Answers, .: to selected exercises for Introductory DE with BVP 3ed

Answers, Hints, and Solutions to Selected Exercises for

Introductory Differential Equations with Boundary Value Problems Third Edition

Martha L. Abell James P. Braselton

AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier

Answers, Hints, and Solutions to Selected Exercises for

Introductory Differential Equations with Boundary Value Problems Third Edition

Martha L. Abell James P. Braselton

AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Academic Press is an imprint of Elsevier

Academic Press is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 Theobald’s Road, London WC1X 8RR, UK Copyright © 2010 by Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions. This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein).

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CHAPTER 0 CHAPTER 1

Contents

CHAPTER 1

Introduction to Differential Equations

1

CHAPTER 2

First-Order Equations

8

CHAPTER 3

Applications of First-Order Differential Equations

46

CHAPTER 4

Higher Order Equations

68

CHAPTER 5

Applications of Higher-Order Differential Equations

103

CHAPTER 6

Systems of Differential Equations

125

CHAPTER 7

Applications of Systems of Ordinary Differential Equations

155

CHAPTER 8

Introduction to the Laplace Transform

171

CHAPTER 9

Eigenvalue Problems and Fourier Series

189

CHAPTER 10 Partial Differential Equations

201

iii

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CHAPTER 1

Introduction to Differential Equations

Exercises 1.1 1. Second-order linear ordinary differential equation. The forcing function is f (x) = x 3 so the equation is nonhomogeneous. 3. Second-order linear partial differential equation 5. This is a first-order ordinary differential equation. It is nonlinear because the deriviatve dy/dx is squared. 7. Second-order linear partial differential equation 9. Second-order nonlinear ordinary differential equation 11. This is a second-order partial differential equation. It is nonlinear because of the product, ∂u uux = u , of functions involving the dependent variable, u = u(x, t). ∂x 13. This is a first-order ordinary differential equation. If we write it as (2t − y)dt/dy − 1 = 0, y is independent, t = t(y) is dependent, and the equation is nonlinear. If we write the equation as dy/dt + y = 2t, t is independent, y = y(t) is dependent, and the equation is linear. 15. This is a first-order ordinary differential equation. It is nonlinear in both x (because of the 2x dx term) and y (because of the −y dy term). 31. Differentiate and collect dy and dx terms: 3x2 dx + 2xy dx + x 2 dy = 0 3x dx + 2y dx + x dy = 0 x dy = (−2y − 3x)dx dy/dx = −2y/x − 3. If x = 1, y = 99.

1

2

CH A P T E R 1:

Introduction to Differential Equations

  1 35. − cos x 2 + C 2 37. Use a u-substitution with u = ln x ⇒ du = 1/x dx. Then,  y=

1 dx = x ln x



1 du = ln|u| + C = ln(ln x) + C. u

39. Use integration by parts with u = x ⇒ du = dx and dv = e−x dx ⇒ v = −e−x .  y=

−x

xe

−x

dx = −xe

 +

41. tan−1 (x) − ln(x + 1) + C x 1 √   43. 6 sin−1 − x 4 − x 2 x 2 − 10 2 4 45. y (x) = 2 e−2 x 47. y (x) =

5 −3 x 7e

+ 27 e4 x

49. y (x) = 13 sin(3 x) 51. y (x) = − 34 − 12 x + 34 e2 x 53. y (t) = −t 7 + t 6 55. y (x) = x 4 − 12 x 2 + 2 x + 1   57. y (x) = −sin x −1 + 2 67. y (x) = c1 x + c2 x 2 69. y (x) = (ex + C)e−2 x 79.

y 2

1

⫺2

⫺1

1

⫺1

⫺2

2

x

e−x dx = −xe−x − e−x + C.

Answers, Hints, and Solutions to Selected Exercises

1 32 (12x − 8 sin 2x + sin 4x)

80. y = y

4

3

2

1

2

4

6

8

10

12

x

1 −x/2 81. y = − 208 e (74ex cos 2x − 74 cos 3x − 111ex sin 2x − 20 sin x)

y 5

1

2

3

4

0.5

1.0

5

⫺5

⫺10

82.

y 1.0

0.5

⫺1.0

⫺0.5 ⫺0.5

⫺1.0

x

6

x

3

4

CH A P T E R 1:

Introduction to Differential Equations

    83. x = 49 e−6t e9t + 8 , y = 89 e−6t e9t − 1 x, y

y

15

15

10

10

5

5

0.2

0.4

0.6

0.8

1.0

t

3

4

5

6

7

8

y 1.0

0.5

⫺1.0

⫺0.5

0.5

1.0

x

⫺0.5

⫺1.0

Exercises 1.2 1. No

2. Yes y

⫺1.0

y

1.0

1.0

0.5

0.5

⫺0.5

0.5

1.0

x

⫺1.0

⫺0.5

0.5

⫺0.5

⫺0.5

⫺1.0

⫺1.0

1.0

x

9

x

Answers, Hints, and Solutions to Selected Exercises

3. Yes

4. No y

⫺1.0

y

1.0

2

0.5

1

⫺0.5

0.5

1.0

x

⫺1

⫺1.0

⫺2

8.

y

2

1

1

⫺1

1

2

x

⫺4

⫺2

⫺1

⫺1

⫺1

⫺2

⫺2

10.

y

6

4

4

2

2 2

4

6

t

2

1

2

x

x

y

6

⫺2

1

y

2

9.

⫺6

⫺1

⫺0.5

7.

⫺2

⫺2

⫺6

⫺4

⫺2

2

⫺2

⫺2

⫺4

⫺4

⫺6

⫺6

4

6

x

5

6

CH A P T E R 1:

Introduction to Differential Equations

11.

12.

y

⫺4

4

4

2

2

⫺2

2

x

4

⫺4

⫺2

2

⫺2

⫺2

⫺4

⫺4

13.

14.

y

⫺1.0

y

y

1.0

1.0

0.5

0.5

⫺0.5

0.5

1.0

x

x

4

⫺1.0

⫺0.5

0.5

⫺0.5

⫺0.5

⫺1.0

⫺1.0

1.0

15. x  = y so y = x  = −4x: {x  = y, y = −4x} 17. {x  = y, y = −13x − 4y} 19. {x  = y, y = −16x + sin t} 21. y = C + 21. (b)

√

π 2

erf (x), where erf (x) =

√2 π

x 0

e−t dt. 2

22. (b)

y

y

2

10

1

5 1

2

3

4

5

x

5

⫺1

⫺5

⫺2

⫺10

10

15

20

25

x

x

Answers, Hints, and Solutions to Selected Exercises

23. (a)

23. (b)

y

⫺1.0

25. (a)

y

1.0

1.0

0.5

0.5

⫺0.5

0.5

1.0

x

⫺1.0

0.5

⫺0.5

⫺0.5

⫺1.0

⫺1.0

25. (b)

y

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0.4

0.6

0.8

1.0

x

0.2

Chapter 1 Review Exercises 1. First-order ordinary linear homogeneous differential equation 3. Second-order linear homogeneous differential equation 5. Second-order nonlinear partial differential equation 17. y = (2 − x 2 ) cos x + 2x sin x  √   √ 19. y = 12 x x 2 − 1 + ln 2x + 2 x 2 − 1 21. y = 14 (x + 4 cos 2x) 23. y = 13 (1 − cos3 x)

1.0

y

1.0

0.2

⫺0.5

0.4

0.6

0.8

1.0

x

x

7

CHAPTER 2

First-Order Equations

Exercises 2.1 1. (a) yes; (b) no; (c) no 3. f (x, y) = y1/5 , so fy (t, y) = 15 y−4/5 is not continuous at (0, 0) and uniqueness is not  5/4 ,y=0 guaranteed. Solutions: y = 45 t  √  √ 2 y, y ≥ 0 y−1/2 , y ≥ 0 5. f (t, y) = 2 |y| = √ , so fy (t, y) = is not continuous at 2 −y, y < 0 −(−y)1/2 , y < 0 (0, 0). Therefore, the hypotheses of the Existence and Uniqueness Theorem are not satisfied. √ dy 1 3/2 7. Yes. = y t ⇒ dy = t 1/2 dt ⇒ ln|y| = 23 t 3/2 + C1 ⇒ y = Ce2t /3 . Application dt y   of the initial conditions yields y = exp 23 (t 3/2 − 1) . 9. Yes. f (t, y) = sin y − cos t and fy (t, y) = cos y are continuous on a region containing (π, 0). 11. y = sec t so y = sec t tan t = y tan t and y(0) = 1. fy (t, y) = t is continuous on −π/2 < t < π/2 and f (t, y) = sec t is continuous on −π/2 < t < π/2 so the largest interval on which the solution is valid is −π/2 < t < π/2.  13. f (t, y) = y2 − 1 and fy (t, y) = 12 ( y2 − 1)−1/2 ; unique solution guaranteed for (a) only. 15. (0, ∞). The solution is y = 14 t −1 (t 4 − 1). 17. (0, ∞), because y = ln t has domain t > 0. 19. (−∞, 1), because y = 1/(t − 1) has domain (−∞, 1) ∪ (1, ∞) and y = 1/(t − 3) has domain (−∞, 3) ∪ (3, ∞). 21. (−2, 2) 23. t > 0, y = t −1 sin t − cos t, −∞ < t < ∞ 8

Answers, Hints, and Solutions to Selected Exercises

21.

23.

y

y

4

6 4

3

2 2 ⫺6

⫺4

t

⫺2

2

4

6

⫺2

1

⫺4 ⫺2

t

⫺1

0

1

⫺6

2

24.

25.

y 2

y

1

0.35 0.30 0.25

⫺2

0.20

t

⫺1

1

0.15 ⫺1

0.10 0.05 ⫺1.5

⫺1.0

⫺0.5

0.5

1.0

1.5

t

⫺2

25. First, we solve the equation (see next section): dy = y2 dt 1 dy = dt y2 −

1 = t+C y y=

−1 . t+C

Applying the initial condition indicates that −1/C = a ⇒ C = −1/a so y = a/(1 − at). This solution is defined for t > 1/a or t < 1/a.

2

9

10

CH A P T E R 2:

First-Order Equations

26.

27.

y

⫺2

y

2

2

1

1

⫺1

1

2

t

⫺2

⫺1

1

⫺1

⫺1

⫺2

⫺2

2

1 2 1 2 27. Separating √ variables (see next section) gives us y dy = −t dt ⇒ 2 y = 2 t √+ C ⇒ y = ± C − t 2 . Applying the initial conditions indicates that C = a2 so y = a2 − t 2 (because y(0) = a is positive). Thus, the interval of definition of the solution is |t| < a.

28.

29. (a)

y

29. (b)

y

y

3

3

2

2

1

1

2 1

21

1 21 22

2

3

4

t

1

2

3

4

5

6

t

1

21

21

22

22

23

23

2

3

4

5

6

t

29. General solution: 12 sin2 t − sin y cos t + 12 sin2 y = C; solution is not unique if x0 = π/2 or 3π/2.

Exercises 2.2 1. Separate variables and integrate: 6 2 x dx 7 1 4 2 y = x3 + C 4 7 8 4 y = x3 + C 7  1/4 8 3 y= x +C . 7

y3 dy =

3. Separate variables and integrate: y−7 dy = 3x −8 dx 1 3 − y−6 = − x −7 + C, 6 7

t

Answers, Hints, and Solutions to Selected Exercises

−7 = C. so y−6 − 18 7 x 9 −7 y =C 5. 3t + 12 t 4 + 52 y − 14

7. sinh 3x − 12 cosh 4y = C 9. y−3/2 − x 3 − 4x 3/2 = C   11. y = sin−1 − 34 cos x + C 13. Separate variables and integrate: dy = −ky dt 1 dy = −k dt y ln|y| = −kt + C y = Ce−kt . In the calculation above, remember that e−kt+C = e−kt eC . C is arbitrary so eC is positive and arbitrary.   1 1 15. y = cosh−1 − 120 sinh 6t − 16 cosh 4t + C 17. 10t − 73 e−3t − ey + 2y4 = C 19.

3 4

1 1 cos θ − 12 cos 3θ + 16 sin 4y − sin y = C

21. y sin 2x + 2xy + 2y2 + 5 + 4Cy = 0  1 1 23. − −64 C cos(2y (x)) −64 C + cos(8x − 2y) + cos(8x + 2y) + 64 cos(2y) + 1  2 cos(8x) + 4 cos(4x − 2y) + 4 cos(4x + 2y) + 8 cos(4x) − 128 = 0 √ 25. − 14 x sin 2x + 14 x 2 − 18 cos 2x − 2 sin y = C  1 1 58 + 6 sin(y) − 2 cos(4x) + sin(y + 4x) − sin(−y + 4x) + 27. 64 −1 + sin(y)  8 cos(2x) − 4 sin(y + 2x) + 4 sin(−y + 2x) − 64 C + 64 C sin(y) = 0 29.

2 3/2 + 1 e3/y 3 (ln x) 3

=C

31. Factor first, then separate, use partial fractions, and simplify: dy/dt = (t 2 + 1)(y2 − 1) 1 dy = (t 2 + 1)dt y2 − 1   1 1 1 − dy = (t 2 + 1)dt 2 y−1 y+1   1  y − 1  1 2 ln = t +t+C 2 y + 1 3

11

12

CH A P T E R 2:

First-Order Equations

2 2 y−1 = Ce 3 t +2t y+1

y= 33. y =

2 2 +2t

1 + Ce 3 t

2 2 +2t

1 − Ce 3 t

.

13 + 12C − 4x(3C + 1) 4C(x − 1) − 3

35.

1 y3 + 1



dy = dt

 1 2−y + dy = dt 3( y + 1) 3( y2 − y + 1)     √  ( y + 1)1/3  2y − 1 =C −3t + 3 tan−1 + 3 ln 2 √ ( y − y + 1)1/6  3 y

22

y

y

2

2

2

1

1

1

21

1

2

t

22

21

1

2

t

22

21

1

21

21

21

22

22

22

FIGURE 1 Direction fields for dy/dt = y3 + y2 , dy/dt = y3 − y2 , and dy/dt = y2 − y3 .

37.

1 

y3 + y

dy = dt

 1 y − 2 dy = dt y y +1

1 ln|y| − ln y2 + 1 = t + C 2 y = Cet  2 y +1 Ce2t 1 − Ce2t  Ce2t y=± 1 − Ce2t

y2 =

2

t

Answers, Hints, and Solutions to Selected Exercises

39.

 1 1 1 − + dy = dt 2( y − 1) y 2( y + 1) 1 1 ln|y − 1| − ln|y| + ln|y + 1| = t + C 2 2 y2 − 1 = Cet y 1 y2 = 1 − Ce2t 1 y=± 1 − Ce2t



41. dy = x 3 dx ⇒ y = 14 x 4 + C. y(0) =

1 4

· 04 + C = 0 ⇒ C = 0 so y(x) =

1 4

x4 .

y 4 3 2 1

22

21

1

x

2

43. Integrating gives us x = sin y + C and applying the initial condition gives us C = 2 so x( y) = sin( y) + 2. x 3.0 2.5 2.0 1.5

26

24

45. y(t) =

2 3

y

22

2

4

6

√

9 + 3 t 3/2 √ 47. y(t) = −1 − −1 + 2 et √ √ 2 x,

49. y(x) = e

√ √ 2 x

y(x) = e−

51. y(x) = arctan(x) + 1 1 53. The solution for (a) is y = esin t , for (b) it is y = , and for (c) it is 1 − sin t   1 4 + r sin t + sin2 t . y= 4

13

14

CH A P T E R 2:

First-Order Equations

y 15

12.5

10

7.5

5

2.5

27.5

25

22.5

2.5

5

7.5

t

22.5

55. y = 2x(x − 2)−1 57. (a) y = −x + 4 − tan(C − x); (b) y = ±

1 [−3(x + C)2 ]1/3 9 x+C

67. Use partial fractions. 1 dt = dt 12 + 4y − y2   1 1 − dy = 8 dt y+2 y−6   y + 2   = 8t + C ln y − 6 y+2 = Ce8t y−6 y=2

3Ce8t + 1 . Ce8t − 1

  √ 71. − 12 x 9 − x 2 + 92 arcsin 13 x − 12 ey(x) cos(y(x)) − 12 ey(x) sin(y (x)) + 12 = 0

Exercises 2.3 1. Use undetermined coefficients. A general solution of the corresponding homogeneous equation y − y = 0 is yh = Cet . The forcing function is f (t) = 10. The associated set of functions for a constant is F = {1}. Because no multiple of 1 is a solution to the corresponding homogeneous equation, we assume that a particular solution takes the form yp = A · 1 = A. Differentiating yp = 0 and substituting into the nonhomogeneous

Answers, Hints, and Solutions to Selected Exercises

equation gives us yp − yp = −A = 10 so A = −10 and yp = −10. Therefore, a general solution of the nonhomogeneous equation is y = yh + yp = −10 + Cet . 3. A general solution of the corresponding homogeneous equation is yh = Cet . The associated set of functions for the forcing function f (t) = −cos t − sin t is F = {cos t, sin t}. Because no element of F is a solution to the corresponding homogeneous equation, we search for a particular solution of the form yp = A cos t + B sin t. Differentiating, yp = B cos t − A sin t and substituting into the nonhomogeneous equation yields yp − yp = B cos t − A sin t − A cos t − B sin t = (−A + B) cos t + (−A − B) sin t = −cos t − sin t. So −A + B = −1 and −A − B = −1, which has solution A = 1 and B = 0. Therefore, yp = cos t and y = yh + yp = cos t + Cet . 5. Use undetermined coefficients. First, a general solution of the corresponding homogeneous equation is yh = Cet . The associated set of functions of the forcing function is F = {et }. Because et is a solution of the corresponding homogeneous equation, multiply F by t, tF = {tet }. Because tet is not a solution of the corresponding homogeneous equation we search for a particular solution of the nonhomogeneous equation of the form yp = Atet with derivative yp = Atet + Aet . Substituting yp into the nonhomogeneous equation yields Atet + Aet − Atet = Aet so A = 1 and yp = tet . Therefore, a general solution of the nonhomogeneous equation is y = yh + yp = Cet + tet . 7. y = 12 t + Ct −1 dy 1 1 + y = e−x to see that p(x) = 1/x. Then, dx x x = x. Multiplying through by the integrating factor

9. First write the equation in standard form

an integrating factor is e and integrating gives us

1/x dx

x

= x ln x

dy + y = e−x dx d (xy) = e−x dx xy = −e−x + C y = Cx −1 − x −1 e−x .

√ 11. y = 4t 2 + 1 + C 4t 2 + 1

13. An integrating factor is μ(t) = e cot t dt = e−ln csc t = sin t. Multiplying through by the integrating factor, integrating the result, and solving for y results in sin t

dy + y cos t = sin t cos t dt d ( y sin t) = sin t cos t dt

15

16

CH A P T E R 2:

First-Order Equations

1 2 sin t + C 2 1 y = sin t + C csc t, 2

y sin t =

which is equivalent to y = −12 cos t cot t + C csc t because sin t cos t dt = −12 cos2 t +C when choosing u = cos t rather than u = sin t when calculating the integral. 15. An integrating factor is   μ(t) = exp −

4t dt 2 4t − 9



  1 1 2 = exp − ln(4t − 9) = √ . 2 4t 2 − 9

Multiplying through by the integrating factor, integrating the result, and solving for y results in 1 dy 4t 1 −√ y= √ 2 4t 2 − 9 dt 4t 2 − 9 4t − 9   d 1 y = √ dt 4t 2 − 9 1 y= √ 4t 2 − 9

√ √

t 4t 2 − 9 t

4t 2 − 9  1 4t 2 − 9 + C 4  1 y = (4t 2 − 9) + C 4t 2 − 9. 4

17. An integrating factor is μ(t) = e2 cot x dx = e−2 ln csc x = sin2 x. Multiplying the equation by the integrating factor, integrating, and solving for y yields sin2 x

19. θ = −1 + Cer

dy + 2y sin2 x cot x = sin2 x cos x dx d (sin2 xy) = sin2 x cos x dx 1 sin2 xy = sin3 x + C 3 1 y = sin x + C csc2 x. 3

2 /2

21. x( y) = −1 − y + Cey 23. x(t) = C/(t 3 − 1) 25. v(s) = se−s + Ce−s 27. Use undetermined coefficients. A general solution of the equation is yh = Ce−t . The associated set of functions for the forcing function f (t) = e−t is F = {e−t }. Because e−t is a solution to the corresponding homogeneous equation, multiply F by t n where n is the smallest integer so that no element of t n F is a solution to the corresponding homogeneous equation. In this case, tF = {te−t } so we assume

Answers, Hints, and Solutions to Selected Exercises

that a particular solution of the nonhomogeneous equation has the form yp = Ate−t . Differentiating yp , yp = −Ate−t + Ae−t , and substituting into the nonhomogeneous equation yields yp + yp = −Ate−t + Ae−t + Ate−t = Ae−t = e−t so A = 1 and yp = te−t . Therefore a general solution of the nonhomogeneous equation is y = yh + yp = Ce−t + te−t . Application of the initial condition yields y = e−t (t − 1). y

1

2

3

4

5

t

20.2 20.4 20.6 20.8 21

2

29. An integrating factor is μ(t) = e 2t dt = et . Multiplying the equation by the integrating factor, integrating, and solving for y yields et

2

dy 2 2 + 2tet y = 2tet dt d  t2  2 e y = 2tet dt 2

2

et y = et + C y = 1 + Ce−t . 2

y 1

Applying the initial condition yields y = 1 − 2e−t . 2

0.5

1

2

3

4

5

t

20.5

21

31. A general solution is y = 2t −1 (t − 1)et + Ct −1 . Applying the initial conditions yields y = (2tet − 2et − 1)/t.

17

18

CH A P T E R 2:

First-Order Equations

y 4 2

2

4

6

8

10

t

22 24

y 4 2

33. y =

2

t 2 − 16 t2 + 4

4

6

8

10

t

22 24

␪ 10 8 6

35. y = te2t

4 2

2

4

6

8

10

t

43. y + y = t has solution y = t − 1 + Ce−t . y(0) = 1 ⇒ C = 2 so y = t − 1 + 2e−t for −1  0 ≤ t < 1. When t = 1, y =  1 − 1 + 2e = 2/e. The solution to y + y = 0, y(1) = 2/e −t t − 1 + 2e , 0 ≤ t < 1 is y = 2e−t . Thus, y(t) = . 2e−t , t ≥ 1  e−2t , 0 ≤ t < 1 45. y(t) = e2−4t , t ≥ 1

Answers, Hints, and Solutions to Selected Exercises

47. y(t) = − 25 cos(2t) − 15 sin(2t) + Cet 49. y(t) = (t + C)e−t 1 51. y(t) = − 25 − 15 t + Ce5 t 1

53. y(t) = −2 cos(t) + 4 sin(t) + 4 et + Ce 2 t 55. y(t) =

2 t 11 e

+ Ce−10 t

57. y(t) = (2t + C)et 59. y(t) = cos(t) + sin(t) + t − 1 + Ce−t 63. y(t) = t − 1 + Ce−t , y(t) = −12 cos(t) + 12 sin(t) + Ce−t , y(t) = 12 cos(t) + 12 sin(t) + Ce−t , y(t) = 12 et + Ce−t 65. y(t) = t − 1 − e−t , y(t) = t − 1, y(t) = t − 1 + e−t , y(t) = t − 1 + 2 e−t , y(t) = t − 1 + 3 e−t

Exercises 2.4 1. My (t, y) = 2y − 12 t −1/2 = Nt (t, y), exact 3. My (t, y) = cos ty − ty sin ty = Nt (t, y), exact 5. My (t, y) = 6ty = 0 = Nt (t, y), not exact 7. My (t, y) = sin 2t = 2 sin 2t = Nt (t, y), not exact 9. My (t, y) = y−1 = Nt (t, y), exact 11. y = C + t 3 13. y = 0, ty2 = C ∂ ∂ (2t + y3 ) = 3y2 = (3ty2 + 4). Let ∂y ∂t F(t, y) have total derivative (2t + y3 )dt + (3ty2 + 4)dy. Then, F(t, y) = (2t + y3 )dt = t 2 + ty3 + g( y). Differentiating F with respect to y, Fy (t, y) = 3ty2 + g ( y) = 3ty2 + 4 ⇒ g ( y) = 4 so g( y) = 4y and F(t, y) = t 2 + ty3 + 4y. A general solution is then t 2 + ty3 + 4y = C or t 2 + ty3 + 4 y(t) = C. ∂ ∂ 17. The equation is exact because (2ty) = 2t = (t 2 + y2 ). Let F(t, y) satisfy Ft (t, y)dt + ∂y ∂t Fy (t, y)dy = 2ty dt + (t 2 + y2 )dy. Then, F(t, y) = 2ty dt = t 2 y + g ( y) = t 2 + y2 so g ( y) = y2 ⇒ g( y) = 13 y3 . Therefore F(t, y) = t 2 + 13 y3 and a general solution of the equation is t 2 + 13 y3 = C. Observe that solving this as a homogeneous equation of degree  next section) results in the following form of the solution:  2 (see the 2 + y2 y 3 t − ln(t) = C. − 13 ln t3

15. Observe that the equation is exact because

∂ ∂ (sin2 y) = 2 sin y cos y = sin 2y = (t sin 2y). ∂y ∂t Let F(t, y) satisfy Ft (t, y)dt + Fy (t, y)dy = sin2 y dt + t sin 2y dy. Then, F(t, y) =

19. The equation is exact because

19

20

CH A P T E R 2:

First-Order Equations

17.

19.

21. y

y

y

6

6

4

4

2

2

4 2

24

22

2

t

4

26

24

22 24

22

2

4

6

t

26

24

22

2

22

22

24

24

26

26

4

6

sin2 y dt = t sin2 y + g( y) so g ( y) = 0 ⇒ g( y) = 0 and F(t, y) = t sin2 y. A general solution is then t sin2 y = C or ln t + 2 ln sin y = C. ∂ t ∂ 21. The equation is exact because (e sin y) = et cos y = (1 + et cos y). Let F(t, y) ∂y ∂t

satisfy Ft (t, y)dt + Fy (t, y)dy = et sin y dt + (1 + et cos y) dy. Then, F(t, y) = et sin y dt = et sin y + g( y) so g ( y) = 1 ⇒ g( y) = t. Thus, F(t, y) = et sin y + t and a general solution of the equation is et sin y + y = C. √ 23. y = 0, y = C sec t 2 + tan t 2 ∂ ∂ (1 + y2 cos ty) = 2y cos ty − ty2 sin ty = (sin ty + ∂y ∂t ty cos ty). Let F(t, y) satisfy Ft (t, y)dt + Fy (t, y)dy = (1 + y2 cos ty) dt + (sin ty +

ty cos ty) dy. Then, F(t, y) = (1 + y2 cos ty) dt = t + y sin ty + g( y) so Fy (t, y) = sin ty + ty cos yt + g ( y) so g ( y) = 0 and g( y) = 0. Then F(t, y) = t + y sin(yt) and a general solution of the equation is t + y sin yt = C.

25. The equation is exact because

27. The equation is exact because

∂ ((3 + t) cos(t + y) + sin(t + y)) = cos(t + y) − (3 + t) ∂y

∂ ((3 + t) cos(t + y)). Let F(t, y) satisfy Ft (t, y)dt + Fy (t, y)dy = ((3 + t) ∂t

cos(t + y) + sin(t + y)) dt + (3 + t) cos(t + y) dy. Then, F(t, y) = ((3 + t) cos(t + y) + sin(t + y)) dt = (3 + t) sin(t + y) + g( y) so Fy (t, y) = (3 + t) cos(t + y) + g ( y) = (3 + t) cos(t + y) ⇒ g ( y) = 0 ⇒ g( y) = 0 so F(t, y) = 3 sin(t + y) + t sin(t + y) and 3 sin(t + y) + t sin(t + y) = C. sin(t + y) =

∂ −2 2 y/t −t y e + 1 = −t −3 yey/t (2t + y) = ∂y



∂ y/t e (1 + y/t) . Let F(t, y) satisfy Ft (t, y)dt + Fy (t, y)dy = −t −2 y2 ey/t + 1 dt + ∂t

ey/t (1 + y/t) dy. Then, F(t, y) = −t −2 y2 ey/t + 1 dt = t + yey/t + g( y). Next, Fy (t, y) = ey/t (1 + y/t) + g ( y) = ey/t (1 + y/t) so g ( y) = 0 ⇒ g( y) = 0. Thus, F(t, y) = yey/t + t and yey/t + t = C.

29. The

equation

is

exact

because

t

Answers, Hints, and Solutions to Selected Exercises

25.

27.

y

26

24

29.

y

y

6

6

6

4

4

4

2

2

2

22

2

4

6

t

26

24

22

2

4

6

t

26

24

22

2

22

22

22

24

24

24

26

26

26

∂ 2

∂ 2

2ty = 4ty = 2t y . Let F(t, y) satisfy ∂y ∂t 2 2 Ft (t, y) dt + Fy (t, y) dy = 2ty dt + 2t y dy. Then, F(t, y) = 2ty2 dt = t 2 y2 + g( y) so Fy (t, y) = 2t 2 y + g ( y) = 2t 2 y, which means that g ( y) = 0 ⇒ g( y) = 0. Therefore F(t, y) = t 2 y2 and a general solution (or, the integral curves) of the differential equation are t 2 y2 = C. Application of the initial condition results in y2 t 2 − 1 = 0.

31. This equation is exact because

Observe that dividing the differential equation by 2ty yields y dt + t dy = 0, which is dy dy 1 + y = 0. This is a first-order linear homogeneous equaequivalent to t + y = 0 or dt dt t tion with integrating factor μ(t) = e 1/t dt = t. Multiplying through by the integrating factor, integrating, and solving for y gives us

t

dy +y = 0 dt d (ty) = 0 dt ty = C y = Ct −1 .

Observe that squaring both sides of the equation and solving for C gives us the same result as that obtained by solving the equation as an exact equation.



∂ ∂ 2 33. The equation is exact because 2ty + 3t 2 = 2t = t − 1 . Let F(t, y)

∂y

∂t satisfy Ft (t, y) dt + Fy (t, y) dy = 2ty + 3t 2 dt + t 2 − 1 dy. Then, F(t, y) =



2ty + 3t 2 dt = t 2 y + t 3 + g( y) so Fy (t, y) = t 2 + g ( y) = t 2 − 1 ⇒ g ( y) = −1 ⇒ g( y) = −y. Therefore, F(t, y) = t 2 y + t 3 − y and the integral curves are t 2 y + t 3 − y = C. Applying the initial condition and solving for y gives us −t 3 − 1 y= . (t − 1) (t + 1)

4

6

t

21

22

CH A P T E R 2:

First-Order Equations

31.

32.

y

24

33.

y

y

4

4

4

2

2

2

22

2

t

4

24

22

2

t

4

24

22

2

22

22

22

24

24

24

t

4

∂ y ∂ y 2

(e − 2ty) = ey − 2t = te − t . Let F(t, y) ∂y ∂t

satisfy Ft (t, y) dt + Fy (t, y) dy = (ey − 2ty) dt + tey − t 2 dy. Then, F(t, y) =

y (e − 2ty) dt = tey − t 2 y + g( y). Differentiating with respect to y, Fy (t, y) = y te − t 2 + g ( y) = tey − t 2 ⇒ g ( y) = 0 ⇒ g( y) = 0 so F(t, y) = tey − t 2 y and the integral curves are tey − t 2 y = C. Applying the initial condition results in tey − t 2 y = 0.

∂ 2 ∂ 37. The equation is exact because y − 2 sin 2t = 2y = (1 + 2ty). Let F(t, y) ∂t ∂y

satisfy Ft (t, y) dt + F(t, y) dy = y2 − 2 sin 2t dt + (1 + 2ty) dy. Then, F(t, y) =

2

y − 2 sin 2t dt = ty2 + cos 2t + g( y). Differentiating with respect to y, Fy (t, y) = 2ty + g ( y) = 1 + 2ty ⇒ g ( y) = 1 ⇒ g( y) = y so F(t, y) = ty2 + cos 2t + y and the integral curves are ty2 + cos 2t + y = C. Applying the initial condition results in ty2 + cos (2t) + y − 2 = 0. 35. The equation is exact because

35.

36.

y

⫺4

37.

y

y

4

4

4

2

2

2

⫺2

2

4

t

⫺4

⫺2

2

4

t

⫺4

⫺2

2

⫺2

⫺2

⫺2

⫺4

⫺4

⫺4

4

  ∂ 1 ∂ 2 39. The equation is exact because − y = −2y = (−2ty). Let F(t, y) 2 ∂y  1 + t ∂t  1 2 satisfy Ft (t, y) dt + Fy (t, y) dy = − y dt − 2ty dy. Then, F(t, y) = 1 + t2

t

Answers, Hints, and Solutions to Selected Exercises

 1 2 − y dt = tan−1 t − ty2 + g( y). Differentiating with respect to y, Fy (t, y) = 1 + t2 −2ty + g ( y) = −2ty ⇒ g ( y) = 0 ⇒ g( y) = 0 so F(t, y) = tan−1 t − ty2 and the integral curves are tan−1 t − ty2 = C. Applying the initial condition results in y2 − arctan(t) = 0. t



38.

39.

y 6

6

4

4

2

2

ä ⫺6

⫺2

⫺4

41.

2

4

6

t

⫺6

⫺2

⫺4

2

⫺2

⫺2

⫺4

⫺4

⫺6

⫺6

42.

y

43.

y 6

6

4

4

4

2

2

2

2

4

6

t

⫺6 ⫺4 ⫺2

2

4

6

t

⫺6 ⫺4 ⫺2

2

⫺2

⫺2

⫺2

⫺4

⫺4

⫺4

⫺6

⫺6

⫺6

C 49. y =

2/3 t e + t2 −t + C =0 t2

53. yt 5 + t 4 y2 + 14 t 4 = C 55. cos (y (t)) t 2 + sin (y) t = C 57. t 2 + y cos ty = C

4

6

t

y

6

⫺6 ⫺4 ⫺2

51. y−1 −

y

4

6

t

23

24

CH A P T E R 2:

First-Order Equations

 √ 1 1 10C 2 t 2 + 10 ; 59. (a) (i) y = −t, t 2 + 2ty + y2 = C; (ii) y = −Ct ± 10 C √ 1  19 1 (iii) y = −39C 2 t 2 + 40 − 20 Ct ± 20 C (i)

(ii)

(iii)

y

⫺1.0

y

y

1.0

1.0

1.0

0.5

0.5

0.5

⫺0.5

0.5

1.0

t

⫺1.0

⫺0.5

0.5

1.0

t

⫺1.0

⫺0.5

0.5

⫺0.5

⫺0.5

⫺0.5

⫺1.0

⫺1.0

⫺1.0

1.0

t

Exercises 2.5 1. This is Bernoulli with n = −1. Let w = y1−(−1) = y2 ⇒ 1 −1 dw y . Then, 2 dt

dw dy dy = 2y ⇒ = dt dt dt

dy 1 − y = ty−1 dt 2 1 −1 dw 1 y − y = ty−1 2 dt 2 dw − y2 = 2t dt dw − w = 2t. dt Use undetermined coefficients to solve for w. A general solution of the corresponding homogeneous equation is wh = Cet . The associated set of functions for the forcing function f (t) = 2t is F = {t, 1}. Because no element of F is a solution to the corresponding homogeneous equation we assume that a particular solution has the form wp = At + B ⇒ wp = A. Substituting wp into the nonhomogeneous equation yields wp − wp = −At + (A − B) = 2t so A = −2 and B = −2 so wp = −2t − 2. A general √ solution is then w = wh + wp = Cet − 2t − 2. Because w = y2 , y = ± Cet − 2t − 2. 3. This is Bernoulli with n = 3 so we let w = y1−3 = y−2 . Then, 1 dw dy − y3 = . Substituting into the equation gives us 2 dt dt dw − y = 2ty3 cos t −ty3 dt dw 1 −2 + y = −2 cos t dt t

dw dy = −2y−3 so dt dt

Answers, Hints, and Solutions to Selected Exercises

dw 1 + w = −2 cos t dt t d (tw) = −2t cos t dt tw = −2 cos t − 2t sin t + C w = −2t −1 cos t − 2 sin t + Ct −1 1 = −2t −1 cos t − 2 sin t + Ct −1 y2 1 y = ±√ −1 −2t cos t − 2 sin t + Ct −1 √ −(2 cos t + 2t sin t − C)t or y = ± . 2 cos t + 2t sin t − C 9 3 5. y3/2 + 20 cos t − 20 sin t − Ce3t = 0

7. Homogeneous of degree 1 9. Not homogenous 11. Homogeneous of degree 0 13. Not homogeneous 15. Not homogeneous 17. The equation is homogeneous of degree 1. Let t = vy. Then, dt = vdy + ydv. Substituting into the equation, separating, and integrating yields 2tdt + (y − 3t)dy = 0 2vy(vdy + ydv) + (y − 3vy)dy = 0 2v(vdy + ydv) + (1 − 3v)dy = 0 (2v2 − 3v + 1)dy = −2vydv 1 −2v dy = 2 dv y 2v − 3v + 1   1 1 1 dy = 2 − dv y 2v − 1 v − 1 ln y = ln(2v − 1) − 2 ln(1 − v) + C 2v − 1 (v − 1)2 (2t − y)y y=C (t − y)2 y=C

(t − y)2 = C. 2t − y 

−y + t Another form of the solution is −2 ln − t





y − 2t + ln t

 − ln(t) = C.

25

26

CH A P T E R 2:

First-Order Equations

19. The equation is homogeneous of degree 2. Observe that either t = vy or y = ut results in an equivalent problem. We choose to use y = ut ⇒ dy = udt + tdu. Then, (ty − y2 )dt + t(t − 3y)dy = 0 (t 2 u − t 2 u2 )dt + t(t − 3ut)(udt + tdu) = 0 (u − u2 )dt + (1 − 3u)(udt + tdu) = 0 2u(1 − 2u)dt + t(1 − 3u)du = 0 2u(1 − 2u)dt = t(3u − 1)du 1 3u − 1 dt = du t 2u(1 − 2u)   1 1 1 1 dt = − + du t 2 u 2u − 1 1 1 ln t = − ln u − ln(2u − 1) + C 2 4 −4 ln t = 2 ln u + ln(2u − 1) + C 1 = Cu2 (2u − 1) t4 y2 (t − 2y) 1 = C t4 t3 ty2 (t − 2y) = C.   1 −t + 2y 1 y − ln − ln(t) = C. Another form of the solution is − ln 4 t 2 t 21. y3 − (3 ln(t) + C) t 3 = 0 23. This is homogeneous of degree 1. (Also, observe that this is a first-order linear equation in y.) Solving it as a homogeneous equation, we let y = ut ⇒ dy = udt + tdu. Then, (t − y)dt + tdy = 0 (t − ut)dt + t(udt + tdu) = 0 (1 − u)dt + udt + tdu = 0 dt = −tdu 1 dt = −du t ln t = −u + C y ln t = − + C t y = t(C − ln t).

Answers, Hints, and Solutions to Selected Exercises

    3y − 2t 1 −t + 2y 2 + ln − ln(t) = C 25. − ln 3 t 2 t 27. The equation is homogeneous of degree 2. Let t = vy ⇒ dt = vdy + ydv. Then, y2 dt = (ty − 4t 2 )dy y2 (vdy + ydv) = (vy2 − 4v2 y2 )dy vdy + ydv = (v − 4v2 )dy ydv = −4v2 dy 4 1 − dy = 2 dv y v 1 −4 ln y = − + C v 4 ln y =

y + C. t

y 1 y − ln − ln(t) = C. 4 t t   √   2  1 t − ty + y2 1 1 (t − 2y) 3 + √ arctan − ln(t) = C 29. y = −t, − ln 2 3 t t2 3 3 Another form of the solution is

31.



−1 1 (−y + t) y−1 e−t/y − ln(y) = C 2

33. y + 2y = t 2 y1/2 is Bernoulli with n = 1/2. Let w = y1−(1/2) = y1/2 . Then, dw 1 −1/2 dy dy y ⇒ = 2y1/2 . Substituting into the equation yields 2 dt dt dt

dw = dt

dy + 2y = t 2 y1/2 dt 2y1/2

dw + 2y = t 2 y1/2 dt

dw 1 + y1/2 = t 2 dt 2 dw 1 + w = t2. dt 2 Using undetermined coefficients, a general solution is w = Ce−t + 12 t 2 − t + 1. Thus, y = (Ce−t + 12 t 2 − t + 1)2 . Applying the initial condition results in

27

28

CH A P T E R 2:

First-Order Equations

y 5

4

3

2

y = 14 (4 − 8t + 8t 2 − 4t 3 + t 4 ).

1

1

35. y =

1 2

√

2

3

4

5

t

2 + 2t 2 t

37. y = − 12 + 12 t 2 √ 39. y = 3 −3 ln(t) + 27t 35.

