 3t -1 y ¢ + 29 t -2 y = 0 . Thus, t0 = 0 is the only singular point. The characteristic equation is l2 - 4 l + 29 = 0 which has roots l1 = 2 + 5i and l 2 = 2 - 5i . Hence, the general solution is y = c1t 2 cos(5 ln t ) + c 2 t 2 sin(5 ln t ), t π 0 . 8. t0 = 0 . The characteristic equation is l2 - 6l + 9 = 0 which has roots l1 = l 2 = 3. Hence, the general solution is y = c1t 3 + c 2 t 3 ln t , t π 0 .

286 • Chapter 10 Series Solutions of Linear Differential Equations

9.

10. 11.

12. 13.

14. 15.

16.

17.

18. 19.

20. 21.

22.

When put in standard form, the differential equation is y ¢¢ + t -1 y ¢ + 9 t -2 y = 0 . Thus, t0 = 0 is the only singular point. The characteristic equation is l2 + 9 = 0 which has roots l1 = 3i and l 2 = -3i . Hence, the general solution is y = c1 cos( 3 ln t ) + c 2 sin( 3 ln t ), t π 0 . t0 = 0 . The characteristic equation is l2 + 2l + 1 = 0 which has roots l1 = l 2 = -1. Hence, the general solution is y = c1t -1 + c 2 t -1 ln t , t π 0 . When put in standard form, the differential equation is y ¢¢ + 3t -1 y ¢ + 17 t -2 y = 0 . Thus, t0 = 0 is the only singular point. The characteristic equation is l2 + 2l + 17 = 0 which has roots l1 = -1 + 4 i and l 2 = -1 - 4 i . Hence, the general solution is y = c1t -1 cos( 4 ln t ) + c 2 t -1 sin( 4 ln t ), t π 0 . t0 = 0 . The characteristic equation is l2 + 10l + 25 = 0 which has roots l1 = l 2 = -5 . Hence, the general solution is y = c1t -5 + c 2 t -5 ln t , t π 0 . Consider the differential equation y ¢¢ + 5 t -1 y ¢ + 40 t -2 y = 0 . We see that, t0 = 0 is the only singular point. The characteristic equation is l2 + 4 l + 40 = 0 which has roots l1 = -2 + 6i and l 2 = -2 - 6i . Hence, the general solution is y = c1t -2 cos(6 ln t ) + c 2 t -2 sin(6 ln t ), t π 0 . t0 = 0 . The characteristic equation is l2 - 3l = 0 which has roots l1 = 0, l 2 = 3. Hence, the general solution is y = c1 + c 2 t 3 , t π 0 . When put in standard form, the differential equation is y ¢¢ - ( t - 1) -1 y ¢ - 3( t - 1) -2 y = 0 . Thus, t0 = 1 is the only singular point. The characteristic equation is l2 - 2l - 3 = 0 which has roots l1 = -3 and l 2 = 1. Hence, the general solution is y = c1 ( t - 1) 3 + c 2 ( t - 1) -1, t π 1 . t0 = 1. The characteristic equation is l2 + 2l + 17 = 0 which has roots l1 = -1 + 4 i, l 2 = -1 - 4 i . Hence, the general solution is y = c1 ( t - 1) -1 cos( 4 ln t - 1 ) + c 2 ( t - 1) -1 sin( 4 ln t - 1 ), t π 1. When put in standard form, the differential equation is y ¢¢ + 6( t + 2) -1 y ¢ + 6( t + 2) -2 y = 0 . Thus, t0 = -2 is the only singular point. The characteristic equation is l2 + 5l + 6 = 0 which has roots l1 = -3 and l 2 = -2 . Hence, the general solution is y = c1 ( t + 2) -3 + c 2 ( t + 2) -2 , t π -2 . t0 = 2 . The characteristic equation is l2 + 4 = 0 which has roots l1 = 2i, l 2 = -2i . Hence, the general solution is y = c1 cos(2 ln t - 2 ) + c 2 sin(2 ln t - 2 ), t π 2 . From the form of the general solution, t0 = -2 and the characteristic equation has roots l1 = 1 and l 2 = -2 . Therefore, the characteristic equation is l2 + l - 2 = 0 . Matching the characteristic equation with the general form given in equation (3), we see that a - 1 = 1 and b = -2 . Thus, the differential equation is ( t + 2) 2 y ¢¢ + 2( t + 2) y ¢ - 2 y = 0 . t0 = 1, l = 0, 0. \ l2 = 0 fi a = 1, b = 0 . From the form of the general solution, t0 = 0 and the characteristic equation has roots l1 = 2 + i and l 2 = 2 - i . Therefore, the characteristic equation is l2 - 4 l + 5 = 0 . Matching the characteristic equation with the general form given in equation (3), we see that a - 1 = -4 and b = 5 . Thus, the differential equation is t 2 y ¢¢ - 3ty ¢ + 5 y = 0 . The characteristic equation has roots l1 = 2 and l 2 = -1 . Therefore, the characteristic equation is l2 - l - 2 = 0 fi a = 0, b = -2 . Thus, the differential equation is t 2 y ¢¢ + ty ¢ - y = g( t) . We can determine the nonhomogenous term g(t) by inserting the given particular solution y P ( t) = 2 t + 1 . Doing so, we obtain t 2 (0) + t(2) - 2(2 t + 1) = -2 t - 2 = g( t) .

Chapter 10 Series Solutions of Linear Differential Equations • 287

23.

24. 25.

26.

27.

28.

29.

30.

31.