37.

y 15 10 5 ⫺4

⫺2

⫺5 ⫺10 ⫺15

39.

y

y

12 250

10

2

4

t

⫺4

⫺2

8

200

6

150

4

100

2

50 2

4

t

1

2

3

4

5

t

41. y4 dt + (t 4 − ty3 )dy = 0 is homogeneous of degree 4. Let t = vy ⇒ dt = vdy + ydv. Then, y4 dt + (t 4 − ty3 )dy = 0 y4 (vdy + ydv) + (v4 y4 − vy4 )dy = 0 (vdy + ydv) + (v4 − v)dy = 0 v4 dy = −ydv 1 1 dy = − 4 dv y v 1 ln y = 3 + C 3v y3 3 ln y = 3 + C. t

Answers, Hints, and Solutions to Selected Exercises

y 1 y3 8 − ln − ln t − + ln 2 = 0. 3 3t t 3 √ √ 43. We need to solve dy/dt = y/t + t/y subject to y( e) = e. The equation is homogeneous of degree 2. To see that this is so, we rewrite the equation: Applying the initial condition results in

y2 + t 2 dy = dt yt yt dy = (y2 + t 2 )dt. Now, we let y = ut ⇒ dy = udt + tdu. Substituting then gives us ut 2 (u dt + t du) = (u2 t 2 + t 2 ) dt u(u dt + t du) = (u2 + 1) dt tu du = dt 1 dt = u2 du t 1 ln t = u3 + C 3 1 y3 + C. ln t = 3 t3 √ √ Now apply the initial condition and solve for y, y = 2 t ln t. 49. Because M(t, y)dt + N(t, y)dy = 0 is homogeneous, we can write the equation in the form dy/dt = F(y, t). If t = r cos θ and y = r sin θ, dt = cos θ dy − r sin θ dθ and dy = sin θ dr + r cos θ dθ. Substituting into the equation gives us dy/dt = F(t, y)   sin θ dr + r cos θ dθ r sin θ =F = F(tan θ) cos θ dr − r sin θ dθ r cos θ sin θ dr + r cos θ dθ = F(tan θ) cos θ dr − F(tan θ)r sin θ dθ F(tan θ)r sin θ dθ + r cos θ dθ = F(tan θ) cos θ dr − sin θ dr F(tan θ) sin θ dθ + cos θ 1 = dr. F(tan θ) cos θ − sin θ r 53. f (t) = t − 1; g(t) = t 2 − t; general solution: (tc − y) − 1 = c2 − c ⇒ y = ct − 1 +  d d   2 c − c2 ; singular solution: (ty − y − 1) = (y ) − y ⇒ ty + y = 2y y − y ⇒ dt dt (t − 2y + 1)y = 0 ⇒ y = 12 (t + 1) ⇒ y = 14 t 2 + 12 t − 34 55. f (t) = 1 − 2t; g(t) = t −2 ; general solution: 1 − 2(tc − y) = c−2 ⇒ y = 12 (c−2 + 2ct − 1); singular solution: y = 12 (3t 2/3 − 1) 59. We see that the equation is a Lagrange equation by rewriting it in the form y = ty2 + (3y2 − 2y3 ) and identifying f (y ) = y2 and g( y) = 3y2 − 2y3 . Differentiating with

29

30

CH A P T E R 2:

First-Order Equations

respect to t yields the equation y = y2 + 2ty y + 6y y − 6y2 y and substituting p = y results in dp dp dp + 6p − 6p2 dt dt dt   dp p − p2 = 2xp + 6p − 6p2 dx p = p2 + 2tp

dx 2xp + 6p − 6p2 = dp p − p2 dx 2p 6p − p2 . + 2 x= dp p − p p − p2 The solution of this linear equation is x =

y = xp2 + 3p2 − 2p3 =

2p3 − 21p2 + 36p + 6C

so 6 p2 − 2p + 1

2p3 − 21p2 + 36p + 6C 2

p + 3p2 − 2p3 . 6 p2 − 2p + 1

61. Differentiating the equation gives us y = (2 − y ) + xy + 4y y . Now, we let p = y and solve for dx/dp: y = (2 − y ) + xy + 4y y dp dp + 4p dx dx dp 2p − 2 = (x + 4p) dx dx x + 4p = dp 2(p − 1) p = 2−p+x

dx 1 4p + x= . dp 2 − 2p 2(p − 1) √ This linear equation has solution x = − 83 + 43 p + 43 p2 + C 2 − 2p so 

  8 4 4 2 y = x(2 − o) + 2p + 1 = − + p + p + C 2 − 2p (2 − p) + 2p2 + 1. 3 3 3 2

63. limt→∞ y(t) =

ky0 k = ay0 a

65. General solution: 2t 2 ln |t| + t 2 = Ct 2 . Initial value problem has two solutions: √  2 y = ± 2 t (ln 4 − ln t).

Answers, Hints, and Solutions to Selected Exercises

64.

65.

y

4

4

2

2

ä ⫺4

y

⫺2

2

4

t

⫺4

t

⫺2

2

⫺2

⫺2

⫺4

⫺4

Exercises 2.6 1. 47.3742, 63.2572 3. 1.8857, 2.09847 5. 79.8458, 123.048 7. 1.95109, 1.95388 9. 83.6491, 88.6035 11. 2.37754, 2.41897 13. 185.34, 206.981 15. 1.95547, 1.95609 17. 90.6405, 90.6927 19. 216.582, 216.992 23. 1.95629, 1.95629 25-27. (a) y(t) = e−t , y(1) = 1/e ≈ 0.367879

25. (Euler’s) 26. (Improved) 27. (4th-Order RK)

h =0.1

h =0.05

h =0.025

0.348678 0.368541 0.429069

0.358486 0.368039 0.414831

0.363232 0.367918 0.36788

29. y(0.5) ≈ 0.566144, 1.12971, 1.68832, 2.23992, 2.78297

4

31

32

CH A P T E R 2:

First-Order Equations

We state the following algorithms that can be used to generate numerical solutions of some initial value problems of the form  y = f (x, y) . y(x0 ) = y0 The following three algorithms implement Euler’s method, the improved Euler’s method, and the Runge-Kutta method of order four. For all three algorithms, first define f (x, y), x0 , y0 , and the desired step size, h. Mathematica Clear[f,h,xe,ye,x0,y0,xr,yr, xi,yi,k1,k2,k3,k4] f[x_,y_]=?; h=?; x0=?; y0=?;

Maple V f:=’f’:h:=’h’:xe:=’xe’:ye:=’ye’: x0:=’x0’:xr:=’xr’:yr:=’yr’: xi:=’xi’:yi:=’yi’: f:=(x,y)->?: h:=?: x0:=?: y0:=?:

The following implements Euler’s method. xe[n_]=x0+n h; xe:=n->x0+n*h: ye[n_]:=ye[n]=h*f[xe[n-1],ye[n-1]]+ye[n-1]; ye:=proc(n) option remember; ye[0]=y0; ye(n-1)+h*f(xe(n-1),ye(n-1)) end: ye(0):=y0:

The following implements the improved Euler’s method. xi[n_]=x0+n h; yi[n_]:=yi[n]=h(f[xi[n-1],yi[n-1]]+ f[xi[n],h*f[xi[n-1],yi[n-1]]+ yi[n-1]])/2+yi[n-1]; yi[0]=y0;

xi:=n->x0+n*h: yi:=proc(n) option remember; yi(n-1)+h/2* (f(xi(n-1),yi(n-1))+f(xi(n-1), yi(n-1)+h*f(xi(n-1),yi(n-1)))) end: yi(0):=y0:

The following implements the fourth-order Runge-Kutta method. xr:=n->x0+n*h: xr[n_]=x0+n h; yr:=proc(n) yr[n_]:=yr[n]=yr[n-1]+h/6(k1[n-1]+ local k1,k2,k3,k4; 2k2[n-1]+2k3[n-1]+k4[n-1]); option remember; yr[0]=y0; k1:=f(xr(n-1),yr(n-1)); k1[n_]:=k1[n]=f[xr[n],yr[n]]; k2:=f(xr(n-1)+h/2, k2[n_]:=k2[n]=f[xr[n]+h/2,yr[n]+h k1[n]/2]; yr(n-1)+h*k1/2); k3[n_]:=k3[n]=f[xr[n]+h/2,yr[n]+h k2[n]/2]; k3:=f(xr(n-1)+h/2, k4[n_]:=k4[n]=f[xr[n+1],yr[n]+h k3[n]] yr(n-1)+h*k2/2); k4:=f(xr(n),yr(n-1)+h*k3); yr(n-1)+h/6*(k1+2*k2+2*k3+k4) end: yr(0):=y0:

Answers, Hints, and Solutions to Selected Exercises

3, 11, and 19. For h = 0.1:

n 0 1 2 3 4 5 6 7 8 9 10

xn

yn (Euler’s method)

0 1 0.1 1 0.2 1.01 0.3 1.03201 0.4 1.06851 0.5 1.12269 0.6 1.19873 0.7 1.30242 0.8 1.44206 0.9 1.63001 1 1.8857

yn (improved Euler’s method)

yn (Runge-Kutta method of order 4)

1 1.005 1.02211 1.05412 1.10465 1.17872 1.28354 1.43008 1.63607 1.93233 2.37754

1 1.00535 1.02298 1.05572 1.10733 1.18299 1.29026 1.44075 1.65355 1.9626 2.43501

For h = 0.05:

n

xn

yn (Euler’s method)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

1 1 1.0025 1.00775 1.01603 1.02764 1.04295 1.06233 1.08626 1.11526 1.14995 1.19107 1.2395 1.29632 1.36284 1.44071 1.53199 1.63934 1.76621 1.91719 2.09847

yn (improved Euler’s method)

yn (Runge-Kutta method of order 4)

1 1.00125 1.00526 1.01231 1.02275 1.03693 1.05529 1.07832 1.10661 1.14084 1.18185 1.23064 1.28845 1.35687 1.43787 1.53404 1.6488 1.78675 1.9543 2.16055 2.41897

1 1.00129 1.00535 1.01247 1.02298 1.03725 1.05572 1.07889 1.10733 1.14176 1.183 1.23208 1.29027 1.35916 1.44077 1.53774 1.65357 1.79301 1.96266 2.17199 2.43514

33

34

CH A P T E R 2:

First-Order Equations

7,15, and 23. For h = 0.1:

n 0 1 2 3 4 5 6 7 8 9 10

xn

yn (Euler’s method)

0 1 0.1 1.08415 0.2 1.17254 0.3 1.26471 0.4 1.36006 0.5 1.45785 0.6 1.55721 0.7 1.6572 0.8 1.75683 0.9 1.85511 1 1.95109

yn (improved Euler’s method)

yn (Runge-Kutta method of order 4)

1 1.08627 1.17664 1.27055 1.36727 1.46596 1.56568 1.66546 1.76429 1.86125 1.95547

1 1.08636 1.17682 1.27082 1.36763 1.4664 1.56621 1.66607 1.76498 1.86201 1.95629

For h = 0.05:

n

xn

yn (Euler’s method)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

1 1.04207 1.08525 1.12947 1.17468 1.2208 1.26777 1.3155 1.36387 1.41281 1.46219 1.51189 1.5618 1.6118 1.66176 1.71155 1.76106 1.81016 1.85873 1.90667 1.95388

yn (improved Euler’s method)

yn (Runge-Kutta method of order 4)

1 1.04262 1.08633 1.13107 1.17677 1.22336 1.27075 1.31884 1.36753 1.41672 1.46629 1.51612 1.56608 1.61605 1.66591 1.71554 1.76481 1.8136 1.86182 1.90934 1.95609

1 1.04263 1.08636 1.13111 1.17682 1.22342 1.27082 1.31892 1.36763 1.41683 1.4664 1.51624 1.56621 1.6162 1.66607 1.7157 1.76498 1.81379 1.86201 1.90954 1.95629

Answers, Hints, and Solutions to Selected Exercises

25–27. (a) y(t) = e−t , y(1) = 1/e ≈ 0.367879 h = 0.1

h = 0.05

h = 0.025

25. (Euler’s) y(1) ≈ 0.348678 y(1) ≈ 0.358486 y(1) ≈ 0.363232 26. (Improved Euler’s) y(1) ≈ 0.368541 y(1) ≈ 0.368039 y(1) ≈ 0.367918 27. (4th-Order Runge-Kutta) y(1) ≈ 0.36788 y(1) ≈ 0.414831 y(1) ≈ 0.429069

25.

Exp[–1]//H 0.367879 Remove[f, x, y] f[x_, y_] = –y; h = .1; x0 = 0; y0 = 1; xe [n_] = x0 + n h; ye [n_] := ye [n] = h f[xe [n – 1], ye [n – 1] + ye [n – 1]; ye [0] = y0 ; ye [10] 0.348678 Remove[f, x, y] f[x_, y_] = –y; h = .05; x0 = 0; y0 = 1; xe [n_] = x0 + n h; ye [n_] := ye [n] = h f[xe [n – 1], ye [n – 1]] + ye [n – 1]; . ye [0] = y0 ; ye [20] 0.358486

26. Remove[f, x, y] f[x_, y_] = –y; h = .1; x0 = 0; y0 = 1; xi [n_] = x0 + n h; yi [n_] := yi [n] = 1 h(f[xi [n – 1], yi [n – 1]] + f[xi[n], 2

35

36

CH A P T E R 2:

First-Order Equations

h f[xi [n − 1], yi [n − 1]] + yi [n – 1]]) + yi [n – 1]; yi [0] = y0 ; yi [10] 0.368541 Remove[f, x, y] f[x_, y_] = –y; h = .05; x0 = 0; y0 = 1; xi [n_] = x0 + n h; yi [n_] := yi [n] = 1 h(f[xi [n − 1], yi [n − 1]] + f[xi [n], ) 2 h f[xi [n − 1], yi [n – 1]] + yi [n – 1]]) + yi [n – 1]; yi [0] = y0 ; yi [20] 0.368039

28.

Clear[x, y] partsol = HDSolve[{y’[x] == Sin[2 x - y[x]], Y[0] == 0.5}, y[x], {x, 0, 15}]; Plot[y[x] /. partsol, {x, 0, 15}]

5 4 3 2 1

2

29.

numericalsols = Map[HDSolve[ {y’[x] == Sin[x y[x]], y[0] == #}, y[x], {x, 0, 7}][[1, 1, 2]]&, {0.5, 1.0, 1.5, 2.0, 2.5}]; Plot[Evaluate[numericalsols], {x, 0, 7}]

4

6

8

10

12

14

3 2.5 2 1.5 1

numericalsols /. X -> 0.5 {0.566144, 1.12971, 1.68832, 2.23992, 2.78297}

0.5 1

2

3

4

5

6

7

Answers, Hints, and Solutions to Selected Exercises

30.

«“Graphics‘PlotField’” dirfield = PlotVectorField[{1, x2 + y2 }, {x, –1, 1}, {y, –1, 1}, Axes → Automatic, AxesOrigin → {0, 0}, HeadWidth → 0, HeadLength → 0, ScaleFunction →(1&), PlotPoints → 30]

1

0.5

⫺1

⫺0.5

0.5

1

0.5

1

⫺0.5

⫺1

1

numsol = HDSolve[{y’[x] == x2 + y[x]2 , y[0] == 0}, y[x], {x, −1, 1}]; Plot[y[x]/-numsol, {x, −1, 1}, PlotRange → {−1, 1}, AspectRatio → 1] numsol/-X → 1 {{y[1] → 0.350247}}

0.75 0.5 0.25

⫺1

⫺0.5 ⫺0.25 ⫺0.5 ⫺0.75 ⫺1

Chapter 2 Review Exercises √ 3 25 t 6 + 125C   3. y = sinh−1 12 cosh(x) + 12 C

1. y =

1 5

5. y4 dy = e5t dt ⇒ 15 y5 = 15 e5t + C ⇒ y5 = e5t + C → y = (e5t + C)1/5 √ −2t 7. y(t) = e± −e +2 C 1 1 y 6 y 9. − 20 cos(10 t) + 18 cos(4 t) − 37 e cos (6 y) − 37 e sin (6 y) = C

37

38

CH A P T E R 2:

First-Order Equations

11. The equation (y − t) dt + (t + y) dy = 0 is homogeneous of degree 1. Either y = ut or t = vy result in an equivalent problem. We choose to use y = ut ⇒ dy = udt + tdu. Then, (y − t) dt + (t + y)dy = 0 (u2 + 2u − 1)dt = −t(u + 1)du 1 u+1 dt = − 2 du t u + 2u − 1  1   ln t = − lnu2 + 2u − 1 + C 2 t −2 = C(u2 + 2u − 1) t −2 = Ct −2 (y2 + 2ty − t 2 ) y2 + 2ty − t 2 = C. √

2C 2 t 2 + 1 √ 13. y = 12 , y = ± −t + C     5 −x + 2t 5 x + 2t 1 x 15. − ln − − ln + ln − ln(t) = C 8 t 8 t 4 t y = −t ± C1

17. 19.

1 3 3 t y − cos t − sin y 1 2 2 t ln y + y = C

21. y = 1 + Ce−t

=C

2 /2

23. r = t −1 (cos t + t sin t + C) 25. y = t −1 (6et − 6tet + 3t 2 et + C)1/3 27. y = −2 + 2 ln(−2/t), y = Ct + 2 ln C  3/2  √ √ 1 29. y = − 13 t 6 t − 6 t 2 − 12 C + 54 6 t − 6 t 2 − 12 C , y=

1 3

 3/2  √ √ 1 t 6 t − 6 t 2 − 12 C − 54 6 t − 6 t 2 − 12 C ,

 3/2  √ √ 1 y = − 13 t 6 t + 6 t 2 − 12 C + 54 6 t + 6 t 2 − 12 C , y=

1 3

 3/2  √ √ 1 t 6 t + 6 t 2 − 12 C − 54 6 t + 6 t 2 − 12 C

31. y + sin(t − y) = π 33. t sin y − y sin t = 0 35. t ln y + y ln t = 0

Answers, Hints, and Solutions to Selected Exercises

37. n

xn

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 2

39. n

xn

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

yn yn (Improved (Euler’s) Euler’s)

yn (Runge-Kutta of order 4)

1 1 1.01118 1.02608 1.04368 1.06345 1.08505 1.10823 1.13281 1.15866 1.18565 1.21368 1.24268 1.27256 1.30328 1.33478 1.36699 1.3999 1.43344 1.46759 1.50232

1 1.00678 1.01921 1.03486 1.05296 1.07309 1.09494 1.11831 1.14303 1.16896 1.19601 1.22407 1.25307 1.28295 1.31364 1.34509 1.37726 1.4101 1.44357 1.47764 1.51229

1 1.00559 1.0181 1.03383 1.052 1.07219 1.0941 1.11752 1.14228 1.16825 1.19533 1.22343 1.25247 1.28237 1.31309 1.34456 1.37675 1.40961 1.44311 1.4772 1.51186

yn yn (Improved (Euler’s) Euler’s)

yn (Runge-Kutta of order 4)

1 1 1.0025 1.0075 1.01503 1.02511 1.03779 1.0531 1.07112 1.09189 1.11548 1.14194 1.17132 1.20364 1.23889 1.27702 1.31791 1.36139 1.40717 1.45488 1.50399

1 1.00125 1.00501 1.01129 1.02013 1.03157 1.04565 1.06245 1.08201 1.10441 1.12971 1.15795 1.18918 1.2234 1.26056 1.30058 1.34329 1.38842 1.43561 1.48438 1.5341

1 1.00125 1.00501 1.01129 1.02012 1.03156 1.04564 1.06242 1.08198 1.10437 1.12966 1.15789 1.1891 1.22329 1.26043 1.30042 1.34309 1.38818 1.43532 1.48403 1.53369

39

40

CH A P T E R 2:

First-Order Equations

Differential Equations at Work A. Modeling the Spread of a Disease 2. (b)

(c)

I

I

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0.0

0

2

4

6

8

10

t

4. (a)

0.0

2

4

6

8

10

2

4

6

8

10

t

(b) I

I

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0.0

0

0

2

4

6

8

10

t

0.0

0

t

B. Linear Population Model with Harvesting

1 (ay0 − h)eat + h ; (d) limt→∞ = ∞; a 1  a  (e) limt→∞ = h/a; (f ) limt→∞ = −∞; y(t) = 0 when t = − ln 1 − y0 a h 2. (a) y = 2; (b) y = (y0 − 2)et/2 + 2; (c) ∞; (d) 2; (f ) −∞; 2 ln 4 1. (a) ay − h = 0 has solution y = h/a; (c) y =

3. y = 12 (2 − et ) = 0 when t = ln 2 ≈ 0.693; y = 12 (4 − 3et/2 ) = 0 when t = 2 ln(4/3) ≈ 0.575 4. y = 12 (2 − et/2 ) = 0 when t = 2 ln 2 ≈ 1.386

√ 5. First solve y = 12 y − 12 , y(0) = 1/2 to obtain y = 1 − 12 et/2 . Then, y(1) = 1 − 12 e ≈ √ 0.176. Then, for year two, solve y = 12 y + 12 , y(1) = 1 − 12 e to obtain y = −1 + √ √ 2e(t−1)/2 − 12 et/2 so y(2) = −1 + 2 e − 12 e ≈ 0.938. y = 12 y + r, y(1) = 1 − 12 e has √ solution y = − 12 et/2 − 2r + (2r + 1)e(t−1)/2 so y(2) = − 12 e − 2r + (2r + 1) e = 12 = √ y(0) when r = 14 ( e − 1) ≈ 0.162. √ e − 12 e ≈ 0.290. y(T ) = e(t−1)/2 − 12 et/2 = 12 = 6. Set r = 0 in the above. Then, y(2)  =  2 ≈ 3.092. y(0) when T = −2 ln −1 + √ e

Answers, Hints, and Solutions to Selected Exercises

7. (a)

(b)

21.0

20.20

20.8

20.15

20.6 20.10 20.4 20.05

20.2 0.2

0.4

0.6

0.8

1.0

(c)

(d)

20.0

20

0.2

0.4

0.2

0.4

0.6

0.8

1.0

19 18

19.5

17 16

19.0

15 14 0.2

0.4

0.6

0.8

1.0

0.6

0.8

1.0

8. (a)

(b)

22.0

22.0

21.5

21.5

21.0

21.0

20.5

20.5

0.2

0.4

0.6

0.8

1.0

(c)

(d)

22.0

22.0

21.5

21.5

21.0

21.0

20.5

20.5

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

41

42

CH A P T E R 2:

First-Order Equations

C. Logistic Model with Harvesting

 √ 1  a ± a2 − 4ch 2c 2. (a) y = 2, y = 5 1. y =

(b) y 7 6 5 4 3 2 1

2

4

6

8

10

t

(c) 5; (d) 5; (e) 0 (extinction) 49 3. (a) 49 − 40h = 0 when h = 40 (c)

(b) y

(d)

y

y

7

7

7

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

2

4

6

8

10

t

2

4

6

8

10

t

2

4

6

8

10

4. No Harvesting

y

Harvesting

y

2.5

1 ⫺2

2.0

⫺4

1.5

⫺6 1.0

⫺8

0.5

⫺10 1

2

3

4

5

t

⫺12

2

3

4

5

t

t

Answers, Hints, and Solutions to Selected Exercises

5. No Harvesting

y

Harvesting

y

12

10

10

8

8 6 6 4 4 2

2 1

2

3

4

t

5

6.

1

2

3

4

5

t

7. y

y

14

14

12

12

10

10

8

8

6

6

4

4

2

2 2

4

6

8

10

12

t

8.

2

4

6

8

10

12

t

9. y

y

14

14

12

12

10

10

8

8

6

6

4

4

2

2 2

4

6

8

10

t

2

4

6

8

10

t

43

44

CH A P T E R 2:

First-Order Equations

D. Logistic Model with Predation 1. a 2. (a)

(b)

W 20

W 20

15

15

10

10

5

5

5

10

15

20

t

5

10

15

20

3. k > 19.3 will work k 519.3 5

10

k 519.4 15

20

5

21

21

22

22

23

23

24

24

25

25

k 519.5 5

10

10

15

20

15

20

k 519.6 15

20

5

21

21

22

22

23

23

24

24

25

25

10

t

Answers, Hints, and Solutions to Selected Exercises

4.

k ⫽ 19.1

k ⫽ 19.2

k ⫽ 19.3

W 20

W 20

W 20

15

15

15

10

10

10

5

5

5

5

10

15

20

t

5

k ⫽ 19.4

10

15

20

t

5

k ⫽ 19.5 W 20

W 20

15

15

15

10

10

10

5

5

5

10

15

20

t

5

k ⫽ 19.7

10

15

20

t

5

k ⫽ 19.8 W 20

W 20

15

15

15

10

10

10

5

5

5

10

15

20

t

6. (a)

5

10

15

20

t

(b)

(c)

W 20

W 20

W 20

15

15

15

10

10

10

5

5

5

5

10

15

20

t

5

10

20

15

20

t

10

t

k ⫽ 19.9

W 20

5

15

k ⫽ 19.6

W 20

5

10

15

20

t

5

10

15

20

5

10

15

20

t

t

45

CHAPTER 3

Applications of First-Order Differential Equations

Exercises 3.1 1. y(t) = 100ekt . With t = 3, y(3) = 100e3k = 200 so e3k = 2, which means that ek = 21/3 and y(t) = 2t/3 . y(30) = 210 = 1024, while y(t) = 2t/3 = 4250 when t = ln(4250)/ ln(2) ≈ 36.16. 3. y(t) = ekt and y(1) = ek = 2/3 so y(t) = (3/2)−t . y(t) = (3/2)−t = 1/3 when t = ln(3)/ ln(3/2) ≈ 2.71. 5. y(1000) = 100e1000k = 50 so ek = 2−1/1000 and y(t) = 100 · 2−t/1000 . y(1) = 100 · 2−1/1000 ≈ 99.93 and y(500) ≈ 70.71. 7. y(t) = 500ekt and y(6) = 500e6k = 600 so ek = (6/5)1/6 and y(t) = 500 · (6/5)t/6 . y(24) = 500 · (6/5)4 = 5184/5 ≈ 1036.8. 500 · (6/5)t/6 = 1000 when t = 6 ln(2)/ ln(6/5) ≈ 22.81. 9. y(5) = e5k = 1/2 so ek = 2−1/5 and y(t) = 2−t/5 . Then y(t) = 2−t/5 = 1/6 when t = 5 ln(6)/ ln(2) ≈ 12.92 and y(15) = 2−3 = 1/8. 11. Let H denote the half-life of the radioactive substance. Then, y(t) = eHk = 1/2 gives us t1 y(t) = (1/2)t/H . Thus, y(t1 ) = (1/2)t1 /H so ln y(t1 ) = − ln 2 and y(t2 ) = (1/2)t2 /H so H t2 ln y(t2 ) = − ln 2. Subtracting these two equations and solving for H gives the result. H 13. We set y(0) = 800. Then, y(5) = 800e5k = 560 ⇒ ek = (7/10)1/5 ⇒ y(t) = 800 · 5 ln(1/2) (7/10)t/5 . To find the half-life solve 800 · (7/10)t/5 = 400 for t: t = ≈ 9.72 ln(7/10) days. 15. The half-life of 226 Ra is approximately 1700 years. Then, y(1700) = e1700k = 1/2 ⇒ ek = (1/2)1/1700 so y(t) = (1/2)t/1700 , y(100) ≈ 0.96 and approximately 96% of the original amount remains.

46

Answers, Hints, and Solutions to Selected Exercises

17. We use the results of Example 2: y(t) = 2−t/5730 . Then, ttool ≈ 3561.13 years old, tfossil ≈ 4222.81 years old, tfossil − tt00 ≈ 661.68 years. No. 19. First, we rewrite the equation as dy/dt − ry = −ay2 . Now, we let w = y1−2 = y−1 . Then, dw/dt = −y−2 dy/dt so −y2 dw/dt = dy/dt. Then, −y2

dw − ry = −ay2 dt

dw + rw = a dt ert

dw + rert w = aert dt

(Divide by −y2 ) (Multiply by the integrating factor ert )

d  rt  e w = aert dt ert w =

a rt e +C r

w=

a + Ce−rt r

y=

1 a/r + Ce−rt

(Integrate) (Solve for w) (Solve for y).

Now apply the initial condition y(0) = y0 to obtain C = (a y0 + r)/(r y0 ). Thus, 1 r y0 y = a ay + r = . 0 a y + (a y0 + r)e−r t 0 + e−r t r r y0  −1 21. y = 100000 1 + 9e−t/100 ; y(25) ≈ 1.25 × 105 ; limt→∞ y(t) = 1,000,000.   −1  3 t ; y(2) ≈ 191; because there is no t so that y = 200, all 23. y = 200 1 + 199 199 students do not theoretically learn of the rumor. 25. y(t) = 1 + 499e−5t ; y(20) ≈ 1; quickly. 27. limm→∞ S0 (1 + k/m)mt = S0 ekt   −1  1 29. (a) −r 1 − y y = 0 ⇒ y = 0, A; (c) y(t) = 2 1 + ert , limt→∞ y(t) = 0; (d) A   −1 , limt→∞ y(t) = 0 y(t) = 6 3 − ert 31. (a) y = 2, semistable; y = 0, unstable; (b) y = 0, unstable; y = 1, unstable; (c) y = 0, semistable; y = 1, unstable; (d) y = 3, stable; y = 0, semistable, y = −3, unstable; (e) y = −1, unstable; y = 0, semistable; (f) y = −1, stable; y = 0, unstable; y = 1, stable.

47

48

CH A P T E R 3:

Applications of First-Order Differential Equations

y

y

3

y 2.0

2.0 1.5 1.0 0.5

2 1 1 2 3 4 5

21

t

1.5 1.0

20.5 21.0

1 2 3 4 5

(a)

0.5

t

0.0

(b)

y

y

y

4

2

1.0 0.5

2 1 2 3 4 5

22

t

24

20.5 21.0 21.5 22.0

t 1 2 3 4 5 (c)

1 2 3 4 5

1

t

21

1 2 3 4 5

t

22

(d)

(e)

(f)

y 2.0 1.5 1.0 0.5 20.5 21.0

1 2 3 4 5

t

(g)

−1  33. y(t) = 100 − 99e−t/100 , −1  y(t) = 2 − e−t/2 .

−1  y(t) = 20 − 19e−t/20 ,

−1  y(t) = 10 − 9e−t/10 ,

1.0

0.8

0.6

0.4

0.2

5

√ 35. (a) P(t) =

4ah − r 2 tan

10

15

20

  √ √ 1 2 2 +r 2 C 4ah − r − t 4ah − r

that P(t) = 0 when t ≈ 3006.

2a

. From the graph we see

Answers, Hints, and Solutions to Selected Exercises

P 300 250 200 150 100 50 500

1000

1500

2000

2500

3000

t

    √ √ 500 4h − 9 −1 With P(t) = 50 4h − 9. tan 0005 200 cos −√ − 1 × 106 h − 156191  √ 1. 4.h − 9.t + 150, h must be slightly smaller than h ≈ 0.1667. P 10 8 6 4 2 1

2

3

4

5

t

22 24

37. y = (r − ay)y ⇒ y = y(ay − r)(2ay − r). The zeros are 0, r/a, and r/(2a). Now make a sign chart to see when y is positive and negative to determine when y is concave up ( y positive) and concave down (y negative).   Ay0 1 y0 39. y(t) = . The denominator is 0 if t = ln . y0 + (A − y0 )ert r y0 − A

Exercises 3.2 1. T = 70ekt + 30 and T (15) = 70e15k + 30 = 80 so 70e15k = 50 and then ek = (5/7)1/15 . Then T (t) = 70 · (5/7)t/15 + 30. Then,  t/15 5 70 + 30 = 50 7  t/15 5 2 = 7 7 15 ln(2/7) ≈ 55.85 min. t= ln(5/7)

49

50

CH A P T E R 3:

Applications of First-Order Differential Equations

3. With T (t) = −35ek + 75, we find ek : T (5) = −35e5k + 75 = 50 −35e5k = −25

 1/5 5 5 so ek = 7 7  t/5 5 T (t) = −35 + 75. 7 e5k =

Next, we solve −35

 5 t/5 7

+ 75 = 60 resulting in t =

5 ln(3/7) ≈ 12.59 min. ln(5/7)

5. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts , observe that t = 0 corresponds to 3:00 p.m. With this convention, T0 = 79, T (3) = 68, and Ts = 60. This means that T (t) = (79 − 60)e−kt + 60 = 19(e−k )t = 19e−kt + 60. Because T (3) = 68, T (3) = 19e−3k + 60 = 68 19e−3k = 8  1/3 8 e−k = 19 

8 so T (t) = 19 19 12:30 p.m.

t/3



8 + 60. Solving T (t) = 19 19

t/3 + 60 = 98.6 yields t ≈ −2.45 hr,

7. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts , observe that T (t) = (90 − 70)e−kt + 70 = 20e−kt + 70. Because T (3) = 20e−3k + 70 = 80, 20e−3k = 10 ⇒ e−3k = 1/2 ⇒ e−k = (1/2)1/3 so T (t) = 20 (1/2)(t/3) + 70. Evaluating at t = 5, T (5) ≈ 76.3◦ F.  t/30 2 + 300; T (−30) = 75◦ F 9. T (t) = −150 3 11. Using Newton’s Law of Cooling, T (t) = (T0 − Ts )e−kt + Ts = (200 − 68)e−kt + 68. T (2) = 132e−2k + 68 = 170 ⇒ 132e−2k = 102 ⇒ e−2k = 17/22 ⇒ e−k = (17/22)1/2 . Thus, T (t) = 132 (17/22)t/2 + 68. Solving T (t) = 132 (17/22)t/2 + 68 = 140 results in t ≈ 4.7 min. 12. This problem is a great class experiment if time allows. Depending upon your assumptions, the problem can be complicated. Is your model confirmed by real data?   13. u(t) = −5(9 + π2 )−1 −8π2 − (2π2 + 27)e−t/4 + 3π sin(πt/12) + 9 cos(πt/12) − 72   15. u(t) = −5(9 + π2 )−1 −14π2 + π2 e−t/4 + 3π sin(πt/12) + 9 cos(πt/12) − 126

1 y, y(0) = 0 and using the integrating factor μ(t) = e 17. (a) Solving y = 8 − 100 2 t/100 , we find that y = 25 + Ce−t/100 . e

1/100 dt

=

Answers, Hints, and Solutions to Selected Exercises

51

(b) First, dV/dt = 4 − 2 = 2 gives us V = 2t + C. V0 = 500 so V (t) = 2t + 500. 1 Now solve y = 4 − y, y(0) = 20 using the integrating factor μ(t) = t + 250

2t 2 + 1000t + C e 1/(t+250)dt = eln(t+250) = t + 250 to obtain y = . Apply the initial t + 250 2t 2 + 1000t + 5000 condition to obtain . t + 250 3 (c) V = −2t + 600; solve y = 4 − y, y(0) = 100; y = −2(t − 300) + 300 − t 1 (t − 300)3 . If R1 = R2 , the volume remains constant, so V (t) = V0 . If 54000 R1 > R2 , V increases. If R1 < R2 , V decreases.

dy 3 19. Solve + y = 8, y(0) = 10. The integrating factor is μ(t) = e 3/(t+200)dt = dt t + 200 e3 ln(t+200) = (t + 200)3 and y = 2t + 400 − 390 · 2003 (t + 200)−3 . V (t) = t + 200 = 400 when t = 200 and y(200) = 605 so the concentration at t = 200 is y(200)/V (200) = 605/400 = 1.5125. 21. (a) u(t) = e−t/4 (−6 + 76et/4 ); (b) u(t) = (9 + π2 )−1 (639 + 71π2 + (81 − π2 )e−t/4 − 90 cos(πt/12) − 30π sin(πt/12)); (c) u(t) = 2(9 + π2 )−1 (333 + 37π2 + (27 − 2π2 ) e−t/4 − 45 cos(πt/12) − 15π sin(πt/12)). u

u

u

76

78

75

76

74

74

76

73

72

74

72

70

72

71

68 5

10

15

20

t

80 78

70 5

(a)

10

15

20

t

5

(b)

u(t) =

4608 + 8π2

15 (c)

23. −4032e−2t ud − 7π2 e−2t ud + 287424e−2t + 489π2 e−2t − 240π sin  πt  + 7π2 ud + 4032ud + 71π2 + 40896 − 5760 cos 12

10

 πt  12

;

the average temperature is      7 1 + 47e48 576 + π2 ud + 9e48 250048 + 433π2  1 24 −489π2 − 287424   u(t) dt = . 24 0 384e48 576 + π2   3 95808 + 4410816e48 + 163π2 + 7661e48 π2    ≈ Solving u(t) = 70 for ud yields ud = 7 1 + 47e48 576 + π2 69.8273.

20

t

52

CH A P T E R 3:

Applications of First-Order Differential Equations

Exercises 3.3 1. m = 1 so solve v = 32 − v, v(0) = 0: v(t) = 32 − 32e−t ; v(2) ≈ 27.67 ft /s. 3. The mass is m = W /g = 1/32 so we solve

1  32 v

= 1 − 2v, v(0) = 8:

dv + 64v = 32 dt e64t

dv + 64e64t v = 32e64t dt d  64t e v = 32e64t dt 1 e64t v = e64t + C 2 1 v = + Ce−64t . 2

Apply the initial condition to obtain v(t) = 12 (1 + 15e−64t ); v(1) ≈ 0.5 ft /s. 5. The ball’s weight is 4 ounces = 1/4 pound so its mass is found by solving 14 = m · 32 giving us m = 1/128 slug. Now solve v = 32 −

8v, v(0) = −64 because the ball is tossed upward. The integrating factor is μ(t) = e 8 dt = e8t . The solution to the ODE is v = 4 + Ce−8t . Applying the initial condition gives us v(t) = 4 − 68e−8t ; v(0) = 0 when t = 18 ln 17 ≈ 0.354 s. 7. dv/dt = 32 − v, v(0) = 0 has solution v(t) = 32 − 32e−t ; ds/dt = v(t), s(0) = 0 has solution s(t) = 32t + 32e−t − 32; s(4) ≈ 96.59 ft < 300 so about 203.41 ft above the ground. 9. We solve 10v = 10 · 9.8 − 10v, v(0) = 0 by separating variables: 49 dv = −v dt 5 1

dv = −dt v − 49 5 49 ln v − = t + C 5 v= Applying the initial condition gives us v(t) =

49 + Ce−t . 5 49 −t 5 (1 − e );

limt→∞ v(t) = 49/5.

11. Because the mass of the object is 100 kg, the weight is mg = 100 · 9.8. Therefore, we solve v = −9.8 − v/1000, v(0) = 100 to obtain v = −9800 + Ce−t/1000 and apply the initial condition resulting in v(t) = −9800 + 9900e−t/1000 ; v(t) = 0 when t ≈ 10.152 s; s(t) = −9800t + 9900000(1 − e−t/1000 ); s(10.152) ≈ 506.76 m. 13. dv/dt = −g, v(0) = v0 has solution v(t) = −gt + v0 ; ds/dt = v(t), s(0) = s0 has solution s(t) = − 12 gt 2 + v0 t + s0 .

Answers, Hints, and Solutions to Selected Exercises

15. Because the object reaches its maximum height when v = −gt + v0 = 0, t = v0 /g and the air resistance is ignored, the object hits the ground when t = 2v0 /g. The velocity at this time is v(2v0 /g) = −g(2v0 /g) + v0 = −v0 . 98 cv0 + 98 −ct/10 17. The solution of 10v = 98 − cv, v(0) = v0 is v(t) = − + e with limit c c 98 limt→∞ v(t) = − c = −19.6 so c = 5. 19. The parachutist’s mass is m = 192/32 = 6 slugs so we solve v = 32 − 12 v2 , v(0) = 60. Here, we use separation of variables: 1 32 − 12 v2 1 8



dv = dt

1 1 + 8+v 8−v

 = dt

ln |8 + v| − ln |8 − v| = 8t + C 8+v = Ce8t . 8−v Now we find v: 8+v = Ce8t 8−v 8 + v = 8Ce8t − Cve8t v + Cve8t = 8Ce8t − 8 v=

8Ce8t − 8 1 + Ce8t

and apply the initial condition v(0) = 60 ⇒

8C − 8 17 = 60 ⇒ C = − C +1 13

17e8t + 13 . The limiting velocity is limt→∞ v(t) = 8. 17e8t − 13 dv dv dr dv 21. (b) = = v ; (c) limt→∞ v2 = v0 2 − 2gR dt dr dt dr √ √ 2 23. g ≈ 32 ft/s ≈ 0.006 mi/s2 ; v0 = 2gR = 2 · .165 · 0.006 · 1080 ≈ 1.46m/s dQ 1 E0 25. In standard form the equation is + Q= . The corresponding homogeneous dt RC R 1 dQ + Q = 0 with general solution Qh = C1 e−t/(RC) . Because E0 /R is a equation is dt RC to see that v(t) = 8

53

54

CH A P T E R 3:

Applications of First-Order Differential Equations

constant and not a solution of the corresponding homogeneous equation, we assume that a particular solution of the nonhomogeneous equation has the form Qp = A, where A is a constant to be determined. Qp = 0 and substituting into the nonhomogeneous equation gives us Qp +

1 E0 Qp = RC R

0+

1 E0 A= RC R A = CE0

so a particular solution of the nonhomogeneous equation is Qp = CE0 . Therefore, a general solution of the nonhomogeneous equation is Q(t) = Qp (t)+ Qh (t) = CE0 + C1 e−t/(RC) . Applying the initial condition gives us Q(t) = E0 C + e−t/(RC) (−E0 C + Q0 ); 1 −t/(RC) (−E0 C + Q0 ). differentiating Q yields I(t) = − e RC √ √ √ 27. The solution of v + 13 v = 16 − 4 3, v(0) = 0 is v(t) = 12(4 − 3) − 12(4 − 3)e−t/3 ; √ √ −t/3 √ x(t) = 12(4 − 3)t + 36(4 − 3)e − 36(4 − 3). 29. v(t) = 32, limt→∞ (32 + Ce−t ) = 32 31. v(t) = −gm/c, limt→∞ (−gm/c + Ce−ct/m ) = −gm/c 33. c = 1/2: v(t) = 64e−t/2 (−1 + et/2 ); c = 1: v(t) = 32e−t (−1 + et ); c = 2: v(t) = 16e−2t (−1 + e2t )

v 50

40

30

20

10 t 0.5

1.0

1.5

2.0

2.5

3.0

Answers, Hints, and Solutions to Selected Exercises

35. Plots of numerical solutions FR ⫽16v3

FR ⫽16 v

v

v

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0.5

1.0

1.5

2.0

t

0.5

1.0

37. The woman’s mass is m = 125/32 slugs. At t = 5, she falls s1 = 272.479 ft. After the parachute opens, she falls approximately 4000 − 272.479300.772 = 3727.52 ft. in approximately 295.772 seconds. The total time of the fall is 295.772 + 5.000 = 300.772 seconds. 5000 4000 3000 2000 1000

100

200

Chapter 3 Review Exercises 1. y = 0 is unstable; y = 1/2 is stable. 3. y = 0 is unstable; y = 4 is stable.

300

400

1.5

2.0

t

55

56

CH A P T E R 3:

Applications of First-Order Differential Equations

Stable at 4; unstable at 0 y 5

1.0

4 0.5 3 y

21

1

2

3

4

5 2

20.5 1 21.0 1

2

3

4

5

6

5

6

t

21

5. y = 0 is stable; y = 2 is unstable; y = 4 is stable. Stable at 4, 0; unstable at 2 y 5 1.5 4 1.0 3

0.5 y

⫺1

1

2

3

4

5

2

⫺0.5 ⫺1.0

1

⫺1.5 1

2

3

4

t

⫺1

7. With y(t) = y0 ekt , y(4) = y0 e4k = 3y0 ⇒ ek = 31/4 so y(t) = y0 · 3t/4 . y0 · 3t/4 = 5y0 ⇒ 14 t ln 3 = ln 5 so t = (4 ln 5)/(ln 3) ≈ 5.86 days. 9. y(t) = y0 ekt ; y(1700) = y0 e1700k = 12 y0 ⇒ ek = (1/2)1/1700 . Then y = y0 · 2−t/1700 , y(50) ≈ 0.9798y0 (97.98% of y0 ).

Answers, Hints, and Solutions to Selected Exercises

11. First we solve y = ky(1000 − y), y(0) = 250: dy = ky(1000 − y) dt 1 dy = k dt y(1000 − y)   1 1 1 + dy = k dt 1000 y 1000 − y ln |y| − ln |1000 − y| = 1000kt + C y = Ce1000kt 1000 − y y=

1000Ce1000kt . 1 + Ce1000kt

Applying the initial condition gives us y(0) =

1000C 1 = 250 ⇒ C = 1+C 3

1000e1000kt 1000e1000k . Then y(1) = = 500 ⇒ e1000k = 3 ⇒ ek = 31/1000 so 3 + e1000kt 3 + e1000k 1000 · 3t = 1000(1 + 31−t )−1 ; y = 750 implies t = (ln 9)/(ln 3) ≈ 2 days. y= 3 + 3t so y =

13. Using Newton’s Law of Cooling, we have T0 = 40 and Ts = 90 so T (t) = 90 − 50ekt . Next T (20) = 90 − 50e20k = 65 ⇒ −50e20k = −25 ⇒ e20k = 1/2 ⇒ ek = 2−1/20 so T (t) = 90 − 50 · 2−t/20 ; T (30) = 90 − 50 · 2−3/2 ≈ 72.3◦ F. 15. We need to solve T  = k(T − 325), T (0) = 100, T (45) = 150 and then determine T (−60). T (t) = Cekt + 325; T (0) = 100 ⇒ C = −225 ⇒ T (t) = −225ekt + 325.  7 1/45 7 k T (45) = 150 ⇒ −225e45k + 325 = 150 ⇒ e45k = 175 . Therefore, 225 = 9 ⇒ e = 9 T (t) = −225 · (7/9)t/45 + 325 so T (−60) ≈ 10.43◦ F. 17. If the object weighs 4 pounds, then its mass is found with F = mg or 4 = m · 32 to be m = 1/8 slug. Therefore, we solve the initial value problem 18 v = 4 − v, v(0) = 0 to find that v(t) = 4 − 4e−8t . Then v(3) = 4 − 4e−24 ≈ 4 ft/s. The distance traveled by the rock after t seconds is found by solving s = 4 − 4e−8t , s(0) = 0, which has solution s(t) = 4t + 12 e−8t − 12 . After 3 seconds, the rock has fallen s(3) ≈ 11.5 ft. 19. dv/dt = −9.8 − v, v(0) = 40 has solution v(t) = 15 (−99 + 299e−t ); v(t) = 0 when t = ln(299/99) ≈ 1.11 s. ds/dt = v, s(0) = 0 has solution s(t) = 15 (−99t − 299e−t + 299); s(ln(299/99)) ≈ 18.11 ft.

57

58

CH A P T E R 3:

Applications of First-Order Differential Equations

19e8t + 11 ; limt→∞ v(t) = 8 ft/s. 19e8t − 11 23. With m = 230 kg, dv/dt = 82/115 − 637v/230000, v(0) = 0, which has solution 164000 230000 41000 v(t) = (1 − e−637t/230000 ); v(t) = 12 when t = ln ≈ 17.23 s. 637 637 39089 164000 37720000000 −637t/230000 dy/dt = v, y(0) = 0 has solution y = t+ (e − 1). 637 405769 H = y(17.23) ≈ 104.17798. 21. dv/dt = 32 − 12 v2 , v(0) = 30 has solution v(t) = 8

25. (a) r 2 = 4 cos2 2θ; (b) r = 0; (c) r = 12 θ 2 ; (d) r = 2(1 + sec θ); (e) r = −3(−1 + cos θ) r

r

0.6

20

0.4 0.2 22

240

21

1

2

220

␪

20

40

60

␪ 80

220

20.2 240

20.4 20.6

260 (c)

(a) r

4

30

r

20 2 10

1

2

3

4

␪

26

25

24

23

22

␪

21

210 22

220 230

24 (d)

    c 1 1 c −0.05t M e p− · dt = lim p− e P= M→∞ 2 · −0.05 0 q 2 q 0      c c = −10 p − lim e−0.05M − 1 = 10 p − . q M→∞ q 

27. (a)

(e)

∞

−0.05t

Answers, Hints, and Solutions to Selected Exercises

(b)

59

1 dx = dt [(q − qE) − ax]x   1 1 a + dx = dt r − qE x (r − qE) − ax 1 (ln |x| − ln |(r − qE) − ax|) = t + C r − qE x = (r − qE)(t + C) ln (r − qE) − ax x = Ke(r−qE)t . (r − qE) − ax x0 . Substitution gives us x(0) = x0 implies that K = (r − qE) − ax0 x x0 e(r−qE)t . Solving for x we find that = (r − qE) − ax (r − qE) − ax0 x0 (r − qE)e(r−qE)t x= . Because h(t) = qEx(t), h(t) = (r−qE)t ax0 e + (r − qE) − ax0 qEx0 (r − qE)e(r−qE)t . ax0 e(r−qE)t + (r − qE) − ax0 x

x

2.0

x

2.0

140 000 130 000

1.8

1.5

120 000

1.6 1.4

5

10

15

20

t

110 000 100 000

0.5

1.2 2

4

6 (c)

8

10

t

90 000

0.0

100 (d)

(d) From the graph, we see that if qE ≈ 0.5, the whale population remains constant. (e) For (i) we obtain approximately 3.819 × 107 while for (ii) we obtain approximately −1.530 × 107 . The negative result in (ii) indicates that the whale population becomes extinct. The more effort that is used to catch whales, the more whales are caught and they become extinct; for moderate efforts, less whales are caught, but they do not become extinct. ⎛ ⎞ 1000 3.28883 × 109 ⎜ 1500 −3.84615 × 109 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2000 −3.84615 × 109 ⎟ ⎜ ⎟ ⎜ 2500 2.95858 × 1015 ⎟ ⎜ ⎟ ⎜ 3000 2.95858 × 1015 ⎟ ⎜ ⎟ ⎜ 3500 2.95858 × 1015 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4000 2.95858 × 1015 ⎠ 4500 2.95858 × 1015

200

300 (f )

400

500

t

60

CH A P T E R 3:

Applications of First-Order Differential Equations

(f) We see that the maximum profit results if E ≈ 2500. Under these conditions, the whale population increases and approaches a limiting population of about 140,000. (g) We see that if there is no harvesting, the limiting population is about 400,000, which is above the optimal stock level of approximately 227,500. What advice would you give the whaling industry? 29. (a) Differentiating y + 2x = c with respect to x gives us y + 2 = 0 ⇒ y = −2 so we must solve y = 1/2. Integrating yields y = x/2 + k, k constant. Observe that the original family is the family of lines with slope −2 so it is no surprise that the family of orthogonal trajectories is the family of lines with slope 1/2. (b) Differentiating the equation yields dy/dx = cecx and solving the original equation for c gives us ln y = cx ⇒ c = (ln y)/x. Therefore, we must solve dy 1 1 x = − cx = − =− . ln y dx ce y ln y y x Separating variables and integrating with integration by parts gives us y ln y dy = −x dx 1 y2 ln y − y2 + x 2 = k. 2 (c) In this case 2yy = 2x so y = x/y and we must solve dy/dx = −y/x. Separating variables and integrating gives us 1 1 dy = − dx ⇒ ln |xy| = k ⇒ xy = k. y x (d) Differentiating the equation gives us 2yy = 2x + c and solving for c yields c = (y2 − x 2 )/x. Then, 2yy = 2x + c y2 − x 2 x 2 2 dy x +y = dx 2xy

2yy = 2x +

so we must solve dy/dx = −2xy/(x 2 + y2 ), which is homogeneous of degree 2. In this case, we let x = vy ⇒ dx = v dy + y dv. Then, 2x dx + (x 2 + y2 )dy = 0 2vy2 ( y dv + v dy) + y2 (v2 + 1)dy = 0 2v( y dv + v dy) + (v2 + 1)dy = 0 2vy dv + (3v2 + 1)dy = 0

Answers, Hints, and Solutions to Selected Exercises

2v 3v2 + 1

1 dv = − dy y

1 ln(3v2 + 1) = − ln |y| + k 3  2  3x + y2 ln = −3 ln |y| + k y2 1 x 2 y + y3 = k. 3 y

210

10

y 20

5

15

25

5

10

x

10

5

25

210

210

25

(a)

5

y

10

10

5

5

25

5

10

x

210

25

5

25

25

210

210

(c)

x

(b)

y

210

10

(d)

10

x

61

62

CH A P T E R 3:

Applications of First-Order Differential Equations

30.

31. y

y

210

10

10

5

5

25

5

10

x

210

25

5

25

25

210

210

33.

10

x

dy y−c dy 2(y − c) dy x = 2k1 x, k1 = 2 ⇒ = ; solve: =− ⇒ 12 y2 − cy = dx dx x dx 2(y − c) x − 14 x 2 + K; multiply by 4: x 2 + 2y2 = 4cy + 4K ⇒ 4c = 1 ⇒ c = 1/4.

35. y2 − x 2 = C 37. (a) Implicitly differentiating y2 = 2cx + 2c2 , we have 2yy = 2c. Using the quadratic  formula with 2c2 + 2xc − y2 = 0, we find that c = 12 −x ± x 2 + 2y2 . If c =   dy 2y 1 2 + 2y2 , then the orthogonal trajectories must satisfy −x − x , =  2 dx x + x 2 + y2 which is a first-order homogeneous equation. With the substitution x = uy ⇒ dx = ydu + udy, we have    2u(u dy + y du) − uy + u2 y2 + 2y2 dy = 0 2y du = √

1 u2 + 2 − u

du =

 1 2 1  2 1 u + u u + 2 + ln u + u2 + 2 = 4 4 2   1 x2 1 x2 1 x x2 + x 2 + 2 + ln + + 2 = 2 2 4y 4 2 y y y

 u2 + 2 − u du 1 dy 2y 1 ln |y| + C 2 1 ln |y| + C. 2

Answers, Hints, and Solutions to Selected Exercises

  dy −x + x 2 + 2y2 , then the orthogonal trajectories satisfy = dx   1 x2 x2 4y 1 x2 1 1 x . Therefore, − x + 2 − ln + + 2 = ln |y| + C.  2 2 2 4y 4 2 y y y x − x 2 + 2y2 2 This family of curves is self-orthogonal. (Both sets of graphs are obtained from the same family.) Similarly, if c =

1 2

A self–orthogonal family of curves y 10

5

210

25

5

10

x

25

210

Differential Equations at Work A. Mathematics of Finance 2. We compute x = 1000e0.08t for the t-values: $1491.82, $2225.54, $3320.12, $4953.03. 4. 157 months

12 000 10 000 8000 6000 4000 2000 ⫺2000

x

50

100

150

t

63

64

CH A P T E R 3:

Applications of First-Order Differential Equations

5. x  − ln(1 + .08/12)x = −599, x(0) = 80000 has solution x = 90149.2 − 10149.2e0.00664454t and x(t) = 0 when t = 329.

80 000

x

60 000 40 000 20 000 50 100 150 200 250 300 350

t

⫺20 000

6. x  − ln(1 + .08/12)x = −599 · 1.0025t , x(0) = 80000 has solution x = 144419e0.00249688t − 64418.7e0.00664454t and x(t) = 0 when t = 195.

x 50 100 150 200 250 300 350 ⫺100 000 ⫺200 000 ⫺300 000

7. x  − ln(1 + .1/12)x = 250, x(0) = 0 has solution x = 30124.8e0.0082988t − 30124.8.

Years 10 20 30

Months

Balance

120 240 360

51424.30 190632 567473

8. x  − ln(1 + .1/12)x = 250 · 1.005t , x(0) = 0 has solution x = 75499.9e0.0082988t − 75499.9e0.00498754t .

Years

Months

Balance

10 20 30

120 240 360

67017.10 303349 1.04302 × 106

Months

Balance

120 240 360

51424.30 190632 567473

9. Years 10 20 30

Years 10 20 30

Months

Balance

35 45 55

133381 345957 897323

t d  −rt  e S = e−rt (I(t) − E(t)); e−rt S = 0 e−ru (I(u) − dt

t E(u))du + C; S(t) = ert 0 e−ru (I(u) − E(u))du + Cert ; applying S(0) = S0 yields C = S0 .

11. S  − rS = I(t) − E(t);

t

Answers, Hints, and Solutions to Selected Exercises

12. Here, E(t) = E(0)eit = 18000e0.03t and   I(0)e jt , 0 ≤ t ≤ T 20000e0.05t , 0 ≤ t ≤ T I(t) = = F + Vei(t−T ) , t > T 0.2 · 18000e0.03T + 0.3 · 18000e0.03T e0.03(t−T ) , t>T  20000e0.05t , 0 ≤ t ≤ T . = 0.03T 0.03t 18000(0.2e + 0.3e ), t > T We will assume that S0 = 0 so the balance of the account    t0 S(t0 ) = ert0 S0 + (I(t) − E(t)e−rt dt 0

t can be rewritten as S(t0 ) = ert0 00 (I(t) − E(t)e−rt dt. Thus, the balance of the account thirty years after retirement is given by  T +30 S(T + 30) = e0.06(T +30) (I(t) − E(t))e−0.06t dt, 0

which is a function of T : 0.06(T +30)

S(T + 30) = e

  20000



T

e−0.01t dt+

0 T +30 

0.2e0.03T + 0.3e0.03t e−0.06t dt−

18000 T



T +30

18000

 e−0.03t dt .

0

(b) By trial and error, we see that if T = 29, the account balance is never zero. S 1.2⫻107

S 2.5⫻106

1.0⫻107 8.0⫻10

2.0⫻106

6

1.5⫻106

6.0⫻106 4.0⫻106

1.0⫻106

2.0⫻106

500 000 10

20

30

40

t 10

20

(a)

B. Algae Growth 1. Note that the equation is separable as well as first-order linear. We solve it by viewing it as a linear equation and using undetermined coefficients. The corresponding homogeneous dN − μN = 0 with characteristic equation r − μ = 0 ⇒ r = μ and a general equation is dt

30 (b)

40

50

60

t

65

66

CH A P T E R 3:

Applications of First-Order Differential Equations

solution of the corresponding homogeneous solution is Nh = Ceμt . A particular solution takes the form Np = B, where B is a constant to be determined. Substituting into the nonhomogeneous equation gives us Np  − μNp = −μB = μA so B = −A and Np = −A. The general solution of the nonhomogeneous equation is then N = Ceμt − A. Applying the initial condition gives us C = A so N(t) = Aeμt − A. 2. μ = 0.693 3. 3e0.693·24 − 3 ≈ 5.033 × 107 and 3e0.693·36 − 3 ≈ 2.062 × 1011

C. Dialysis 1. Use a computer algebra system to perform the algebra and calculus. A general solution of z = −αz is z = Ce−αx . Because z = u − v, u − v = Ce−αx . Using QB u = −k(u − v) yields QB u = −kCe−αx so dividing by QB and integrating results in u=

Ck + αC1 QB eαx Ck + C = . 1 αQB eαx αQB eαx

Solving for v and substituting results in v=

 1 1  (QB u + ku) = Ck − αCQB + αC1 QB eαx . αx k αQB e

Applying the initial conditions u(0) = u0 and v(L) = 0 leads to the system of equations

1 αQB eαL

1 (Ck + C1 αQB ) = u0 αQB   Ck − αCQB + αC1 QB eαL = 0,

u0 (αQB − k) ku0 QB eαL and C1 = αL . Substituting these αL ke + αQB − k ke + αQB − k values and simplifying the results into the formulas for u and v yields which has solutions C =

u = u0

QB eαx − QD eαL   eαx QB − QD eαL

and

v = u0

eαL − eαx . QD  αL eαx e −1 QB

2. In this case, u(x) = 50.0623e−0.5625x − 14.2623. After dialysis has been performed once, waste levels are approximately u(1) ≈ 14.2623, which are within the average range. eαL − 1 and αQB + keαL − k eαL − 1 , which are the same. QB (U0 − u(L)) = u0 kQB αQB + keαL − k QB eαL − 1 eαL − 1 CL = (u0 − u(L)) = kQB = Q 4. B αQB − k u0 αQB + keαL − k + eαL k eαL − 1 1 − eαL = QB = QB αQB αQB −αL − + eαL 1− e QD QD

3.

L 0

k(u(x) − v(x))dx = u0 kQB

Answers, Hints, and Solutions to Selected Exercises

D. Antibiotic Production 1. X =

Xmax eμmax t Xmax − 1 + eμmax t

2. Hours 4 8 12 16 20 24

Mass 2.69487 5.50521 8.02624 9.3104 9.78178 9.93326

X 10

8

6

4

2

5

3. P(t) = 4.

1200

10

15

20

600

800

t

 1  10EeKd t (eKd t − e24Kd ) Kd

P

1000 800 600 400 200

200

400

1000

t

67

CHAPTER 4

Higher Order Equations

Exercises4.1

  t 4t − 1   = 1; linearly independent 1. W (S) =  1 4   −6t   e e−4t  3. W (S) =  = 2e−10t ; linearly independent −6e−6t −4e−4t    −1 t t −2   = −2t −4 + t − 4 = −t −4 5. Linearly independent; W (S) =  −2 −t −2t −3  7. Linearly independent; W (S) = 0 9. W ({et , e−t }) = −2

11. W ({t −1/2 , t 3 }) = 72 t 3/2 13. y(t) = −2e2t + e−t 15. y = −t −1/3 + 2t 3 17. y = sin t + cos t + t sin t   d d 1 t d d −t ) = 1 (et − e−t ) = sinh t; 19. (a) (cosh t) = (e + e (b) dt dt 2 2 dt (sinh t) = dt   1 t 1 t −t −t (c) cosh2 t − sinh2 t = 14 (e2t + 2 + e−2t ) − 2 (e − e ) = 2 (e + e ) = cosh t; − 2 + e−2t ) = 1; (d) W ({cosh t, sinh t}) = cosh2 t − sinh2 t = 1    cos 4t sin 4t   21. W (S) =  = 4 cos2 4t + 4 sin2 4t = 4 −4 sin 4t 4 cos 4t   −1  t t   23. W (S) =  −2 = 2t −1 −t 1   5 dt  5t  25. e e 3t 3t 3t y2 = e dt = e dt = e e−t dt = −e2t [e3t ]2 e6t 1 2t 4 (e

68

Answers, Hints, and Solutions to Selected Exercises

Because every linear combination (including constant multiples) of solutions to a linear homogeneous equation is a solution, this means that a second linearly independent solution of the equation is y2 = e2t . Therefore, a fundamental set of solutions is S = {e2t , e3t } and a general solution of the linear homogeneous equation is y = c1 e2t + c2 e3t . 27. Here p(t) = −4 and y1 = e2t so a second linearly independent solution is  y2 = e

2t



e− −4 dt dt = e2t [e2t ]2



e4t dt = te2t . e4t

This means that a fundamental set of solutions is S = {e2t , te2t } and a general solution of the homogeneous equation is y = e2t (c1 + c2 t). Applying the initial conditions yields y = te−2t . 29. S = {cos 3t, sin 3t}, y = c1 cos 3t + c2 sin 3t, y = cos 3t − 43 sin 3t 31. S = {t −4 , t}, y = c1 t −4 + c2 t sin t 33. For t > 0, y2 (t) = √ so a general solution is y = t −1/2 (c1 cos t + c2 sin t). t 35. Substituting y = e−t/2 and y = et/3 into the equation y + (b/a) y + (c/a) y = 0 results in the system of equations 2 b/a − 4 c/a = 1 and 3 b/a − 9 c/a = 1 so b/a = 5/6 and c/a = 1/6. Hence, the equation is 6y + 5y + y = 0. 37. Substitution of either function into the differential equation and equating coefficients gives us a − b = −3 and 2a = 4 so a = 2 and b = 5. 

   1 − p(t) dt − p(t) dt 39. W f (t), e dt = e ( f (t))2 41. First, we calculate d d   W ({y1 , y2 }) = y1 y2 − y1 y2 dt dt = y1 y2 + y1 y2 − y1 y2 − y1 y2 = y1 y2 − y1 y2 . Because y1 and y2 are both solutions of the equation, y1 = −p(t)y1 − q(t)y1 and y2 = −p(t)y2 − q(t)y2 so d W ({y1 , y2 }) = y1 y2 − y1 y2 dt = y1 [−p(t)y2 − q(t)y2 ] − y2 [−p(t)y1 − q(t)y1 ] = −p(t)[y1 y2 − y1 y2 ] = −p(t)W ({y1 , y2 }). Therefore, W ({y1 , y2 }) satisfies the first-order linear separable equation W  +  dW 1 p(t)W = 0 ⇒ = −p(t)W ⇒ dW = −p(t)dt with solution ln |W | = − p(t) dt dt W  or W ({y1 , y2 }) = Ce− p(t) dt .

69

70

CH A P T E R 4:

Higher Order Equations

43. y = c1 + c2 tan(c3 + c4 ln t) is a solution of the equation if (a) y = c1 (note that if c2 = 0 or c4 = 0, y is a constant function); or (b) y = − 12 + c2 tan(c3 + c2 ln t). The Principle of Superposition does not hold if c1  = 0 and y = c1 + c2 tan(c3 + c2 ln t), 2 ty − 2yy = − c1 c22 sec2 (c3 + c2 ln t)  = 0. t 45. y = −πt −1 cos 4t 47. y = t −3 (a cos t + b sin t); y(π) = −aπ−3 = 0 implies a = 0; y(2π) = implies that a = 0. y = t −3 sin t, C arbitrary.

18 −3 a π

=0

49. We solve y1 + p(t)y1 + q(t)y1 = 0, y2 + p(t)y2 + q(t)y2 = 0 for p(t) and q(t): p(t) = y1 y2 − y1 y2 y y − y1 y2 and q(t) = 1 2 . W ({y1 , y2 }) W ({y1 , y2 })

Exercises 4.2 1. y = 0 has characteristic equation r 2 = 0 so r = 0 has multiplicity two. Two linearly independent solutions to the equation are y1 = 1 and y2 = t; a fundamental set of solutions is S = {1, t}, and a general solution is y = c1 + c2 t. 3. y + y = 0 has characteristic equation r 2 + r = 0, which has solutions r1 = 0 and r2 = −1. Two linearly independent solutions to the equation are y1 = 1 and y2 = e−t ; a fundamental set of solutions is S = {1, e−t }, and a general solution is y = c1 + c2 e−t . 5. y = c1 e−6t + c2 e−2t 7. y = c1 e−t/4 + c2 e−t/2 9. The characteristic equation is r 2 + 16 = 0 with roots r1,2 = ±4i so a fundamental set of solutions is S = {cos 4t, sin 4t} and a general solution is y = c1 cos 4t + c2 sin 4t. √ 2 11. The characteristic equation √ is r + 7√= 0 with roots r1,2 = ±i 7 so a fundamental √ set 7t), sin( 7t)} and a general solution is y = c cos( 7t) + of solutions is S = {cos( 1 √ c2 sin( 7t). 13. The characteristic equation is 7r 2 + 4r − 3 = (r + 1)(7r − 3) = 0 with solutions r1 = −1 and r2 = 3/7. Thus, a fundamental set of solutions is S = {e−t , e3t/7 } and a general solution is y = c1 e−t + c2e3t/7 . 15. The characteristic equation is r 2 − 6r + 9 = (r − 3)2 = 0 with solutions r1,2 = 3. Then, one solution is y1 = e3t and a second linearly independent solution found through reduction of order is y2 = te3t . Thus, a fundamental set of solutions is S = {e3t te3t } and a general solution is y = c1 e3t + c2 te3t . 17. General: y = c1 + c2 et/3 ; IVP: y = −21(1 − et/3 ) 19. General: y = c1 e3t + c2 e4t ; IVP: y = 14e3t − 11e4t 21. General: y = c1 e5t + c2 e2t ; IVP: y = e5t 23. General: y = c1 cos 10t + c2 sin 10t; IVP: y = cos 10t + sin 10t 25. General: y = e−2t (c1 + c2 t); IVP: y = e−2t (1 + 5t) 27. General: y = e−2t (c1 cos 4t + c2 sin 4t); IVP: y = e−2t (2 cos 4t + sin 4t)

Answers, Hints, and Solutions to Selected Exercises

31. y = (1 + C + Ct)/((1 + Ct) sin t) 33. y = 12 t(−5 + 2t 5 + 10C)/(t 5 + 5C) 35. y =

C ln (1 + sin t) sin t − C ln (−1 + sin t) sin t − sin t − 2 C cos t (−C ln (−1 + sin t) + C ln (1 + sin t) − 1)

37. (a) y = c1 t 2/3 + c2 t; (b) y = t(c1 + c2 ) ln t 39. The characteristic equation of ay + 2by + cy = 0 is ar 2 + 2br + c = 0, which has √ −b ± b2 − ac solutions r1,2 = . With b2 − ac > 0, two linearly independent solutions a are

y1 = exp and y2 = exp

−b +



√  √ 

b2 − ac b b2 − ac t = exp − t exp t a a a

−b −



√  √ 

b2 − ac b b2 − ac t = exp − t exp − t . a a a

Two other linearly independent solutions are given by

√ 

√ 

 1 b 1 b2 − ac b2 − ac (y1 + y2 ) = exp − t · exp t + exp − t 2 a 2 a a

√  

b b2 − ac = exp − t cosh t a a and

√ 

√  

1 b b2 − ac b2 − ac 1 (y1 − y2 ) = exp − t · exp t − exp − t 2 a 2 a a

√ 

 b2 − ac b t . = exp − t sinh a a 41. y = Ce−t sin 2t, y = 0, y = e−t cos 2t         43. y = 32 a + 12 b e−t + − 12 a− 12 b e−3t so y = − 32 a + 12 b e−t −3 − 12 a − 12 b e−3t ;

 √ √ 1 3 (b + a) (3 a + b)  y = 0 if t = − ln ± ; for none, (b + a)(3a + b) ≤ 0 while 3 b+a for one, (b + a)(3a + b) > 0 45. (a) No; (b) To be a general solution, a fundamental set for the equation is S = {t cos t, t sin t}. Now substitute each of these functions into the differential equation and set the result equal to zero. Solve the resulting system for p(t) and q(t) to obtain p(t) = −2/t and q(t) = (t 2 + 2)/t 2 .

71

72

CH A P T E R 4:

Higher Order Equations

Exercises 4.3 1. F = {t, 1} 3. F1 = {e−t }, F2 = {1} 5. F1 = {e−t }, F2 = {t 4 , t 3 , t 2 , t, 1} 7. F1 = {cos 2t, sin 2t}, F2 = {e−4t } 9. F1 = {cos 3t, sin 3t}, F2 = {cos 2t, sin 2t} 11. F1 = {e−t cos 2t, e−t sin 2t}, F2 = {1} 13. yp = Ae2t (y = c1 cos t + c2 sin t + 85 e2t ) 15. yp = Ate3t (y = c1 e2 + c2 e3t + te3t ) 17. yp = At 2 + Bt + C (y = et (c1 + c2 t) + t 2 + 4t + 6) 19. yp = A cos 2t + B sin 2t (y = c1 cos t + c2 sin t − 13 cos 2t) 21. yp = At cos 2t + Bt sin 2t + Ct + D (y = c1 cos 2t + c2 sin 2t + 14 t sin 2t + 14 t) 1 6 23. yp = At 6 + Bt 5 + Ct 4 + Dt 3 + Et 2 (y = c1 + c2 t + 10 t − 13 t 3 )

25. A fundamental set of solutions for the corresponding homogeneous equation is S{e−2t , et } so a general solution of the corresponding homogeneous equation is yh = c1 e−2t + c2 et . The associated set of functions is F = {1}. Because no element of F is a solution of the corresponding homogeneous equation, we assume a particular solution takes the form yp = A · 1 = A with derivatives yp = yp = 0. Substituting yp into the nonhomogeneous equation gives us yp + yp − 2yp = −2A = −1 so A = 1/2 and a particular solution of the nonhomogeneous equation is yp = 1/2. Thus, y = yh + yp = c1 et + c2 e−2t + 1/2. 27. y = c1 e4t + c2 e−2t + 1 − 4t 29. The characteristic equation of the corresponding homogeneous equation is r 2 + 2r + 26 = 0 with solutions r1,2 = −1 ± 5i. A fundamental set of solutions for the corresponding homogeneous equation is S = {e−t cos 5t, e−t sin 5t} and a general solution is yh = e−t (c1 cos 5t + c2 sin 5t). The associated set of functions is F = {t, 1}. Because no element of F is a solution to the corresponding homogeneous equation, we assume that a particular solution has the form yp = At + B with derivatives yp = A and yp = 0. Substituting yp into the nonhomogeneous equation gives us yp + 2yp + 26yp = 26At + (2A + 26B) = −338t so 26A = −338 ⇒ A = −13 and 2A + 26B = 0 ⇒ B = 1 so a particular solution of the nonhomogeneous equation is yp = −13t + 1. Thus, y = yh + yp = c1 e−t sin (5t) + c2 e−t cos (5t) + 1 − 13t. 31. y = 280 − 60t + 5t 2 + c1 e−1/2t + c2 e−1/4t 33. S = {1, e2t }, yh = c1 + c2 e2t , F = {cos 3t, sin 3t}, yp = A cos 3t + B sin 3t, y = −4 sin (3t) + 83 cos (3t) + c1 e2t + c2 54 35. y = c1 e−3t + c2 e3 t − 216 169 cos (2t) − 13 sin (2t) t

37. S = {e−2t , te−2t }, yh = e−2t (c1 + c2 t), F = {t 2 cos 2t, t cos 2t, cos 2t, t 2 sin 2t, t sin 2t, sin 2t}, yp = (At 2 + Bt + C) cos 2t + (Et 2 + Ft + G) sin 2t, y = c1 e−2t + c2 e−2t t + (−4t + 3) cos (2t) − 4t sin (2t) (t − 1)

Answers, Hints, and Solutions to Selected Exercises

 39. y = c1 e−t + c2 e5t + −36t 3 + 18t 2 − 6t e5t 41. The corresponding homogeneous equation is y + 4y = 0 with  characteristic equation k 2 + 4k = k(k + 4) = 0 → k = 0 or k = −4 so S = 1, e−4t is a fundamental set for −4t the corresponding homogeneous h =  c1 + c2 e . The forcing function is  4t equation and  y−4t 4t −4t . An element of F2 is a solution to the f (t) = 8e − 4e so F1 = e and F2 = e   corresponding homogeneous equation so we multiply F2 by t giving us tF2 = te−4t . No element of F1 or tF2 is a solution to the corresponding homogeneous equation so we assume that a particular solution has the form yp = Ae4t + Bte−4t . Differentiating twice gives us yp = 4Ae4t − 4Bte−4t + Be−4t and yp = 16Ae4t + 16Bte−4t − 8Be−4t and substituting into the nonhomogeneous equation gives us yp + 4yp = 32Ae4t − 4Be−4t = 8e4t − 4e−4t . Equating coefficients we see that 32A = 8 so A = 1/4 and −4B = −4 so B = 1. Thus, yp = 14 e4t + te−4t and y = yh + yp = c1 + c2 e−4t + 14 e4t + te−4t . 43. The corresponding homogeneous equation is y + 4y = 0 with  characteristic equation k 2 + 4k = k(k + 4) = 0 → k = 0 or k = −4 so S = 1, e−4t is a fundamental set for the corresponding homogeneous equation and yh = c1 + c2 e−4t . The forcing function is  f (t) = −24t − 6 − 4te−4t + e−4t so F1 = {t, 1} and F2 = te−4t , e−4t . We multiply F1   by t and F2 by t so that no element of tF1 = t 2 , t is a solution of the corresponding   homogeneous equation and so that no element of tF2 = t 2 e−4t , te−4t is a solution of the corresponding homogeneous equation. We now assume that a particular solution has the form yp = At 2 + Bt + Ct 2 e−4t + Dte−4t . Then, yp = 2At + B − 4Ct 2 e−4t + (2C − 4D)te−4t + De−4t and yp = 2A + 16Ct 2 e−4t + (−16C + 16D)te−4t + (2C − 8D)e−4t . Substituting into the nonhomogeneous equation yields yp + 4yp = 8At + (2A + 4B) − 8Cte−4t + (2C − 4D)e−4t = −24t − 6 − 4te−4t + e−4t and equating coefficients gives us 8A 2A +4B

−8C 2C

−4D

= = = =

−24 −6 −4 1.

Thus, A = −3, B = 0, C = 1/2 and D = 0 so yp = −3t 2 + 12 t 2 e−4t . A general solution is then y = yh + yp = c1 + c2 e−4t − 3t 2 + 12 t 2 e−4t with derivative y = −6t − 2t 2 e−4t + te−4t − 4c2 e−4t . 45. S = {1, t}, yh = c1 + c2 t, F1 = {t 2 , t, 1}, t 2 F1 = {t 4 , t 3 , t 2 }, F2 = {et }, F3 = 1 4 t + {cos t, sin t}, yp = At 4 + Bt 3 + Ct 2 + Eet + F cos t + G sin t, y = c1 t + c2 + 12 t e − sin t 47. S = {e2t , e−2t }, yh = c1 e2t + c2 e−2t , F = {1}, yp = A, y = 2e−t + 2et − 4

73

74

CH A P T E R 4:

Higher Order Equations

49. The corresponding homogeneous equation is y + 2y − 3y = 0, which has characteristic equation r 2 + 2r − 3 = (r + 3)(r − 1) ⇒ r1 = −3, r2 = 1. Then, a fundamental set of solutions for the corresponding homogeneous equation is S = {e−3t , et } and a general solution of the corresponding homogeneous equation is yh = c1 e−3t + c2 et . The associated set of functions for the forcing function, f (t) = −2, is F = {1}. Because y = 1 is not a solution to the corresponding homogeneous equation, we assume that a particular solution of the nonhomogeneous equation has the form yp = A · 1 = A. Differentiating, yp = yp = 0 and substituting into the nonhomogenous equation gives us −3A = −2 ⇒ A = 2/3 so yp = 2/3 and a general solution of the nonhomogeneous equation is y = yh + yp = c1 e−3t + c2 et + 2/3. Application of the initial conditions yields y = 2et − 2e−3t + 2/3. 51. y = e−4t + 4e−4t t + 1/4 53. The corresponding homogeneous equation is y + 6y 25y = 0, which has characteristic equation r 2 + 6r + 25 = 0 ⇒ r1,2 = −3 ± 4i. Then, a fundamental set of solutions for the corresponding homogeneous equation is S = {e−3t cos 4t, e−3t sin 4t} and a general solution of the corresponding homogeneous equation is yh = e−3t (c1 cos 4t + c2 sin 4t). The associated set of functions for the forcing function, f (t) = −1, is F = {1}. Because y = 1 is not a solution to the corresponding homogeneous equation, we assume that a particular solution of the nonhomogeneous equation has the form yp = A · 1 = A. Differentiating, yp = yp = 0 and substituting into the nonhomogenous equation gives us 25A = −1 ⇒ A = −1/25 so yp = −1/25 and y = yh + yp = e−3t (c1 cos 4t + c2 sin 4t) − 1/25. Application of the initial conditions yields y = 7/4 e−3t sin (4t) − 1/25.

49. y 6

51. y 1.2

5

1.0

0.5

4

0.8

0.4

3

0.6

0.3

2

0.4

0.2

1

0.2 0.2

0.4

0.6

0.8

1.0

t

53.

y

0.6

0.1 0.5

1.0

1.5

2.0

t

⫺0.1

0.5

1.0

1.5

2.0

2.5

3.0

t

55. S = {1, et }, F1 = {t, 1}, tF1 = {t 2 , t}, F2 = {t 2 e2t , te2t , e2t }, yp = At 2 + Bt + Ct 2 e2t + Ete2t + Fe2t , y = −2t 2 e2t + 6e2t t − 7e2t + 3/2t 2 + 5et + 3t − 3/2 57. The corresponding homogeneous equation is y + 4y = 0 with  characteristic equation k 2 + 4k = k(k + 4) = 0 → k = 0 or k = −4 so S = 1, e−4t is a fundamental set for function is the corresponding homogeneous equation and yh = c1 + c2 e−4t . The forcing  f (t) = −24t − 6 − 4te−4t + e−4t so F1 = {t, 1} and F2 = te−4t , e−4t . We multiply F1   by t and F2 by t so that no element of tF1 = t 2 , t is a solution of the corresponding   homogeneous equation and so that no element of tF2 = t 2 e−4t , te−4t is a solution of the corresponding homogeneous equation. We now assume that a particular solution has the form yp = At 2 + Bt + Ct 2 e−4t + Dte−4t . Then, yp = 2At + B − 4Ct 2 e−4t + (2C − 4D)te−4t + De−4t and yp = 2A + 16Ct 2 e−4t + (−16C + 16D)te−4t + (2C − 8D)e−4t .

Answers, Hints, and Solutions to Selected Exercises

Substituting into the nonhomogeneous equation yields yp + 4yp = 8At + (2A + 4B) − 8Cte−4t + (2C − 4D)e−4t = −24t − 6 − 4te−4t + e−4t and equating coefficients gives us 8A 2A +4B −8C 2C

−4D

= = = =

−24 −6 −4 1.

Thus, A = −3, B = 0, C = 1/2 and D = 0 so yp = −3t 2 + 12 t 2 e−4t . A general solution is then y = yh + yp = c1 + c2 e−4t − 3t 2 + 12 t 2 e−4t with derivative y = −6t − 2t 2 e−4t + te−4t − 4c2 e−4t . Applying the initial conditions gives us y(0) = c1 + c2 = 0 y (0) = −4c2 = 0 so c1 = c2 = 0 and the solution to the initial value problem is y = −3t 2 + 12 t 2 e−4t . 59. The corresponding homogeneous equation is y + 9y = 0 with general solution yh = c1 cos 3t + c2 sin 3t. For 0 ≤ t < π, a particular solution has the form yp = At + B and we find that yp = 29 t and for 0 ≤ t < π, the solution to the initial value problem is y = 2 2 2 2  9 t − 27 sin 3t. Evaluating y = 9 t − 27 sin 3t and y at t = π shows us that y(π) = 2π/9   and y (π) = 4/9 so for t ≥ π, we solve y + 9y = 0, y(π) = 2π/9, y (π) = 4/9, which 2 t − 2 sin 3t, 0 ≤ t < π 4 sin 3t. Thus, y = 9 2 27 . has solution y = − 29 cos 3t − 27 4 sin 3t, t ≥ π − 9 cos 3t − 27 y 0.6 0.4 0.2 ⫺0.2 ⫺0.4 ⫺0.6

1

2

3

4

Note: A CAS was used to construct the solution. 61.

⎧ ⎪ ⎪sin 2t, 0 ≤ t < π ⎨ y = 12 (5 − 5 cos 2t + 2 sin 2t) , π ≤ t < 2π ⎪ ⎪ ⎩ sin 2t, t ≥ 2π

5

6

t

75

76

CH A P T E R 4:

Higher Order Equations

y 1

2

4

6

8

t

⫺1

⫺2

⫺3

⫺4

⫺5

Note: A CAS was used to construct the solution. 63.

(a) Let yp (t) be a particular solution to the nonhomogeneous equation. The characteristic equation of the corresponding homogeneous equation is ar 2 + br + c √ −b ± b2 − 4ac , where our convention is that r1 correwith solutions r1,2 = 2a sponds to the “+” solution and r2 corresponds to the “−” solution. Note that because b is assumed to be positive, −b is negative. If b2 − 4ac > 0, r1,2 are both negative (See Exercise 36 in Section 4.2), there are constants c1 and c2 so that y1 = c1 er1 t + c2 er2 t + yp (t), and there are constants b1 and b2 so that y2 = b1 er1 t + b2 er2 t + yp (t). Then,  lim (y2 − y2 ) = lim (b1 − c1 )er1 t + (b2 − c2 )er2 t = 0 − 0 = 0.

t→∞

t→∞

If b2 − 4ac = 0, r = r1,2 is negative, there are constants c1 and c2 so that y1 = er1 t (c1 + c2 t) + yp (t), and there are constants b1 and b2 so that y2 = er1 t (b1 + b2 t) + yp (t). Subtracting and taking the limit as t → ∞ yields 0. Finally, if b2 − 4ac < 0, −b/(2a) < 0, there are constants c1 and c2 so that

−bt/(2a)

y1 = e

c1 cos

√

√   4ac − b2 4ac − b2 t + c2 sin t + yp (t), 2a 2a

and there are constants b1 and b2 so that

√ 

√  2 2 4ac − b 4ac − b y2 = e−bt/(2a) b1 cos t + b2 sin t + yp (t). 2a 2a Subtracting and taking the limit as t → ∞ yields 0.