From the form of the general solution, the characteristic equation has roots l1 = 2 and l 2 = 3 . Therefore, the characteristic equation is l2 - 5l + 6 = 0 . Matching the characteristic equation with the general form given in equation (3), we see that a - 1 = -5 and b = 6 . Thus, the differential equation is t 2 y ¢¢ - 4 ty ¢ + 6 y = g( t) . We can determine the nonhomogenous term g(t) by inserting the given particular solution y P ( t) = ln t . Doing so, we obtain t 2 y P¢¢ - 4 ty P¢ + 6 y P = g( t) or t 2 (- t -2 ) - 4 t( t -1 ) + 6 ln t = g( t) . Thus, g( t) = -5 + 6 ln t . Under the change of variable t = e z , the differential equation transforms into Y ¢¢ ( z) - Y ¢ ( z) - 2Y ( z) = 2 . The general solution is Y ( z) = c1e - z + c 2e 2 z - 1 fi y = c1t -1 + c 2 t 2 - 1. Under the change of variable t = e z , the differential equation t 2 y ¢¢ - ty ¢ + y = t -1 transforms into Y ¢¢ ( z) - 2Y ¢ ( z) + Y ( z) = (e z ) -1 or Y ¢¢ ( z) - 2Y ¢ ( z) + Y ( z) = e - z . Solving this constant coefficient equation using the techniques of Chapter 4, we find the general solution Y ( z) = c1e z + c 2 ze z + 0.25e - z . Since z = ln t , the solution can be converted to y ( t) = c1t + c 2 t ln t + 0.25 t -1 . Under the change of variable t = e z , the differential equation transforms into Y ¢¢ ( z) + 9Y ( z) = 10e z . The general solution is Y ( z) = c1 cos( 3z) + c 2 sin( 3z) + e z fi y = c1 cos( 3 ln t) + c 2 sin( 3 ln t) + t . Under the change of variable t = e z , the differential equation t 2 y ¢¢ - 6 y = 10 t -2 - 6 transforms into Y ¢¢ ( z) - Y ¢ ( z) - 6Y ( z) = 10(e z ) -2 - 6 or Y ¢¢ ( z) - Y ¢ ( z) - 6Y ( z) = 10e -2 z - 6 . Solving this constant coefficient equation using the techniques of Chapter 4, we find the general solution Y ( z) = c1e 3 z + c 2e -2 z - 2 ze -2 z + 1. Since z = ln t , the solution can be converted to y ( t) = c1t 3 + c 2 t -2 - 2 t -2 ln t + 1. Under the change of variable t = e z , the differential equation transforms into 1 5 Y ¢¢ ( z) - 5Y ¢ ( z) + 6Y ( z) = 3z . Therefore, Yc = c1e 2 z + c 2e 3 z , Yp = Az + B = z + . 2 12 1 5 1 5 The general solution is Y ( z) = c1e 2 z + c 2e 3 z + z + fi y = c1t 2 + c 2 t 3 + ln t + . 2 12 2 12 Under the change of variable t = e z , the differential equation t 2 y ¢¢ + 8 ty ¢ + 10 y = 36( t + t -1 ) transforms into Y ¢¢ ( z) + 7Y ¢ ( z) + 10Y ( z) = 36(e z + e - z ) . Solving this constant coefficient equation using the techniques of Chapter 4, we find the general solution Y ( z) = c1e -5 z + c 2e -2 z + 2e z + 9e - z . Since z = ln t , the solution can be converted to y ( t) = c1t -5 + c 2 t -2 + 2 t + 9 t -1 . The complementary solution is yC ( t) = c1t -1 + c 2 t 3 . For a particular solution, use y P ( t) = At + B . Then, the general solution is y ( t) = c1t -1 + c 2 t 3 - 2 t - 2 . Imposing the initial conditions, we obtain y (1) = c1 + c 2 - 2 - 2 = 1 and y ¢ (1) = -c1 + 3c 2 - 2 = 3. Solving, we find 5 5 the solution of the initial value problem is y ( t) = t -1 + t 3 - 2 t - 2 . The interval of existence 2 2 is 0 < t < • . Consider the initial value problem t 2 y ¢¢ - 5 ty ¢ + 5 y = 10, y (1) = 4, y ¢ (1) = 6 . The complementary solution is yC ( t) = c1t 5 + c 2 t . By inspection, a particular solution is y P ( t) = 2 . Thus, the general solution is y ( t) = c1t 5 + c 2 t + 2 . Imposing the initial conditions, we obtain y (1) = c1 + c 2 + 2 = 4 and y ¢ (1) = 5c1 + c 2 = 6 . Solving, we find the solution of the initial value problem is y ( t) = t 5 + t + 2 . The interval of existence is the entire t-axis.

288 • Chapter 10 Series Solutions of Linear Differential Equations

32.

33.

34.

The complementary solution is yC ( t) = c1t -1 + c 2 t -1 ln(- t) . For a particular solution, use y P ( t) = At + B . Then, the general solution is yC ( t) = c1t -1 + c 2 t -1 ln(- t) + 2 t + 9 . Imposing the initial conditions, we obtain y (-1) = -c1 - 2 + 9 = 1 and y ¢ (-1) = -c1 + c 2 + 2 = 0 . Solving, we find the solution of the initial value problem is y ( t) = 6 t -1 + 4 t -1 ln(- t) + 2 t + 9 . The interval of existence is -• < t < 0 . Consider the initial value problem t 2 y ¢¢ + 3ty ¢ + y = 2 t -1, y (1) = -2, y ¢ (1) = 1 . The complementary solution is yC ( t) = c1t -1 + c 2 t -1 ln t . Using the change of variable t = e z as in Example 2, we find a particular solution y P ( t) = t -1 (ln t) 2 . Thus, the general solution is y ( t) = c1t -1 + c 2 t -1 ln t + t -1 (ln t) 2 . Imposing the initial conditions, we obtain y (1) = c1 = -2 and y ¢ (1) = -c1 + c 2 = 1. Solving, we find the solution of the initial value problem is y ( t) = -2 t -1 - t -1 ln t + t -1 (ln t) 2 . The interval of existence is the positive t-axis. 1 dy 1 d 2 y 1 1 Ê d 2 y dy ˆ dy dy dz 1 dy d 2 y = = = + = Á - ˜. ; dt dz dt t dz dt 2 t 2 dz t dz 2 t t 2 Ë dz 2 dz ¯ d3y 2 Ê d 2 y dy ˆ 1 Ê d 3 y d 2 y ˆ 1 Ê d 3 y d2y dy ˆ = 3 + = ˜ ˜ . Therefore, 3 3 Á 2 3 Á 3 2˜ 3 Á 3 2 +2 dt t Ë dz dz ¯ t Ë dz dz ¯ t Ë dz dz dz ¯ Ê d 2Y dY ˆ d 3Y d 2Y dY Ê dY ˆ 3 2 t y ¢¢¢ + a t y ¢¢ + b ty ¢ + gy = 3 - 3 2 + 2 + aÁ 2 ˜ + b ÁË ˜¯ + gY = 0 dz dz dz dz ¯ dz Ë dz d 3Y d 2Y dY + ( a 3 ) + gY = 0 . 3 2 + (b - a + 2) dz dz dz Consider the differential equation t 3 y ¢¢¢ + 3t 2 y ¢¢ - 3ty ¢ = 0 . Assuming a solution of the form y ( t) = tl , we obtain the characteristic equation l3 - 4 l = 0 . The roots are l1 = 0, l 2 = 2 and l 3 = -2 . The general solution is y ( t) = c1 + c 2 t 2 + c 3 t -2 , t π 0. a = 0, b = 1, g = -1 fi Y ¢¢¢ - 3Y ¢¢ + 3Y ¢ - Y = 0 . The characteristic equation is l3 - 3l2 + 3l - 1 = (l - 1) 3 = 0 . The roots are l1 = l 2 = l 3 = 1. Therefore, Y = c1e z + c 2 ze z + c 3 z 2e z fi y = c1t + c 2 t ln t + c 3 t(ln t) 2 . Consider the differential equation t 3 y ¢¢¢ + 3t 2 y ¢¢ + ty ¢ = 8 t 2 + 12 . Using the change of variable t = e z as suggested in Exercise 34, the differential equation transforms to Y ¢¢¢ ( z) = 8e 2 z + 12 . The general solution is Y ( z) = c1 + c 2 z + c 3 z 2 + e 2 z + 2 z 3 . Using the fact that z = ln t , the general solution becomes y ( t) = c1 + c 2 ln t + c 3 (ln t) 2 + t 2 + 2(ln t) 3 , t > 0 . a = 6, b = 7, g = 1 fi Y ¢¢¢ + 3Y ¢¢ + 3Y ¢ + Y = 0 . The characteristic equation is (l + 1) 3 = 0 . The roots are l1 = l 2 = l 3 = -1 . Therefore, Yc = c1e - z + c 2 ze - z + c 3 z 2e - z , Yp = Az + B fi Y = c1e - z + c 2 ze - z + c 3 z 2e - z + z - 1 fi