Answers, Hints, and Solutions to Selected Exercises

(b) No. For example both y1 (t) = 12 t sin t and y2 (t) = sin t + 12 t sin t are solutions of y + y = cos t and limt→∞ (y2 (t) − y1 (t)) = limt→∞ sin t does not exist. (c) Refer to Section 4.2. Every solution, yh (t), of ay + by + cy = 0 has the property that limt→∞ yh (t) = 0. A particular solution of ay + by + cy = 0 is yp = k/c so a general solution is y = yh (t) + k/c. Thus, limt→∞ (yh (t) + k/c) = 0 + k/c = k/c. (d) y = c1 + c2 e−bt/a + kt/b; limt→∞ y(t) = ∞ if k > 0 and limt→∞ y(t) = −∞ if k < 0. (e) y = c1 + c2 t + 12 kt 2 ; limt→∞ y(t) = ∞ if k > 0 and limt→∞ y(t) = −∞ if k < 0. 67. ω = 2 because a general solution of y + 4y = 0 is yh = c1 cos 2t + c2 sin 2t. 69. Resonance occurs when c = 3. For c  = 3, x(t) = c1 cos 3t + c2 sin 3t − c21−9 sin ct. If c = 3, x(t) = c1 cos 3t + c2 sin 3t − 16 t cos 3t.

Exercises 4.4 1. A fundamental set of solutions for the corresponding homogeneous equation is S = {cos 2t, sin 2t} and a general solution is yh = c1 cos 2t + c2 sin 2t. Assuming that a particular solution  has the form yp = u1 y1 + u2 y2 , we first compute W (S) =  cos 2t sin 2t   −2 sin 2t 2 cos 2t  = 2. Next, 

 1 0 W (S) 1



 1  cos 2t W (S) −2 sin 2t

u1 = and u2 =

  1 1 sin 2t  dt = − sin 2t dt = cos 2t 2 cos 2t  2 4   1 1 0 dt = cos 2t dt = sin 2t. 1 2 4

Therefore, a particular solution is yp = u1 y1 + u2 y2 = 14 cos2 2t + 14 sin2 2t = 14 and a general solution is y = yh + yp = c1 sin (2t) + c2 cos (2t) + 1/4. (Undetermined coefficients are easier.) 3. y = c1 e5t + c2 e2t − 12 e3t 5. y = c1 e−2t sin (4t) + c2 e−2t cos (4t) + 18 te−2t 7. S = {cos (4t) , sin (4t)}, yh = c1 cos (4t) + c2 sin (4t), W (S) = 4. yp = u1 y1 + u2 y2 ⇒     1  0 1 1 sin 4t  u1 = dt = −1 dt = − t W (S) csc 4t 4 cos 4t  4 4 and  u2 =

 1  cos 4t W (S) −4 sin 4t

  1 1 1 0  dt = cot 4t dt = − ln csc 4t = ln sin 4t. csc 4t  4 16 16

1 Thus, y = yh + yp = c1 sin (4t) + c2 cos (4t) + 16 ln (sin (4t)) sin (4t) − 14 t cos (4t).

77

78

CH A P T E R 4:

Higher Order Equations

9. y (t) =

1 −t 49 e

(c1 sin (7t) + c2 cos (7t) + ln (sin (7t)) sin (7t) − 7t cos (7t))

11. S = {et cos 5t, et sin 5t}, yh = et (c1 cos 5t + c2 sin 5t), W (S) = 5e2t . yp = u1 y1 + u2 y2 ⇒    1  0 et sin 5t  dt t t  W (S) e (sec 5t + csc 5t) e (5 cos 5t + sin 5t)  1 =− sin 5t (sec 5t + csc 5t) dt 5 

 1 1 1 =− (tan 5t + 1) dt = − ln sec 5t + t 5 5 5 

u1 =

and    1  et cos 5t 0  dt t t  W (S) e (cos 5t − 5 sin 5t) e (sec 5t + csc 5t) 



 1 1 1 1 1 = (1 + cot 5t) dt = t − ln csc 5t = t + ln sin 5t . 5 5 5 5 5 

u2 =

Therefore, y = yh + yp = c1 et sin (5t) + c2 et cos (5t) 

1 1 1 − − cos (5t) ln (cos (5t)) − sin (5t) ln (sin (5t))+t (− sin (5t) + cos (5t)) et . 5 5 5 13. S = {e3t cos 5t, e3t sin 5t}, yh = e3t (c1 cos 5t + c2 sin 5t), W (S) = 5e6t . yp = u1 y1 + u2 y2 ⇒     0 e3t sin 5t 1   u1 =  3t  dt W (S) e tan 5t e3t (5 cos 5t + 3 sin 5t) =−

1 5

1 =− 5

 sin 5t tan 5t dt = − 

1 5



sin2 5t 1 dt = − cos 5t 5



1 − cos2 5t dt cos 5t

1 (ln(sec 5t + tan 5t) − sin 5t) 25

(sec 5t − cos 5t) dt = −

 

1 1 + sin 5t =− ln − sin 5t 25 cos 5t

and  1  e3t cos 5t u2 = 3t W (S) e (3 cos 5t − 5 sin 5t)  1 1 sin 5t dt = − cos 5t. = 5 25 

  0  dt e3t tan 5t 

Answers, Hints, and Solutions to Selected Exercises

1 + sin (5t) cos (5t) ln . cos (5t) 

Therefore, y = c1

e3t

sin (5t) + c2

e3t

cos (5t) −

1 3t 25 e

15. y = c1 e6 t sin (t) + c2 e6t cos (t) + e6t (t sin (t) + ln (cos (t)) cos (t)) 17. S = {e3t , e−3t }, yh = c1 e3t + c2 e−3t , W (S) = −6. yp = u1 y1 + u2 y2 ⇒   1  0 e−3t  dt W (S) (1 + 3et )−1 −3e−3t    e−3t 1 et 1 dt = dt 6 1 + 3et 6 e4t (1 + 3et )

  3 9 27 1 1 81 − + − + dt (u = et ⇒ du = et dt) 6 u 1 + 3u u4 u3 u2 

1 1 3 9 − 2 + 2 − − 27 ln u + 27 ln(1 + 3u) 6 u 3u 2u 

1 1 3 − e−2t + e−2t − 9e−t − 27t + 27 ln(1 + 3et ) 6 3 2

 u1 = = = = = and

    1  e3t 1 e3t 0  dt u2 = dt = − 3t t −1   (1 + 3e ) W (S) 3e 6 1 + 3et  1 u2 du (u = et ⇒ du = et dt) =− 6 1 + 3u

  1 1 1 1 1 =− − + u+ du 6 9 3 9 1 + 3u 

1 1 1 1 =− − u + u2 + ln(1 + 3u) 6 9 6 27 

1 1 1 t 1 2t t =− − e + e + ln(1 + 3e ) . 6 9 6 27 

 2t  1 Therefore, y = c1 e3t + c2 e−3t − 324 9e − 6et + 2 ln 1 + 3et + 18e3t + 486e5t − 1458   ln 1 + 3et e6t − 81e4t + 1458 ln et e6t e−3t .   19. y = c1 et + c2 e−t + 12 e−t t + 12 cosh (2t) − 12 sinh (2t) e2t − 12 cosh (2t) − 12 + t  − 12 sinh (2t) 21. y = c1 e2t + c2 e2t − e2t (ln (t) + 1) 23. S = {e−3t , te−3t }, W (S) = e−6t , yh = e−3t (c1 + c2 t). yp = u1 y1 + u2 y2 ⇒  u1 =

 1  0 W (S) t −1 e−3t

  te−3t  dt = −1 dt = −t e−3t (1 − t)

79

80

CH A P T E R 4:

Higher Order Equations

and

 u2 =

 1  e−3t W (S) −3e−3t

  0  dt = t −1 dt = ln t. t −1 e−3t 

Therefore, y = c1 e−3t + c2 e−3t t + t (−1 + ln (t)) e−3t ⎛ ⎞    2

tan 12 et ⎜ ⎟ −t −t ⎜ ⎟  25. y = ⎝2   2 − c1 e + c2 ⎠ e 1 t t 1 + tan 2 e e

27. S = {et , tet }, yh = et (c1 + c2 t), W (S) = e2t . yp = u1 y1 + u2 y2 ⇒  u1 =

 1  √ 0 W (S) et 1 − t 2

   1 tet  2 dt = − 3 1 − t 2 1 − t dt = − t t e (1 + t) 3

and  u2 =

     1  dt = √0 1 − t 2 dt = t 1 − t 2 + sin−1 t . t  2 e 1−t 2

 1 et W (S) et

√ √ Therefore, y = yh + yp = c1 et + c2 et t + 16 et (2 1 − t 2 + 1 − t 2 t 2 + 3t arcsin (t)).   29. y = c1 e2t + c2 e2t t + 12 e2t arctan (t) t 2 + t − arctan (t) − t ln t 2 + 1 31. S = {cos(t/2), sin(t/2)}, yh = c1 cos(t/2) + c2 sin(t/2), W (S) = 1/2. Assuming that yp = u1 y1 + u2 y2 ⇒      

   

1 0 sin  1    1 2 t     t + 1 dt u1 = tan  dt = −2 1 W (S) sec 1 t + csc 1 t 2 cos 1 t  2

 = −2 2 ln sec



2



1 t +t 2

2

2



and  u2 =

    1 cos  1  2t   W (S) − 1 sin 1 t 2 2

  

   1     dt = 2 1 + cot t dt 1 1  2 sec 2 t + csc 2 t  0

 

 

1 1 = 2 t − ln csc t = 2 2 + ln sin t . 2 2 Therefore, y = yh + yp = c1 sin( 12 t) + c2 cos( 12 t) + cos( 12 t) ln(cos( 12 t)) + sin( 12 t) ln 1 1 (sin( 2 t)) − 2 t(− sin( 12 t) + cos( 12 t)). 33. y = c1 cos 3t + c2 sin 3t + 19 cos 3t ln cos 3t + 13 t sin 3t

   cos 2t sin 2t  35. yh = c1 cos 2t + c2 sin 2t and S = {cos 2t, sin 2t} so W (S) =  = −2    sin 2t 2 cos 2t   0  cos 2t sin 2t  0  = − 12 sin 2t tan 2t and u2 = 12  = 2. Then, u1 = 12  tan 2t 2 cos 2t  −2 sin 2t tan 2t 

Answers, Hints, and Solutions to Selected Exercises

sin t cos t so u1 = − 14 (ln (sec 2t + tan 2t) − sin 2t) and u2 = 12 cos2 2t so yp = − 14 cos 2t (ln (sec 2t + tan 2t) − sin 2t) + 12 sin 2t cos2 2t and y = yh + yp = c1 cos 2t + c2 sin 2t − 14 cos 2t (ln (sec 2t + tan 2t) − sin 2t) + 12 sin 2t cos2 2t.    cos 2t sin 2t   37. yh = c1 cos 2t + c2 sin 2t and S = {cos 2t, sin 2t} so W (S) =  =   −2 sin 2t 2 cos   2t  0  cos 2t sin 2t  0  2. Then, u1 = 12  = − sin2 t and u2 = 12  = sin t  tan t 2 cos 2t −2 sin 2t tan t    cos t so u1 = 14 (sin 2t − 2t) and u2 = 12 ln cos t − 12 cos 2t so yp = 14 cos 2t   (sin 2t − 2t) + 12 sin 2t ln cos t − 12 cos 2t and y = yh + yp = c2 cos 2t + c2 sin 2t +   1 1 1 (sin cos 2t 2t − 2t) + sin 2t ln cos t − cos 2t . 4 2 2    cos 2t sin 2t   39. yh = c1 cos 2t + c2 sin 2t and S = {cos 2t, sin 2t} so W (S) =   = 2.    −2 sin 2t 2 cos 2t    cos 2t  0 sin 2t  0 = Then, u1 = 12  = − 12 tan2 2t and u2 = 12   sec 2t tan 2t 2 cos 2t −2 sin 2t sec 2t tan 2t  1 1 1 2 tan 2t so u1 = 4 (2t − tan 2t) and u2 = 4 ln sec 2t so yp = =

1 1 cos 2t (2t − tan 2t) + sin 2t ln sec 2t 4 4 1 1 t cos 2t + sin 2t ln sec 2t 2 4

because − 14 cos 2t tan 2t = − 14 sin 2t is a solution of the corresponding homogeneous equation. Hence, a general solution is y = yh + yp = c1 cos 2t + c2 sin 2t + 12 t cos 2t + 1 4 sin 2t ln sec 2t.    cos 2t sin 2t   41. yh = c1 cos 2t + c2 sin 2t and S = {cos 2t, sin 2t} so W (S) =  = 2. −2 sin 2t 2 cos 2t      0  cos 2t sin 2t  0  Then, u1 = 12  2 = − 12 sec 2t tan 2t and u2 = 12  =  sec 2t 2 cos 2t −2 sin 2t sec2 2t  1 1 1 1 1 2 sec 2t so u1 = − 4 sec 2t and u2 = 4 ln (sec 2t + tan 2t) so yp = − 4 + 4 sin 2t ln (sec 2t + tan 2t) and y = yh + yp =c1 cos 2t + c2 sin 2t − 14 + 14 sin 2t ln (sec 2t + tan 2t) with first derivative y = 12 [cos 2t (4c2 + ln (sec 2t + tan 2t)) − 4c1 sin 2t + tan 2t]. Application of the initial conditions results in 1 − + c1 = 0 4 2c2 = 1 so c1 = 1/4 and c2 = 1/2 and the solution to the initial value problem is y = 14 cos 2t + 1 1 1  2 sin 2t − 4 + 4 sin 2t ln (sec 2t + tan 2t). The following figure shows the graph of y + 2  4y = sec 2t, y(0) = 0, y (0) = c for various values of c.

81

82

CH A P T E R 4:

Higher Order Equations

2

y

1.5

1

0.5

⫺0.75

⫺0.5

⫺0.25

0.25

⫺0.5

⫺1

⫺1.5

0.5

0.75

t

Answers, Hints, and Solutions to Selected Exercises

43. S = {cos t, sin t}, yh = c1 cos t + c2 sin t, W (S) = 1. yp = u1 y1 + u2 y2 ⇒    1  0 sin t  dt = − sin t tan2 t dt W (S) tan2 t cos t      2 = − sin t sec t − 1 dt = − (sec t tan t − sin t) dt 

u1 =

= − sec t − cos t and   sin2 t 0  u2 = dt dt = 2  tan t cos t   1 − cos2 t = dt = (sec t − cos t) dt cos t 

 1  cos t W (S) − sin t

= ln(sec t + tan t) − sin t. Therefore, by y = y u2 y2 = c1 cos(t) + h + yp= yh + u1 y  solution  is given 

1 +

 a general t t t t − sin + log sin + cos − 2. Applying sin(t) c2 − log cos 2 2 2 2 the initial conditions gives us y = 2 cos t + sin t − 2 + sin t − ln(sec t + tan t).  √  √ √ 1 1 π cos 3t + 36 2(6 + 2 ln 2) sin 3t − 13 t cos 3t + 19 sin 3t ln sin 3t 45. y = − 16 2 + 36      t t 51. y = 1k 0 f (z) cos kz dz sin kt − 0 f (z) sin kz dz cos kt      t −kz t 1 53. y = 2k f (z) dz ekt − 0 ekz f (z) dz e−kt 0e 55. S = {t −1 , t −1 ln t}, yh = t −1 (c1 + c2 ln t), W (S) = t −3 . Assuming that yp = u1 y1 + u2 y2 ⇒  u1 = =−

 1  0 W (S) ln t

  t −1 ln t  dt = − t 2 (ln t)2 dt t −2 (1 − ln t)

2 3 2 2 1 t + t ln t − (ln t)2 27 9 3

and  u2 =

 1  t −1 W (S) −t −2

  0  dt = t 2 ln t dt ln t 

1 1 = − t 3 + t 3 ln t. 9 3 Therefore, y = yh + yp = c1 t −1 + c2 t −1 ln t + ln t − 2. 5 57. y = c1 t 6 + c2 t −1 − 13 ln t + 18

83

84

CH A P T E R 4:

Higher Order Equations

59. If y = eu(t) = eu , then y = eu u and y = eu (u )2 + eu u . Substitution into the equation and dividing by e2u gives us   e−2t eu · (eu (u )2 + eu u ) − e2u (u )2 − 2t(1 + t)e2u = 0   e−2t ((u )2 + u ) − (u )2 − 2t(1 + t) = 0 u = 2t(1 + t)e2t . Integrating twice gives us u = 12 t 2 e2t − 12 te2t + 14 e2t + c1 t + c2  exp 12 t 2 e2t − 12 te2t + 14 e2t + c1 t + c2 .

so

y=

61. (a) y = t 2 (c1 cos t + c2 sin t) + t; (b) y = t 2 (c1 cos t + c2 sin t) + t (infinitely many solutions) 63. (a) y = t −1/2 (c1 cos 2t + c2 sin 2t) + t −1/2 ; (b) No solution; (c) y = t −1/2 (1 − cos 2t) 65. A general√solution is y = c1 t + c2 sin t + cos t and the solution to the initial value problem 4 2 is y = π−4 t + π+4 π−4 sin t + cos t.

Exercises 4.5  2   3t t 2t − 2t 2     1. W (S) =  6t 1 2 − 4t  = 0, linearly dependent    6 0 −4    −t   e e3t te3t     −t 3t 3t 3t  3. W (S) =  −e 3e e + 3te  = 16e5t , linearly independent    e−t 9e3t 6e3t + 9te3t     1 t t2 t3 t 4      0 1 2t 3t 2 4t 3      2  5. W (S) =  0 0 2 6t 12t  = 288, linearly independent   0 0 0 6 24t     0 0 0 0 24  7. 4th order, y = (c1 + c2 t + c3 t 2 )e−2t + c4 e2t 9. 4th order, y = c1 + c2 t + c3 cos 3t + c4 sin 3t 11. 6th order, y = e−3t [(c1 + c2 t) cos 4t + (c3 + c4 t) sin 4t] + c5 e−5t + c6 e−t/3 13. All are solutions of y = 0 but the Wronskian of the set is 0 so they cannot be a fundamental set. 15. Yes: y(4) − 14 y − 2y − 72 y + y = 0

  17. k = 0 with multiplicity 3 is the solution of the characteristic equation so S = 1, t, t 2 is a fundamental set of solutions and, thus, y = c0 + c1 t + c2 t 2 is a general solution.

Answers, Hints, and Solutions to Selected Exercises

19. y = c1 + c2 e5t + c3 e5t t 21. The characteristic equation is k 4 + 16k 2 = 0 and the left-hand side of the equation factors resulting in k 2 (k 2 + 16) = 0 so k1,2 = 0 is a 0 of multiplicity 2 and k 2 + 16 = 0 has roots k3,4 = ±4i. Thus, a fundamental set of solutions is S = {1, t, cos 4t, sin 3t} and a general solution of the equation is y = c1 + c2 t + c3 cos 4t + c4 sin 4t. 23. The characteristic equation is 3k 3 − 4k 2 − 5k + 2 = 0. Using synthetic division, we see that k = −1 is a zero of 3k 3 − 4k 2 − 5k + 2: −1| 3 −4 −5 2 3 −7 2 0 This means that 3k 3 − 4k 2 − 5k + 2 = (k + 1)(3k 2 − 7k + 2) and the quadratic factors quickly so we see that 3k 3 − 4k 2 − 5k + 2 = (k + 1)(3k 2 − 7k + 2) = (k + 1)(3k − 1)(k − 2) = 0 so k1 = −1 has corresponding solution y1 = e−t , k2 = 1/3 has corresolution y3 = e2t . A funsponding solution y2 = et/3 , and k3 = 2 has corresponding  −t t/3 2t damental set of solutions is then S = e , e , e and a general solution is y = c1 e−t + c2 et/3 + c3 2t. 25. The characteristic equation is k 3 − 5k + 2 = 0. Using synthetic division, we see that k = 2 is a zero of k 3 − 5k + 2: 2| 1 0 −5 2 1 2 −1 0 This means that k 3 − 5k + 2 = (k − 2)(k 2 − 5k + 2) and although k 2 + 2k − 1 does not “factor√easily,” using the quadratic formula we see that k 2 + 2k − 1 =√0 if k2,3 = −1 ± 2. Thus, k1 = 2 has corresponding solution y1 = e2t , k2 = −1 − 2 has cor√ √ (−1− 2)t responding √solution y2 = e , and k3 = −1 + 2  has corresponding solution √ √ (−1+ 2)t 2t (−1− 2)t (−1+ 2)t so a fundamental set of solutions is S = e , e ,e y3 = e √

and a general solution is y = c1 e2t + c2 e(−1−

2)t

√

+ c3 e(−1+

2)t .

27. The characteristic equation is k 4 + k 3 = k 3 (k + 1) = 0 so k1,2,3 = 0 is azero of multi plicity 3 and k4 = −1 is a zero of multiplicity 1. A fundamental set is S = 1, t, t 2 , e−t ; a general solution is y = c1 + c2 t + c3 t 2 + c4 e−t. 29. The characteristic equation is r 4 − 16 = (r 2 + 4)(r 2 − 4) = 0 with solutions r1 = 2, r2 = −2, and r3,4 = ±2i. Therefore, four linearly independent solutions to the equation are y1 = e2t , y2 = e−2t , y3 = cos 2t, and y4 = sin 2t. A general solution is then y = c1 e−2t + c2 e2t + c3 sin (2t) + c4 cos (2t). 31. The characteristic equation is r 4 + 7r 3 + 6r 2 − 32r − 32 = 0. Observe that the first three terms have a common factor of r 2 and the last two have a common factor of −32 so the equation can be rewritten as r 2 (r 2 + 7r + 6) − 32(r + 1) = 0. r 2 + 7r + 6 = (r + 6)(r + 1) so the characteristic equation further factors as r 2 (r + 6)(r + 1) − 32(r + 1) = (r + 1)[r 2 (r + 6) − 32] = (r + 1)(r − 2)(r + 4)2 = 0. The solutions of the characteristic equation are r1 = −1 with corresponding solution y1 = e−t , r2 = 2 with corresponding solution y2 = e2t , and r3,4 = −4 with solutions y3 = e−4t , and t4 = te−4t . Therefore, a general solution is y = c1 e2t + c2 e−t + c3 e−4t + c4 e−4t t.

85

86

CH A P T E R 4:

Higher Order Equations

33. The characteristic equation is k 5 + 4k 4 = k 4 (k + 4) = 0 so k1,2,3,4 = 0 is a zero of multiplicity 4 and k5 = −4 is a zero of multiplicity 1. Four linearly independent solutions corresponding to k1,2,3,4 = 0 are y1 = 1, y2 = t, y3 = t 2 , and y4 = t 3 . One solution corresponding to k5 = −4 is y5 = e−4t . Therefore, a fundamental set is S =   2 3 −4t 1, t, t , t , e ; a general solution is y = c1 + c2 t + c3 t 2 + c4 t 3 + c5 e−4t . 35. The characteristic equation is k 5 + 3k 4 + 3k 3 + k 2 = k 2 (k 3 + 3k 2 + 3k + 1) = k 2 (k + 1)3 = 0 so k1,2 = 0 is a zero of multiplicity 2 with corresponding solutions y1 = 1 and y2 = t and k3,4,5 = −1 is a zero of multiplicity 3 with corresponding solutions y3 = e−t ,  −t 2 −t −t y4 = te , and y5 = t e . Thus, a fundamental set is S = 1, t, e , te−t , t 2 e−t and a general solution is y = c1 + c2 t + c3 e−t + c4 te−t + c5 t 2 e−t . 37. The characteristic equation is r 4 + 8r 2 + 16 = (r 2 + 4)2 = 0, which has solution r1,2,3,4 = −2i so four linearly independent solutions are y1 = cos 2t, y2 = t cos 2t, y3 = sin 2t, and y4 = t sin 2t. Thus, a general solution is y = (c1 + c2 t) cos 2t + (c3 + c4 t) sin 2t. 39. The solutions of r 6 + 12r 4 + 48r 2 + 64 = 0 are r1,2,3,4,5,6 = ±2i. Thus, six linearly independent solutions are y1 = cos 2t, y2 = t cos 2t, y3 = t 2 cos 2t, y4 = sin 2t, y5 = t sin 2t, and y6 = t 2 sin 2t. Thus, a general solution is y = (c1 + c2 t + c3 t 2 ) cos 2t + (c4 + c5 t + c6 t 2 ) sin 2t. 41. The characteristic equation of the corresponding homogeneous equation is r 3 − 1 = (r − 1)(r 2 + r + 1) = 0. One solution of the characteristic equation is r1 = 1 with corresponding solution of the differential equation y1 =et . Using the quadratic formula to  √ solve r 2 + r + 1 = 0, we obtain r1,2 = 12 −1 ± 3i so two more linearly independent √  √  solutions are y2 = e−t/2 cos 23 t and y3 = e−t/2 sin 23 t . A general solution is then  √   √  y = c1 et + e−t/2 c2 cos 23 t + c3 sin 23 t . Applying the initial conditions results   √ √  √  in y = et − 3e−1/2t sin 1/2 3t − e−1/2t cos 1/2 3t 43. A general solution of the equation is y = (c1 + c2 t)e2t + (c3 + c4 t)e−2t . Applying the initial conditions yields y = t(e2t − e−2t ). 45. The characteristic equation is k 4 − 5k 2 + 4 = (k 2 − 4)(k 2 − 1) = (k + 2)(k − 2)(k + 1)(k − 1) = 0. A solution corresponding to k1 = −2 is y1 = e−2t , a solution corresponding to k2 = 2 is y2 = e2t , a solution corresponding to k3 = −1 is y3 = a solution corresponding to k4 = 1 is y4 = et . Thus, a fundamental e−t , and  −2t  2t −t t set is e , e , e , e and a general solution is y = c1 e−2t + c2 e2t + c3 e−t + c4 et with derivatives y = −2c1 e−2t + 2c2 e2t − c3 e−t + c4 et , y = 4c1 e−2t + 4c2 e2t + c3 e−t + c4 et , and y = −8c1 e−2t + 8c2 e2t − c3 e−t + c4 et . Applying the initial conditions gives us c1 + c2 + c3 + c4 = −1 −2c1 + 2c2 − c3 + c4 = 3 4c1 + 4c2 + c3 + c4 = −7 −8c1 + 8c2 − c3 + c4 = 15, which has solution c1 = −2, c2 = 0, c3 = 1, and c4 = 0 so the solution to the initial value problem is y = e−t − 2e−2t .

Answers, Hints, and Solutions to Selected Exercises

47. The characteristic equation for the corresponding homogeneous equation is 8k 5 + 4k 4 + 66k 3 − 41k 2 − 37k = k(8k 4 + 4k 3 + 66k 2 − 41k − 37) = 0. This shows us that k1 = 0 is a solution of the characteristic equation with multiplicity 1; a solution of the differential equation corresponding to k1 = 0 is y1 = 1. Now, by inspection we see that k2 = 1, with corresponding solution y2 = et , is a zero of 8k 4 + 4k 3 + 66k 2 − 41k − 37 and using synthetic division, 1| 8 8

4 66 −41 12 78 37

−37 , 0

shows us that 8k 4 + 4k 3 + 66k 2 − 41k − 37 = (k − 1)(8k 3 + 12k 2 + 78k + 37). By the Rational Root theorem, the possible rational roots of 8k 3 + 12k 2 + 78k + 37 are ±1, ±1/2, ±1/4, ±1/8, ±37, ±37/2, ±37/4, and ±37/8. By inspection we note that ±1 are not roots of 8k 3 + 12k 2 + 78k + 37. Testing 1/2, 1/2| 8 8

12 16

78 86

37 , 80

shows us that k = 1/2 is not a root of 8k 3 + 12k 2 + 78k + 37. On the other hand, testing −1/2, −1/2|

8 8

12 8

78 74

37 , 0

shows us that k = −1/2 is a root of 8k 3 + 12k 2 + 78k + 37 and that 8k 3 + 12k 2 + 78k + 37 = (k − 1/2)(8x 2 + 8x + 74) = (2k − 1)(4k 2 + 4k + 37). A solution corresponding to k3 = 1/2 is y3 = et/2 . We use the quadratic formula to solve 4k 2 + 4k + 37 = 0 and obtain k4,5 = − 12 ± 3i with corresponding solutions y4 = e−t/2 cos 3t and y5 = e−t/2 sin 3t. Thus, a fundamental set is  t S = 1, e , e−t/2 , e−t/2 cos 3t, e−t/2 sin 3t and a general solution is y = c1 + c2 et + c3 e−t/2 + e−t/2 (c4 cos 3t + c5 sin 3t), with derivatives y = c2 et − 12 c3 e − t/2 +      e − t/2 − 12 c4 + 3c5 cos 3t + − 3c4 − 12 c5 sin 3t , y = c2 et + 14 c3 e − t/2 + e − t/2      35 − 35 4 c4 − 3c5 cos 3t + 3c4 − 4 c5 sin 3t ,      107 99 99 107 c − c cos 3t + c + c sin 3t , 4 5 4 5 8 5 4 8 y(4)

y = c2 et − 18 c3 e − t/2 + e − t/2 and

1 = c2 − 16 c3 e − t/2 + e−t/2    1081 105 105 16 c4 + 2 c5 cos 3t + − 2 c4

et

  + 1081 16 c5 sin 3t . Applying the initial conditions results in the system y(0) = c1 + c2 + c3 + c4 = 4 1 1 y (0) = c2 − c3 − c4 + 3c5 2 2 1 35 y (0) = c2 + c3 − − 3c5 4 4 1 107 99 y (0) = c2 − c3 + c4 − c5 8 8 4 1 1081 105 (4) y (0) = c2 + c3 + c4 + c5 16 16 2

= −14 = −14 = 139 =−

29 , 4

87

88

CH A P T E R 4:

Higher Order Equations

which has solution c1 = 0, c2 = 0, c3 = 1, c4 = 3, and c5 = −4. Thus, the solution to the initial value problem is y = e−t/2 (1 + 3 cos 3t − 4 sin 3t), which is graphed in the following plot. y 4

3

2

1

2

4

6

8

10

12

t

⫺1 ⫺2

Note: A CAS was used to assist in constructing the solution. 49. The characteristic equation is r 5 + 8r 4 = r 4 (r + 8) = 0 with solutions r1,2,3,4 = 0 and r5 = −8. For the first four roots, solutions are y1 = 1, y2 = t, y3 = t 2 , and y4 = t 3 . For the fifth, we obtain y5 = e−8t . Therefore a general solution is y = c1 + c2 t + c3 t 2 + c4 t 3 + c5 e−8 . Application of the initial conditions results in y = 8t 3 + 4t + 9. 53. (a) 3; (b) t − 1; (c) 2; (d) 0

59.

  e−3t     −3 e−3t  W (S) =   9 e−3t     −27 e−3t −12t 1080e

e−t

e−4t cos (3t)

−e−t

−4 e−4t cos (3t) − 3 e−4t sin (3t)

e−t

7 e−4t cos (3t) + 24 e−4t sin (3t)

−e−t

44 e−4t cos (3t) − 117 e−4t sin (3t)

     −4t −4t −4 e sin (3t) + 3 e cos (3t)  =  7 e−4t sin (3t) − 24 e−4t cos (3t)    44 e−4t sin (3t) + 117 e−4t cos (3t)  e−4t sin (3t)

   f (t)  t f (t) t 2 f (t)   d   = 2( f (t))3 f (t) + t dtd f (t) 2t f (t) + t 2 dtd f (t) 61. (b) W (S) =  dt f (t)   2   d f (t) 2 d f (t) + t d2 f (t) 2 f (t) + 4t d f (t) + t 2 d2 f (t) dt dt dt 2 dt 2 dt 2 63. (a) y = − 0.09090909091 e − 3.0 t +0.5142594770 sin (1.414213562t) +0.09090909091 cos (1.414213562t); (b) y = 2.020725943 e2.0 t sin (1.732050808 t ) + e2.0 t cos (1.732050808 t) − 0.5773502693 e2.0 t sin (1.732050808 t) t − 3.500000000 e2.0 t cos (1.732050808 t) t; (c) y = 0.0002889303734 e−36.99601821 t − 1.061192194e0.4334929696t sin (0.5236492683t) − 1.000288930 e0.4334929696t cos (0.5236492683t)

Answers, Hints, and Solutions to Selected Exercises

65. y = 0.05460976246 y0 e0.4712630674t + 0.8550147286 y0 e−0.2076539530 t sin (1.126208087 t) − 0.05460976246 y0 e−0.2076539530 t cos (1.126208087 t) 67. y(t) = c2 cos2 ( 12 (2c1 − t))

Exercises 4.6 1. y = c1 + c2 t + c3 e−t + 12 et 3. The characteristic equation for the corresponding homogeneous equation is r 5 − r 4 = r 4 (r − 1) = 0 with roots r1,2,3,4 = 0 and r5 = 1. Therefore, a fundamental set of solutions is S = {1, t, t 2 , t 3 , et } and a general solution of the corresponding homogeneous equation is yh = c1 + c2 t + c3 t 2 + c4 t 3 + c5 et . For the forcing function, the associated set of functions is F = {1}. Because the function 1 is a solution of the corresponding homogeneous equation, we multiply F by t n where n is the lowest positive integral power so that no element of t n F is a solution of the corresponding homogeneous equation. In this case t 4 F = {t 4 } and we see that t 4 is not a solution of the corresponding homogeneous equation. Therefore, we assume that a particular solution takes the form yp = At 4 with derivatives yp = 4At 3 , yp = 12At 2 , and so on. Substituting yp into the nonhomoge(5)

(4)

4

1 neous equation results in yp − yp = −24A = 1 ⇒ A = − 24 so a particular solution of the nonhomogeneous equation is yp (t) = −1/24. Therefore, a general solution of the 1 4 equation is y = yh + yp = c1 + c2 t + c3 t 2 + c4 t 3 + c5 et − 24 t .

5. y = c1 + c2 t + c3 e−3t + c4 e3t + 16 te3t 7. The corresponding homogeneous equation has general solution yh = c1 e3t + c2 e2t + c3 et . For the forcing function, we have F1 = {e−3t }. Because e−3t is a solution of the corresponding homogeneous equation we multiply F1 by t n where n is the smallest positive integer so that no element of t n F1 is a solution of the corresponding homogeneous equation. In this case, t f1 = {te−3t }. For the second term, F2 = {te−t , e−t }. No element of F2 is a solution to the corresponding homogeneous equation. Combining F1 and F2 , we search for a particular solution of the nonhomogeneous equation of the form yp = Ate−3t + Bte−t + Ce−t . Computing the derivatives and substituting into the nonhomogeneous equation, solving for A, B, and C, and forming yh and then y = yh + yp gives us the general solution of the equation: y = e−3t t + 34 te−t − 78 e−t + 32 e−3t − 14 e−t t 2 + c1 e−3t + c2 e−2t + c3 e−t . 9. y = et + c1 e4t + c2 e−5t cos (t) + c3 e−5t sin (t) 11. The corresponding homogeneous equation is y + 4y = 0, which has general solution yh = c1 + c2 cos 2t + c3 sin 2t. A fundamental set is S = {1, cos 2t, sin 2t} with Wronskian   1 cos 2t sin 2t   W (S) = 0 −2 sin 2t 2 cos 2t  = 8. 0 −4 cos 2t −4 sin 2t  Next,

  0  1 u1 =  0 8 tan 2t

cos 2t −2 sin 2t −4 cos 2t

 sin 2t  1 2 cos 2t  = tan 2t −4 sin 2t  4

89

90

CH A P T E R 4:

Higher Order Equations

so u1 =

1 8

ln sec 2t;

 1  1 u2 = 0 8 0

 sin 2t  2 cos 2t  = sin2 2t −4 sin 2t 

0 0 tan 2t

so u2 = 18 (4t − sin 4t); and  1  1  u3 = 0 8 0

cos 2t −2 sin 2t −4 cos 2t

 0  1 0  = − tan 2t 4 sec 2t 

so u3 = − 18 ln sec 2t. Thus, yp =

1 1 1 ln sec 2t + cos 2t(4t − sin 4t) − sin 2t ln sec 2t 8 8 8

and a general solution is y = yh + yp = c1 + c2 cos 2t + c3 sin 2t + 18 ln sec 2t + 1 1 8 cos 2t(4t − sin 4t) − 8 sin 2t ln sec 2t. 13. The corresponding homogeneous equation is y(4) + 4y = 0 with characteristic equation k 4 + 4k 2 = k 2 (k + 4) = 0 so a fundamental set is S = {1, t, cos 2t, sin 2t} and a general solution of the corresponding homogeneous equation is yh = c1 + c2 t + c3 cos 2t + c4 sin 2t. Next,   1 t cos 2t sin 2t   0 1 −2 sin 2t 2 cos 2t  W (S) =   = 32. 0 0 −4 cos 2t −4 sin 2t  0 0 8 sin 2t −8 cos 2t  Next,

so u1 =

  0   0 1   u1 = 32  0 sec2 2t

cos 2t −2 sin 2t −4 cos 2t 8 sin 2t

 sin 2t  1 2 cos 2t  = − t sec2 2t −4 sin 2t  4 −8 cos 2t 

1 16 (ln sec 2t − 2t tan 2t);

 1  0 1   u2 = 32 0 0 so u2 =

t 1 0 0

1 8

0 0 0 sec2 2t

cos 2t −2 sin 2t −4 cos 2t 8 sin 2t

 sin 2t  2 cos 2t  1 = sec2 2t −4 sin 2t  4 −8 cos 2t 

tan 2t;  1  0 1   u3 = 32 0 0

t 1 0 0

0 0 0 sec2 2t

 sin 2t  2 cos 2t  1 = sec 2t tan 2t −4 sin 2t  8 −8 cos 2t 

Answers, Hints, and Solutions to Selected Exercises

so u3 =

1 16

sec 2t; and  1 t  1 0 1  u4 =  32 0 0  0 0

cos 2t −2 sin 2t −4 cos 2t 8 sin 2t

 0  0  1  = − sec 2t 0  8  sec2 2t 

1 so u4 = − 16 ln(sec 2t + tan 2t). Hence,

1 1 1 (ln sec 2t − 2t tan 2t) + t tan 2t + cos 2t sec 2t 16 8 16 1 − sin 2t ln(sec 2t + tan 2t) 16 1 (ln sec 2t − sin 2t ln(sec 2t + tan 2t)) = 16

yp =

and a general solution is y = yh + yp = c1 + c2 t + c3 cos 2t + c4 sin 2t +

1 (ln sec 2t − sin 2t ln(sec 2t + tan 2t)). 16

15. The corresponding homogeneous equation is y + 9y = 0, which has characteristic equation r 3 + 9r = r(r 2 + 9) = 0 with solutions r1 = 0 and r2,3 = ±3i. Therefore, a fundamental set of solutions is S = {1, cos 3t, sin 3t} and W (S) =  1 cos(3t) sin(3t)     0 −3 sin(3t) 3 cos(3t)  = 17. A general solution of the corresponding homoge  0 −9 cos(3t) −9 sin(3t) neous equation is yh = c1 + c2 cos 3t + c2 sin 3t. Next,   cos(3t) sin(3t)    0  1 1 1   0 −3 sin(3t) 3 cos(3t) sec2 3t dt = tan 3t, u1 = dt =    27  2 9 27 sec (3t) −9 cos(3t) −9 sin(3t)   0 sin(3t)   1 1   0 3 cos(3t)  dt u2 = 0  27  0 sec2 (3t) −9 sin(3t)  1 1 sec(3t) dt = − ln(sec(3t) + tan(3t)), =− 9 27 and

  cos(3t) 0   1  1 1 1   0  dt = − u3 = sec(3t) tan(3t) dt = − sec(3t). 0 −3 sin(3t)  27  9 27 0 −9 cos(3t) sec2 (3t)

91

92

CH A P T E R 4:

Higher Order Equations

Therefore, y=

1 [(c1 + ln(cos(3t/2) − sin(3t/2)) − ln(cos(3t/2) + sin(3t/2))) cos 3t+ 27 c2 sin 3t + c3 ] .

17. A general solution of the corresponding homogeneous equation y + 4y = 0 is yh = c1 + c2 cos 2t + c3 sin 2t. Using variation of parameters to find a particular solution of the nonhomogeneous equation yields yp = 18 (−2t cos(2t) − log(cos(t) − sin(t)) + log(sin(t) + cos(t)) + sin(2t) log(cos(2t))). Therefore, y = yh + yp = 18 (c1 + c2 cos2 t − 2t cos 2t − ln(cos t − sin t) + ln(cos t + sin t) + c3 sin 2t + ln(cos 2t) sin 2t). 19. The characteristic equation for the corresponding homogeneous equation is r 3 − 3r 2 + 3r − 1 = (r − 1)3 = 0 with solutions r1,2,3 = 1. Therefore, a fundamental set of solutions for the corresponding homogeneous equation is S = {et , tet , t 2 et } and a general solution of the corresponding homogeneous equation is yh = et (c1 + c2 t + c3 t 2 ). With variation of parameters, a particular solution of the nonhomogeneous equation is found to be yp = 14 et t 2 (2 log(t) − 3). Therefore, a general solution of the nonhomogeneous equation 1 2 t 2 t t t 2 t is y = yh + yp = −3 4 t e + 2 ln (t) t e + c1 e + c2 te + c3 t e . 21. Use undetermined coefficients: y =

1 −3t 6e

+ c1 e−4t + c2 e−2t + c3 e3 t .

23. The characteristic equation for the corresponding homogeneous equation is r 3 + 3r 2 + 2r = r(r 2 + 3r + 2) = r(r + 1)(r + 2) = 0 so a fundamental set of solutions for the corresponding homogeneous equation is S = {1, e−t , e−2t } and a general solution of the corresponding homogeneous equation is yh = c1 + c2 e−t + c3 e−2t . Using undetermined coefficients, assume that a particular solution has the form yp = A cos t + B sin t. Substitute yp into the nonhomogeneous equation to determine yp . Then a general solution of the 1 3 sin (t) − 10 cos (t) + c1 e−2t + nonhomogeneous equation is given by y = yh + yp = 10 −t c2 e + c3 25. y = − 12 t 2 + c1 + c2 cos t − ln(− cos(t/2) − sin(t/2)) − ln(− cos(t/2) + sin(t/2)) − (c3 − ln(cos(t/2) − sin(t/2)) + ln(cos(t/2) + sin(t/2))) sin t 27. y = 1 + t − cos t − ln(cos t) − 2 tanh−1 (tan(t/2)) sin t sin(t) cos(t) − . Apply the 2 2 initial conditions to solve the initial value problem: y = 2 − 2 cos t − 12 t sin t.

29. A general solution of the equation is y = c1 e−t + c3 t + c2 −

1  1  y + 2 y = t ln t t ln t  1  1  2 the forcing function is f (t) = 2 . Next, W ({1, t , t ln t}) =  0  t ln t 0

31. First write the equation in standard form: y −

−2 ln(t). Now,  u1 = −

  1   2 ln t  

0 0 1 t 2 ln t

t2 2t 2

1 to see that t 2 ln t  t ln(t)  t2 2t ln(t) + 1  = 1   2 t

 t ln(t)  ln(t) + 1  dt = t , 1  2 ln t  t

Answers, Hints, and Solutions to Selected Exercises

 u2 = −

 1 1  0  2 ln t  0

and  u3 = −

 t ln(t)   ln(t) + 1  dt = − 1 ,  1  2t ln t  t

0 0 1 2 t ln t

 1 1  0  2 ln t  0

t2 2t 2

0 0 1 t 2 ln t

   1  .  dt =  ln t 

Therefore, y = yh + yp = c1 + c2 t 2 + c3 t ln t + t. √ 33. A general solution is y = c1 t + c2 t ln t + c3 t − t 2 ; the solution of the initial value problem is y = t − t 2 + 2t ln t. 35. A general solution is y = c1 + c2 t + c3 t ln t + c4 t 2 + 2t −1/2 ; the solution of the initial value problem is y = 18 −41 + 16t −1/2 + 12t + 13t 2 − 30t ln t .

Exercises 4.7 1. Substituting y = x r into the equation and simplifying gives us 4r(r − 1) − 8r + 5 = (2r − 5)(2r − 1) = 0 so r1 = 5/2 and r2 = 1/2. A fundamental set of solutions is S = {x 1/2 , x 5/2 }; a general solution is y = c1 x 1/2 + c2 x 5/2 . 3. y = c1 x + c2 x 4 5. Substituting y = x r into the equation and simplifying gives us 4r(r − 1) + 17 = 4r 2 − 1 4r + 17 = 0. Using the quadratic formula to √solve for r gives √ us r1,2 = 2 ± 2i. Therefore, a fundamental √ set of solutions is S = { x cos(2 ln x), x sin(2 ln x)} and a general solution is y = x(c1 cos(2 ln x) + c2 sin(2 ln x)). 7. y = x(c1 cos(3 ln x) + c2 sin(3 ln x)) 9. Substituting y = x r into the equation and simplifying gives us 4r(r − 1) + 8r + 1 = with solutions r1,2 = −1/2. Therefore, a fundamental set of solutions is 4r 2 + 4r√+ 1 = 0√ S = {1/ x, ln x/ x} and a general solution is y = x −1/2 (c1 + c2 ln x). (We are assuming that x > 0.) 11. y = x 3 (c1 + c2 ln x) 13. Substituting y = x r into the equation and simplifying gives us r(r − 1)(r − 2) + 22r(r − 1) + 124r + 140 = r 3 + 19r 2 + 104r + 140 = (r + 2)(r + 7)(r + 10) = 0 so r1 = −10, r2 = −7, and r3 = −2. A fundamental set of solutions is then S = {x −10 , x −7 , x −2 } and a general solution is y = c1 x −10 + c2 x −7 + c3 x −2 . 14. y = c1 x −5 + c2 x 2 + c3 x 10 16. y = c1 x −1 + c2 cos(2 ln x) + c3 sin(2 ln x) 17. Substituting y = x r into the equation x 3 y + 2xy − 2y = 0 and simplifying gives us r(r − 1)(r − 2) + 2r − 2 = (r − 1)(r 2 − 2r + 2) = 0 with solutions r1 = 1 and r2,3 = 1 ± i. Therefore, a fundamental set of solutions is S = {x, x cos(ln x), x sin(ln x)} and a general solution is y = c1 x + x(c2 cos(ln x) + c3 sin(ln x)). 19. y = x −1 (c1 + c2 ln x + c3 (ln x)2 )

93

94

CH A P T E R 4:

Higher Order Equations

21. A fundamental set of solutions for the corresponding homogeneous equation is S = {x −2 , x −2 ln x} with Wronskian W (S) = x −5 . In standard form, the equation is y + 5  4 y + 2 y = x −7 , which shows us that the forcing function is f (x) = x −7 . Now, we x x use variation of parameters to form a particular solution of the form yp = u1 y1 + u2 y2 :     log(x)  0      2 x  dx = − ln x dx = 1 + ln x u1 = x 5  1 2 log(x)  x4 9x 3 3x 3  1 −  7  x x3 x3 and  u2 =

  1  5  x2 x  2 −  x3

After forming yp , y = yh + yp =

0 1 x7

    1   dx = x −4 dx = − 3 .  3x 

c1 c2 ln (x) 1 −5 + +9x x2 x2

23. y = c1 sin (ln (x)) + c2 cos (ln (x)) + 15 x −2 25. With the substitution x = et , the equation becomes d 2 y/dt 2 + dy/dt − 6y = 2et . The corresponding homogeneous equation has solution yh = c1 e2t + c2 e−3t = c1 x 2 + c2 x −3 . Using undetermined coefficients, we assume that a particular solution of the transformed equation takes the form yp = Aet . Substituting yp into the transformed nonhomogeneous equation gives us Aet + Aet − 6Aet = −4Aet = 2et so A = − 12 and yp = −2et = − 12 x. c1 Therefore, a general solution of the nonhomogeneous equation is y = yh + yp = −3 + x c2 x 2 − 12 x. 27. With the substitution x = et , the equation becomes d 2 y/dt 2 + 4y = 8. The corresponding homogeneous equation has general solution yh = c1 cos 2t + c2 sin 2t = c1 sin (2 ln (x)) + c2 cos (2 ln (x)). We now assume that a particular solution of the transformed equation has the form yp = A. Substituting into the nonhomogeneous equation gives us yp + 4yp = 4A = 8 so A = 2 an yp = 2. Therefore, a general solution of the nonhomogeneous equation is y = yh + yp = c1 sin (2 ln (x)) + c2 cos (2 ln (x)) + 2. c2 1 −3 29. y = 25 x + c1 x 2 + 4 + c3 x 2 ln (x) x 9 √ 1 2 3 31. y = 5 x + 5 x 33. A general solution of the equation is y = c1 cos(2 ln x) + c2 sin(2 ln x) and the solution to the initial value problem is y = cos (2 ln (x)). 35. A general solution of the equation is y = c1 x −10 + c2 x + c3 x 2 and the solution to the 1 −10 8 x + 11 x initial value problem is y = − 34 x 2 + 44 x + 12 x 2 sin (ln (x)) − 32 x 2 cos (ln (x)) √ 39. A general solution of the equation is y = c1 x + c2 x −1 + 15 x −2 and the solution to the √ 1 −2 initial value problem is y = 22 − 53 x −1 . 15 x + 5 x

37. y =

3 2

Answers, Hints, and Solutions to Selected Exercises

√ √ 1 3 x and the solution to 41. A general solution of the equation is y = c1 x + c2 x ln x + 25 √ √ 24 8 1 3 the initial value problem is y = 25 x − 5 x ln (x) + 25 x .    x r1 x r2   43. W (S) =   = −x r1 +r2 −1 (−r2 + r1 )  x r1 −1 r1 x r2 −1 r2  49. y = c1 + c2 cos(6 ln x) + c3 sin(6 ln x) 51. y = x(c1 + c2 ln x + c3 (ln x)2 ) 53. (c) y = c1 sin(2 arctan(x)) + c2 cos(2 arctan(x)); (d) y = c1 sin(2 arctan(x)) + c2 cos(2 arctan(x)) + 14 arctan(x); (e) y = 12 sin(2 arctan(x)); (f) y = 38 sin(2 arctan(x)) + 14 arctan(x). 106 −2 14 sin (5 ln (x)) 56 cos (5 ln (x)) x + − 25 5 25 x2 x2 √ 67. y = ± 2 c1 ln (x) + 2 c2 65. y =

Exercises 4.8 1. x = 0, R ≥ 1 3. x = ±2, R ≥ 1

      91 671 455 2 3 4 5. y = a0 + a1 x + 11 2 a1 − 15 a0 x + 6 a1 − 55 a0 x + 24 a1 − 4 a0 x +   4651 671 5 120 a1 − 4 a0 x + · · ·       5 5 a1 + 14 a0 + 24 x4 + 7. y = a0 + a1 x + 12 a1 + a0 + 12 x 2 + 12 a1 + 13 a0 + 13 x 3 + 24   11 1 1 5 120 a1 + 12 a0 + 12 x + · · · 3 9 9. y = a0 + a1 x − 14 a1 x 3 + 16 a1 x 4 − 80 a1 x 5 + · · ·       5 11. y = a0 + a1 x + 12 a1 − a0 x 2 + − 13 a1 − 13 a0 x 3 + − 24 a1 + 16 a0 x 4 +   1 1 − 120 a1 + 15 a0 x 5 + · · · 1 5 1 13. y = a0 + a1 x + 34 a0 x 2 + 12 a1 x 3 − 32 a0 x 4 − 32 a1 x 5 + · · · 1 8 1 x + 1386 x 12 + · · · 15. y = 1 + 13 x 4 + 42 5 15 17. y = a1 + a2 (x − 1) + 18 a2 (x − 1)2 + 96 a2 (x − 1)3 + 512 a2 (x − 1)4 +  39 5 6 2048 a2 (x − 1) + O (x − 1) 1 5 1 7 37 19. y = 1 + 16 x 3 − 30 x + 240 x − 90720 x9 + · · · 7 6 11 7 89 8 79 9 21. y = x + 12 x 2 − 18 x 4 − 18 x 5 − 120 x + 1680 x + 2688 x + 2880 x +··· 1 7 1 23. (a) y = a1 + a2 x − 12 a1 x 2 + 16 a2 x 3 − 18 a1 x 4 + 24 a2 x 5 − 240 a1 x 6 + 112 a2 x 7 − 11 13 3 1 1 8 9 2 3 4 1920 a1 x + 8064 a2 x + · · · ; (b) y = a1 + a2 x − 2 a1 x − 6 a2 x − 8 a1 x − 1 1 1 3 11 5 6 a2 x 7 − 896 a1 x 8 − 17280 a2 x 9 + · · · ; y = a1 + a2 x − 40 a2 x −  48 a1 x − 240    1 1 1 1 1 2 3 2 4 2 ka1 x + 3 a2 − 6 ka2 x + − 6 ka1 + 24 k a1 x +

95

96

CH A P T E R 4:

Higher Order Equations

   1 1 1 1 2 1 2 − 15 ka2 + 120 k 2 a2 + 10 a2 x 5 + 60 k a1 − 720 k 3 a1 − 45 ka1 x 6 +   23 1 1 1 − 1260 ka2 + 280 k 2 a2 + 42 a2 − 5040 k 3 a2 x 7 +   1 1 11 1 − 1680 k 3 a1 + 40320 k 4 a1 + 2520 k 2 a1 − 105 ka1 x 8 +   43 1 11 1 1 2 3 4 9 45360 k a2 − 11340 k a2 − 2835 ka2 + 362880 k a2 + 216 a2 x + · · · 

25. 2, 23 , 25 , 27 , 29 ,

2 11

1 4 1 5 1 6 17 7 53 11 9 41 27. y = 1 − 12 x 2 + 13 x 3 − 24 x − 24 x + 40 x − 5040 x − 20160 x 8 + 7560 x − 302400 x 10 − 3097 8219 3431 11 12 13 19958400 x + 119750400 x − 2075673600 x + · · ·

Exercises 4.9 1. x = 0, regular 3. x = 4, regular   1 2 1 3 1 3 x + 286 x + 5720 x 4 + 486200 x5 + · · · + 5. y = c1 x 7/2 1 + 13 x + 22   3 2 3 3 9 4 9 5 c2 1 − 35 x + 10 x − 10 x − 40 x − 200 x +···   3 2 29 3 13 251 7. y = c1 1 + 14 x − 104 x − 6864 x + 65472 x 4 + 11348480 x 5 + · · · x −5/9 +   1 13 2 59 29 53 c2 1 + 14 x − 644 x − 61824 x 3 + 247296 x 4 + 12364800 x5 + · · ·   1 2 1 3 1 1 x − 360 x + 8640 x 4 − 302400 x5 + · · · + 9. y = c1 x 1 − 13 x + 24    1 4 1 5 25 4 c2 ln (x) x 2 − 13 x 3 + 24 x − 360 x + · · · x −1 + (−2 − 2 x + 49 x 3 − 288 x +  157 5 −1 21600 x + · · · )x    √  9 4 3 2 9 11. y = c1 1 + 32 x 2 − 40 x + · · · x −2 + c2 3 x 1 − 26 x + 1976 x4 + · · ·   77 2 77 3 209 4 4807 5 x + 384 x + 3072 x + 245760 x +··· + 13. y = c1 x 7/4 1 + 78 x + 160    3 + 105 x 4 + 231 x 5 + · · · x −5/4 + (12 + 15 x + 15 x 2 − 13 x 3 − c2 ln (x) 15 x 8 64 256 4 2  1741 4 4141 5 −5/4 256 x − 1024 x + · · · )x 15. y = c1 x −7 + c2 x 17. y = c1 x 2 + c2 x 2 ln x

   

√ √ −1 + x 2 + ln −1/2 + x + −x + x + c2 x (−1 + x)−3/2 21. y = c1 −2 √ x (−1 + x) 23. y = c1 (1 − 2 x) + c2 ((−1 + 2 x) ln (x) − 2 + (1 − 2 x) ln (−1 + x)) 27. (a) = x−1/2 (c1 cos x + c2 sin x); (b) y = c1 J5 (4x) + Y5 (4x) 29 4 5 33. y = 1 + (1 − x) + 38 (−1 + x)2 − 13 (−1 + x)3 + 101 384 (−1 + x) − 128 (−1 + x) + · · ·

Answers, Hints, and Solutions to Selected Exercises

Chapter 4 Review Exercises

 5t  e 1  1. Both are solutions of y − 25y = 0 and W ({e5t , 1}) =   5 e5t 0 linearly independent    t t ln (t)   3. W (S) =   = t; linearly independent  1 ln (t) + 1    t cos (t)   5. All three are solutions of y + y = 0 and W (S) =  1 − sin (t)   0 − cos (t) linearly independent

    = −5e5t  = 0; 

 sin (t)   cos (t)  = t;  − sin (t) 

11. First, write the equation in standard form, y + (2/t)y + y = 0, to see that p(t) = 2/t. Using the reduction of order formula (4.4), y2 (t) = v(t)f (t) where  v(t) =

 1 e− p(t) dt dt = 2 (f (t))



 1 e− 2/t dt dt = −1 2 (t sin t)

 csc2 t dt = − cot t.

Then, a second linearly independent solution of the equation is y2 (t) = t −1 sin t cot t = t −1 cos t and a general solution is y = t −1 (c1 cos t + c2 sin t). 13. The characteristic equation is 6r 2 + 5r − 4 = (2r − 1)(3r + 4) = 0 with solutions r1 = 1/2 and r2 = −4/3. Therefore, a fundamental set of solutions is S = {et/2 , e−4t/3 } and a general solution is y = c1 e1/2t + c2 e−4/3t . 15. y = c1 e−2t + c2 e−t 17. y = c1 e1/2t + c2 e2t 19. y = c1 e−1/4t + c2 e1/5t 21. The characteristic equation is 2r 3 + 3r 2 + r = r(r + 1)(2r + 1) = 0 with solutions r1 = 0, r2 = −1, and r3 = −1/2. Therefore, a fundamental set of solutions is S = {1, e−t , e−t/2 } and a general solution is y (t) = c1 + c2 e−1/2t + c3 e−t . 23. The characteristic equation is 9r 3 + 12r 2 + 13r = r(9r 2 + 12r + 13) = 0 with solutions r1 = 0 and r2,3 = −2/3 ± i. Therefore, a fundamental set of solutions is S = {1, e−2t/3 cos t, e−2t/3 sin t} and a general solution is y = c1 + c2 e−2/3 t sin (t) + c3 e−2/3t cos (t). 25. A fundamental set of solutions for the corresponding homogeneous equation is S = {1, e−5t } and a general solution of the corresponding homogeneous equation is yh = c1 + c2 e−5t . To find a particular solution, we use undetermined coefficients. The associated set of functions is F = {t 2 , t, 1}. Because 1 is a solution of the corresponding homogeneous equation we multiply F by t giving us tF = {t 3 , t 2 , t}. No element of tF is a solution of the corresponding homogeneous equation so we assume that a particular solution has the form yp = At 3 + Bt 2 + Ct. Substituting yp into the nonhomogeneous equation 2 t. A general solution of the and solving for A, B, and C gives us yp = 13 t 3 − 15 t 2 + 25 1 2 1 3 2 t + c2 . nonhomogeneous equation is then y = yh + yp = − 5 t + 3 t + c1 e−5t + 25

97

98

CH A P T E R 4:

Higher Order Equations

3 27. y = c1 e−t sin (2t) + c2 e−t cos (2t) + 17 sin (2t) − 12 17 cos (2t)

29. A fundamental set of solutions for the corresponding homogeneous equation is S = {1, e2t } and a general solution of the corresponding homogeneous equation is yh = c1 + c2 e2t . To find a particular solution, we use variation of parameters. First, we compute W (S) = 2e2t . Assuming that yp = u1 y1 + u2 y2 ,    1  1 e2t 0 e2t  −2t dt = − e · dt u1 = 2t 2t W (S) 1/(1 + e ) 2e  2 1 + e2t  

1 1  =− t − ln 1 + e2t 2 2 

and     1 1 1 1 0  dt dt = e−2t W (S) 0 1/(1 + e2t ) 2 1 + e2t  

1 1 −2t 1  2t = − e + ln 1 + e . 2 2 2 

u2 =

Therefore, y = yh + yp = + c1 e2 t + c2 .

1 4

 ln 1 + e2t +

1 2t 4e

 ln 1 + e2t −

1 4

−

1 2

t−

1 2

t e2t

3 −2t e 31. y = e3t (c1 cos 2t + c2 sin 2t) + 29

33. Use a CAS to help you with the algebra. A fundamental set of solutions for the corresponding homogeneous equation is S = {e−4t , e−3t } and a general solution is yh = c1 e−3t + c2 e−4t . To find a particular solution, we use undetermined coefficients. The associated set of functions is F = {t 2 e−4t , te−4t , e−4t }. Multiplying F by t gives us tF = {t 3 e−4t , t 2 e−4t , te−4t } and we assume that a particular solution takes the form equation yp = At 3 e−4t + Bt 2 e−4t + Cte−4t . Substituting yp into the nonhomogeneous and solving for A, B, and C gives us yp = −e−4t t 3 + 3t 2 + 6t + 6 . Therefore a general solution of the nonhomogeneous equation is y = yh + yp = c1 e−3t + c2 e−4t −  −4t 2 t 6 + 3t + t e . 35. Use a CAS to help you with the algebra. A fundamental set of solutions for the corresponding homogeneous equation is S = {e−2t , te−2t , e4t } and a general solution of the corresponding homogenous equation is yh = (c1 + c2 t)e−2t + c3 e4t . We use undetermined coefficients to find a particular solution of the nonhomogeneous equation. For the first term in the forcing function, the associated set of functions is F1 = {e4t }. Multiplying F1 by t gives us tF1 = {te4t }. For the second term, the associated set of functions is F2 = {e−2t }. Multiplying F2 by t 2 gives us t 2 F2 = {t 2 e−2t }. Therefore, we assume that a particular solution of the nonhomogeneous equation has the form yp = Ate4t + Bt 2 e−2t . Substituting yp into the nonhomogeneous equa 2 1 −2t e 18t + 6e6t t + 6t − 2e6t + 1 . tion and solving for A and B results in yp = 216 Therefore, a general solution of the nonhomogeneous equation is y = yh + yp =   1 36

1 10 t 1 2 4t 1 1 4t te10 t − 108 e + 12 t e + 36 te4t + 216 e e−6t + c1 e−2 t + c2 e4t + c3 e−2t t.

4 8 1 2 − 125 t + 25 t + c1 e−t cos (2t) + c2 e−t sin (2t) + c3 e−t cos (2t) t + 37. y = 625 −t c4 e sin (2t) t

Answers, Hints, and Solutions to Selected Exercises

39. The characteristic equation is r 2 + 10r + 16 = (r + 8)(r + 2) = 0 with solutions r1 = −8 and r2 = −2 so a fundamental set of solutions is S = {e−8t , e−2t } and a general solution is y = c1 e−8t + c2 e−2t . Applying the initial conditions results in y = 23 e−2t − 23 e−8 t . 41. The characteristic equation is r 2 + 25 = 0 with solutions r1,2 = ±5i. Therefore, a fundamental set of solutions is S = {cos 5t, sin 5t} and a general solution is y = c1 cos 5t + c2 sin 5t. Applying the initial conditions results in y = cos (5t). 43. y =

19 t 25 e

45. y =

1 2

−4t + 1 tet − 19 25 e 5

sin (t) t

47. The characteristic equation for the corresponding homogeneous equation is r 2 + 1 = 0 with solutions r1,2 = ±i so a fundamental set of solutions for the corresponding homogeneous equation is S = {cos t, sin t} with W (S) = 1 and a general solution is yh = c1 cos t + c2 sin t. Using variation of parameters, we assume that a particular solution takes the form yp = u1 y1 + u2 y2 . Using the variation of parameters formulas,  u1 =

and

 1  0 W (S) csc t

  sin t  dt = − dt = −t cos t 

   1  cos t 0  u2 = dt = cot t dt = − ln csc t = ln sin t. W (S) − sin t csc t  Therefore, y = yh + yp = c1 sin (t) + c2 cos (t) + ln (sin (t)) sin (t) − t cos (t).  1 4t 5 49. y = 20 e t + 20t  51. y = 14 et−1 8 − e − 4t + 4et − 3et 2 + 2et 2 ln t 

55. (a) y = −e−t−1 + 12 et−1 57. Assuming that y = x r and substituting into the equation yields r(r − 1) − 4r + 6 = (r − 2)(r − 3) = 0 with solutions r1 = 2 and r2 = 3. Therefore, a fundamental set of solutions if S = {x 2 , x 3 } and a general solution of the equation is y = c1 x 2 + c2 x 3 . 59. y = −x −1 + 2x −1/2 61. Assuming that y = x r and substituting into the equation yields r(r − 1) − 7r + 25 = 0 with solutions r1,2 = 4 ± 3i. Therefore, a fundamental set of solutions is S = {x 4 cos(3 ln x), x 4 sin(3 ln x)} and a general solution is y = x 4 (c1 cos(3 ln x) + c2 sin(3 ln x)).     63. y = a1 + a2 x + (2 a2 − 2 a1 ) x 2 + 2 a2 − 83 a1 x 3 + 43 a2 − 2 a1 x 4 +   2 16 a − a x5 + · · · 2 1 3 15 65. y = a1 + a2 x − 32 a1 x 2 − 16 a2 x 3 − 58 a1 x 4 − 18 a2 x 5 + · · · 1

67. y = c1 x −2 + c2 x 2

99

100

CH A P T E R 4:

Higher Order Equations

     69. y = c1 (−1 + x)4 x − 16 + c2 −6 (−1 + x)4 x − 16 ln (−1 + x) +    59 2 197 3 − 6 x4 6 (−1 + x)4 x − 16 ln (x) + 101 x − x − + 22 x 6 2 60

Differential Equations at Work B. Modeling the Motion of a Skier 1. With v = s = ds/dt, s = k 2 − h2 (s )2 becomes v = k 2 − h2 v2 . Then, 1

dv = dt 

1 1 1 + = dt 2k k + hv k − hv k 2 − h2 v 2

1 (ln(k + hv) − ln(k − hv)) = t + C 2hk 

k + hv ln = 2hkt + C k − hv k + hv = Ce2hkt k − hv k Ce2hkt − 1 h Ce2hkt + 1  

k hv0 . v = tanh hkt + tanh−1 h k v=

Applying the initial condition gives us v(0) = k(C − 1)/(Ch + h) = v0 so C = (k + hv0 )/(k − hv0 ) and then v=

k hv0 − k + (hv0 + k)e2hkt . h k − hv0 + (hv0 + k)e2hkt

2. Integrating gives us     ln −2 hv0 e2hkt − 1 + k e2hkt + 1 kt − . s =C+ 2 h h Then, s(0) = C + ln(−4k)/h2 = 0 so C = − ln(−4k)/h2 . ⎛ ⎞ 3. 30 4.39078 0.001376 ⎜ 35 5.13939 0.001376 ⎟ ⎜ ⎟ ⎜ 40 5.84888 0.001376 ⎟ ⎜ ⎟ ⎝ 45 6.51387 0.001376 ⎠ 50 7.12928 0.001376

Answers, Hints, and Solutions to Selected Exercises

7. ␣ ⫽ 30 55 50 45 40 35 30 25

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CH A P T E R 4:

Higher Order Equations

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10. Students should notice that for longer races, drag affects the skier the most. Minimizing body area by wearing tight clothing can help skiers increase their speed substantially.

C. The Schrödinger Equation 1. In this problem, remember that e represents the elementary charge rather than the unique number whose natural logarithm has value one. For the given values, a0 = 2 4π0 h2 11 , 2a = 1.05893 × 1010 . E = − h = 5.29465 × 10 = 2.17869 × 1018 . 0 μe2 2μa0 2 5. u(r) = c1 r  ((1 − 2

( + 1)p 4 p2 + 6 p2 + 2 2 p2 − 2 Z − 2 Z 2 r +2 r − −2  − 2 (−2  − 2)(−6 − 4 )

p(6 p2 + 11 p2 + 6 2 p2 − 3 Z2 − 9 Z − 6 Z + 3 p2 ) 3 r + · · · ))+ (−2  − 2)(−6 − 4 )(−12 − 6 ) p r+ c2 r −−1 ((1 + 2 2  + 4  + 1 − (− − 1)2 8

4 p2 + 6 (− − 1)p2 + 2 (− − 1)2 p2 + Z2 +Z − 3 Z(− − 1) − 2 Z − Z(− − 1)2 2 2 r 2 − · · · )) ( + 4  + 1 − (− − 1)2 )(−1 + 2 + 6  − (− − 1)2 ) 6. Z Z2 r + 40000 r 2 + · · · ))+ −2  − 2 (−2  − 2)(−6 − 4 ) Z c2 r −−1 ((1 + 200 2 r+  + 4  + 1 − (− − 1)2

R(r) = c1 r  ((1 + 200

40000

Z2 r 2 + · · · )) (2 + 4  + 1 − (− − 1)2 )(−1 + 2 + 6  − (− − 1)2 )

CHAPTER 5

Applications of Higher-Order Differential Equations

Exercises 5.1 1. m = 4 slugs, k = 9 lb/ft; released 1 ft above equilibrium with zero initial velocity 3. m = 1/4 slug, k = 16 lb/ft; released 3/4 ft (8 in.) below equilibrium with an upward initial velocity of 2 ft/s 5. A general solution of x  + x = 0 is x(t) = c1 cos t + c2 sin t. Application of the initial conditions yields x(t) = 3 cos t − 4 sin t = 5 cos(t − φ), φ = −cos−1 (3/5) ≈ 0.93 rads; period = 2π, amplitude = 5. 7. A general solution of x  + 16x = 0 is c1 cos 4t + c2 sin 4t. Then, x  (t) = −4c1 sin 4t + 4c2 cos 4t; x(0) = c1 = −2, x  (0) = 4c2 = 1 ⇒ c2 = 1/4 so x(t) = −2 cos 4t + 1 4 sin 4t, which we can rewrite as  √  2 1 65 2 cos(4t − φ) = cos(4t − φ), x(t) = (−2) + 4 4 √ √ where φ = cos−1 (−8/ 65) ≈ 3.02 rads; period = π/2; amplitude = 65/4. 9. A general solution of x  + 9x = 0 is x(t) = c1 cos 3t + c2 sin 3t.√Application of the initial conditions yields x(t) = 13 cos 3t − 13 sin 3t. Thus, x(t) = 32 cos(3t − φ), φ = √ √ −cos−1 (1/ 2) = −π/4 rads; period = 2π/3; amplitude = 2/3. 11. Here, 16 = k · 1/2 ⇒ k = 32 and 16 = m · 32 ⇒ m = 1/2 so we solve 12 x  + 32x = 0, x(0) = 1, x  (0) = 0. A general solution of x  + 64x = 0 is x(t) = c1 cos 8t + c2 sin 8t and application of the initial conditions yields x(t) = cos 8t, maximum displacement = 1 ft when −8 sin 8t = 0 or 8t = nπ so that t = nπ/8, n = 0, 1, 2, . . . . 13. We first determine the spring constant k and the mass m: F = ks ⇒ 6 = 12 · k ⇒ k = 12. 3  Next, F = mg ⇒ 6 = m · 32 ⇒ m = 3/16. Thus, we solve 16 x + 12x = 0 or, equiva lently, x + 64x = 0. A general solution of this equation is x(t) = c1 cos 8t + c2 sin 8t. If the object is lifted 3 in. = 1/4 ft above the equilibrium position and released, we find the solution that satisfies the initial conditions x(0) = −1/4 and x  (0) = 0: x(0) = c1 = −1/4, x  (0) = 8c2 = 0 ⇒ x(t) = − 14 cos 8t; t = π/16 sec; x(5) ≈ 0.167 ft; x(t) = 18 sin 8t; t = π/8 sec.

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x 0.2

0.1

0.5

1.0

1.5

2.0

2.5

3.0

t

⫺0.1 ⫺0.2

15. General solutions are x = c1 cos t + c2 sin t, x = c1 cos 2t + c2 sin 2t, and x = c1 cos 3t + c2 sin 3t. Applying the initial conditions yields x = −cos t, x = −cos 2t, and x = −cos 3t, respectively. As k increases, the frequency at which the spring-mass system passes through equilibrium increases. x 1.0

0.5

1

2

3

4

⫺0.5

⫺1.0

16.

x 1.0

0.5

1

⫺0.5

⫺1.0

2

3

4

5

6

t

5

6

t

Answers, Hints, and Solutions to Selected Exercises

1  x + 8x = 0, 17. Here, 1 = k · 1/8 ⇒ k = 8 and 1 = m · 32 ⇒ m = 1/32 so we solve 32   x(0) = b, x (0) = −1. A general solution of x + 256x = 0 is x(t) = c1 cos 16t + 1 c2 sin 16t and application of the initial conditions yields x(t) = b cos 16t − 16 sin 16t.    1 1 1 The amplitude of the solution is A = b2 + = b2 + and b2 + =2⇒ k/m 256 256 √ b = 1023/16.

x 2

1

0.5

1.0

1.5

t

⫺1

⫺2

√ √ 19. T = 2π m/k ⇒ 2π m/32 = π/2 ⇒ m = 2 slugs   2 β β2 21. First, x(t) = α2 + 2 cos(ωt − φ) ⇒ v(t) = x  (t) = −ω α2 + 2 . The maximum ω ω   2 2 β β value of −ω α2 + 2 sin(ωt − φ) is ω α2 + 2 . ω ω 23. y(t) = 1.61 cos 3.5t − 0.856379 sin 3.5t; maximum displacement =  2 2 (1.61) + (0.856379) = 1.8236 ft 25. First, we find the mass of the cylinder: 512 = m · 32 ⇒ m = 16 slugs. Also, h = mg/(πR2 ρ) = 512/(62.5π) ≈ 2.61 ft ⇒ Equilibrium position is 4 − 2.61 ≈ 1.39 ft and π · 62.5 y(0) = 3 − 1.39 ≈ 1.61 ft so we solve the initial value problem y + y = 0, 16 π · 62.5 y(0) = 1.61, y (0) = −3. A general solution of y + y = 0 is y(t) = c1 cos 3.5t + 16 c2 sin 3.5t and applying the initial conditions results in y(t) = 1.61 cos 3.5t − 0.856379 sin 3.5t. Maximum displacement is 1.612 + 0.85644442 ≈ 1.8236 ft. 27. For (a), we solve x  + 4x = 0, x(0) = α, x  (0) = 0 for α = 1, 4, −2. α = 1: x = cos 2t; α = 4: x = 4 cos 2t; α = −2: x = −2 cos 2t. From the graph we see that varying α affects the amplitude of the displacement but not the times at which the mass passes through the equilibrium position. For (b), we solve x  + 4x = 0, x(0) = 0, x  (0) = β for β = 1, 4, −2. β = 1: x = 12 sin 2t; β = 4: x = 2 sin 2t; β = −2: x = −sin 2t. From the graph, we see that varying β not only

105

106

CH A P T E R 5:

Applications of Higher-Order Differential Equations

affects the initial velocity but also the amplitude of the displacement. The time at which the object passes through the equilibrium position is not affected.

(a)

(b) x

x

4

2

2

1

2

4

6

t

8

2

22

21

24

22

28.

4

6

8

t

x 0.6 0.4 0.2 2

4

6

t

8

⫺0.2 ⫺0.4 ⫺0.6 1 (−14 cos 2t + 5 sin 2t) and for (b) the solution is 29. For (a) the solution is x = 24  √ √ √  1 x = 72 −42 cos 6t + 5 6 sin 6t . For (a), the period is π and the amplitude is √ √ 1√ 1 2/3π and the amplitude is 12 199 ≈ 6 137/2 ≈ 1.37941. For (b) the period is 1.17556.

(a)

(b)

x

0.6

0.6

0.4

0.4

0.2

0.2 2

4

6

8

t

x

2

20.2

20.2

20.4

20.4

20.6

20.6

4

6

8

t

Answers, Hints, and Solutions to Selected Exercises

30.

x

0.5

2

4

6

8

t

⫺0.5

31. The solution is x=

     √ 1 1√ 1√ 15t + 8 15 sin 15t . 45 cos 60 2 2

√ √ The period is 4π/ 15 and the amplitude is 14 199/15 ≈ 0.910586. x

0.5

2

4

6

8

t

⫺0.5

Exercises 5.2 1. m = 1, c = 4, k = 3; released from equilibrium with an upward initial velocity of 4 ft/s. 3. m = 1/4, c = 2, k = 1; released 6 inches above equilibrium with a downward initial velocity of 1 ft/s. 5. Ageneral solution of x  + 4x  + 13x = 0 is x = e−2t (c1 cos 3t + c2 sin 3t).Application of the initial conditions yields   √c1 = 1, −2c1 + 3c2 = −1 ⇒ c1 =√1, c2 = 1/3 and x(t) = 1 −2t e cos 3t + 3 sin 3t = 310 e−2t cos(3t − φ), φ = cos−1 (3/ 10) ≈ 0.32 rads, Q.P.: 2π/3, t ≈ 0.63. 7. The characteristic equation of x  + 2x  + 26x = 0 is r 2 + 2r + 26 = 0 with solutions r1,2 = −1 ± 5i so a general solution is x = e−t (c1 cos 5t + c2 sin 5t). Application

107

108

CH A P T E R 5:

Applications of Higher-Order Differential Equations

of the yields c1 = √ 1, −c1 + 5c2 = 1 ⇒ c1 = 1, c2 = 2/5 so x =  initial conditions  √ 2 −t e cos 5t + 5 sin 5t . Thus, x(t) = 529 e−t cos(5t − φ), φ = cos−1 (5/ 29) ≈ 0.38 rads, Q.P.: 2π/5, t ≈ 0.39. 9. A general solution of x  + 8x  + 15x = 0 is x = c1 e−5t + c2 e−3t and application of the initial conditions yields c1 + c2 = 0, −5c1 − 3c2 = 1 ⇒ c1 = −1/2, c2 = 1/2 ⇒ x(t) = − 12 e−5t + 12 e−3t ; overdamped; does not pass through equilibrium;   x 12 ln(5/3) ≈0.093.   11. The characteristic equation is r 2 + 32 r + 12 = r + 12 (r + 1) = 0 so a general solution is given by x = c1 e−t + c2 e−t/2 . Application of the initial conditions yields the system of equations c1 + c2 = −1, −c1 − 12 c2 = 2 ⇒ c1 = −3, c2 = 2. Thus, x(t) = −3e−t + 2e−t/2 ; overdamped; x(t) = 0 implies t = 2 ln(3/2) ≈ 0.811; maximum displacement = 1 at t = 0. 13. A general solution of x  + 8x  + 16x = 0 is x = c1 e−4t + c2 te−4t and application of the initial conditions yields c1 = 4, −4c1 + c2 = −2 so c1 = 4 and c2 = 14 so x(t) = 4e−4t + 14te−4t ; critically damped; does not pass through equilibrium; maximum displacement = 4 at t = 0. 15. The characteristic equation of x  + 10x  + 25x = 0 is r 2 + 10r + 25 = (r + 5)2 = 0 with solutions r1,2 = −5 so a general solution of the ODE is x = e−5t (c1 + c2 t). Application of the initial conditions yields c1 = −5 and −5c1 + c2 = 1 so c1 = −5 and c2 = −24 and x(t) = −5e−5t − 24te−5t ; critically damped; does not pass through equilibrium; maximum displacement = 5 at t = 0. 17. Because the spring-mass system is critically damped, we have c2 − 4mk = 0 ⇒ c2 − 4 · 1 · 1 = 0 ⇒ c2 = 4 ⇒ c = 2. 19. The object’s mass is given by F = mg ⇒ 32 = m · 32 ⇒ m = 1. We solve the initial value problem x + 10x  + 24x = 0, x(0) = −1/2, x  (0) = 0. The characteristic equation is r 2 + 10r + 24 = 0 ⇒ r1 = −4, r2 = −6 so a general solution of the differential equation is x = c1 e−4t + c2 e−6t . Application of the initial conditions yields the system of equations c1 + c2 = −1/2, −4c1 − 6c2 = 0, which has solutions c1 = −3/2 and c2 = 1 so x(t) = − 32 e−4t + e−6t ; does not pass through equilibrium; maximum displacement = 1/2 at t = 0. 21. First we find the spring constant: m = 70 ⇒ F = mg = 70 · 9.8 = 686 ⇒ k = F/s = 686/−.25 = 2744. We solve the initial value problem 70x  + 280x  + 2744x = 0, x(0) = −3, x  (0) = 0. A general solution of the equation is 



11 11 t + c2 sin 4 t . x = e−2t c1 cos 4 5 5 Application  of the initial conditions yields the system of √equations c1 = −3, = −3 and c2 = − 32 5/11 so x(t) = e−2t −2c1 + 4 11 5 c2 = 0, which has solution c1   √ √ √ √ 21 cos(4 11/5t − φ), where −3 cos(4 11/5t) − 32 5/11 sin(4 11/5t) or x(t) = √ 2 11

Answers, Hints, and Solutions to Selected Exercises

  √ √ φ = cos−1 (−2 11/7) ≈ 2.81646; x(t) = 0 when t = 14 5/11 12 (2n + 1)π + φ , n an integer or t ≈ 0.739471, 1.26899, 1.7985, 2.32802, 2.85753, . . . 23. The motion is critically damped if c2 − 4mk = 0 ⇒ c2 − 4 · 4 · 64 = 0 ⇒ c = 32. The motion is underdamped if 0 < c < 32. 25. The quasiperiod is 1 π 13 = = √ 2 6 1 4km − c 4 · 13 · − c2 13 4π

4πm

⇒ c = 10/13. 27. The solution of mx  + cx  + kx = 0, x(0) = α, x  (0) = 0 is −

αe

t

√

c2 −4km+c



2m

x=

  √   √  √ t c2 −4km t c2 −4km 2 m m c e − 1 + c − 4km e +1 √ 2 c2 − 4km

while the solution of mx  + cx  + kx = 0, x(0) = 0, x  (0) = β is −

mβe x=

t

√  c2 −4km+c 2m

√

 √  t c2 −4km m e −1

c2 − 4km

Use the Principle of Superposition to form the result. 29. x(t) = c1 e−pt + c2 te−pt , ρ = c/(2m); x(0) = α, x  (0) = β implies that x(t) = αe−ρt (1 + ρt); x(t) = 0 implies t = −1/ρ < 0 31. The mass passes through the equilibrium when cos(μt − φ) = 0, which occurs when μt −  1 1 π + nπ + φ . The difference between φ = 12 π + nπ, n = 0, ±1, ±2, dots or t = 2 μ two consecutive items is     1 1 1 1 π π + (n + 1)π + φ − π + nπ + φ = , μ 2 μ 2 μ which is exactly one-half of the quasiperiod. 33. Suppose that two successive minima or maximum occur when tn = and tn+2 =

 1  −1 tan (−ρ/μ) + φ + nπ μ  1  −1 tan (−ρ/μ) + φ + (n + 2)π . μ

Then, cos(μtn − φ) = cos(μtn+2 − φ) and    x(tn ) = exp 2cπ/ 4mk − c2 . x(tn+2 )

109

110

CH A P T E R 5:

Applications of Higher-Order Differential Equations

√ √ √ 35. Note that c = 2 6 results in critical damping, c = 4 6 in overdamping, =− 6 √ √ √ and c √ √ √ 1 −3 2(1+ 3)t 6t 2 in underdamping. c = 2 6: x = te−t 6 ; c = 4 6: x = √ e (e − 1); 6 2   √ √ √ c = 6: x = 13 2e−t 3/2 sin √3 t . 2

x 0.20

0.15

0.10

0.05

0.5

36.