35.

36.

37.

38.

fi y = c1t -1 + c 2 t -1 ln t + c 3 t -1 (ln t) 2 + ln t - 1.

Section 10.4 1.

When put in standard form, the differential equation is y ¢¢ + t -1 (cos t) y ¢ + t -1 y = 0 . Thus, t = 0 is the only singular point. The coefficient functions are p( t) = t -1 (cos t) and q( t) = t -1 . Clearly tp( t) = cos t and t 2q( t) = t are analytic. Therefore, t = 0 is a regular singular point.

Chapter 10 Series Solutions of Linear Differential Equations • 289

2. 3.

4.

5.

6. 7.

sin t 1 sin t t2 t4 t6 and . Since q ( t ) = = = 1 + - + ... and t 2q( t) = 1 are both tp ( t ) t2 t2 t 3! 5! 7! analytic at t = 0 , then t = 0 is a regular singular point. When put in standard form, the differential equation is y ¢¢ + ( t + 1) -1 y ¢ + ( t 2 - 1) -1 y = 0 . Thus, t = 1 and t = -1 are singular points. The coefficient functions are p( t) = ( t + 1) -1 and q( t) = ( t 2 - 1) -1. Clearly ( t - 1) p( t) = ( t - 1)( t + 1) -1 and ( t - 1) 2 q( t) = ( t - 1)( t + 1) -1 are analytic at t = 1. Therefore, t = 1 is a regular singular point. Similarly, t = -1 is also a regular singular point. 1 t +1 1 p( t) = 2 and q( t) = . 2 = 2 2 ( t - 1) ( t - 1) ( t + 1) ( t - 1) ( t + 1) 2 1 1 1 1 At t = -1, ( t + 1) p( t) = and ( t + 1) 2 q( t) = as t Æ -1. Therefore, t = -1 2 Æ 2 Æ 4 4 ( t - 1) ( t - 1) is a regular singular point. 1 At t = 1, lim( t - 1) p( t) = lim does not exist.. Therefore, t = 1 is an irregular singular t Æ1 t Æ1 ( t - 1)( t + 1) point. When put in standard form, the differential equation is y ¢¢ + t -2 (1 - cos t) y ¢ + t -2 y = 0 . Thus, t = 0 is the only singular point. The coefficient functions are p( t) = t -2 (1 - cos t) and q( t) = t -2 . t t3 t5 Using a Maclaurin series, tp( t) = t -1 (1 - cos t) = - + - L is analytic at t = 0 as is 2! 4! 6! t 2q( t) = 1. Therefore, t = 0 is a regular singular point. t2 t 1 p( t) = q( t) = . Since neither tp( t) = nor t 2q( t) = are analytic at t = 0, there is an irregular t t t singular point at t = 0 . When put in standard form, the differential equation is y ¢¢ + (1 - e t ) -1 y ¢ + (1 - e t ) -1 y = 0 . Thus, t = 0 is the only singular point. The coefficient functions are p( t) = (1 - e t ) -1 and q( t) = (1 - e t ) -1 . Using a Maclaurin series, p( t) =

-1

8.

9.

-1

Ê ˆ Ê ˆ t2 t3 t t2 tp( t) = t(1 - e ) = tÁ - t - - - L˜ = Á -1 - - - L˜ is analytic at t = 0 as is t 2q( t).. 2! 3! 2! 3! Ë ¯ Ë ¯ Therefore, t = 0 is a regular singular point. -1 t+2 1 1 = p( t) = and q( t) = . 2 2 = 2 (2 - t)(2 + t) ( t - 2) (4 - t ) ( t - 2) ( t + 2) 2 1 1 -( t + 2) Æ 0 and ( t + 2) 2 q( t) = as t Æ -2 . Therefore, At t = -2, ( t + 2) p( t) = 2 Æ 16 ( t - 2) ( t - 2) t = -2 is a regular singular point. 1 1 At t = 2, ( t - 2) p( t) = -1 and ( t - 2) 2 q( t) = as t Æ 2 . Therefore, t = 2 is a regular 2 Æ 16 ( t + 2) singular point. When put in standard form, the differential equation is y ¢¢ + (1 - t 2 ) -1/ 3 y ¢ + (1 - t 2 ) -1/ 3 ty = 0 . Thus, t = 1 and t = -1 are singular points. The coefficient functions are p( t) = (1 - t 2 ) -1/ 3 and q( t) = t(1 - t 2 ) -1/ 3 . Neither of the functions ( t ± 1) p( t) or ( t ± 1) 2 q( t) is analytic at t = ±1 . Therefore, t = 1 is an irregular singular point as is t = -1. t -1