1.0

1.5

2.0

38.

x

2.5

0.3

10

0.2

5

t

3.0

x

0.1 2 0.5

1.0

1.5

2.0

2.5

3.0

4

6

8

t ⫺5

⫺0.1 ⫺10

39. c = 2: x = e−t (2t − 1); c =

√ √ √ √ √ 8: x = (1/ 2 − 2)e−(1+ 2)t + (1 − 1/ 2)e(1− 2)t

x 1.5 1.0 0.5

1 ⫺0.5 ⫺1.0

2

3

4

5

6

t

10

12

t

Answers, Hints, and Solutions to Selected Exercises

Exercises 5.3 1. First, we determine the mass and the value of the spring constant. F = ks ⇒ 8 = 1 · k ⇒ k = 8 and F = mg ⇒ m = F/g = 16/32 = 1/2. Then we solve the initial value problem x + 16x = 2 cos 3t, x(0) = 0, x  (0) = 2. A general solution of the corresponding homogeneous equation is xh = c1 cos 4t + c2 sin 4t. Using the method of undetermined coefficients we assume that a particular solution of the nonhomogenous equation has the form xp = A cos 3t + B sin 3t. Substitution of xp into the nonhomogeneous equation yields 7A cos 3t + 7B sin 3t = 2 cos 3t ⇒ A = 2/7, B = 0 so a general solution of the nonhomogeneous equation is x(t) = c1 cos 4t + c2 sin 4t + 27 cos 3t. Application of the initial conditions yields 2/7 + c1 = 0, 4c2 = 2 ⇒ c1 = −2/7, c2 = 1/2 so x(t) = − 27 cos 4t + 12 sin 4t + 27 cos 3t. The natural frequency of the system is ω = 4. 3. First, we determine the mass and the value of the spring constant. F = 16 ⇒ m = F/g = 16/32 = 1/2 and F = 16 ⇒ k = F/s = 16/(2/3) = 24 (s = 8 inches = 2/3 foot). Then, we solve the initial value problem x + 48x = 4 cos t, x(0) = 1/3, x  (0) = 0. The characteristic equation of the corresponding homogeneous equation x  + 48x = 0 is √ r 2 + 48 = 0 ⇒ r1,2 = ±4 of the corresponding homogeneous  √3i so  a general  solution √  equation is xh = c1 cos 4 3t + c2 sin 4 3t . Using the method of undetermined coefficients, we assume that there is a particular solution of the nonhomogeneous equation of the form xp = A cos t + B sin t. Substitution of this function and its derivatives into the nonhomogeneous equation yields 47A cos t + 47B sin t = 4 cos t so A =4/47 and √ B = 0. Thus, a general solution of the nonhomogeneous equation is x = c1 cos 4 3t +  √  4 c2 sin 4 3t + 47 cos t. Applying the initial conditions results in the system of equa√ tions c1 + 4/47 = 1/3, 4 3c2 = 0, which has solution c1 = 35/141 and c2 = 0. Thus, √ 35 4 x(t) = 141 cos(4 3t) + 47 cos t. 5. We solve x  + 9x = 4 cos ωt, x(0) = 0, x  (0) = 0. A general solution of the corresponding homogeneous equation is xh = c1 cos 3t + c2 sin 3t. If ω  = 3, we assume that a particular solution of the nonhomogeneous equation is sp = A cos ωt + B sin ωt. Substitution of this function into the nonhomogenous equation yields A(9 − ω2 ) cos ωt + B(9 − ω2 ) sin ωt = 4 cos ωt ⇒ A = 4/(9 − ω2 ), B = 0 so a general solution of the nonhomogeneous equation is x = c1 cos 3t + c2 sin 3t + 4 (cos 3t − cos ωt), ω  = 3 4 ω2 −9 cos ωt. x(t) = ; resonance occurs if ω = 3. 2 9 − ω2 t sin 3t, ω = 3 3

1.0 0.5

0.5

20.5

5 10 15 20 25

1.5 1.0 0.5 20.5 21.0 21.5

5 10 15 20 25

5 10 15 20 25

21 22 23 2

5 10 15

20 25

10 15 20 25

21 22

15 5 10 15 20 25

10

25 210

5 5 10 15 20 25

5

10 15 20 25

20.5 21.0

5 25

1.0 0.5

1 5

20.5 21.0 10 5

2 1

4 2 22 24

20.5 21.0

1.0 0.5

210 5 10 15 20 25

215

10

15

20

25

111

112

CH A P T E R 5:

Applications of Higher-Order Differential Equations

7. We solve x  + 8x  + 25x = cos t − sin t, x(0) = x  (0) = 0. A general solution of the corresponding homogeneous equation is xh = e−4t (c1 cos 3t + c2 sin 3t). Using the method of undetermined coefficients, there is a particular solution of the form xp = A cos t + B sin t. Substitution of xp into the nonhomogeneous equation yields 1 1 xp = 20 cos t − 40 sin t so a general solution of the nonhomogeneous equation is 1 1 −4t x = e (c1 cos 3t + c2 sin 3t) + 20 cos t − 40 sin t. Application of the initial conditions   1 7 1 −4t yields x(t) = −e 20 cos 3t + 120 sin 3t 40 (2 cos t − sin t). 9. We solve x  + 4x = cos t, x(0) = x  (0) = 0; xh = c1 cos 2t + c2 sin 2t; xp = A cos t + B sin t = 13 cos t ⇒ x(t) = c1 cos 2t + c2 sin 2t + 13 cos t. Application of the initial   conditions yields x(t) = 13 (cos t − cos 2t); the envelope functions are ± 23 sin 12 t . If the external force is changed to f (t) = cos(t/2), we solve x  + 4x = cos(t/2), x(0) = 4 x  (0) = 0. xp = A cos(t/2) + B sin(t/2) = 15 cos(t/2). Application of the initial condi4 4 8 tions yields x(t) = 15 cos(t/2) − 15 cos 2t; the envelope functions are ± 15 sin 34 t. The maximum displacement decreases. External force = f (t) = cos(t)

External force = f (t) = cos(t/2) x

x 0.6

0.4

0.4

0.2

0.2 20.2 20.4 20.6

5

10

15

20

25

30

t

5

10

15

20

25

30

t

20.2 20.4

 11. Solve: 4x  +4x  +26x = 250 sin t, x(0) = x  (0) = 0; x(t)= e−t/2 c1 cos( 52 t) + c2 sin( 52 t)

 − 2 cos t + 11 sin t; x(t) = e−t/2 2 cos( 52 t) − 4 sin( 52 t) − 2 cos t + 11 sin t; transient:

 e−t/2 2 cos( 52 t) − 4 sin( 52 t) ; steady-state: −2 cos t + 11 sin t 13. First, write the equation as x  + (k/m)x = (F/m) sin ωt. Both k and m are √ √ positive so xh = c1 cos( k/mt) + c2 sin( k/mt). xp = A cos ωt + B sin ωt = √ √ F Fω √ F √ sin ωt. (a) x(t) = α cos( k/mt) − m/k sin( k/mt) + k−mω m/k 2 2 2 k−mω k−mω  √ √ Fω F √ sin ωt; (b) x(t) = β − k−mω2 m/k sin( k/mt) + k−mω2 m/k sin ωt; (c) x(t) = √  √ √ Fω F √ α cos( k/mt) + β − k−mω m/k sin( k/mt) + k−mω m/k sin ωt. 2 2 15. First solve x  + x = 1, x(0) = x  (0) = 0 ⇒ x(t) = 1 − cos t. Because x(π) = 2 and   x  (π) = 0 next solve x + x = 0, x(π) = 2, x (π) = 0 ⇒ x(t) = −2 cos t. 1 − cos t, 0 ≤ t ≤ π Thus, x(t) = . −2 cos t, t > π 17. First solve x  + x = t, x(0) = x  (0) = 0 ⇒ x(t) = t − sin t. x(1) = 1 − sin 1 and x  (1) = 1 − cos 1 so next solve x  + x = 2 − t, x(1) = 1 − sin 1, x  (1) = 1 − cos 1 ⇒

Answers, Hints, and Solutions to Selected Exercises

2 1 x(t) = (sin ⎧ 2 − 2 sin 1) cos t + 4 cos 1 sin 2 sin t. Thus, ⎪ ⎪ ⎨t − sin t, 0 ≤ t ≤ 1 x(t) = −2 sin 1 cos t + (2 cos 1 − 1) sin t + 2 − t, 1 < t ≤ 2 . ⎪ ⎪ ⎩(sin 2 − 2 sin 1) cos t + 4 cos 1 sin2 1 sin t, t > 2 2

√ 2 2 19. The roots of the characteristic √ equation √ = 0 are r1,2 = −λ ± λ − ω . −λt Therefore, xh (t) = e (c1 cos ω2 − λ2 t + c2 sin ω2 − λ2 t), which can be written as   ω 2 − λ2 t − φ xh (t) = Ae−λt sin       ω2 − λ2 t cos φ − cos ω2 − λ2 t sin φ , = Ae−λt sin r 2 + 2λr + ω2

where A =

 c12 + c22 , sin φ = c1 /A, and cos φ = c2 /A. A particular solution of x  +

2λx  + ω2 x = F sin γt is

xp (t) =

 F ω2 − γ 2 sin(γt) − 2Fγλ cos(γt)

 . γ 4 + γ 2 4λ2 − 2ω2 + ω4

With sin(A + B) = sin A cos B + cos A sin B, we have F sin (γt + θ), xp (t) = 2 2 (ω − γ )2 + 4λ2 γ 2 −2λγ ω2 − γ 2 sin θ =  and cos φ =  . (ω2 − γ 2 )2 + 4λ2 γ 2 (ω2 − γ 2 )2 + 4λ2 γ 2 21. b = 0: x = 12 t sin t; b = 1: x = 12 (2 sin t + t sin t). As λ → 0, the motion approaches resonance. 6

22.

x

x 10

4 5

2 t 2

⫺2

4

6

8

10

12

20

⫺6

t

80

⫺10

23. (a) x(t) =

100 39 (cos(19t/10) − cos 2t);

(b) x(t) =

100 41 (cos 2t − cos(21t/10))

24.

x

3

4

2

2

1 40

60

80

x

t

t 20 ⫺4

60

⫺5

⫺4

⫺2

40

⫺1 ⫺2 ⫺3

1

2

3

4

5

6

113

114

CH A P T E R 5:

Applications of Higher-Order Differential Equations

25. x(t) =

⎧ ⎪ t − sin t + a cos t + b sin t, 0 ≤ t ≤ 1 ⎪ ⎪ ⎨

−t − sin t + 2 sin(t − 1) + 2 + a cos t + b sin t, 1 < t ≤ 2 ⎪ ⎪ ⎪ ⎩−sin t + 2 sin(t − 1) − sin(t − 2) + a cos t + b sin t, t > 2

√ 5000 27. x(t) = − 89733775561 (cos(299581t/50) − cos(20 15t))

Exercises 5.4 1. Solve 2Q + 32Q = 220 or Q + 16Q = 110, Q(0) = Q (0) = 0. A general solution is Q = c1 cos 4t + c2 sin 4t + 55/8. Application of the initial conditions yields Q(t) = 55 55  8 (1 − cos 4t). I(t) = Q (t) = 2 sin 4t. 3. Solve 14 Q + 64Q = 16t or Q + 256Q = 64t, Q(0) = Q (0) = 0. A general solution of the equation is Q = c1 cos 16t + c2 sin 16t + t/4 and application of the initial conditions 1 yields Q(t) = 64 (16t − sin 16t). I(t) = Q (t) = 14 (1 − cos 16t).  + 125Q + 5000Q = 0, Q(0) = 0, Q (0) = 4. A general solution is Q(t) = 5. Solve Q  √   √  −125t/2 7t + c2 sin 25 7t . Application of the initial conditions yields e c1 cos 25 2 2  √  √ Q(t) = 81757 e−125t/2 sin 252 7 t ; max.: 0.0225 at t ≈ 0.0147.

5.

Q 0.020

3. 6

0.015

5

0.010 0.005

4 0.05 0.10 0.15 0.20 0.25

t

3 4

2

I

3

1

2

1

2

3

4

5

6

1 0.05

0.10

0.15

0.20

0.25

t

 7. Solve Q + 125Q 5000Q = 630 t 5000 sin t. A general solution is Q(t) =  +  cos √  √ 2  25 12185 −125t/2 c1 cos 2 7t + c2 sin 25 7t + 12502813 cos t + 62526875 e 2 12502813 sin t. Applica  √  12185 tion of the initial conditions yields Q(t) = e−125t/2 − 12502813 cos 252 7 t + 26554371√ 312570325 7  √  25 7 12185 62526875 sin + 12502813 cos t + 12502813 sin t; steady-state charge: limt→∞ Q(t) = 2 t 12185 12502813

12185 62526875 cos t + 62526875 12502813 sin t; steady-state current: − 12502813 sin t + 12502813 cos t.

Answers, Hints, and Solutions to Selected Exercises

8.

115

Q 0.04 0.03 0.02 0.01

4 ⫺0.01

2 2

4

6

8

10

0.2

0.4

0.6

0.2

0.4

0.6

0.8

1.0

t

I

12 4

⫺2

3

⫺4

2 1 ⫺1

0.8

1.0

t

⫺2

9. (a) s(x) = 20 3 3 x

1 4 1 2 1 3 300 x + 3 x − 15 x ; (b) s(x)

=

1 4 10 2 2 3 30 x + 3 x − 3 x ; (c) s(x)

2 = 13 x 4 + 100 3 x −

d4s = 8, s(0) = s (0) = dx 4 1 4 1 3 1 4 100 x + 10 0, s(10) = s (10) = 0. (a) s(x) = 300 3 x − 15 x ; (b) s(x) = 30 x + 3 x − 2 3 1 4 1000 20 3 3 x ; (c) s(x) = 3 x + 3 x − 3 x ; simple support leads to larger maximum displacement.

11. In each case, we consider the boundary value problem EI

14 and 22. (a) s(x) =

1 6 10000 125 3 1 6 1250 125 3 360 x + 9 x − 9 x ; (d) s(x) = 360 x + 3 x − 18 x 480000π−4 sin(πx/10); (b) no solution; (c) s(x) = −80π−4

14 and 23. (a) s(x) =

 −600πx − 300π3 x − +π3 x 3 − 6000 sin(πx/10) ; (d) s(x) = 240π−4

 −100πx + πx 3 + 2000 sin(πx/10) 21.

22.

s 20 000 15 000 10 000 5000 25000 210 000

23.

s 80 000 60 000 40 000

2

4

6

8

10

x

20 000 2

4

6

8

10

x

    1 1 1 15. Q = 0 is Q = c1 cos √ t + c2 sin √ t . AppliC LC√ LC cation of the initial conditions yields Q(t) = Q0 cos(t/ LC); differentiation yields √ √ √ I(t) = −Q0 / LC sin(t/ LC); maximum Q = Q0 ; maximum I = Q0 / LC. A general solution of LQ +

s 60 000 50 000 40 000 30 000 20 000 10 000 2

4

6

8

10

x

116

CH A P T E R 5:

Applications of Higher-Order Differential Equations

17. If CR2 − 4L < 0, a general solution of the corresponding homogeneous equation is √

√

  R 4L − R2 C 4L − R2 C Qh = − exp − t c1 cos t + c2 sin t . √ √ 2L 2L C 2L C A particular solution of the nonhomogeneous equation is Qp = −

  CE0 2 CωR cos ωt + (CLω − 1) sin ωt 1 − 2CLω2 + C 2 ω2 (L 2 ω2 + R2 )

and Ip = E0 ((Lω − 1/(Cω)2 + R2 ))−1 (R sin ωt − (Lω − 1/(Cω) cos ωt).  E0 19. Ip = (Lω − 1/(Cω))2 + R2 cos(ωt − φ), cos φ = 2 2 (Lω − 1/(Cω))  +R  − (Lω − 1/(Cω))/ (Lω − 1/(Cω))2 + R2 . Let g(ω) = (Lω − 1/(Cω))2 + R2 ; √ g (ω) = 0 ⇒ ω = 1/ LC. The motion stops in (a) because limt→∞ θ(t) = 0; it eventually oscillates in (b). 1 5 3 x 6 − 36 x + 100 25. (a) s(x) = 36000 9 x; (b) no solution; (c) s(x) = 1 5 3 6 s(x) = 36000 x − 72 x + 25 x 6

1 5 3 6 36000 x − 9 x + 150x; (d)

Exercises 5.5 1. g/L = 32/2 = 16 ⇒ we solve θ  + 16θ = 0 subject to the indicated conditions. The characteristic equation is r 2 + 16 = 0 ⇒ r1,2 = ±4i so a general solution is θ(t) = 1 1 c1 cos 4t + c2 sin 4t. (a) θ(t) = 20 cos 4t; (b) θ(t) = 20 cos 4t + 14 sin 4t; (c) θ(t) = √ 1 1 20 cos 4t − 4 sin 4t; maximum displacement: (a) 1/20; (b) and (c) 26/20 ≈ 0.255. ␪

␪

2.

0.2

0.4

0.1

0.2 1

2

3

4

5

6

t

1

20.1

20.2

20.2

20.4

2

3

4

5

6

t

√ 3. The characteristic equation of θ  + 2 7θ  + 16θ = 0 has roots r1,2 =  √  √ 1 2 −2 7 ± 6i = − 7 ± 3i. Therefore, a general solution of the equation is θ(t) = √ √ √ 1 −t 7 e ( 7 sin 3t + 3 cos 3t); (b) θ(t) = e−t 7 (c1 cos 3t + c2 sin 3t). (a) θ(t) = 60   √  √ √  √ 1 −t 7 1 −t 7 ( 7 + 20) sin 3t + 3 cos 3t ; (c) θ(t) = 60 e ( 7 − 20) sin 3t + 3 cos 3t . 60 e 5. θ  (t) = −θ0 ω sin ωt + v0 cos ωt; θ  (t) = −θ0 ω2 cos ωt − v0 ω sin ωt; θ  + ω2 θ = −θ0 ω2 cos ωt − v0 ω sin ωt + θ0 ω2 cos ωt + v0 ω sin ωt = 0; θ(0) = θ0 , θ  (0) = v0 7. Because θ = θ0 cos ωt + (v0 /ω) sin ωt and θ = A cos(ωt − φ) = A cos ωt cos φ + A sin ωt sin φ, we have A cos φ = θ0 and A sin φ = v0 /ω. Therefore, cos φ = θ0 /A

Answers, Hints, and Solutions to Selected Exercises

9. 11. 13. 15.

and  sin φ = v0 /(ωA), so (θ0 /A)2 + (v0 /(ωA))2 = 1. Solving for A, we find that A = θ02 + v02 /ω2 . The phase angle is determined by finding the angle φ that satisfies cos φ = θ0 /A and sin φ = v0 /(ωA). √ T = 2π 2/9.8 ≈ 2.83 s √ T = 2π 8/32 ≈ 3.14 s √ 2π L/9.8 = 1 if L = 0.248 m  Maximum displacement: θ02 + v02 /ω2 when cos(ωt − φ) = ±1 ⇒ ωt − φ = nπ, (n =    √ 0, ±1, ±2, . . . ) ⇒ t = (1/ω)(nπ + φ); ω = g/L, φ = cos−1 θ0 / θ02 + v02 /ω2

17. The characteristic equation is Lr 2 + br + g = 0 so that the roots are r1,2 =   1  −b ± b2 − 4Lg . Thus, if b2 − 4Lg > 0, overdamped with solution 2L

   2 − 4Lg b −b + 1 t + θ=  (bθ0 + b2 − 4Lgθ0 + 2Lv0 ) exp 2L 2 b2 − 4Lg

   2 − 4Lg −b − b (−bθ0 + b2 − 4Lgθ0 − 2Lv0 ) exp t . 2L If 

b2 −  4Lg = 0,

critically

damped

with

solution

θ = θ0 e−bt/(2L) +

bθ0 + v0 te−bt/(2L) ; if b2 − 4Lg < 0, underdamped with solution 2L    b2 − 4Lg −bt/(2L) θ = Ce F cos Wt + sin Wt 4Lg − b2   bγ + C −F cos γt + 2 sin γt , Lγ − g

1  2 Lγ 2 − g and S = b − 4gL. The pendulum eventu2 2 2 4 2 2 2L b γ + L γ − 2Lγ g + g ally oscillates. where C =

19. x(t) = A cos ωt ⇒ x  (t) = −Aω sin ωt. Substitution into the nonlinear term yields   dx  x2 − 1 = −A3 ω cos2 ωt sin ωt + Aω sin ωt. dt However, cos2 ωt = 1 − sin2 ωt, so we have  dx  = −A3 ω sin ωt sin ωt + A3 ω sin3 ωt + Aω sin ωt.  x2 − 1 dt With the identity sin3 ωt = − 14 sin 3ωt + 34 sin ωt, it follows that   dx  1  1 = − A3 ω + Aω sin ωt − A3 ω sin 3ωt.  x2 − 1 dt 4 4

117

118

CH A P T E R 5:

Applications of Higher-Order Differential Equations

21. (a) θ(t) = 2 sin t; (b) θ(t) = 2 cos t; (c) θ(t) = −2 cos t; (d) θ(t) = −sin t; (e) θ(t) = −2 sin t; (f ) θ(t) = cos t − sin t; (g) θ(t) = −cos t + sin t

2

␪

1

2

4

6

8

t

⫺1

⫺2

√

√

15t 15t 1 −t/4 √ 22 and 23. θ(t) = e 15 sin + 15 cos , 15 4 4

√

15t 4



√

√

4e−t/4 sin 15t 15t 1 −t/4 √ − e 15 sin + 15 cos , , √ 15 4 4 15 √

15t √

√

4e−t/4 sin 4 1 −t/4 √ 15t 15t e 15 sin + 3 cos , − , √ 3 4 4 15 √

√

15t 15t 1 −t/4 √ − e 15 sin − 5 cos , 5 4 4 √

√

15t 15t 1 −t/4 √ 15 sin e − 5 cos , 5 4 4 √

√

15t 15t 1 −t/4 √ 15 sin − e + 3 cos , 3 4 4 √

√

15t 15t 1 −t/4 √ e 3 15 sin + 5 cos , 5 4 4 √

√

√ 15t 15t 1 −t/4 e 13 15 sin + 15 cos , 15 4 4 √

√

15t 15t 1 −t/4 √ e 7 15 sin − 15 cos , 15 4 4 √

√

√ 15t 15t 1 −t/4 e 11 15 sin − 15 cos 15 4 4

Answers, Hints, and Solutions to Selected Exercises

2

u

2

1

u

2

1

5

10

15

1

t

5

10

15

t

⫺1

⫺1

⫺1

⫺2

⫺2

⫺2

2

u

2

1

u

2

1

5

10

15

t

5

10

15

⫺1

⫺2

⫺2

⫺2

2

1

u

2

1

5

10

15

5

10

15

⫺1

⫺1

⫺2

⫺2

⫺2 u

2

1

1 5

10

15

t

5

10

15

5

10

15

5

10

15

t

t

u

t

⫺1

u

15

1

t

2

10

u

t

⫺1

u

5

1

⫺1

2

u

t

u 2 1 5

10

15

t

⫺1

⫺1

⫺1

⫺2

⫺2

⫺2

Chapter 5 Review Exercises 1. First, we find m = W /g = 32/32 = 1. Next, we find the spring constant with F = k · s ⇒ 32 = 1/2 · k ⇒ k = 64. Now, we solve the initial value problem x + 64x = 0, x(0) = 1/3, x  (0) = 0. A general solution is x = c1 cos 8t + c2 sin 8t and a solution of the initial value problem is x(t) = 13 cos 8t; maximum displacement = 1/3; t = π/16, π/8. 3. Solve 5x  + 20x  + 65x = 0, x(0) = 0, x  (0) = −1. A general solution is x = e−2t (c1 cos 3t + c2 sin 3t). Application of the initial conditions yields x(t) = 1 −2t −  3 e sin 3t, lim  t→∞ x(t) = 0; quasiperiod = 2π/3; maximum displacement =  1  x( 3 tan−1 (3/2)) ≈ 0.144; t = π/3.

t

119

120

CH A P T E R 5:

Applications of Higher-Order Differential Equations

0.5

1.0

1.5

2.0

2.5

3.0

⫺0.05 ⫺0.10

5. Solve 4x  + 16x = 4 or x  + 4x = 1, x(0) = x  (0) = 0. The differential equation has general solution x = c1 cos 2t + c2 sin 2t + 14 and application of the initial conditions results in x(t) = 14 (1 − cos 2t); maximum displacement is 1/2 and occurs when t = π/2. 0.5

6.

0.4

10 5

0.3 0.2

10

20

30

40

50

⫺5

0.1 2

4

6

8

10

⫺10

12

7. Solve 4x  + 16x = 4 cos t or x  + 4x = cos t, x(0) = x  (0) = 0. A general solution is x(t) = c1 cos 2t + c2 sin 2t + 13 cos 2t and application of the initial conditions yields x(t) = 13 cos t − 13 cos 2t; beats; the envelope functions are y = ± 23 sin(t/2). 8.

x 0.6 0.4

x 1.0 0.5

0.2 ⫺0.2 ⫺0.4 ⫺0.6

5

10

15

20

25

t

2

4

6

8

10

12

t

⫺0.5 ⫺1.0

9. Solve 4Q + 80Q + 436Q = 100, Q(0) = Q (0) = 0. A general solution is Q(t) = e−10t (c1 cos 3t + c2 sin 3t + 25/109) and an application of the initial conditions 25 −10t sin 3t; lim yields Q(t) = 327 [3 − 3e−10t (cos 3t − 10 sin 3t)]; I(t) = 25 t→∞ Q(t) = 3 e 25/109; limt→∞ I(t) = 0. 11. Solve Q + 104 Q = 220, Q(0) = Q (0) = 0. A general solution is Q = c1 cos 100t + 11 c2 sin 100t + 11/500 and application of the initial conditions yields Q(t) = 500 (1 − 11 cos 100t); I(t) = 5 sin 100t; limt→∞ Q(t) and limt→∞ I(t) do not exist. 12. Solve Q + 104 Q = 100 sin 10t, Q(0) = Q (0) = 0. A general solution of the equation 24 is Q = c1 cos 100t + c2 sin 100t + 2499 sin 2t and application of the initial conditions   1 1 50 results in Q = 2499 − 2 sin 100t + 25 sin 2t ; I = 2499 (−cos 100t + cos 2t); the limits do not exist.

Answers, Hints, and Solutions to Selected Exercises

125 3 1 5 1 6 2 13. s(4) = 0, s(0) = s (0) = s(10) = s (10) = 0 ⇒ s(x) = 250 3 x − 9 x + 12 x − 360 x

15. We solve s(4) = x(10 − x), s(0) = s (0) = s (10) = s (10) = 0 with solution s(x) = 1250 2 250 3 1 5 1 6 3 x − 9 x + 12 x − 360 x . 17. Solve 12 θ  + 32θ = 0, θ(0) = 1, θ  (0) = 0 ⇒ θ(t) = cos 8t; maximum displacement = 1; t = π/16 19. Solve 12 θ  + 8θ  + 32θ = 0, θ(0) = 1, θ  (0) = 0 ⇒ θ(t) = c1 e−8t + c2 te−8t . Application of the initial conditions yields θ(t) = e−8t + 8te−8t ; motion is not periodic. 21. θ(t) = 0.2168 cos 3.7t + 0.0471622 sin 3.7t 23. The solution to y + y = 1 + t 2 , y(0) = y (0) = 0 is y = 3 − t 2 − 3 cos t. Then, y(1) = 2 − 3 cos 1 and y (1) = −2 + 3 sin 1 and the solution to y + y = 0, y(1) = 2 − 3 cos 1, y (0) = −2 + 3 sin 1 is y = (−3 + 2 cos 1 + 2 sin 1) cos t − 2(cos 1 − sin 1) sin t. Thus, 1 − t2, 0 ≤ t < 1 the solution to y + y = , y(0) = y (0) = 0 is 0, t ≥ 1 3 − t 2 − 3 cos t, 0 ≤ t ≤ 1 y(t) = . (−3 + 2 cos 1 + 2 sin 1) cos t − (2 cos 1 − 2 sin 1 + sin 2) sin t, t > 1  √  25. x(t) = 25 cos 7255 t 27. x(t) = A sin ωn t + B cos ωn t + F/k, x(t) = (x0 − F/k) cos ωn t + F/k

Differential Equations at Work A. Rack-and-Gear Systems 1. Natural frequency: ωn 2

kr 2 = so ωn = 2I¯ + (W /g)r 2 + m3 r 2 /3



kr 2 2I¯ + (W /g)r 2 + m3 r 2 /3

2. θ(t) = θ0 sin ωn t θ0 ωn cos ωn t. The maximum value of θ˙ is θ0 ωn , so the maxi3. θ(t) = θ0 sin ωn t so θ˙ (t) =   1 W 2 m3 r 2 (θ0 ωn )2 ; the maximum value of θ(t) is mum value of T is Tmax = I¯ + r + 2 g 2·3 θ0 , so the maximum value of U is Umax = 12 kr 2 θ02 . The value for ωn that was found in (1) is found with Tmax = Umax also. 4. The natural frequency approaches a finite value found with L’Hopital’s Rule to be   kr 2 k = . lim 2 2 r→∞ 2I¯ + (W /g)r + m3 r /3 W /g + m3 /3 The method for finding the natural frequency of a mechanical system by solving Tmax = Umax for ωn is called Rayleigh’s Energy Method. Rayleigh discovered he did not need the exact shape of the displacement function to obtain a good approximation of the natural frequency. In fact, we can test this with the spring-mass system without damping modeled by mx  + kx = 0. Assume that the displacement is x(t) = x0 sin ωn t and use Rayleigh’s √ Energy Method to show that the natural frequency is ωn = k/m as expected. Notice that the term x02 appears in both Tmax and Umax so they cancel in the calculation of ωn . Therefore,

121

122

CH A P T E R 5:

Applications of Higher-Order Differential Equations

ωn does not depend on the amplitude of x(t) as is the case with any linear ordinary differential equation. This means that Rayleigh’s Energy Method can be used to approximate the natural frequency of the system without actually solving a differential equation.

B. Soft, Hard, and Aging Springs Soft Springs 10 2 5 2

4

6

8

10

22 24

25

26

210

0.5

1.0

1.5

2.0

2.5

3.0

0.5

1.0

1.5

2.0

2.5

3.0

10

2

4

6

8

10

5

20.5 21.0 21.5

25

22.0 210 40

20

220

210

10

220

240

20

Answers, Hints, and Solutions to Selected Exercises

Hard Springs

4

4

2

2

2

4

6

8

10

⫺2

⫺2

⫺4

⫺4

2

4

6

8

10

2

4

6

8

10

Aging Springs 4 5 2

2

4

6

8

10 ⫺2

⫺5 ⫺4

C. Bodé Plots 1. x(t) = 2.

1 −t 12 e

 √ √ √  −3et cos 2t + 9 cos 3t + 7 3 sin 3t

1.0

0.5

2 ⫺0.5

⫺1.0

4

6

8

123

124

CH A P T E R 5:

Applications of Higher-Order Differential Equations

3. M(2) ≈ .25; φ(2) ≈ 90 4. 10⫺9 10⫺13 10⫺17 10⫺21

0.2

0.5

2

⫺50

⫺100

⫺150

1.0

4

2.0

6

5.0

8

10.0

10

CHAPTER 6

Systems of Differential Equations

Exercises 6.1 1. Remember that x and y are functions of t: x = x(t) and y = y(t). The solution of x = 6  or dx/dt = 6 is x = x(t) = 6t + c1 because 6 dt = 6t + c1 . For the second equation,  y = cos t or dy/dt = cos t, and integrating gives us cos t dt = sin t + c2 . Thus, y = sin t + c2 . 1 3. x = 0 ⇒ x = c1 . y = −2y ⇒ dy = −2 dt ⇒ y = c2 e−2t . If you had solved the proby lem in a different order than is shown here, you might have arrived at the solution   x(t) = c2 , y(t) = c1 e−2t . 5. Refer back to Chapter 2 for help with solving first-order equations. For the first equation, x1 = −3x1 , x1 (0) = −1 has solution x1 = −e−3t . Next, x2 = 1, x2 (0) = 1 has solution x2 = t + 1. 7. Differentiating the x equation gives us x = −3x + 6y while solving the x equation for y gives us y = 16 (x + 3x). Now, substitute the y and y equations into the x equation and set the equation equal to 0: x = −3x + 6y   x = −3x + 6 4x − y x = −4x + 21x x + 4x − 21x = 0. This second-order equation with constant  coefficients has characteristic equation 2 r + 4r − 21 = 0 ⇒ r1 = −7, r2 = 3 so x(t) = − 32 c1 e−7 t + c2 e3t , y(t) =  c1 e−7 t + c2 e3t .  9.

x(t) = c1 sin(t) + c2 cos(t), y(t) = − 12 c1 cos(t) + 12 c2 sin(t) − 12 c1 sin(t) −  1 c cos(t) 2 2 125

126

CH A P T E R 6:

Systems of Differential Equations

11. Differentiating the x equation and substituting the y equation into the result gives us x = y x = −x + 1 x + x = 1. For x + x = 1, the corresponding homogeneous equation is x + x = 0, which has general solution xh = c1 cos t + c2 sin t. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation. A lot of practice and a thoughtful guess gives us xp = 1. Therefore, x = c1 cos t + c2 sin t + 1. Choosing a different linear combination gives us {x(t) = −c1 cos(t) + c2 sin(t) + 1, y(t) = c1 sin(t) + c2 cos(t)}. 13. {x = y, y = −4x + 3y} 15. {x = y, y = −16x + t sin t} 17. {y = x, x = z, z  = −6x − 3y − 3z + t} 21. Using D = d/dt, in operator notation the system becomes Dx = −2x − 2y + 4, Dy = (D + 2)x + 2y = 4 −5x + y or . Applying D − 1 to the first equation and −2 to the 5x + (D − 1)y = 0 (D − 1)(D + 2)x + 2(D − 1)y = (D − 1)(4) second equation gives us . Adding these −10x + −2(D − 1)y = 0 equations gives us [(D − 1)(D + 2) − 10](x) = −4 or x + x − 12x = −4, which has solution x = c1 e4t + c2 e3t + 13 . From the x equation, x = −2x − 2y + 4, y = 12 (4 − 2x − x ) = c1 e4t − 52 c2 e3t + 53 . 23. x(t) = e−3t (2c2 t + c1 ) −

et 2 + , y(t) = c2 e−3t + 13 4 9

25. x(t) = et (c1 (1 − 2t) − (c3 + c4 )t + c2 + c4 ) + (c1 + c3 ) sin(t) − (c2 + c4 ) cos(t), y(t) = 12 e−it (2e(1+i)t (c1 (2t − 1) + c3 t + c4 (t − 1) − c2 ) + (1 + i)((c1 − i(c2 + ic3 + c4 ))e2it − ic1 + c2 − ic3 + c4 )) or x = (c2 + c1 ) sin t + c1 cos t − c3 et − c4 tet , y = c2 cos t + (−c2 − 2c1 ) sin t + c3 et + c4 tet 3 1 27. x = − 10 cos t − 10 sin t − 2c1 e2t + (3c3 − 2c2 )e−t − 2c3 te−t , y = 3 4c3 )e−t + 3c3 te−t , z = 25 cos t + 10 sin t + c1 e2t + c2 e−t + c3 te−t

1 2

sin t + (3c2 −

29. In operator notation, the system is (D − 3)x + 2y = 0 −2x + (D + 1)y = 10

⇒

(D − 3)(D + 1)x + 2(D + 1)y = 0 4x − 2(D + 1)y = −20

Adding, we have (D2 − 2D + 1)x = −20 with general solution x = c1 et + c2 et − 20. Solving the first differential equation for y, we have y = − 12 (x − 3x) = c1 et − 12 c2 tet − 30. Application of the initial conditions yields x(0) = c1 − 20 = 0, y(0) = c1 − 12 c3 − 30 = 0 with solution c1 = 20, c2 = −20. Therefore x = 20et − 20tet − 20, y = 30et − 20tet − 30.

Answers, Hints, and Solutions to Selected Exercises

3

0.2

0.4

0.6

0.8

2

1.0

⫺5

⫺10

1

⫺20

⫺15

⫺10

⫺5 ⫺1

⫺15 ⫺2 ⫺20



D + 1 e t



31.

= (D + 1)(2et ) − (D + 1)(et ) = et  = 0

D + 1 2et









D2

t2

D2 D2

D2

t 2





, 2 33. 2 = 0,

4t 2 4D2

= 0,

4D2 4t 2

= 0 4D2 x + 4D2 y = 4t 2 4D 4D









D+1

t −D

D+1 −D

t







35. k = −4;

= 0;

= 0 ⇔ k = −4;

=0⇔

−4(D + 1) 4D

kt 4D

−4(D + 1) kt

k = −4





D+1

D+1 −c

−c



37. c  = 4:

= −8 + 2c = 0 ⇒ c = 4. If c = 4,

 = 0.

2(D + 1) −8

2(D + 1) −8



D+1 e−t



However,

= 0.

2(D + 1) 2e−t

D2 x + D2 y = t 2

39. (a) x1 = x2 , x2 = −3x + y = −3x1 + y1 , y1 = y2 , y2 = −x1 − y1 ; (b) x1 = x2 , x2 = −x + 2y + cos 2t = −x1 + 2y1 + cos 2t, y1 = y2 , y2 = −2x1 + y1 x = 2x − xy x = 2x − xy 41. For (a), while for (b), . In each case, the solution that y = −3y + xy y = 3y − xy satisfies x(0) = 3, y(0) = 2 is x(t) = 3, y(t) = 2.

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y

y

6

6

5

5

4

4

3

3

2

2

1

1

1

2

3 (a)

4

5

6

x

1

2

3 (b)

Exercises 6.2  −3 5 1. 3 2

 −5 21 3. 1 −3 

 −1 −11 −2 5. AB = ; BA = −2 −28 −16 ⎛

22 ⎜ 7. AB = ⎝ 22 −31

−5 −24 −4

−3 −27 ⎛

⎞ 1 −13 ⎜−25 ⎟ ⎜ −17⎠ and BA = ⎜ ⎝ 0 37 −27

9. 17 11. −3 13. 1 

1 1 2 15. −1 0 ⎛ ⎞ −1 −3 5 −2 ⎜0 0 1 0⎟ ⎜ ⎟ 17. ⎜ ⎟ 1 −3 1 ⎠ ⎝1 1 2 −4 2   1 1 ; λ2 = −4, v1 = 19. λ1 = −5, v1 = 1 2

9 20 −9 −11

⎞ 16 −35 20 5 ⎟ ⎟ ⎟ −8 −28⎠ −22 22

4

5

6

x

Answers, Hints, and Solutions to Selected Exercises

21. λ1,2 = −2 ± 2i, v1,2

 1 ± 2i = 5

23. The eigenvalues are found by solving



−1 − λ −1 −2





2−λ 2 = −2 + 3λ − λ3 = −(λ + 2)(λ − 1)2 = 0,

2



−2 −1 −1 − λ

which shows us that ⎛ ⎞λ1 = −2 has multiplicity one and λ2,3 = 1 has multiplixi ⎜ ⎟ city two. Let vi = ⎝yi ⎠ denote an eigenvector corresponding to λi . For λ1 = −2, zi ⎛ ⎞ 1 −1 −2 ⎜ ⎟ (A − λ1 I)v1 = 0 has augmented matrix ⎝ 2 4 2 ⎠, which reduces to −2 −1 1 ⎛ ⎞ ⎛ ⎞ 1 0 −1 1 ⎜ ⎟ ⎜ ⎟ ⎝0 1 1 ⎠. Thus, x1 = z1 and y1 = −z1 . Choosing z1 = 1 gives us v1 = ⎝−1⎠. For 0 0 0 1 ⎛ ⎞ −2 −1 −2 ⎜ ⎟ λ2,3 = 1, (A − λ2,3 I)v2,3 = 0 has augmented matrix ⎝ 2 1 2 ⎠, which reduces −2 −1 −2 ⎛ ⎞ 1 1/2 1 ⎜ ⎟ to ⎝0 0 0⎠. Thus, y2,3 and z2,3 are free and x2,3 = − 12 y2,3 − z2,3 . More specifically, 0 0 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −1 −1/2 −1 2 s−t ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ for s, t ∈ R, v2,3 = ⎝ s ⎠ = ⎝ 1 ⎠s + ⎝ 0 ⎠t are eigenvectors corresponding t 0 1 to v2,3 . In this case, we see that choosing s = 2 and t = 0 and then s = 0⎛and⎞t = 1 −1 ⎜ ⎟ gives us two linearly independent eigenvectors corresponding to λ2,3 , v2 = ⎝ 2 ⎠ and 0 ⎛ ⎞ −1 ⎜ ⎟ v3 = ⎝ 0 ⎠. 1 25. The eigenvalues are found by solving



−3 − λ 0 −1





−1 −1 − λ −3 = −10 − 16λ − 7λ2 − λ3 = −(λ + 1)(λ2 + 6λ + 10) = 0



1 0 −3 − λ

resulting in λ1 = −1 and λ2,3 = −3 ± i.

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⎛ ⎞ xi ⎜ ⎟ Let vi = ⎝yi ⎠ denote an eigenvector zi ⎛ −2 ⎜ (A − λ1 I)v1 = 0 has augmented matrix ⎝−1 1

27.

29.

31.

33.

corresponding to λi . For λ1 = −1,

⎞ ⎛ ⎞ 0 −1 1 0 0 ⎟ ⎜ ⎟ 0 −3⎠, which reduces to ⎝0 0 1⎠ 0 −2 0 0 0 ⎛ ⎞ 0 ⎜ ⎟ so x1 = y1 = 0 and z1 is free. Choosing z1 = 1 gives us v1 = ⎝0⎠. λ2 = −3 + i, 1 ⎛ ⎞ −2 − 3i 0 −1 ⎜ ⎟ (A − λ2 I)v2 = 0 has augmented matrix ⎝ −1 −3i −3 ⎠, which reduces 1 0 −2 − 3i ⎛ ⎞ 1 0 −i ⎜ ⎟ to ⎝0 1 −1 − i⎠ so x2 = iz2 and y2 = (1 + i)z2 . Choosing z2 = 1 gives us v2 = 0 0 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −i 0 −1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝1 + i⎠ = ⎝1⎠ + ⎝ 1 ⎠i. Since eigenvectors of complex conjugate eigenvalues are 1 1 0 ⎛ ⎞ ⎛ ⎞ 0 −1 ⎜ ⎟ ⎜ ⎟ also complex conjugates, v3 = ⎝1⎠ − ⎝ 1 ⎠i. 1 0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 1 −1 ⎜−2⎟ ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ λ1 = −2 and λ2,3,4 = −1; v1 = ⎜ ⎟, v2 = ⎜ ⎟, and v3,4 = ⎜ ⎟. ⎝1⎠ ⎝0⎠ ⎝0⎠ 1 1 0   1 − 2 e−2t −2e−2t ; −5e−5t − 15 e−5t   −sin t cos t − t sin t sin t cos t + t sin t ; cos t t cos t + sin t −cos t sin t − t cos t ⎞ ⎛ ⎞ ⎛ 1 4t e 4e4t ⎜ ⎟ ⎜ 14 ⎟ ⎝−3 sin 3t ⎠; ⎝ 3 sin 3t ⎠ 3 cos 3t − 13 cos 3t

37. ⎛ For⎞ A, the eigenvalues ⎛ ⎞ are λ1,2 = −1 and λ3 = 0 with corresponding eigenvectors v1,2 = 0 1 √ ⎜ ⎟ ⎜ ⎟ ⎝1⎠ and v3 = ⎝6⎠. For B, the eigenvalues are λ1,2 = 3 ± 2 11 and λ3 = 3 with 1 5 ⎛ ⎛ ⎞ √ ⎞ 1 − 15 (3 ± 2 11) −10 ⎜ ⎟ ⎜ ⎟ corresponding eigenvectors v1,2 = ⎝ −1 ⎠ and v3 = ⎝ 25 ⎠. 1 19 39. λ1,2 = 1 ± ik; λ1,2 = k ± i

Answers, Hints, and Solutions to Selected Exercises

Exercises 6.3 1.

3.

5.

7.

    x 1 4 x = y 2 −1 y      x 1 0 x et = + y 1 1 y 0    t(t 3 − 1)x 1 −t x = t(t 3 − 1)y −2t 2 t 3 + 1 y ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ x 1 1 1 x cos t ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ 1 ⎠⎝y⎠ + ⎝ 0 ⎠ ⎝y⎠ = ⎝2 0 z 0 −2 −1 z sin t

9. −5, linearly independent 11. 2 − 4 cos2 2t, linearly independent 13. 2e2t , linearly independent ⎛ ⎞  c1 c1 ⎜ ⎟ 15–22. (t)C where C = for 2 × 2 or ⎝c2 ⎠ for 3 × 3 c2 c3   23. x(t) = e−t + 2e4t , y(t) = −e−t + 3e4t 25. {x(t) = sin(t), y(t) = − sin(t) + cos(t)} 27. x = 4t −1 − 1, y = 4 − t 2 29. {x(t) = t, y(t) = sin(3t) + cos(3t), z(t) = − cos(3t) + sin(3t)}  t e−t e 33. (t) = 2et e−t

Exercises 6.4

    x c Remember that C denotes a constant vector. For X = , C has the form C = 1 while y c2 ⎛ ⎞ ⎛ ⎞ x c1 for X = ⎝y⎠, C has the form C = ⎝c2 ⎠. z c3  e4t e2t C 1. X = 4t e 3e2t  3. X =

6e−4t

0

e−4t

e2t

 e2 t 5. X = 2t e

C

e2t t C e2 t t + e2t

131

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CH A P T E R 6:

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et sin(2t)

7. X =

et cos(2t)

− 12 et sin(2t) + 12 et cos(2t) ⎛ ⎞ 0 e3t 0 ⎜ ⎟ 9. X = ⎝ 0 e2t e2t t ⎠C 2t 0 0 e ⎛ ⎞ sin(t) cos(t) 0 ⎜ ⎟ ⎟ 11. X = ⎜ ⎝−cos(t) sin(t) 0 ⎠C 0

0



− 12 et cos(2t) − 12 et sin(2t)

C

e2 t

13. The eigenvalues of A are λ1 = 15 and λ2 = −4 with corresponding eigenvectors    −4t e15 t e −5 2 v1 = and v2 = so X = C 1 −4t 7 15 t 7 1 e − e 2 5  6 −1 15. The eigenvalues and corresponding eigenvectors of A = are λ1 = 5 and 5 0   1 1 λ2 = 1 with corresponding eigenvectors v1 = and v2 = . Thus, a general 1 5  t e5 t e solution is X = C. 5et e5 t  7 0 17. The eigenvalues and corresponding eigenvectors of A = are λ1 = −8 and 5 −8   0 3 and v2 = . Thus, a general λ2 = 7 with corresponding eigenvectors v1 = 1 1  0 3e7 t solution is X = C. e−8 t e7 t  −5 3 19. The eigenvalues and corresponding eigenvectors of A = are λ1 = −11 and 2 −10   −1 3 and v2 = . Thus, a general λ2 = −4 with corresponding eigenvectors v1 = 2 1  −4t e e−11 t solution is X = C. 1 −4t −2 e−11 t 3e  −6 −4 21. The eigenvalues and corresponding eigenvectors of A = are λ1 = −12 and −3 −10   2 −2 and v2 = . Thus, a general λ2 = −4 with corresponding eigenvectors v1 = 3 1  −4t e e−12t C. solution is X = − 12 e−4t 32 e−12 t

Answers, Hints, and Solutions to Selected Exercises

 −1 . A general solution is then 23. λ1,2 = −8 and there is only one eigenvector v1 = 1  −8 t e−8 t t e X= C. −e−8 t −e−8 t t + 12 e−8 t 25. The eigenvalues of the coefficient matrix are λ1,2 = ±4i and the corresponding   sin (4t) cos (4t) ±2i eigenvectors are v1,2 = . Thus, X = C. 1 1 −1 2 cos (4t) − 2 sin (4t) ⎛ ⎞ ⎛ ⎞ 1 0 ⎜ ⎟ ⎜ ⎟ 27. The eigenvalues are 4, −2, and −1 with corresponding eigenvalues of ⎝0⎠, ⎝1⎠, and 0 0 ⎛ 4t ⎞ ⎛ ⎞ −t e e 0 −1 ⎜ ⎟ ⎜ ⎟ 0 e−2t ⎟ ⎝ 0 ⎠ so a general solution is X = ⎜ ⎝0 ⎠C. 5 0 −5e−t 0 ⎛ ⎞ −7 ⎜ ⎟ 29. The eigenvalues are λ1 = 3 and λ2,3 = −1 with corresponding eigenvectors v1 = ⎝ 4 ⎠ 6 ⎛ ⎞ ⎛ −t 3t −t t ⎞ e e e −1 ⎜ ⎟ ⎜ ⎟ and v1,2 = ⎝ 0 ⎠. A general solution is X = ⎝ 0 − 47 e3t −e−t ⎠C. 2 −2e−t − 6 e3t −2e−t t 7

31. The are λ1 = 1⎞and λ2,3 = 1 ± 2i with corresponding eigenvectors v1 = ⎛ ⎞eigenvalues ⎛ 1 −1 ∓ 32 i ⎜ ⎟ ⎜ ⎟ ⎝0⎠ and v2,3 = ⎝ ∓i ⎠; a general solution is 0 1 ⎛ t ⎞ t e cos(2t) et sin(2t) e ⎜ ⎟ 6 t 4 t 6 t 4 t ⎟ X=⎜ 13 e cos(2t) − 13 e sin(2t) 13 e sin(2t) + 13 e cos(2t) ⎠C. ⎝0 0

4 t 6 t − 13 e cos(2t) − 13 e sin(2t)

6 t 13 e

4 t cos (2t) − 13 e sin(2t)

eigenvectors 33. The eigenvalues are⎛λ1,2 3,4 = 0 with corresponding ⎛ ⎞ ⎞ = −1 and ⎛ λ⎞ ⎛ ⎞ 0 1 −2 0 ⎜−1⎟ ⎜2⎟ ⎜−2⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ v1 = ⎜ ⎟, v2 = ⎜ ⎟, v3 = ⎜ ⎟, and v4 = ⎜ ⎟. A general solution is ⎝0⎠ ⎝1⎠ ⎝0⎠ ⎝1⎠ 2 0 1 0 ⎞⎛ ⎞ ⎛ −2et −2et 1 4 c1 ⎜ ⎟ ⎜ 0 −3et 1 1⎟ ⎟ ⎜c2 ⎟ ⎜ X=⎜ ⎟ ⎜ ⎟. 3et 0 2⎠ ⎝c3 ⎠ ⎝ 0 3et 0 1 0 c4   35. A general solution is x(t) = c1 et + c2 et et − 1 , y(t) = c2 e2t and the solution of the IVP   is x = 4et −1 + et , y = 4e2t . 37. A general solution is x(t) = c1 e4t , y(t) = 2c1 e4t t + c2 e4t and the solution of the IVP is x = 8e4t , y = 16e4t t.

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39. x(t) = c1 cos(4t) − c2 sin(4t), y(t) = c1 sin(4t) + c2 cos(4t); x = −8 sin 4t, y = 8 cos 4t 41. x(t) = c1 e2t (2et − 1) + c3 e2t (et − 1), y(t) = c3 (−et )(et − 1)2 − 2c1 e2t (et − 1) + c2 et , z(t) = −2c1 e2t (et − 1) − c3 e2t (et − 2); x = −e2t (−1 + 2et ), y = 2et (1 − et + e2t ), z = 2e2t (−1 + et ) 43. x(t) = c1 (−e−t )(3et − 4) + c2 e−t (et − 1) + 8c3 e−t (et − 1) − 3c4 e−t (et − 1), y(t) = −3c4 e−t (et − 1)2 − 2c1 e−t (−3et + e2t + 2) + c2 e−t (−et + e2t + 1) + 2c3 e−t (−7et + 3e2t + 4), z(t) = − 12 c2 e−t (et − 1)2 + 32 c4 e−t (et − 1)2 + c1 e−t (−3et + e2t + 2) − c3 e−t (−8et + 3e2t + 4), w(t) = 2c1 (et − 1) + c2 (1 − et ) − 6c3 (et − 1) + c4 (3et − 2); x = 9e−t (−1 + et ), y = e−t (9 − 15et + 7e2t ), z = − 12 e−t (9 − 18et + 7e2t ), w = −7(−1 + et ) √   √  √  √ 5−5 t 1 − 12 e 54. (a) x(t) = 10 5 − 3 5 e 5t + 5 + 3 5 , √     √ 1 √ √  5−5 t 1 −2 y(t) = 10 e − 5 − 5 e 5t + 5 + 5 ; (b) x(t) = e2t , y(t) = e2t (3t + 1); √  √  √  √  3 sin 7t 5 sin 7t (c) x(t) = + cos 7t , y(t) = cos 7t − √ √ 7 7  1 −3t 55. x(t) = 13 e 6(c1 − 3c2 + 4c3 )e3t sin(2t) + 3(3c1 + 4c2 − c3 )e3t cos(2t) + 4c1   1 −3t (c1 − 16c2 + 17c3 ) e3t sin(2t) + 2 (4c1 + c2 + 3c3 ) e3t −12c2 + 3c3 , y(t) = 26 e   1 −3t cos(2t) − 8c1 + 24c2 − 6c3 , z(t) = 13 e −2 (3c1 + 4c2 − c3 ) e3t sin(2t)  3t +4(c1 − 3c2 + 4c3 )e cos(2t) − 4c1 + 12c2 − 3c3  17t/3  1 −11t/3 57. x(t) = 17 e 14e x0 − 7e17t/3 y0 + 3x0 + 7y0 , y(t) =   1 −11t/3 − 17 e 6e17t/3 x0 − 3e17t/3 y0 − 6x0 − 14y0 y 1.0

0.5

21.0

20.5

0.5

20.5

21.0

1.0

x

Answers, Hints, and Solutions to Selected Exercises

Exercises 6.5  2e−t 1. X = e−t

  1 c1 −et + 1 c2 et

3. x = −c1 e−t − 2c2 et − t + 1, y = c1 e−t + 3c2 et − 2t    e−2t 3e−t c1 −cos t 5. X = −2t + e 2e−t c2 sin t − cos t    −e−2t −2e−t c1 −te−2t + 7. X = 2e−2t 3e−t c2 2te−2t − e−2t    3e−2t e−t c1 0 9. X = + 2e−2t e−t c2 −2e−t    2te−t + e−t −2te−t c1 e−t 11. X = + −2t −t −t 2te −2te + e c2 0   1 −6 x  X + F(t), where X = and 13. In matrix form, the system is X = 1 4 y  f1 (t) F(t) = . f2 (t)  1 −6 For (a), has eigenvalues λ1 = −2 and λ2 = −1 with corresponding 1 4   2 3 and v2 = . Hence, a general solution is eigenvectors v1 = 1 1   2 −2t 3 −t X = c1 e + c2 e 1 1   2e−2t 3e−t c1 = . e−2t e−t c2 This shows us that a fundamental matrix for the system is  2e−2t 3e−t . = e−2t e−t Because we started with x, y notation we end with x, y notation: x = 2c1 e−2t + 3c2 e−t y = −c1 e−2t + c2 e−t . For (b), we know solution of the corresponding homogeneous a general  that from (a) 2e−2t 3e−t c1 . Because ae2t is not a solution for any vector system is Xh = e−2t e−t c2

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  a1 a1 2t a= , we assume that a particular solution has the form Xp = ae2t = e , a2 a2 where A is to be determined. Substituting Xp into the nonhomogeneous equation gives us x − (x − 6y) = 6a2 et = 0 y − (x − 4y) = −a1 et + 5a2 et = −2e−t . From the first equation, we see that a2 = 0 and from the second it follows that a1 = 2 so  0 2t e . As in (a), because we started with x, y notation we end with x, y notation: Xp = 2 x = 2c1 e−2t + 3c2 e−t + 2et y = −c1 e−2t + c2 e−t . For (c), we know solution of the corresponding homogeneous  that from (a) a general 2e−2t 3e−t c1 system is Xh = . Because ae−t is a solution for some vectors a = e−2t e−t c2   a1 a1 −t , we assume that a particular solution has the form Xp = ae−t = e , where a2 a2 a is linearly independent of v2 . Substituting Xp into the nonhomogeneous equation gives us x − (x − 6y) = −2(a1 − 3a2 )e−t = −2e−t y − (x − 4y) = −(a1 − 3a2 )e−t = −e−t  1 −t so a1 = 1 + 3a2 . Choosing a2 = 0 gives us Xp = e and as in (a) and (b) because 0 we started with x, y notation we end with x, y notation: x = 2c1 e−2t + 3c2 e−t + e−t y = −c1 e−2t + c2 e−t .  2 3 15. The eigenvalues of A = are λ1,2 = −1 and there is only one linearly inde−3 −4  −1 pendent eigenvector corresponding to this eigenvalue, v = . Thus, one solution 1 of the system is X1 = ve−t . We now look for a second linearly independent solution of the form X2 = w1 te−t + w2 e−t . Differentiating, X2 = −w1 te−t + (w1 − w2 )e−t , and substituting into the system gives us X2 = −w1 te−t + (w1 − w2 )e−t = AX2 = A(w1 te−t + w2 e−t ) = Aw1 te−t + Aw2 e−t .

Answers, Hints, and Solutions to Selected Exercises

Equating coefficients gives us Aw1 = −w1 . This means that w1 is an eigenvector of A with corresponding eigenvalue λ = −1. We choose w1 = v. Thus, w1 − w2 = v − w2 = Aw2 , which shows us that w2 satisfies (A + I) w2 = v:    3 3 w2,1 −1 = −3 −3 w2,2 1    1 1 w2,1 −1/3 = . 0 0 w2,2 0 Then w2,1 + w2,2 = −1/3. We choose w2,1 = −1/3, which gives us   −1 −t −1/3 −t te + e . X2 = 1 0 Combining X1 and X2 , a fundamental matrix for the system is  −e−t −te−t − 13 e−t = e−t te−t   −e−t −te−t − 13 e−t c1 and a general solution of the system is X = . e−t te−t c2 When dealing with repeated eigenvalues for which there is only one linearly independent eigenvector, one must be especially carefully in choosing the form of Xp . For (b), we assume that Xp = at 2 e−t + bte−t + ce−t ,    a1 b1 c1 , b= , and c = are constant vectors to be determined. where a = a2 b2 c2 Substituting Xp into the nonhomogeneous equation gives us  −3(a1 + b1 )t 2 + (2a1 − 3b1 − 3b2 )t + (b1 − 3c1 − 3c2 ) −t  Xp − AXp = e 3(a1 + b1 )t 2 + (5b1 + 3b2 ) + (b2 + 3c1 + 3c2 )  1 = e−t −1 and equating coefficients we find that a1 + a2 = 0 2a1 − 3b1 − 3b2 = 0 b1 − 3c1 − 3c2 = 1 a1 + a2 = 0 2a2 + 3b1 + 3b2 = 0 b2 + 3c1 + 3c2 = −1,

Remember that when the equation Av = λv is satisfied, λ is an eigenvalue of A with corresponding eigenvector v.

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which has solution a 1 = 0, a 2 = 0, b1 =1 + 3c1 + 3c2 , b2 = −1 − 3c 1 −3c2 . Choosing te−t −e−t −te−t − 13 e−t c1 te−t and X = + . c1 = c2 = 0, Xp = −te−t e−t te−t c2 −te−t For (c), we proceed in much the same way as in (b). We assume that Xp = at 2 e−t + bte−t + ce−t ,    a1 b1 c1 , b= , and c = are constant vectors to be determined. where a = a2 b2 c2 Substituting Xp into the nonhomogeneous equation gives us Xp − AXp

 −3(a1 + b1 )t 2 + (2a1 − 3b1 − 3b2 )t + (b1 − 3c1 − 3c2 ) −t = e 3(a1 + b1 )t 2 + (5b1 + 3b2 ) + (b2 + 3c1 + 3c2 )  2 = e−t −1

and equating coefficients we find that a1 + a2 = 0 2a1 − 3b1 − 3b2 = 0 b1 − 3c1 − 3c2 = 2 a1 + a2 = 0 2a2 + 3b1 + 3b2 = 0 b2 + 3c1 + 3c2 = −1, which has solution a1 = 3/2, a2 = −3/2, b1 = 2 + 3c1 + 3c2 , b2 = −1 −  3 2 −t −t 2 t e + 2te 3c1 − 3c2 . Choosing c1 = c2 = 0 gives us Xp = and − 32 t 2 e−t − te−t   3 2 −t  −t −t −te−t − 13 e−t c1 −e 2 t e + 2te + . X= c2 e−t te−t − 32 t 2 e−t − te−t 17. x(t) = (c1 + t) cos(t) − sin(t)(c2 + log(cos(t))), y(t) = (c1 + t) sin(t) + cos(t) (c2 + log(cos(t)))        19. x(t) = c1 cos(t) + sin(t) c2 − 2 log sin 2t + 2 log cos 2t , y(t) = −c1 sin(t)+        cos(t) c2 − 2 log sin 2t + 2 log cos 2t −2  3t  1 −4t 6t 3t 21. x(t) = 9 e (c1 20e − 4e − 7 + 4(e − 1)(2(c2 + c3 )e3t − 2c2 + c3 )) − t 2 + 11t 83 1 −4t (32(5c1 − 4c2 − c3 )e3t − 48(c1 − 2(c2 + c3 ))e6t − 16(7c1 − 4 − 16 , y(t) = 96 e 1 −4t 4t 8c2 + 4c3 ) − 3e (4t(8t − 15) + 125)), z(t) = 144 e (16 (−5c1 + 4c2 + c3 ) e3t − 32 (c1 − 2 (c2 + c3 )) e6t + 16(7c1 − 8c2 + 4c3 ) + e4t (99 − 108t))

Answers, Hints, and Solutions to Selected Exercises

⎛ ⎞ x(t) ⎜ ⎟ 23. In matrix form, we are solving X = AX + F(t), where X = ⎝y(t)⎠, A = z(t) ⎛ ⎞ ⎛ ⎞ −2 1 −1 f1 (t) ⎜ ⎟ ⎜ ⎟ ⎝ 5 −2 5 ⎠, and F(t) = ⎝ f2 (t) ⎠. A has eigenvalues λ1 = −1 and λ2 = −1 ± 2i 2 −1 1 f3 (t) ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −1 −1 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ with corresponding eigenvectors v1 = ⎝ 0 ⎠ and v2,3 = ⎝ 1 ⎠ ± ⎝−2⎠ i. One solu1 2 0 ⎞ ⎛ −t −e ⎟ ⎜ tion of the homogeneous equation is X1 = ⎝ 0 ⎠. Two more linearly independent e−t solutions are ⎡⎛ ⎞ ⎛ ⎞ ⎤ ⎛ ⎞ −1 0 −e−t cos 2t ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ X1 = e−t ⎣⎝ 1 ⎠ cos 2t − ⎝−2⎠ sin 2t ⎦ = ⎝e−t (cos 2t + 2 sin 2t)⎠ 2e−t cos 2t 2 0 and

⎡⎛ ⎞ ⎛ ⎞ ⎤ ⎛ ⎞ −1 0 −e−t sin 2t ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ X2 = e−t ⎣⎝ 1 ⎠ sin 2t + ⎝−2⎠ cos 2t ⎦ = ⎝e−t (−2 cos 2t + sin 2t)⎠. 2 0 2e−t sin 2t

Thus, a general solution of the system is ⎛ −t −e ⎜ X=⎜ ⎝ 0 e−t

⎞⎛ ⎞ ⎟ ⎜c1 ⎟ e−t (cos 2t + 2 sin 2t) e−t (−2 cos 2t + sin 2t)⎟ ⎠ ⎝c2 ⎠. c3 2e−t cos 2t 2e−t sin 2t −e−t cos 2t

−e−t sin 2t

For (b), we assume that a particular solution takes the form Xp = aet + b cos 2t + c sin 2t, ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ b1 c1 a1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ where the vectors a = ⎝a2 ⎠, b = ⎝b2 ⎠, and c = ⎝c2 ⎠ are to be determined. a3 b3 c3 Substituting into the nonhomogeneous system and simplifying the results gives us ⎛

⎞ (3a1 − a2 + a3 )et + (2b1 − b2 + b3 + 2c1 ) cos 2t + (−2b1 + 2c1 − c2 + c3 ) sin 2t ⎜ ⎟ Xp − AXp = ⎝(−5a1 + 3a2 − 5a3 )et + (−5b1 + 2b2 − 5b3 + 2c2 ) cos 2t + (−2b2 − 5c1 + 2c2 − 5c3 ) sin 2t ⎠ (−2a1 + 2a2 )et + (−2b1 + 2b2 − b3 + 2c3 ) cos 2t + (−2b3 − 2c1 + 2c2 − c2 ) sin 2t

139

140

CH A P T E R 6:

Systems of Differential Equations

⎛

⎞

3et − cos 2t + sin 2t

⎜ ⎟ = ⎝−5et + 2 cos 2t − 7 sin 2t ⎠ −2et + 4 cos 2t − sin 2t and equating coefficients results in the system a1 − a2 + a3 = 3 2b1 − b2 + b3 + 2c1 = −1 −2b1 + 2c1 − c2 + c3 = 1 −5a1 + 3a2 − 5a3 = −5 −5b1 + 2b2 − 5b3 + 2c2 = 2 −2b2 − 5c1 + 2c2 − 5c3 = −7 −2a1 + 2a2 = −2 −2b1 + 2b2 − b3 + 2c3 = 4 −2b3 − 2c1 + 2c2 − c2 = −1, which has solution a1 = 1, a2 = 0, a3 = 0, b1 = 0, b2 = 1, b3 = 0, c1 = 0, c2 = 0, and c3 = 1. Thus, a particular solution of the nonhomogeneous system is ⎛

⎞ et ⎜ ⎟ Xp = ⎝cos 2t ⎠ sin 2t and a general solution of the nonhomogeneous system is X = Xh + Xp ⎛ ⎞⎛ ⎞ ⎛ ⎞ −e−t −e−t cos 2t −e−t sin 2t c1 et ⎜ ⎟⎜ ⎟ ⎜ ⎟ =⎝ 0 e−t (cos 2t + 2 sin 2t) e−t (−2 cos 2t + sin 2t)⎠ ⎝c2 ⎠ + ⎝cos 2t ⎠. e−t 2e−t cos 2t 2e−t sin 2t c3 sin 2t 25. x(t) = − 12 cos(2t)c2 + 12 sin(2t)c1 + 18 cos(2t) + 14 sin(2t)t + c3 , y(t) = 12 c2 cos(2t) − 12 c2 sin(2t) − 18 cos(2t) − 14 sin(2t)t − 2c3 , z(t) = sin(2t)c2 + cos(2t)c1 + 12 t cos(2t) + 3c3 27. x(t) = −1 + t + e−t , y(t) = 0 1 5t e + 14 et 29. x(t) = − 23 e−t + 16 e5 t − 12 et , y(t) = 23 e−t + 12   2t 1 sin(3t) − 7 cos(3t) , 31. x(t) = 20 13 + e 39 13   1 2t 62 8 (3t) − e sin − 13 cos(3t) y(t) = − 30 − 13 2 39

Answers, Hints, and Solutions to Selected Exercises

 1 + sin(2t) sin(2t) + cos(2t) − sin(2t) ln − 33. x(t) = cos(2t)   1 + sin(2t) 1 + 54 cos(2t) ln(cos (2t)) + 52 sin(2t)t + 12 , 4 cos(2t) ln cos(2t)   1 + sin(2t) y(t) = − 14 sin(2t) − 14 cos (2t) ln − 12 ln(cos(2t)) sin(2t) + cos(2t) 1 1 4 cos(2t) ln (cos(2t)) + cos(2t)t + 2 sin(2t)t 

− 14

35. x =

1 2

1 2

1 (5 ln(cos(t) − sin(t)) cos(2t) −5 ln(cos(t) + sin(t)) cos(2t) + 4 cos(2t) −3 sin(2t)) 4

and y=

1 −cos2 (2t) + 2 ln(cos(t) − sin(t)) cos(2t) − 2 ln(cos(t) + sin(t)) cos(2t) 2 + cos(2t) − sin2 (2t) − ln(cos(t) − sin(t)) sin(2t)  + ln(cos(t) + sin(t)) sin(2t) − 2 sin(2t) x,y

⫺0.75 ⫺0.5 ⫺0.25

y

8

8

6

6

4

4

2

2

0.25

0.5

0.75

t

⫺1

⫺0.5

0.5

1

1.5

2

x

√   t c1 t + t c2 1   1 f (t) c1 f (t) 39. X = + f (t) 1 c2 f (t)  1 37. X = t 

453 −8 t 3 −t 2 3 −t −t + 14 − 14 e t + 49 te−t , y(t) = 391 41. (a) x(t) = − 199 343 e + 1372 e 343 e + 151 −8 t 1 179 −2t t t − 14 + 49 te−t + 37 e−t t 2 ; (b) x(t) = 260 + 10 1372 e 81 e − 81 e 9 te − 5 2 t 20 −2t 10 2 −2t 7 −2t 3 173 t 173 −2t 10 t t − 9 t e − 9 e t , y(t) = − 81 e + 81 e − 9 te + 76 t 2 et + 79 3 t e − 27 e 14 3479 10 −t t 2 e−2t + 27 e−2t t + 79 e−2t t 3 ; (c) x(t) = 279 289 cos(4t) − 4624 sin(4t) + 289 e + 1 1 5 1321 300 11 −t −t 2 sin(4t)t − 4 cos(4t)t + 17 te , y(t) = 2312 sin(4t) + 289 cos(4t) − 289 e + 3 1 −t 17 te − 2 cos(4t)t

141

142

CH A P T E R 6:

Systems of Differential Equations

  43. Homogeneous: x(t) = e−7t 3 − 2e2t , y(t) = −e−7t ; nonhomogeneous:   210t + 3600e−7t − 2548e−5t + 173 1 7t − 48e−7t − 1 , y(t) = 49 1225 Homogeneous

x,y 1.0

x(t) =

Nonhomogeneous x,y 1.0

0.5

0.5

0.5

1.0

1.5

2.0

t

0.5

⫺0.5

⫺0.5

⫺1.0

⫺1.0

1.0

1.5

2.0

t

45. Homogeneous: x(t) = e3t (3 sin(2t) + cos(2t)), y(t) = −4e3t sin(2t); nonhomogeneous: x(t) = 14 e3t ((2t + 15) sin(2t) − (6t + 1) cos(2t) + 5), y(t) = 12 e3t (−9 sin(2t) + (4t + 3) cos(2t) − 3) Nonhomogeneous

Homogeneous x,y

x,y

2 ⫻ 106 5 ⫻ 106 1 ⫻ 106

1

2

3

4

5

6

⫺1 ⫻ 106

1

t

⫺5 ⫻ 106

⫺2 ⫻ 106

Exercises 6.6 1. λ1,2 = 1, −11; saddle point, unstable 3. λ1,2 = 2; deficient node, unstable 5. λ1,2 = −8, −12; improper node, asymptotically stable 7. λ1,2 = −1; star node, asymptotically stable √ 9. λ1,2 = 2 ± 5i 2; spiral point, unstable √ 11. λ1,2 = ±i 41; center, stable

2

3

4

5

6

t

Answers, Hints, and Solutions to Selected Exercises

y

13.

⫺10

10

5

5

⫺5

5

10

x

⫺10

⫺5

⫺5

⫺5

⫺10

⫺10

y

15.

⫺10

y

14.

10

10

5

5

5

10

x

⫺10

10

5

10

x

y

16.

10

⫺5

5

⫺5

⫺5

⫺5

⫺10

⫺10

x

 0 1    17. x − x − 2x = 0, X = X, saddle point, unstable 2 1  0 1    19. x − 6x + 9x = 0, X = X, deficient node, unstable −9 6 y

21.

⫺10

10

5

5 x

⫺5

5

10

⫺10

x

⫺5

⫺5

⫺5

⫺10

⫺10

y

23.

⫺10

y

22.

10

10

5

5 x 5

10

⫺10

10

5

10

y

24.

10

⫺5

5

x

⫺5

⫺5

⫺5

⫺10

⫺10

27. A1 = K3 C1 /K1 , B1 = K3 C1 /K2 , C1 arbitrary.

143

144

CH A P T E R 6:

Systems of Differential Equations

Exercises 6.7 1. (0, 0), saddle point, unstable; (1, 0), inconclusive–center or spiral point 3. (0, 0), saddle point, unstable; (1, −1), spiral point, unstable 5. (0, 0), inconclusive–center or spiral point y

1.

y

2.

3 2

⫺3 ⫺2 ⫺1 ⫺1

1

2

3

x

4

2

2

⫺3

y

1

⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺4

3

1

1

4.

y

3.

3

1

2

3

4

x

⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺2

⫺3

⫺3

⫺4

⫺4

y

5.

1

4

4

2

2

2

2

4

x

⫺4

⫺2

2

x

4

3

4

x

y

6.

4

⫺2

2

⫺4

⫺2

2

⫺2

⫺2

⫺2

⫺4

⫺4

⫺4

x

4

7. (0, 0), node or spiral point, unstable 9. (0, 0), improper node, asymptotically stable; (0, 4), saddle point, unstable; (2, 0), saddle point, unstable; (3, 1), spiral point, unstable 7.

8.

y 4 2

⫺4

⫺2

2 ⫺2 ⫺4

4

x

9.

y

y

4

6

3

5

2

4

1

3

⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2

3

2

x

1 ⫺1 ⫺1

11. (−4, 4), saddle point, unstable; (1, −1), saddle point, unstable 13. (2, −1), λ1,2 = −6, 1, saddle; (8, −1), λ1,2 = 6, 1, unstable node √ 15. (0, 0), λ1,2 = ± 2, saddle; (1, 1), λ1,2 = −1 ± i, stable spiral

1

2

3

4

5

x

Answers, Hints, and Solutions to Selected Exercises

11.

10. y 6

y

12.

y

5

6

5

4

4

4

3

3

2

2

1

1 ⫺1 ⫺1

1

2

3

4

5

6

x

2

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

x

2

⫺6 ⫺4 ⫺2 ⫺2

14.

4

6

2

3

4

x

⫺4 ⫺6

⫺2

13. y

2

15.

y

y 4

2

4 2

4

6

8

10

3

x

2

2

⫺2

1

⫺4

⫺6

⫺4

⫺2

2

⫺6

x ⫺2 ⫺1

1

x

⫺1

⫺2

⫺8

⫺2

17. (0, 2), λ1,2 = −6, 1, unstable node; (0, −2), λ1,2 = −2, −4, stable node; (2, 0), λ1,2 = √ √ ±2 2, saddle; (−2, 0), λ1,2 = ±2 2, saddle √ 19. (0, 0), λ1,2 = ±2, saddle; (−1, −1), λ1,2 = 1 ± i 3, unstable spiral y

16.

y

17.

3

6 4

2 2

2

1

⫺3 ⫺2 ⫺1

1

2

3

x

⫺1

⫺4

⫺2

2

⫺6

y

3

2

2

1

1 1

2

3

x

⫺3 ⫺2 ⫺1

⫺8 y

20.

3

⫺1

⫺4

⫺4

⫺3

⫺3 ⫺2 ⫺1

x ⫺12 ⫺10 ⫺8 ⫺6 ⫺4 ⫺2 4 ⫺2

⫺2

⫺2

19.

y

18.

4

1

⫺1

⫺2

⫺2

⫺3

⫺3

2

3

x

2

x

145

146

CH A P T E R 6:

Systems of Differential Equations

21. (0, 0), λ1,2 = a1 , −a2 , saddle; (a1 /b1 , 0), λ1,2 = −a 1 , a1 c2 /b1 − a2 , stable node;   a2 b1 a2 b1 2 4a2 (a1 c2 + a2 b1 ) λ1,2 = ± − , (−a2 /c2 , (a1 c2 + a2 b1 )/(c1 c2 )), c2 c2 b1  2 a2 b1 4a2 (a1 c2 + a2 b1 ) unstable node if − ≥ 0; unstable spiral if c2 b1 2  4a2 (a1 c2 + a2 b1 ) a2 b1 − <0 c2 b1   25. H (x, y) = (2y + x−1 y−2 ) dy + g(x); g(x) = − −2x dx so H (x, y) = x2 − (xy)−1 + y2 . y

24.

25. y 10

6 4

8

2 6 ⫺6

⫺4

⫺2

2

4

6

x 4

⫺2

2

⫺4 ⫺6

2

4

6

8

  27. H (x, y) = (x cos xy − sin 2y) dy + g(x); g(x) = − sin 2x dx = 1 1 2 cos 2y + 2 cos 2x + sin xy. y

26.

⫺4

cos 2x so H (x, y) =

1.5

4

1

2

0.5

⫺2

x

y

27.

6

⫺6

1 2

10

2

4

6

x

⫺1.5

⫺1

⫺0.5

0.5

⫺2

⫺0.5

⫺4

⫺1

⫺6

⫺1.5

29. x(t) = c1 et , y(t) = − 13 c1 2 e2t + c2 et

1

1.5

x

Answers, Hints, and Solutions to Selected Exercises

y 1

0.5

⫺1

x

⫺0.5

0.5

1

⫺0.5 ⫺1

31. y

y

y

y

1

1

1

1

0.5

0.5

0.5

0.5

⫺1 ⫺0.5

0.5

1

⫺1

⫺1 ⫺0.5

0.5

1

⫺

13 15

⫺1 ⫺0.5

0.5

1

⫺

11 15

⫺1 ⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺1

⫺1

⫺1

⫺1

y

y

y

1

1

1

0.5

0.5

0.5

0.5

0.5

1

⫺

7 15

⫺1 ⫺0.5

0.5

1

⫺

1 3

⫺1 ⫺0.5

0.5

1

⫺

1 5

⫺1 ⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺1

⫺1

⫺1

⫺1

y

y

y

1

1

1

0.5

0.5

0.5

0.5

0.5

1

1 15

⫺1 ⫺0.5

0.5

1

1 5

⫺1 ⫺0.5

0.5

1

1 3

⫺1 ⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺1

⫺1

⫺1

⫺1

y

y

1 0.5 ⫺1 ⫺0.5

y

1

0.5

1

⫺1 ⫺0.5

0.5

1

⫺1 ⫺0.5

0.5

1

0.5

1

⫺

1 15

7 15

1

0.5 11 15

1

3 5

y

1

0.5 3 5

0.5

⫺

y

1

⫺1 ⫺0.5

1

y

1

⫺1 ⫺0.5

0.5

0.5

0.5

1

13 15

⫺1 ⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺0.5

⫺1

⫺1

⫺1

⫺1

1

147

148

CH A P T E R 6:

Systems of Differential Equations

32.

y 6 5 4 3 2 1 1

2

3

4

5

6

x

Exercises 6.8 1. x(1) ≈ 8.64479, y(1) ≈ 8.29263 3. x(1) ≈ −7.2362, y(1) ≈ −1.5998 5. x(1) ≈ −0.113115, y(1) ≈ −1.06576 7. x(1) ≈ −326.204, y(1) ≈ 608 9. x(1) ≈ 0.164251, y(1) ≈ −0.504587 11. x(1) ≈ 1.52319, y(1) ≈ 0.419975 13. x = y, y = −2x − 3y, x(0) = 0, y(0) = −3; Euler’s method yields x(1) ≈ −0.723913, y(1) ≈ 0.40179; exact solution is x(t) = 3e−2t − 3e−t so x(1) = 3e−2 (1 − e) ≈ −0.697632. 15. x = y, y = −9x, x(0) = 0, y(0) = 3; Euler’s method yields x(1) ≈ 0.346313, y(1) ≈ −4.49743; exact solution is x(t) = sin 3t so x(1) = sin 3 ≈ 0.14112. 17. x = y, y = −t −2 (16x + ty), x(1) = 0, y(1) = 4; Euler’s method yields x(2) ≈ 0.354942, y(2) ≈ −2.90834; exact solution is x(t) = sin(4 ln t) so x(2) = sin(4 ln 2) ≈ 0.360687. 19. (See 13) Runge-Kutta yields x(1) ≈ −0.697621, y(1) ≈ 0.291602 21. (See 15) Runge-Kutta yields x(1) ≈ 0.141307, y(1) ≈ −2.96975 23. (See 17) Runge-Kutta yields x(1) ≈ 0.360845, y(1) ≈ −1.86541 25. x(t) = 13 e−2t (e3t − 1), y(t) = 13 e−2t (e3t + 2) 26. x(t) = − 13 e−2t (3 cos 3t + sin 3t), y(t) = 13 e−2t (3 cos 3t + 11 sin 3t)

Answers, Hints, and Solutions to Selected Exercises

27. x(t) = 2et (t + 1), y(t) = −2tet 25. y 50

y

26.

40

y

27.

1

1

0.5

0.5

30 ⫺1

20

⫺0.5

10 10

20

30

40

50

x

0.5

2 x 4

6

22 24 26

Chapter 6 Review Exercises  

0.5

⫺1

4

2

⫺0.5

⫺1

6

22

⫺1

⫺0.5

y

24

x

⫺0.5

28.

26

1

1 −2 1. λ1 = 11, λ2 = −4, v1 = , v2 = 2 1  1 3. λ1,2 = −4, v = 1 ⎛ ⎞ ⎛ ⎞ −1 ± 12 i −2 ⎜ ⎟ ⎜ ⎟ 5. λ1,2 = ±3i, λ3 = −1, v1,2 = ⎝ −1/2 ⎠, v3 = ⎝−1⎠ 1 1

1

x

149

150

CH A P T E R 6:

Systems of Differential Equations

⎛ ⎞ 0 ⎜ ⎟ 7. λ1,2,3 = 0, v = ⎝1⎠ 1        9. x = 15 et 4 + e5t c1 − 4 −1 + e5t c2 , y = 15 et −e5t c1 + c1 + 4e5t c2 + c2 11. x = −e−2t + 2et , y = 3e−2t − 3et , x(0) = 1, y(0) = 0 13. x = 2 sin 4t, y = cos 4t + sin 4t 15. x =

17 −2t 4 e

sin 4t, y = 14 e−2t (4 cos 4t − sin 4t)

17. x = e−t (2tc1 + c1 + 2tc2 ), y = e−t (c2 − 2t (c1 + c2 )) 19. x = e−t ((5 − 4et )c1 + 2(−1 + et )(3c2 − 2c3 )), y = (−2 + 2e−2t )c1 + (3 − 2e−2t )c2 + 2(−1 + e−2t )c3 , z = e−2t (−3c2 (−1 + et )2 + (3 − 5et + 2e2t )c1 + (3 − 4et + 2e2t )c3 ) 1 21. x = cos 3t + 11 3 sin 3t, y = 3 (−5 + 5 cos 3t + sin 3t), z = − cos 3t + 5 sin 3t

23. x(t) = −e7 t + c2 e−t , y(t) = 79 e7 t + c2 e−t + c1 e−2t 25. x(t) = c2 e−t + c1 e5 t + 1, y(t) = −c2 e−t + 12 c1 e5 t − 1 27. x(t) = c2 sin(t) + c1 cos(t) − cos(t) − t sin(t), y(t) = −c2 cos(t) + c1 sin(t) + t cos(t) − sin(t) 45 2 1 2 2 29. x(t) = 15 2 t ln(t) − 4 t − t ln(t) + t + c1 t + c2 , y(t) = t ln(t) − t + 9 c1 + 5 t ln(t) − 15 2 2 2 2 t + 3 c1 t + 3 c2      1 1 1 31. x = 210 140t 3/2 + 16t 7/2 , y = 15 −10t 3/2 − 12t 5/2 , z = 210 140t 3/2 + 56t 5/2 +  3/2 16t

32.