290 • Chapter 10 Series Solutions of Linear Differential Equations 1

7

p( t) = 1, q( t) = t 3 . Since tp( t) = t is analytic at t = 0, but t 2q( t) = t 3 is not, there is an irregular singular point at t = 0 . 11. For this problem, p( t) = (sin 2 t) / P ( t) . Since we know there are singular points at t = 0 and t = ±1, we know that P ( t) must be zero at those points. Since tp( t) is analytic at t = 0 and since (sin 2t) / t tends to 2 as t Æ 0 , it follows that t 2 is a factor of P ( t) . Similarly, ( t - 1) p( t) is not analytic at t = 1 and thus ( t - 1) 2 must be a factor of P ( t) . The same argument applies at t = -1 and thus ( t + 1) 2 must be a factor of P ( t) . In summary, P ( t) = t 2 ( t - 1) 2 ( t + 1) 2 = t 2 ( t 2 - 1) 2 . 12. P ( t) = 1 . 13. For this problem, p( t) = [ tP ( t)]-1 . Since we know there are singular points at t = ±1 , we know that P ( t) must be zero at t = ±1 . Since t 2q( t) = 1 / t , it follows [without any assumptions on P ( t) ] that t = 0 is an irregular singular point. Since, ( t - 1) p( t) is not analytic at t = 1 it follows that ( t - 1) 2 must be a factor of P ( t) . The same argument applies at t = -1 and thus ( t + 1) 2 must be a factor of P ( t) . In summary, P ( t) = ( t - 1) 2 ( t + 1) 2 = ( t 2 - 1) 2 . 14(a). t = 0 is a regular singular point if n = 1. 14(b). t = 0 is an irregular singular point if n ≥ 2 . 15. For this problem, tp( t) = t / (sin t) and t 2q( t) = 1 / t n - 2 . Since t / (sin t) is analytic at t = 0 , it follows that t = 0 is a regular singular point if n = 0, 1, 2 and an irregular singular point if n > 2. 1 t +1 1 16 (a). tp( t) = - and t 2q( t) = Æ as t Æ 0 . Thus, t = 0 is a regular singular point. 2 2 2 10.

•

16 (b). Substituting the series y = Â an tl + n into the differential equation, we obtain n =0 •

[2l (l - 1) - l + 1]a0 tl + Â [(2(l + n )(l + n - 1) - (l + n ) + 1) an + an -1 ] tl + n = 0 . Therefore, the n =1

indicial equation is F (l ) = 0 where F (l ) = 2l2 - 3l + 1. The roots of the indicial equation are 1 l1 = and l 2 = 1. 2 - an -1 - an -1 = , n = 1, 2, K 16 (c). an = 2 F (l + n ) 2(l + n ) - 3(l + n ) + 1 - an -1 , n = 1, 2,K.. For l 2 = 1, the recurrence relation is an = 2 2(1 + n ) - 3(1 + n ) + 1 È t2 t3 ˘ + L˙ . 16 (d). y ( t) = a0 Ít - + Î 3 30 ˚ 17 (a). For this problem, tp( t) = 1 and t 2q( t) = ( t - 1) / 4 . Thus, t = 0 is a regular singular point. •

17 (b). Substituting the series y = Â an tl + n into the differential equation 4 t 2 y ¢¢ + 4 ty ¢ + ( t - 1) y = 0 , n =0 •

we obtain ( 4 l2 - 1) a0 tl + Â [( 4 (l + n ) 2 - 1) an + an -1 ] tl + n = 0 . Therefore, the indicial equation n =1

is F (l ) = 0 where F (l ) = 4 l2 - 1. The roots of the indicial equation are l1 = -1 / 2 and l 2 = 1 / 2 .

Chapter 10 Series Solutions of Linear Differential Equations • 291

- an -1 - an -1 = , n = 1, 2, K F (l + n ) 4 (l + n ) 2 - 1 For l = 1 / 2 , the recurrence relation is an = - an -1 /[ 4 ( n + 0.5) 2 - 1], n = 1, 2,K .. 17 (d). y ( t) = a0 [ t1/ 2 - (1 / 8) t 3 / 2 + (1 / 192) t 5 / 2 - L] . t 3 and t 2q( t) = . Both limits exist as t Æ 0 . Thus, t = 0 is a regular singular point. 18 (a). tp( t) = 16 16 17 (c). an =

•

18 (b). Substituting the series y = Â an tl + n into the differential equation, we obtain n =0 •

[16l (l - 1) + 3]a0 tl + Â [(16(l + n )(l + n - 1) + 3) an + (l + n - 1) an -1 ] tl + n = 0 . Therefore, the n =1

indicial equation is F (l ) = 0 where F (l ) = 16l2 - 16l + 3. The roots of the indicial equation 1 3 are l1 = and l 2 = . 4 4 -(l + n - 1) an -1 -(l + n - 1) an -1 = , n = 1, 2, K 18 (c). an = F (l + n ) 16(l + n )(l + n - 1) + 3 -( 3 / 4 + n - 1) an -1 3 , n = 1, 2, K, .. For l 2 = , the recurrence relation is an = 16( 3 / 4 + n )( 3 / 4 + n - 1) + 3 4 7 11 4 È 3 t4 ˘ t 7 18 (d). y ( t) = a0 Ít 4 + + L˙, t > 0 . 32 10240 ÍÎ ˙˚ 19 (a). For this problem, tp( t) = 1 and t 2q( t) = t - 9 . Thus, t = 0 is a regular singular point. •

19 (b). Substituting the series y = Â an tl + n into the differential equation t 2 y ¢¢ + ty ¢ + ( t - 9) y = 0 , we n =0 •

obtain (l2 - 9) a0 tl + Â [((l + n ) 2 - 9) an + an -1 ] tl + n = 0 . Therefore, the indicial equation is n =1

F (l ) = 0 where F (l ) = l2 - 9 . The roots of the indicial equation are l1 = -3 and l 2 = 3 . - an -1 - an -1 = , n = 1, 2, K 19 (c). an = F (l + n ) (l + n ) 2 - 9 For l = 3 , the recurrence relation is an = - an -1 /[( n + 3) 2 - 9], n = 1, 2,K .. 19 (d). y ( t) = a0 [ t 3 - (1 / 7) t 4 + (1 / 112) t 5 - L] . 20 (a). tp( t) = t + 2 and t 2q( t) = - t . Both limits exist as t Æ 0 . Thus, t = 0 is a regular singular point. •

20 (b). Substituting the series y = Â an tl + n into the differential equation, we obtain n =0 •

[l (l - 1) + 2l ]a0 tl -1 + Â {[(l + n + 1)(l + n ) + 2(l + n + 1)]an +1 + (l + n - 1) an }tl + n = 0 . n =0

Therefore, the indicial equation is F (l ) = 0 where F (l ) = l2 + l . The roots of the indicial equation are l1 = -1 and l 2 = 0 . -(l + n - 1) an , n = 0,1, 2, K 20 (c). an +1 = (l + n + 2)(l + n + 1) -( n - 1) an , n = 0,1, 2, K,.. For l 2 = 0 , the recurrence relation is an = ( n + 2)( n + 1)

292 • Chapter 10 Series Solutions of Linear Differential Equations

È t˘ 20 (d). y ( t) = a0 Í1 + ˙ . Î 2˚ 21 (a). For this problem, tp( t) = 3 and t 2q( t) = 2 t + 1. Thus, t = 0 is a regular singular point. •

21 (b). Substituting the series y = Â an tl + n into the differential equation t 2 y ¢¢ + 3ty ¢ + (2 t + 1) y = 0 , n =0 •

we obtain (l2 + 2l + 1) a0 tl + Â [((l + n ) 2 + 2(l + n ) + 1) an + 2 an -1 ] tl + n = 0 . Therefore, the n =1

indicial equation is F (l ) = 0 where F (l ) = l2 + 2l + 1. The roots of the indicial equation are l1 = l 2 = -1. -2 an -1 -2 an -1 = , n = 1, 2, K 21 (c). an = F (l + n ) ((l + n ) + 1) 2 For l = -1 , the recurrence relation is an = -2 an -1 / n 2 , n = 1, 2,K .. 21 (d). y ( t) = a0 [ t -1 - 2 + t - L]. 22 (a). Both limits exist as t Æ 0 . Thus, t = 0 is a regular singular point. •

22 (b). Substituting the series y = Â an tl + n into the differential equation, we obtain n =0 •

[l (l - 1) - l - 3]a0 tl + Â {[(l + n ) 2 - 2(l + n ) - 3)]an + (l + n - 1) an -1}tl + n = 0 . Therefore, the n =1

indicial equation is F (l ) = 0 where F (l ) = l2 - 2l - 3 . The roots of the indicial equation are l1 = -1 and l 2 = 3 . -(l + n - 1) an -1 -(l + n - 1) an -1 = , n = 1, 2, K 22 (c). an = F (l + n ) (l + n ) 2 - 2(l + n ) - 3 -( n + 2) an -1 , n = 1, 2, K,.. For l 2 = 3, the recurrence relation is an = n(n + 4) È ˘ 3t 4 t 5 + + L˙ . 22 (d). y ( t) = a0 Ít 3 5 5 Î ˚ 23 (a). For this problem, tp( t) = t - 2 and t 2q( t) = t . Thus, t = 0 is a regular singular point. •

23 (b). Substituting the series y = Â an tl + n into the differential equation ty ¢¢ + ( t - 2) y ¢ + y = 0 , we n =0

•

obtain (l2 - 3l ) a0 tl -1 + Â (l + n + 1)[(l + n - 2) an + an -1 ] tl + n = 0 . Therefore, the indicial n =0

equation is F (l ) = 0 where F (l ) = l2 - 3l . The roots of the indicial equation are l1 = 0 and l 2 = 3 . - an -(l + n + 1) an -(l + n + 1) an = , n = 0,1, 2, K = 23 (c). an +1 = (l + n + 1)(l + n - 2) (l + n - 2) F (l + n ) For l = 3 , the recurrence relation is an = - an -1 / ( n + 1), n = 0, 1,K.. 23 (d). y ( t) = a0 [ t 3 - t 4 + (1 / 2) t 5 - L] . 2 sin t 24 (a). tp( t) = Æ -2 as t Æ 0 and t 2q( t) = 2 + t Æ 2 as t Æ 0 . Thus, t = 0 is a regular singular t point.

Chapter 10 Series Solutions of Linear Differential Equations • 293

[

]

24 (b). t 2 y ¢¢ - 2 sin ty ¢ + (2 + t) y = l (l - 1) a0 tl + (l + 1)la1tl +1 + (l + 2)(l + 1) a2 tl + 2 + L 3

È t ˘ -2 Ít - + ...˙ la0 tl -1 + (l + 1) a1tl + (l + 2) a2 tl +1 + L + (2 + t) a0 tl + a1tl +1 + a2 tl + 2 + ... = 0 . Î 3! ˚ l For t : l (l - 1) a0 - 2la0 + 2 a0 = (l2 - 3l + 2) a0 = (l - 1)(l - 2) a0 = 0 . For tl +1: l (l + 1) a1 - 2(l + 1) a1 + 2 a1 + a0 = [(l + 1)(l - 2) + 2]a1 + a0 = 0 . 2 For tl + 2 : (l + 2)(l + 1) a2 - 2(l + 2) a2 + la0 + 2 a2 + a1 = 0 . 3! Therefore, the indicial equation is F(l ) = (l - 1)(l - 2) = 0 . The roots of the indicial equation are l1 = 1 and l 2 = 2 . t3 t4 2 24 (c). y ( t) = a0 [ t - - - L] 2 6 25 (a). For this problem, tp( t) = 4 and t 2q( t) = te t . Thus, t = 0 is a regular singular point.

[

]

[

]

•

25 (b). Given the series y = Â an tl + n , we have ty ¢¢ = l (l - 1) a0 tl -1 + (l + 1)la1tl + L , n =0

-4 y ¢ = la0 t + (l + 1) a1tl + L , and e t y = [1 + t + (1 / 2!) t 2 + L][ a0 tl + a1tl +1 + L] = a0 tl + ( a1 + 1) tl +1 + L . Therefore, substituting the series into the differential equation ty ¢¢ - 4 y ¢ + e t y = 0 , we obtain l (l - 5) a0 tl -1 + [(l + 1)(l - 4 ) + a0 ] tl + L = 0 . Therefore, the indicial equation is l2 - 5l = 0 . The roots of the indicial equation are l1 = 0 and l 2 = 5 . 25 (c). y ( t) = a0 [ t 5 - (1 / 6) t 6 - (5 / 84 ) t 7 - L] t t2 26 (a). tp( t) = Æ -1 as t Æ 0 and t 2q( t) = Æ 0 as t Æ 0 . Thus, t = 0 is a regular singular sin t sin t point. 26 (b). (sin t) y ¢¢ - y ¢ + y = È t3 t5 ˘ l -2 l -1 l l +1 Ít - 3! + 5! ...˙ l (l - 1) a0 t + (l + 1)la1t + (l + 2)(l + 1) a2 t + (l + 3)(l + 2) a3 t + L Î ˚ l -1

[

]

[

] [

]

- la0 tl -1 + (l + 1) a1tl + (l + 2) a2 tl +1 + L + a0 tl + a1tl +1 + a2 tl + 2 + L = 0 .. l -1

2

For t : l (l - 1) a0 - la0 = (l - 2l ) a0 = l (l - 2) a0 = 0 . For tl : l (l + 1) a1 - (l + 1) a1 + a0 = (l + 1)(l - 1) a1 + a0 = 0 . 1 1 For tl +1: (l + 2)(l + 1) a2 + (l + 2) a2 - l (l - 1) a0 + a1 = (l + 2) 2 a2 + a1 - l (l - 1) a0 = 0 . 6 3! Therefore, the indicial equation is F(l ) = l (l - 2) = 0 . The roots of the indicial equation are l1 = 0 and l 2 = 2 . t3 t4 2 26 (c). y ( t) = a0 [ t - + + L] 3 24 27 (a). For this problem, tp( t) = t / (2 - 2e t ) and t 2q( t) = t 2 / (1 - e t ) . Thus, t = 0 is a regular singular point. •

27 (b). Given the series y = Â an tl + n , we have n =0

(1 - e ) y ¢¢ = -l (l - 1) a0 tl -1[-0.5l (l - 1) a0 - (l + 1)la1 ]tl + L, 0.5 y ¢ = 0.5[la0 tl -1 + (l + 1) a1tl + L]. t

294 • Chapter 10 Series Solutions of Linear Differential Equations

Therefore, substituting the series into the differential equation (1 - e t ) y ¢¢ + (1 / 2) y ¢ + y = 0 , we obtain -l (l - 1.5) a0 tl -1 + [-(l + 1)(l - 0.5) a1 + 0.5(-l2 + l + 2) a0 ] tl + L = 0 . Therefore, the indicial equation is l2 - 1.5l = 0 . The roots of the indicial equation are l1 = 0 and l 2 = 1.5 . 27 (c). y ( t) = a0 [ t 3 / 2 + (1 / 2) t 5 / 2 - (17 / 96) t 7 / 2 + L]

Section 10.5 1 (a). When put in standard form, the differential equation is y ¢¢ - (2 t) -1 (1 + t) y ¢ + t -1 y = 0 . Therefore, t = 0 is a regular singular point. •

1 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

(2l2 - 3l ) a0 tl -1 + Â [(l + n + 1)(2(l + n ) - 1) an +1 - (l + n - 2) an ] t n +l = 0 . n =0

1 (c). 1 (d). 1 (e). 2 (b).

Therefore, the exponents at the singularity are l1 = 0 and l 2 = 1.5 . The recurrence relation is an +1 = [(l + n - 2) an ] /[(l + n + 1)(2l + 2 n - 1)], n = 0, 1,K . For l1 = 0 , y = a0 [1 + 2 t - t 2 ] is a polynomial solution. For l 2 = 3 / 2 , y = a0 [ t 3 / 2 - (1 / 10) t 5 / 2 - (1 / 280) t 7 / 2 - L] . Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the second series found in part (d) converges for 0 < t. Substituting the series into the differential equation, we obtain •

[2l (l - 1) + 5l ]a0 tl -1 + [2l (l + 1) + 5(l + 1)]a1tl + Â [2(l + n + 1)(l + n + 5 / 2)an +1 + 3an -1] t n +l n =1

2 (c). 2 (d). 2 (e). 3 (a).

3 = 0 . Therefore, F(l ) = 2l (l + 3 / 2) fi l1 = - , l 2 = 0 . 2 -3an -1 , n = 1, 2,... and (l + 1)(2l + 5) a1 = 0 The recurrence relation is an +1 = 2(l + n + 1)(l + n + 5 / 2) 3 For l1 = - , y = a0 [ t -3 / 2 - ( 3 / 2) t1/ 2 + (9 / 40) t 5 / 2 + L] . 2 For l 2 = 0 , y = a0 [1 - ( 3 / 14 ) t 2 + (9 / 616) t 4 - L] . The series converges for 0 < t. When put in standard form, the differential equation is y ¢¢ - ( 3t) -1 y ¢ + ( 3t 2 ) -1 (1 + t) y = 0 . Therefore, t = 0 is a regular singular point. •

3 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

( 3l2 - 4 l + 1) a0 tl + Â {[ 3(l + n )(l + n - 1) - l - n + 1]an + an -1} t n +l = 0 . n =1

Therefore, the exponents at the singularity are l1 = 1 / 3 and l 2 = 1. 3 (c). The recurrence relation is an = - an -1 /[ 3(l + n )(l + n - 1) - l - n + 1], n = 1, 2,K . 3 (d). For l1 = 1 / 3, y = a0 [ t1/ 3 - t 4 / 3 + (1 / 8) t 7 / 3 + L]. For l 2 = 1, y = a0 [ t - (1 / 5) t 2 + (1 / 80) t 3 + L]. 3 (e). Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the series found in part (d) converge for 0 < t.

Chapter 10 Series Solutions of Linear Differential Equations • 295

4 (b). Substituting the series into the differential equation, we obtain •

[6l (l - 1) + l + 1]a0 t

l

+ Â {[6(l + n )(l + n - 1) + (l + n ) + 1]an - an -1}t n +l = 0 . Therefore, n =1

1 1 F(l ) = 6l2 - 5l + 1 fi l1 = , l 2 = . 3 2 4 (c). The recurrence relation is an =

an -1 , n = 1, 2,... 6(l + n )(l + n - 1) + (l + n ) + 1

1 , y = a0 [ t1/ 3 + (1 / 5) t 4 / 3 + (1 / 110) t 7 / 3 + L]. 3 1 For l 2 = , y = a0 [ t1/ 2 + (1 / 7) t 3 / 2 + (1 / 182) t 5 / 2 + L]. 2 4 (e). The series converges for 0 < t. 5 (a). When put in standard form, the differential equation is y ¢¢ - 5 t -1 y ¢ + t -2 (9 + t 2 ) y = 0 . Therefore, t = 0 is a regular singular point.

4 (d). For l1 =

•

5 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0

l

(l - 6l + 9) a0 t + [(l + 1)l - 5(l + 1) + 9]a1tl +1 + 2

•

 {[(l + n)(l + n - 1) - 5(l + n) + 9]a

n

+ an -1} t n +l = 0

.

n =2

5 (c). 5 (d). 5 (e). 6 (b).

Therefore, the exponents at the singularity are l1 = l 2 = 3. The recurrence relation is an = - an -2 / (l + n - 3) 2 , n = 2, 3,K . For l1 = 3 , y = a0 [ t 3 - (1 / 4 ) t 5 + (1 / 64 ) t 7 + L] . Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the series found in part (d) converges for 0 < t. Substituting the series into the differential equation, we obtain •

[4l (l - 1) + 8l + 1]an tl + Â {[4(l + n) 2 + 4(l + n) + 1]an - 2an -1} t n +l = 0 . Therefore, n =1

6 (c). 6 (d). 6 (e). 7 (a).

1 F(l ) = 4 l2 + 4 l + 1 fi l1 = l 2 = - . 2 2 an -1 , n = 1, 2,... The recurrence relation is an = (2(l + n) + 1) 2 1 For l1 = - , y = a0 [ t -1/ 2 + (1 / 2) t1/ 2 + (1 / 8) t 3 / 2 + L] . 2 The series converges for 0 < t. When put in standard form, the differential equation is y ¢¢ - 2 t -1 y ¢ + t -2 (2 + t) y = 0 . Therefore, t = 0 is a regular singular point. •

7 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

(l2 - 3l + 2) a0 tl + Â {[(l + n ) 2 - 3(l + n ) + 2]an + an -1} t n +l = 0 . n =1

Therefore, the exponents at the singularity are l1 = 1 and l 2 = 2 . 7 (c). The recurrence relation is an = - an -1 /[(l + n - 1)(l + n - 2)], n = 1, 2,K . 7 (d). For l 2 = 2 , y = a0 [ t 2 - (1 / 2) t 3 + (1 / 12) t 4 + L] .

296 • Chapter 10 Series Solutions of Linear Differential Equations

7 (e). Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the series found in part (d) converges for 0 < t. 8 (b). Substituting the series into the differential equation, we obtain •

[l (l - 1) + 4l ]a0 tl + [l (l + 1) + 4(l + 1)]a1tl +1 + Â {[(l + n + 1)(l + n + 4)]an +1 - 2an -1} t n +l = 0 n =1

8 (c). 8 (d). 8 (e). 9 (a).

Therefore, F(l ) = l2 + 3l fi l1 = -3, l 2 = 0 . 2 an -1 , n = 1, 2,... and (l + 1)(l + 4 ) a1 = 0 The recurrence relation is an +1 = (l + n + 1)(l + n + 4 ) For l 2 = 0 , y = a0 [1 + (1 / 5) t 2 + (1 / 70) t 4 + L] . The series converges for 0 < t. When put in standard form, the differential equation is y ¢¢ + t -1 y ¢ - t -2 (1 + t 2 ) y = 0 . Therefore, t = 0 is a regular singular point. •

9 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

(l2 - 1) a0 tl + [(l + 1) 2 - 1]a1tl +1 + Â {[(l + n ) 2 - 1]an - an - 2 } t n +l = 0 . n =2

9 (c). 9 (d). 9 (e). 10 (b).

Therefore, the exponents at the singularity are l1 = -1 and l 2 = 1. The recurrence relation is an = an -2 /[(l + n ) 2 - 1] , n = 2, 3,K . For l 2 = 1, y = a0 [ t + (1 / 8) t 3 + (1 / 192) t 4 + L]. Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the series found in part (d) converges for 0 < t. Substituting the series into the differential equation, we obtain [l (l - 1) + 5l + 4]a0 tl + [l (l + 1) + 5(l + 1) + 4]a1tl +1 •

+ Â {[(l + n )(l + n + 4 ) + 4 ]an - an - 2 } t n +l = 0 . Therefore, F(l ) = l2 + 4 l + 4 fi l1 = l 2 = -2 . n =2

an -2 , n = 2, 3,... and (l + 1)(l + 5) a1 = 0 (l + n + 2) 2 10 (d). For l = -2 , y = a0 [ t -2 + (1 / 4 ) + (1 / 64 ) t 2 + L]. 10 (e). The series converges for 0 < t. 11 (a). When put in standard form, the differential equation is y ¢¢ + t -1 y ¢ - t -2 (16 + t) y = 0 . Therefore, t = 0 is a regular singular point.

10 (c). The recurrence relation is an =

•

11 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

(l2 - 16) a0 tl + Â {[(l + n ) 2 - 16]an - an -1} t n +l = 0 . n =1

Therefore, the exponents at the singularity are l1 = -4 and l 2 = 4 . 11 (c). The recurrence relation is an = an -1 /[(l + n ) 2 - 16] , n = 1, 2,K . 11 (d). For l 2 = 4 , y = a0 [ t 4 + (1 / 9) t 5 + (1 / 180) t 6 + L]. 11 (e). Note that tp( t) and t 2q( t) are analytic everywhere. Thus, see equations (18)-(21), the series found in part (d) converges for 0 < t.

Chapter 10 Series Solutions of Linear Differential Equations • 297

12 (b). Substituting the series into the differential equation, we obtain

[8l

2

]

•

- 2l - 1 a0 t + Â {[8(l + n ) 2 - 2(l + n ) - 1]an + an -1} t n +l = 0 . Therefore, l

n =1

1 1 F(l ) = 8l2 - 2l - 1 fi l1 = - , l 2 = . 4 2 12 (c). The recurrence relation is an =

- an -1 , n = 1, 2,... ( 4 (l + n ) + 1)(2(l + n ) - 1)

1 12 (d). For l1 = - , y = a0 [ t -1/ 4 - (1 / 2) t 3 / 4 + (1 / 40) t 7 / 4 + L] . 4 1 For l 2 = , y = a0 [ t1/ 2 - (1 / 14 ) t 3 / 2 + (1 / 616) t 5 / 2 + L] .. 2 12 (e). The series converges for 0 < t. 13 (a). When put in standard form, the differential equation is y ¢¢ - t -1 ( t 2 + 1) -1 (1 + t) y ¢ + t -1 ( t 2 + 1) -1 y = 0 . Therefore, t = 0 is a regular singular point and all other points are ordinary points. •

13 (b). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

•

 (l + n - 1)(l + n - 2)an -1 t n +l +

 (l + n + 1)(l + n - 1)a

n =1

n =-1

n +1

t n +l

•

- Â (l + n - 1) an t n +l = 0 n =0

14 (a).

14 (b).

15 (a). 16 (a).

16 (b).

Therefore, indicial equation is l2 - 2l = 0 . The exponents at the singularity are l1 = 0 and l2 = 2 . sin 3t Æ 3 as t Æ 0 and t 2q( t) = cos t Æ 1 as t Æ 0 . Thus, t = 0 is a regular singular tp( t) = t point. Ê ˆ Ê t2 ˆ ( 3t) 3 + ...˜ y ¢ + Á1 - + ...˜ y = 0 . t 2 y ¢¢ + Á 3t 3! Ë ¯ Ë 2! ¯ 2 Therefore, indicial equation (l + 1) = 0 fi l1 = l 2 = -1. When put in standard form, the differential equation is y ¢¢ - ( t 2 - 4 ) -2 y ¢ + ( t 2 - 4 ) -2 y = 0 . Therefore, t = 2 and t = -2 are irregular singular points. All other points are ordinary points. 1 1 2 tp( t) = 1 Æ 1 as t Æ 0 and t q( t) = 1 Æ -1 as t Æ 0 . Thus, t = 0 is a regular (1 - t) 3 (1 - t) 3 singular point. Neither ( t - 1) p( t) nor ( t - 1) 2 q( t) are analytical at t = 1, so t = 1 is an irregular singular point. 1 1 1 1 Ê 1 ˆ (1 - t) 3 = 1 - t - t 2 + ... fi t 2 Á1 - t - t 2 + ...˜ y ¢¢ + ty ¢ - y = 0 . Ë 3 9 ¯ 3 9 2 Therefore, indicial equation l - 1 = 0 fi l1 = -1, l 2 = 1.

298 • Chapter 10 Series Solutions of Linear Differential Equations •

17 (a). We need to substitute the series y = Â an ( t - 1) n +l into the differential equation. Before doing n =0 •

so, let us make the change of variable t = t - 1. We now substitute the series y = Â ant n +l into n =0

the transformed equation, -t (t + 2) y ¢¢ - 2(t + 1) y ¢ + a (a + 1) y = 0 , obtaining •

-2l2 a0t l -1 + Â {[-(l + n ) 2 - (l + n ) + a (a + 1)] an - 2(l + n + 1) 2 an +1t l + n = 0 . n =0

Thus, the exponents at the singularity are l1 = l 2 = 0 . 17 (b). For l = 0 , the recurrence relation is an +1 = [- n 2 - n + a (a + 1)]an /[2( n + 1) 2 ] . a (a + 1)[-2 + a (a + 1)] È a (a + 1) ˘ Thus, y ( t) = a0 Í1 + ( t - 1) + ( t - 1) 2 + L˙ . 16 2 Î ˚ 17 (c). When a = 1, y ( t) = a0 t . 18 (a). (1 - t) 2 = -( t - 1)( t + 1) = -( t - 1)(( t - 1) + 2), t = ( t - 1) + 1. Let t = t - 1. We now substitute the series into the transformed equation, -t (t + 2) y ¢¢ - (t + 1) y ¢ + a 2 y = 0 , obtaining •

-[2l (l - 1) + l ]a0t l -1 + Â {-[2(l + n + 1)(l + n ) + (l + n + 1)]an +1 + [-(l + n ) 2 + a 2 ] an }t l + n . n =0

1 Thus, F(l ) = 2l2 - l = 0 and the exponents at the singularity are l1 = 0 and l 2 = . 2 2 2 - n + a an 18 (b). For l1 = 0 , the recurrence relation is an +1 = . ( n + 1)(2 n + 1) È ˘ a 2 (a 2 - 1) 2 y ( t ) = a 1 + a ( t 1 ) + ( t - 1) 2 + L˙ . and 0Í 6 Î ˚

[

]

[

]

-( n + 1 / 2) 2 + a 2 an 1 For l 2 = , the recurrence relation is an +1 = . ( n + 3 / 2)(2 n + 2) 2 1 3 5 È ˘ (a 2 - 14 ) (a 2 - 14 )(a 2 - 49 ) 2 2 ( t - 1) + ( t - 1) 2 + L˙, t - 1 > 0 . and y ( t) = a0 Í( t - 1) + 3 30 Î ˚ -(n + l ) 2 + a 2 an +1 1 = = lim n Æ• ( n + l + 1)(2 n + 2l + 1) n Æ• a 2 n 1 fi convergence for t < 1 or t - 1 < 2 \ R = 2 . 2 1 1 1 18 (d). When a = , one solution (with l = ) reduces to y ( t) = a0 ( t - 1) 2 . 2 2

18 (c). By the Ratio Test, lim

•

19 (a). Substituting the series y = Â an t n +l into the differential equation, we obtain n =0 •

l2 a0 tl -1 + Â {(l + n + 1) 2 an +1 - (l + n - a ) an } t n +l = 0 . n =0

19 (b). The recurrence relation is an +1 = ( n - a ) an / ( n + 1) 2 . For a = 5 , the solution is y ( t) = a0 [1 - 5 t + 5 t 2 - (5 / 3) t 3 + (5 / 24 ) t 4 - (1 / 120) t 5 ] . 19 (c). y ( t) is neither an even nor an odd function. Theorem 10.2 does not apply. 20. The indicial equation is l (l - 1) + al + b = l2 + (a - 1)l + b = 0 . Since l1 = 1, l 2 = 2 , then l2 + (a - 1)l + b = (l - 1)(l - 2) = l2 - 3l + 2 fi a = -2, b = 2 .

Chapter 10 Series Solutions of Linear Differential Equations • 299

21. 22.

The indicial equation is l2 + (a - 1)l + b = 0 . In order to have l1 = 1 + 2i and l 2 = 1 - 2i , we need (l - l1 )(l - l 2 ) = l2 - (l1 + l 2 )l + l1l 2 = l2 - 2l + 5 . Therefore, a = -1 and b = 5 . The indicial equation is l (l - 1) + al + 2 = 0 has l = 2 as a root. Therefore, 2(1) + 2a + 2 = 0 fi a = -2 . Therefore, •

•

t 2 y ¢¢ - 2 ty ¢ + (2 + bt) y =  {(l + n )(l + n - 1) - 2(l + n ) + 2}an t n +l + b  an -1 t n +l = 0 n =0

n =1 •

fi [l (l - 1) - 2l + 2]a0 tl + Â {[(l + n ) 2 - 3(l + n ) + 2]an + ban -1}t n +l = 0 . n =1

[

]

For l = 2 , the recurrence relation becomes ( n + 2) 2 - 3( n + 2) + 2 an + ban -1 = 0, n = 1, 2,...

[

]

2

2

Therefore, n + 4 n + 4 - 3n - 6 + 2 an + ban -1 = ( n + n ) an + ban -1 = 0 fi b = -4 . 23. The indicial equation is l2 = 0 and the corresponding recurrence relation is ( n + 1) 2 an +1 + anan + ban -1 = 0 . Therefore, a = -1 and b = 3. 24 (a). p( t) is odd and q( t) is even, so we expect even and odd solutions. 24 (b). The indicial equation is l (l - 1) + l - u 2 = 0 or F (l ) = l2 - u 2 fi l1 = -u, l 2 = u . For the Bessel equation, l (l - 1) + l - u 2 = 0 or F (l ) = l2 - u 2 . The indicial equation and exponents at the singularity are the same for both equations. • 2

2

l

2

2

24 (c). [l - u ]a0 t + [(l + 1) - u ]a1t

l -1

+ Â {[(l + n ) 2 - u 2 ]an - an - 2 } t n +l = 0 n =2

an - 2 fi an = , n = 2, 3,... (l + n ) 2 - u 2 - an -2 For Bessel’s equation, an = , n = 2, 3,... The minus sign creates a “term-to-term” (l + n ) 2 - u 2 change of sign in the series solution. This sign alteration is not present in the series solutions of the modified Bessel equation.