33.

y

y 2

2.0 1.5

1 1.0 0.5

21.0 20.5

22 0.5

1.0

1.5

2.0

21

1

x 21

20.5 21.0

22

2

x

Answers, Hints, and Solutions to Selected Exercises

Differential Equations at Work A. Modeling a Fox Population in Which Rabies Is Present 1. K52

K51 Susceptible

Susceptible

Infected

1.000 0.999 0.998 0.997 0.996

4. 3 10212 2. 3 10212 22. 3 10212 24. 3 10212

10 20 30 40 50

25. 3 10213 21. 3 10212

25. 3 10214

20

30

40

50

8. 3 10215 6. 3 10215 4. 3 10215 2. 3 10215 24. 3 10215 26. 3 10215

0.506 0.505 0.504 0.503 0.502 10 20 30 40 50 0.501 10 20 30 40 50

Susceptible

Infected 1. 3 10210 5. 3 10211

10

20

30 40 Rabid

25. 3 10211 21. 3 10210

50

21. 3 10210

10 20 30 40 50

10 20 30 40 50

22. 3 10210

40

50

K58 Infected 6. 3 10212 4. 3 10212 2. 3 10212 10 20 30 40 50 Rabid 3. 3 10211 2. 3 10211 1. 3 10211 10 20 30 40 50

21. 3 10212

2. 3 10

Susceptible 0.1280 0.1275 0.1270 0.1265 0.1260 0.1255

1. 3 10212

22. 3 10212 24. 3 10212 30

22. 3 10212 24. 3 10212 0.1280 0.1275 0.1270 0.1265 0.1260 0.1255

6. [(σ + α)(α + β)]/(σβ) = 1.00103

10 20 30 40 50

Total Population

10 20 30 40 50

Infected

2. 3 10212

4. 3 10212 212

20

K54

10 20 30 40 50 Rabid

0.339 0.338 0.337 0.336 0.335 10

21. 3 10211 22. 3 10211 23. 3 10211

0.255 0.254 0.253 0.252 0.251

Total Population

2. 3 10210 1. 3 10210

Total Population

Rabid

K53 Susceptible 0.339 0.338 0.337 0.336 0.335

10 20 30 40 50

10 20 30 40 50

1.000 0.999 0.998 10 20 30 40 50 0.997 10

Infected 1. 3 10213 5. 3 10214

Total Population

Rabid 1.5. 3 10212 1. 3 10212 5. 3 10213

10 20 30 40 50

0.506 0.505 0.504 0.503 0.502 0.501

10 20 30 40 50

Total Population

0.255 0.254 0.253 10 20 30 40 50 0.252 0.251 10 20 30 40 50

151

152

CH A P T E R 6:

Systems of Differential Equations

B. Controlling the Spread of a Disease γ 1. If S0 < γ/λ, I  (0) = λS0 I0 − γI0 < λ · I0 − γI0 = 0. Thus, the rate of infection immeλ γ diately begins to decrease. If I  (0) = λS0 I0 − γI0 > λ · I0 − γI0 = 0, the rate of λ infection first increases. 2. The equation is separable. dI /dS = −1 + ρ/S becomes dI = (−1 + ρ/S)dS, which has solution I = −S + ρ ln S + C. Applying the initial condition results in I0 = −S0 + ρ ln S0 + C so C = I0 + S0 − ρ ln S0 and thus I + S − ρ ln S = I0 + S0 − ρ ln S0 . 3. The maximum value of I occurs when dI /dS = −1 + ρ/S = (ρ − S)/S = 0 ⇒ S = ρ so the maximum is   ρ −1 . I = −ρ + ρ ln ρ + I0 + S0 − ρ ln S0 = I0 + S0 + ρ · ln S0 4. We solve the system λSI + μ − μS = 0, λSI − γI − μI = 0 for S and I . λSI − γI − μI = 0 ⇒ I (λS − γ − μ) = 0 ⇒ I = 0 or λS − γ − μ = 0 with solution S = (γ + μ)/λ. If I = 0, S = 1; if S = (γ + μ)/λ, −λ ·

γ +μ γ +μ I +μ−μ· =0 λ λ

γ +μ −μ λ γ +μ μ· −μ μ λ − (γ + μ) λ I =− = . γ +μ λ γ +μ 5 and 6. To have a “meaningful” equilibrium point, we must have −(γ + μ)I = μ ·

S +I ≤ 1 γ + μ μ λ − (γ + μ) + ≤1 λ λ γ +μ γ 2 + γμ + λμ ≤1 λ(γ + μ) γ 2 = γμ + λμ ≤ λ(γ + μ) γ[(γ + μ) − λ] ≤ 0 ⇒ (γ + μ) − λ ≤ 0. Otherwise, there is no “meaningful” equilibrium point. If there is an equilibrium point, it must be a stable spiral.

Answers, Hints, and Solutions to Selected Exercises

7.

λ

Disease Measles Chickenpox Mumps Scarlet Fever

8.

Disease

4.44 2.69 1.58 1.7

Minimum R-value

Measles 0.933186 Chickenpox 0.911505 Mumps 0.876544 Scarlet Fever 0.882354

9. Chickenpox

Measles I

I

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

S 0.2

0.4

0.6

0.8

S

1.0

0.2

Mumps

0.4

0.6

0.8

1.0

0.8

1.0

Scarlet fever

I

I

1.0

1.0

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

S 0.2

0.4

0.6

0.8

1.0

S 0.2

0.4

0.6

153

154

CH A P T E R 6:

Systems of Differential Equations

10. A variety of responses are possible here. Some responses might include discussions of immunization programs, isolation programs, or simply letting the epidemic run its course.

C. Fitzhugh-Nagumo Model 1.0 0.5 0.0

1.5

20.5 21.0 1.0

1.0

0.5 0.5 0.0 20.5

20

21.0 21.0

20.5

40

60

80

100

20.5 0.0 0.5 1.0 1.0

0.5 0.0 20.5 21.0 1.0

1.0 0.5

0.5 2

0.0

20.5

20.5 21.0 21.0

21.0 20.5

21.5 0.0 0.5 1.0

4

6

8

10

12

14

CHAPTER 7

Applications of Systems of Ordinary Differential Equations

Exercises 7.1 1. (a) We solve the system ⎞  ⎧ ⎧ ⎧  ⎛ dQ dQ  0 1 ⎪ ⎪ ⎪ Q Q ⎪ ⎪ ⎪ = I = I ⎪ ⎪ ⎪ ⎝ ⎪ ⎪ ⎪

⎠ ⎪ ⎪ ⎪ ⎨ dt ⎨ dt ⎨ I  = −10 3 −10 3 I dI 1 10 ⇔ dI 10 10 ⇔  :    = − Q − I = − Q − I ⎪ ⎪ ⎪ Q(0) 0 ⎪ ⎪ ⎪ dt 3 · 0.1 3 dt 3 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = ⎪ ⎪ ⎪ ⎩ ⎩ ⎩ I(0) 1 Q(0) = 0, I(0) = 1 Q(0) = 0, I(0) = 1 √  ⎧ √ 3 5 −5t/3 5 ⎪ ⎪ ⎪ Q(t) = e sin t ⎪ ⎨ 5 3 √  √  ⎪ √ −5t/3 ⎪ 5 5 ⎪ ⎪ sin t + e−5t/3 cos t ⎩I(t) = − 5e 3 3 (b) We solve the system ⎧ ⎧ dQ dQ ⎪ ⎪ ⎪ ⎪ = I =I ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ dt ⎨ dt dI dI 1 10 1 10 10 1 =− Q − I + e−t ⇔ ⎪ = − Q − I + e−t ⎪ ⎪ ⎪ dt 3 · 0.1 3 3 dt 3 3 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ Q(0) = 0, I(0) = 1 Q(0) = 0, I(0) = 1

⎧     ⎪ Q 0 1 Q ⎪ ⎪ ⎪ =



⎪ ⎪ ⎪ I −10 3 −10 3 I ⎪ ⎪   ⎪ ⎪ 0 ⎨ + e−t : ⇔⎪ 1 3 ⎪ ⎪ ⎪     ⎪ ⎪ ⎪ Q(0) 0 ⎪ ⎪ ⎪ = ⎪ ⎩ I(0) 1

√  √  ⎧ √ ⎪ 1 −t 7 5 −5t/3 5 1 −5t/3 5 ⎪ ⎪ sin t − e cos t ⎪ ⎨Q(t) = 3 e + 15 e 3 3 3    √ √ √  ⎪ 1 2 5 5 5 4 ⎪ ⎪ ⎪I(t) = − e−t − e−5t/3 sin t + e−5t/3 cos t ⎩ 3 3 3 3 3 155

156

CH A P T E R 7:

Applications of Systems of Ordinary Differential Equations

⎧ dQ ⎪ ⎪ = −Q − I2 ⎪ ⎪ ⎪ dt ⎪ ⎪ ⎨ 5. (a) We solve dI = Q − 3I : 2 ⎪ ⎪ dt ⎪ ⎪ ⎪ ⎪ ⎪ ⎩Q(0) = 10−6 , I (0) = 0 2 ⎧ ⎪ Q(t) = 10−6 e−2t (1 + t) ⎪ ⎨ ⎪ I (t) = 10−6 te−2t ⎪ ⎩2

⇒ I(t) =

dQ = −10−6 e−2t (1 + 2t) dt

⇒ I1 (t) = I(t) + I2 (t) = −10−6 e−2t (1 + t) ⎧ dQ ⎪ ⎪ ⎪ dt = −Q − I2 + 90 ⎪ ⎪ ⎨ (b) We solve dI = Q − 3I : 2 ⎪ ⎪ dt ⎪ ⎪ ⎪ ⎩ Q(0) = 0, I2 (0) = 0 ⎧ 135 135 −2t ⎪ ⎪ Q(t) = − e − 45te−2t ⎪ ⎨ 2 2 ⇒ I(t) = 90e−2t (1 + t) 45 45 −2t ⎪ −2t ⎪ ⎪ ⎩I2 (t) = 2 − 2 e − 45te ⇒ I1 (t) = ⎧ dQ ⎪ = −Q − I2 + 90 ⎪ ⎪ ⎪ dt ⎪ ⎪ ⎪ ⎪ dI2 ⎨ = Q − I2 + I3 9. (a) We solve dt : ⎪ ⎪ dI3 ⎪ ⎪ = I2 − I3 − 90 ⎪ ⎪ dt ⎪ ⎪ ⎩ Q(0) = 0, I2 (0) = 1, I3 (0) = 0 ⎧ −t −t ⎪ ⎪Q(t) = 90 − te − 90e ⎪ ⎨ I2 (t) = e−t ⎪ ⎪ ⎪ ⎩ I3 (t) = −90 + te−t + 90e−t

45 −2t e (3 + e2t + 2t) 2

Answers, Hints, and Solutions to Selected Exercises

⎧ dQ ⎪ ⎪ = −Q − I2 + 90 sin t ⎪ ⎪ dt ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dI2 ⎪ ⎪ ⎨ = Q − I2 + I3 (b) We solve dt : ⎪ ⎪ ⎪ dI 3 ⎪ ⎪ = I2 − I3 − 90 sin t ⎪ ⎪ ⎪ dt ⎪ ⎪ ⎪ ⎪ ⎩ Q(0) = 0, I2 (0) = 1, I3 (0) = 0 ⎧ −t −t ⎪ ⎪Q(t) = −te + 45e + 45 sin t − 45 cos t ⎪ ⎨ I2 (t) = e−t ⎪ ⎪ ⎪ ⎩ I3 (t) = te−t − 45e−t − 45 sin t + 45 cos t ⎧ dx

⎪ ⎪

−λ 1 ⎨ dt = y

= λ2 + 9 = 0 ⇒ λ = ±3i; undamped; center 13. ,

⎪ dy ⎪

−9 −λ ⎩ = −9x dt dx dy d2 x . Then, x  + 10x  + 50x = 0 ⇒ = 2 = −50x − 10x  = 17. We let y = dt ⎧ dt dt dx

⎪ =y ⎪

−λ ⎨ 1

dt

= λ2 + 10λ + 50 = 0 ⇒ ,

−50x − 10y so

⎪ ⎪ dy

−50 −10 − λ ⎩ = −50x − 10y dt λ1 = −5 + 5i, λ2 = −5 − 5i; underdamped, stable spiral ⎧ 9 −t 1 −9t ⎧ ⎪ ⎪ ⎪ ⎨x(t) = cos 3t ⎨x(t) = 8 e − 8 e 21. (13) ; (15) ; (19) Because ⎩y(t) = −3 sin 3t ⎪ ⎪ 9 −t 9 −9t ⎪ ⎩y(t) = − e + e 8 8     1 1 , X1 (t) = e−5t and λ1 = λ2 = −5 with corresponding eigenvector v = −5 −5       1 0 1 −5t X2 (t) = t+ e . A general solution is X(t) = c1 e−5t + −5 1 −5         1 0 c 1 1 c2 t+ e−5t . X(0) = = implies that c1 = 1 −5c1 + c2 −5 1 0 and c2 = 5.

157

158

CH A P T E R 7:

25.

Applications of Systems of Ordinary Differential Equations

Mathematica

Maple

sys={q’[t]==i[t],i’[t]==-1/(l c)q[t]r/l’i[t]+e[t]/l, q[0]==q0,i[0]==i0} l=1; r=40; c=4/1000; e[t_]=120Sin[t]; q0=0; i0=0; sol1=DSolve[sys,{q[t],i[t]},t] <{GrayLevel[0], GrayLevel[.4]}] ParametricPlot[{q[t],i[t]}/.sol1, {t,0,7},AspectRatio->1, PlotRange->{{-1,1},{-1,1}}, Compiled->False]

sys:=’sys’:q:=’q’:i:=’i’: sys:={diff(q(t),t)=i(t), diff(i(t),t)=-1/(l*c)*q(t)r/l*i(t)+e(t)/l, q(0)=q0,i(0)=i0}; l:=1; r:=40; c:=4/1000; e:=t->120*sin(t); q0:=0; i0:=0; sol1:=dsolve(sys,{q(t),i(t)}); plot(subs(sol1,{q(t),i(t)}),t=0..7); plot(subs(sol1,[q(t),i(t),t=0..7])); 1 0.75 0.5

0.4 0.25 0.2 ⫺1 ⫺0.75 ⫺0.5 ⫺0.25 1

2

3

4

5

6

7

0.25

0.5

0.75

⫺0.25

⫺0.2 ⫺0.4

⫺0.5 ⫺0.75 ⫺1

30.

Clear[x,y,k,m,c] sys={x’[t]==y[t],y’[t]==-k/m x[t]c/m y[t], x[0]==x0,y[0]==y0} m=1; c=1; k=1/2; sol=DSolve[sys,{x[t],y[t]},t] <10 Cos[s],y0->10 Sin[s]}, {s,0,2Pi-2Pi/9,2Pi/9}];

Spr_2:=dsolve({diff(x(t),t)=y(t), diff(y(t),t)=-1/2*x(t)y(t)},{x(t),y(t)}); to_plot:=proc(pair) subs({_C1=pair[1],_C2=pair[2]}, [rhs(Spr_2[2]),rhs(Spr_2[1]), t=-Pi/2..2*Pi]) end: pairs:=[[0,0],[1,1],[-1,1], [1,-1],[1,2],[1,-2], [-1,2],[-1,-2],[2,1], [2,-1],[3,1],[3,-1], [3,2],[-2,-3],[-2,3],[-2,-1], [-3,1],[-3,-1],

1

Answers, Hints, and Solutions to Selected Exercises

[-4,1],[-4,-2],[-4,3],[-4,-4], [-4,5]]; to_graph:=map(to_plot,pairs); plot(convert(to_graph,set), view=[-5..5,-5..5]);

Short[toplot] grays=Table[GrayLevel[i], {i,0,.5,.5/8}]; ParametricPlot[Evaluate[toplot], {t,0,10},PlotStyle->grays, PlotRange->{{-10,10},{-10,10}}, AspectRatio->1]

10 7.5 5 2.5 ⫺10 ⫺7.5 ⫺5 ⫺2.5

2.5

5

7.5

10

⫺2.5 ⫺5 ⫺7.5 ⫺10

Stable Spiral Point.

31.

m=1; c=2; k=3/4; sol=DSolve[sys,{x[t],y[t]},t] toplot=Table[{sol[[1,1,2]], sol[[1,2,2]]}/. {x0->10 Cos[s],y0->10 Sin[s]}, {s,0,2Pi-2Pi/9,2Pi/9}]; Short[toplot] ParametricPlot[Evaluate[toplot], {t,-10,10},PlotStyle->grays, PlotRange->{{-10,10},{-10,10}}, AspectRatio->1]

Spr_2:=dsolve({diff(x(t),t)=y(t), diff(y(t),t)=-3/4*x(t)y(t)},{x(t),y(t)}); to_plot:=proc(pair) subs({_C1=pair[1],_C2=pair[2]}, [rhs(Spr_2[2]),rhs(Spr_2[1]), t=-Pi/2..2*Pi]) end: pairs:=[[0,0],[1,1],[-1,1], [1,-1],[1,2],[1,-2], [-1,2],[-1,-2],[2,1], [2,-1],[3,1],[3,-1], [3,2],[-2,-3],[-2,3],[-2,-1], [-3,1],[-3,-1], [-4,1],[-4,-2],[-4,3],[-4,-4], [-4,5]]; to_graph:=map(to_plot,pairs); plot(convert(to_graph,set), view=[-5..5,-5..5]);

159

160

CH A P T E R 7:

Applications of Systems of Ordinary Differential Equations

10 7.5 5 2.5 ⫺10 ⫺7.5 ⫺5 ⫺2.5

2.5

7.5

5

10

⫺2.5 ⫺5 ⫺7.5 ⫺10

Stable Node.

Exercises 7.2

⎧ 1 ⎪  ⎪ ⎪x = y − x ⎪ 2 ⎪ ⎨ 1 3. We solve y = x − y : ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎩ x(0) = 0, y(0) = 4 ⎧ 8 8 −3t/2 ⎪ ⎪   ⎨x(t) = 3 − 3 e 8 4 , lim (x(t), y(t)) = , 4 8 ⎪ 3 3 ⎪y(t) = + e−3t/2 t→∞ ⎩ 3 3

⎧ 1 ⎪  ⎪ ⎪x = 2 − x ⎪ 5 ⎪ ⎪ ⎨ 1 1 : 7. We solve y = x − y ⎪ 5 5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩x(0) = 0, y(0) = 0 

x(t) = 10 − 10e−t/5 , y(t) = 10 − 10e−t/5 − 2te−t/5



lim x(t) = lim y(t) = 10, x(t)

t→∞

t→∞

⎧ 1 ⎪ ⎪ x  = 15 − x ⎪ ⎪ ⎪ 20 ⎨ 11. (a) We solve y = 1 x − 1 y : ⎪ ⎪ 20 10 ⎪ ⎪ ⎪ ⎩x(0) = 0, y(0) = 0   x(t) = 300 − 300e−t/20 , y(t) = 150 + 150e−t/10 − 300e−t/20 ; (b) lim x(t) = 300, lim y(t) = 150 t→∞

t→∞

Answers, Hints, and Solutions to Selected Exercises

⎧ dx ⎪ ⎪ = 2y − 2x + 1 ⎨ dt 15. System: ⎪ dy ⎪ ⎩ = −3y + x + 1 dt 

 5 23 −4t 8 −t 3 23 −4t 4 −t 5 x(t) = − e + e , y(t) = + e + e ; lim x(t) = , t→∞ 4 12 3 4 12 3 4

lim y(t) =

t→∞

3 ; maximum is x = 4, y ≈ 1 4

      2 3 5 x(t) 21. We solve = X, X(0) = : X(t) = 0 −1 10 y(t)   2t −t 15e − 10e = ; lim x(t) = 0 t→∞ 10e−t ⎛ ⎞ ⎛ ⎞ 1 5 1 8 ⎜ ⎟ ⎜ ⎟ 25. We solve X = ⎝2 −6 2⎠ X, X(0) = ⎝2⎠: X(t) = 4 8 4 0 ⎛ ⎞ 43 46 7t 1 −8t ⎟ ⎛ ⎞ ⎜ 7 + 21 e − 3 e ⎜ ⎟ x(t) ⎜ ⎟ 4 7t 2 −8t ⎜ ⎟ ⎜ ⎟ e + e ⎟ ; z(1) ≈ 7095.86 ⎝y(t)⎠ = ⎜ 3 3 ⎜ ⎟ ⎜ 43 136 z(t) 1 −8t ⎟ ⎝ ⎠ 7t − + e − e 7 21 3 X



−a − λ 0 0

−a − λ 0

= (−a − λ)(−a − λ)(−λ) = 0; λ1 = λ2 = −a, λ3 = 0; 29. a

0 a −λ ⎛ ⎞ 0 ⎜ ⎟ −at a general solution is X = c1 X1 (t) + c2 X2 (t) + c3 X3 (t) where X1 (t) = ⎝−1⎠ e , 1 ⎡⎛ ⎞ ⎛ ⎞⎤ ⎛ ⎞ ⎛ ⎞ 0 −1/a 0 x0 ⎢⎜ ⎟ ⎜ ⎟⎥ ⎜ ⎟ ⎜ ⎟ X2 (t) = ⎣⎝−1⎠ t + ⎝ 1/a ⎠⎦ e−at , X3 (t) = ⎝0⎠. X(0) = ⎝y0 ⎠ yields 1 0 1 z0  c  c2 2 − = x0 , −c1 + = y0 , c1 + c3 = z0 with solution c1 = −(x0 + y0 ), c2 = −ax0 , a a and c3 = x0 + y0 + z0 : ⎧ −at ⎪ ⎨x(t) = x0 e −at y(t) = y0 e + ax0 te−at ; lim x(t) = 0,lim y(t) = 0, t→∞ t→∞ ⎪ ⎩ z(t) = x0 + y0 + z0 − (x0 + y0 )e−at − ax0 te−at lim z(t) = x0 + y0 + z0 t→∞

161

162

CH A P T E R 7:

Applications of Systems of Ordinary Differential Equations

⎧ ⎪ x  = −x + 10 ⎪ ⎪ ⎨y = x − y 39. We solve  : ⎪ z =y ⎪ ⎪ ⎩x(0) = 8, y(0) = 2, z(0) = 2

⎧ −t ⎪ ⎨x(t) = 10 − 2e −t ; y(t) = 10 − 8e − 2te−t ⎪ ⎩z(t) = 10t − 8 + 10e−t + 2te−t

lim x(t) = 10, lim y(t) = 10, lim z(t) = ∞

t→∞

41.

t→∞

t→∞

x:=’x’:y:=’y’: sol:=dsolve({diff(x(t),t)= p*(y(t)/v2-x(t)/v1), diff(y(t),t)=p*(x(t)/v1-y(t)/v2), x(0)=a,y(0)=b},{x(t),y(t)});

Clear[x,y,a,b,p,v2,v1] sol=DSolve[{x’[t]==p(y[t]/v2x[t]/v1), y’[t]==p(x[t]/v1-y[t]/v2), x[0]==a,y[0]==b},{x[t],y[t]},t] sol=Simplify[sol] Apart[sol[[1,1,2]]] Apart[sol[[1,2,2]]]

V1 (a + b) V2 (a + b) and lim y(t) = . t→∞ V1 + V2 V1 + V2 V1 (a + b) V2 (a + b) > ⇒ V1 > V2 . (b) We must have V1 + V2 V1 + V2 (c) (a) The results indicate that lim x(t) = t→∞

D[sol[[1,1,2]],t]//Apart D[sol[[1,2,2]],t]//Apart

Thus, for x(t) to be an increasing function we must have bV1 − aV2 > 0 and for y(t) to be an increasing function we must have −bV1 + aV2 > 0. Thus, it is not possible for both x(t) and y(t) to be increasing functions. (d)

Clear[x,y,a,b,p,v2,v1] sol=DSolve[{x’[t]==p(y[t]/v1x[t]/v1), y’[t]==p(x[t]/v1-y[t]/v1), x[0]==a,y[0]==a},{x[t],y[t]},t] sol=Simplify[sol]

They are both constant.

sol:=dsolve({diff(x(t),t)= p*(y(t)/v1-x(t)/v1), diff(y(t),t)=p*(x(t)/v1-y(t)/v1), x(0)=a,y(0)=a},{x(t),y(t)});

Answers, Hints, and Solutions to Selected Exercises

42.

(a) Clear[x,y,a,b,p,v2,v1] sol=DSolve[{x’[t]==p(y[t]-x[t]), y’[t]==p(x[t]-y[t]), x[0]==1,y[0]==2},{x[t],y[t]},t] sola=sol /. p->0.25 Plot[Evaluate[{x[t],y[t]} /. sola],{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]}, PlotRange->{0,2}] ParametricPlot[{x[t],y[t]} /. sola,{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]},Compiled->False, AxesOrigin->{0,0}, PlotRange->{{0,5},{0,5}}, AspectRatio->1]

sol:=dsolve({diff(x(t),t)= p*(y(t)-x(t)), diff(y(t),t)=p*(x(t)-y(t)), x(0)=1,y(0)=2},{x(t),y(t)}); sola:=subs(p=0.25,sol); plot(subs(sola,{x(t),y(t)}),t=0..10); plot(subs(sola,[x(t),y(t),t=0..10])); 5 4

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solb=sol /. p->0.5 Plot[Evaluate[{x[t],y[t]} /. solb],{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]}, AxesOrigin->{0,0}, PlotRange->{0,2}] ParametricPlot[{x[t],y[t]} /. solb,{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]},Compiled->False, AxesOrigin->{0,0}, PlotRange->{{0,5},{0,5}}, AspectRatio->1]

4

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solb:=subs(p=0.5,sol); plot(subs(solb,{x(t),y(t)}),t=0..10); plot(subs(solb,[x(t),y(t),t=0..10]));

0.75 0.5 0.25 0

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(c) solc=sol /. p->1 Plot[Evaluate[{x[t],y[t]} /. solc],{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]}, AxesOrigin->{0,0}, PlotRange->{0,2}] ParametricPlot[{x[t],y[t]} /. solc,{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]},Compiled->False, AxesOrigin->{0,0}, PlotRange->{{0,5},{0,5}}, AspectRatio->1] 2

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solc:=subs(p=1.0,sol); plot(subs(solc,{x(t),y(t)}),t=0..10); plot(subs(solc,[x(t),y(t),t=0..10]));

0.5 0.25 0

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sold=sol /. p->2 Plot[Evaluate[{x[t],y[t]} /. sold],{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]}, AxesOrigin->{0,0}, PlotRange->{0,2}] ParametricPlot[{x[t],y[t]} /. sold,{t,0,10}, PlotStyle->{GrayLevel[0], GrayLevel[.4]}, Compiled->False, AxesOrigin->{0,0}, PlotRange->{{0,5},{0,5}}, AspectRatio->1]

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sold:=subs(p=2.0,sol); plot(subs(sold,{x(t),y(t)}),t=0..10); plot(subs(sold,[x(t),y(t),t=0..10]));

0.75 0.5 0.25 0

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Answers, Hints, and Solutions to Selected Exercises

   dx dt a1 − a2

43. In matrix form the system is = a2 dy dt   a1 − a2 b1 are a2 b1 − b2 λ1,2 =

b1 b1 − b2

  x . The eigenvalues of y

" ! 1 a1 − a2 + b1 − b2 ± (a1 − a2 + b1 − b2 )2 − 4(a1 b1 − 2a2 b1 − a1 b2 + a2 b2 ) . 2

capa={{a1-a2,b1},{a2,b1-b2}} eigsa=Eigenvalues[capa]//Simplify DSolve[{x’[t]==(a1-a2)x[t]+b1 y[t], y’[t]==a2 x[t]+ (b1-b2)y[t]},{x[t],y[t]},t]

capa:=array([[a1-a2,b1],[a2,b1-b2]]); linalg[eigenvals](capa); dsolve({diff(x(t),t)=(a1-a2)*x(t)+ b1*y(t), diff(y(t),t)= a2*x(t)+(b1-b2)*y(t), x(0)= x0,y(0)= y0},{x(t),y(t)});

Note: The problem may be solved as a system or the system may be written as a second1 order ODE where x  = (a1 − a2 )x  + b1 y (derivative of first equation) or y = (x  − b1 1    (a1 − a2 )x ). Substitution of y and y = (x − (a1 − a2 )x) into the second equation and b1 simplification yields x  − [(a1 − a2 ) + (b1 − b2 )]x  + [(b1 − b2 )(a1 − a2 ) − a2 b1 ]x = 0. We consider the characteristics of this equation (same as the eigenvalues of the system). Periodic if a1 + b1 = a2 + b2 and a1 b1 − 2a2 b1 − a1 b2 + a2 b2 > 0. For example, if a1 = a2 = −1, b1 = b2 = 1, x0 = 1, and y0 = 0, then {x(t) = cos t, y(t) = −sin t}. Exponential Decay if (b2 − b1 ) > (a1 − a2 ) and (b1 − b2 )(a1 − a2 ) − a2 b1 ≥ 0. For example, if a1 = −1, a2 = 0, b1 = b2 = 1, x0 = 1, and y0 = 0, then {x(t) = e−t , y(t) = 0}. Exponential Growth if (a1 − a2 ) > (b2 − b1 ) (will have at least one positive eigenvalue). For example, if a1 = 2, a2 = 0, b1 = b2 = 1, x0 = 1, and y0 = 0, then {x(t) = e2t , y(t) = 0}.

Exercises 7.3

#   I − 2C = 0 1 −2 3. (a) ⇒ (I, C) = (2k, k). The eigenvalues of are λ1,2 = 2(I − C − k) = 0 2 −2 # √ 1 I − 2C = 0 ⇒ (I, C) = (2k, k). The − ± i 7 so (2k, k) is a stable spiral. (b) I −C −k = 0 2   1 −2 eigenvalues of are λ1,2 = ±i so (2k, k) is a center. 1 −1

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⎧ dx ⎪ ⎪ =y ⎨ $ dy/dt k2x 1 dy dt 7. (a) = =− ; (c) ydy = − ∫ k 2 xdx ⇒ y2 = ; (b) ⎪ dy dx dx/dt y 2 ⎪ ⎩ = −k 2 x dt k2 2 − x + C; (d) (0,0), center; solution is a family of ellipses as expected. 2 $x dx dy d2 x dy dx = y − F(x) = y − 0 f (u)du and = −g(x), then = − f (x) = 2 dt dt dt dt dt dx d2 x dx −g(x) − f (x) , which is equivalent to 2 + f (x) + g(x) = 0. dt dt dt ⎧ ⎪ √ ⎪ du = y ⎨ 1 ± 1 − 4kα dθ 2 ; y = 0 and α − u − ku = 0 ⇒ u = 15. . We choose ⎪ dy 2k ⎪ 2 ⎩ = α − u − ku dθ √ √ 1 − 1 − 4kα 1 + 1 − 4kα because it is closer to u = 0 than u = , and u= 2k   √ 2k 1 + 1 − 4kα we are considering a small change in the orbit. J , 0 = 2k   √ 0 1 √ ⇒ λ2 + 1 − 4kα = 0; center. − 1 − 4kα 0 11. If

1 2 1 1 x , then the paths in the phase plane are x 2 + y2 = C. 2 2 2 ⎧ dx ⎪ ⎪ =y ⎨ dt Equilibrium point of is (0,0) and is classified as a center. ⎪ dy ⎪ ⎩ = −V  (x) = −x dt ⎧ dθ ⎪ ⎪ =y ⎨ dθ dt ; equilibrium Points: (θ, 0); if > 0, then we integrate 23. ⎪ dy FR dt ⎪ ⎩ = − sgn(y) dt  I   dθ d dθ 1 dθ 2 I = = −FR with respect to θ to obtain the parabolas I dt dθ dt 2 dt dθ dθ − FRθ + C, > 0. Similar calculations follow for < 0. dt dt 19. If V (x) =

Answers, Hints, and Solutions to Selected Exercises

26.

with(DEtools): eqpts:=solve({x-xˆ3/3-y=0, x+1-y=0}); evalf(eqpts); DEplot2([x-xˆ3/3-y,x+1-y],[x,y], 0..15,x=-2..1,y=-2..1); DEplot2([x-xˆ3/3-y,x+1-y],[x,y], 0..15,{[0,-2,0],[0,0,-2], [0,-0.5,-2],[0,-2,-2], [0,0,0.75],[0,-1,0.75], [0,-2,-1]},x=-2..1,y=-2..1, stepsize=0.1);

Clear[f,g] f[x_,y_]=x-xˆ3/3-y; g[x_,y_]=x+1-y; Solve[{f[x,y]==0,g[x,y]==0}] N[%] <GrayLevel[.4], PlotPoints->30,Axes->Automatic, AxesOrigin->{0,0}, ScaleFunction->(1&)] Clear[numsol] numsol[{x0_,y0_}]:=NDSolve[{ x’[t]==x[t]-x[t]ˆ3/3-y[t], y’[t]==x[t]+1-y[t], x[0]==x0,y[0]==y0},{x[t],y[t]}, {t,0,15}] inits={{-2,0},{0,-2},{-0.5,-2}, {-2,-2},{0,.75},{-1,.75},{-2,-1}}; toplot=Map[numsol,inits]; paremplots=ParametricPlot[ Evaluate[{x[t],y[t]} /. toplot], {t,0,15},PlotStyle->GrayLevel[0], Compiled->False, DisplayFunction->Identity]; Show[pvf,paremplots, PlotRange->{{-2,1},{-2,1}}]

1 0.5

22

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21

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1

20.5 21 21.5 22

27. We guess x0 = 1, graph the resulting solution parametrically, and then use the graph to obtain x0 = 2.

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n1=NDSolve[{x’[t]==y[t], y’[t]==-x[t]-(x[t]ˆ2-1)y[t], x[0]==1,y[0]==0},{x[t],y[t]}, {t,0,20}] ParametricPlot[{x[t],y[t]} /. n1, {t,0,20}, Compiled->False, PlotRange->{{-3,3},{-3,3}}, AspectRatio->1] n2=NDSolve[{x’[t]==y[t], y’[t]==-x[t]-(x[t]ˆ 2-1)y[t], x[0]==2,y[0]==0},{x[t],y[t]}, {t,0,20}] ParametricPlot[{x[t],y[t]} /. n2, {t,0,20}, Compiled->False, PlotRange->{{-3,3},{-3,3}}, AspectRatio->1] 3

DEplot2([y,-x-(xˆ2-1)*y], [x,y],0..20,{[0,1,0]},x=-3..3, y=-3..3,arrows=’NONE’, stepsize=0.1); DEplot2([y,-x-(xˆ2-1)*y], [x,y],0..20,{[0,2,0]},x=-3..3, y=-3..3,arrows=’NONE’, stepsize=0.1);

3 2 1

⫺3

⫺2

2

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⫺1

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3

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1

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⫺1 ⫺2 ⫺3

Chapter 7 Review Exercises ⎧ dQ ⎪ ⎪ = −Q − I2 ⎪   ⎪ dt ⎨ −1 −1 dI2 3. (a) We solve : has eigenvalues and = Q − I2 ⎪ 1 −1 ⎪ ⎪ dt ⎪ ⎩ Q(0) = 10−6 , I2 (0) = 0   i corresponding eigenvectors λ1 = −1 + i, v1 = and λ2 = −1 − i, v2 = 1 ⎧ 1 ⎪   ⎪ e−t cos t ⎨Q(t) = −i 1000000 . , which results in ⎪ 1 1 ⎪ ⎩I2 (t) = e−t sin t 1000000

Answers, Hints, and Solutions to Selected Exercises

⎧ dQ ⎪ ⎧ ⎪ = −Q − I2 + 120 ⎪ 59999999 −t ⎪ ⎪ ⎨ dt ⎨Q(t) = 60 + 60e−t sin t − e cos t 1000000 (b) We solve dI2 : 59999999 −t = Q − I2 ⎪ ⎪ ⎪ ⎩I2 (t) = 60 − e sin t − 60e−t cos t ⎪ dt ⎪ 1000000 ⎩ −6 Q(0) = 10 , I2 (0) = 0 ⎧ dx 1 ⎪ ⎪ = (y − x) ⎪   ⎪ 2 ⎨ dt −1/2 1/2 1 : has eigenvalues and 7. We solve dy = (x − y) ⎪ 1/2 −1/2 ⎪ ⎪ dt 2 ⎪ ⎩ x(0) = 5, y(0) = 10     1 −1 corresponding eigenvectors λ1 = 0, v1 = and λ2 = −1, v2 = ; 1 1   15 5 −t 15 5 −t x(t) = − e , y(t) = + e . 2 2 2 2 ⎧ dx ⎪ = −3x + 4y + 8z ⎪ ⎪ ⎪ dt ⎪ ⎛ ⎞ ⎪ ⎪ ⎪ −3 4 8 ⎨ dy = x − 3y + 4z ⎜ ⎟ : ⎝ 1 −3 11. We solve 4 ⎠ has eigenvalues dt ⎪ ⎪ dz ⎪ 4 5 −10 ⎪ ⎪ = 4x + 5y − 10z ⎪ ⎪ ⎪ ⎩ dt x(0) = 4, y(0) = 4, z(0) = 8 ⎛ ⎞ ⎛ ⎞ 2 8 ⎜ ⎟ ⎜ ⎟ and corresponding eigenvectors λ1 = 3, v1 = ⎝1⎠, λ2 = −14, v2 = ⎝ 4 ⎠, 1 −13 ⎧ 5 128 139 ⎪x(t) = − e−5t − e−14t + e3t ⎪ ⎛ ⎞ ⎪ ⎪ 3 51 17 ⎪ ⎪ 10 ⎨ 7 64 139 3t ⎜ ⎟ λ3 = −5, v3 = ⎝−7⎠; y(t) = e−5t − e−14t + e ⎪ 6 51 34 ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎩z(t) = − 1 e−5t + 208 e−14t + 139 e3t 6 51 34   c a−k 13. (0, 0), saddle if a > k; stable node (star) if a ≤ k. , , center if a > k; d b saddle if a ≤ k.

Differential Equations at Work B. Food Chains < (0.5&),Axes->Automatic,AxesOrigin->{2,8}, PlotPoints->30] numsol=NDSolve[{x1’[t]==a*x1[t]-b*x1[t]ˆ2-c*x1[t]*x2[t], x2’[t]==-d*x2[t]+e*x1[t]*x2[t], x1[0]==3,x2[0]==2},{x1[t],x2[t]}, {t,0,15}]

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pxy=Plot[Evaluate[{x1[t],x2[t]}/.numsol],{t,0,8}, PlotStyle->{GrayLevel[0],GrayLevel[0.5]},AspectRatio->1, PlotRange->{0,11}] pp=ParametricPlot[{x1[t],x2[t]}/.numsol,{t,0,8},Compiled->False, AspectRatio->1]

C. Chemical Reactor (1) and (2) are direct. (3)

cval=Solve[cin-c==v/q c kt,c] step2=h v/(q cp)ckt /. cval[[1]]

cval:=solve(cin-c=v/q*c*kt,c); step2:=subs(c=cval,h*v/(q*cp)*c*kt);

(4) FindRoot (Mathematica) or fsolve (Maple V) may be useful.

CHAPTER 8

Introduction to the Laplace Transform

Exercises 8.1

 1. We use integration by parts once ( ueu du = eu (u − 1) + C) to evaluate the   M   ∞ −st 21 −st improper integral: L{21t} = 0 e · 21t dt = limM→∞ − 2 e (st + 1)  = s 0 1 −Ms −2 21 limM→∞ 2 (1 − e (Ms + 1)) = 21s s  ∞ ∞ 2 (1−s)t M 2 3. L{2et } = 0 e−st · 2et dt = 2 0 et−st dt = limM→∞ e  = s−1 1−s 0  u 1 u 5. Use integration by parts twice or use the formula e sin u du = 2 e (−cos u + sin u). ∞ 2 Then, L{2 sin 2t} = 0 e−st · 2 sin 2t dt = limM→∞ − 2 e−st (2 cos 2t + s +4 M  4 s sin 2t) = 2 s +4 0 M  ∞ 1 7. L{ f (t)} = 1 e−st · 1 dt = limM→∞ − e−st  = e−s s−1 s 1 π/2   π/2 −st 1 s + e−πs/2 −st 9. L{ f (t)} = 0 e · cos t dt = 2 e (−s cos t + sin t) = s +1 s2 + 1 0 3  3 1 11. L{ f (t)} = 0 e−st · (3 − t) dt = 2 e−st (1 + s(t − 3)) = (3s − 1 + e−3s )s−2 s 0 10 M    10 ∞ 1 1 13. L{ f (t)} = 0 e−st · 1 dt + 10 e−st · −1 dt = − e−st  + limM→∞ e−st  = s s 0 10 (1 − 2e−10s )s−1 M  ∞ 1 −st (k cos kt + s sin kt) = 15. L{sin kt} = 0 e−st sin kt dt = limM→∞ − 2 e  s + k2 0 2 2 k/(s + k ) 171

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CH A P T E R 8:

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21. (s − 1)/[(s − 1)2 + 9] 23. 5/[(s + 1)2 + 25] 25. 16(2s + 1)−3 27. −18(s − 3)−1 29.

s2 − 25 (s2 + 25)2

31.

1 5 + 2 s s + 25

33.

s2 + 2s − 3 (s2 + 2s + 5)2

35. (s + 1)−8 37. s−6 − s−4 + s−2 39.

24(5s4 − 10s2 + 1) (s2 + 1)5

41.

6s (s2 − 9)2

43. 45. 47.

14s (s2 + 49)2 2(s3 + 27s) (s2 − 9)3 2 s2 + 2s − 3

49.

s+2 (s + 2)2 + 16

51.

4 (s + 5)2 + 16

s2 + 2k 2 2k 2 ; (b) s3 + 4k 2 s s3 + 4k 2 s 61. No to all. 55. (a)

Exercises 8.2 1.

1 7 5040 t

3. e−5t 5.

1 2 −6t 2t e

7.

1 3

sin 3t

9. 8e−8t − 7e−7t

Answers, Hints, and Solutions to Selected Exercises

11.

2 −2t 9e

13.

1 2

+ 79 e7t

+ 12 e4t

15. − 18 e−2t + 18 e6t 17. 2 − sin t 19. 2 + e2t 21. t + cos 2t 23. 1 + t + t 2 25. t − 12 t 2 + 13 t 3 − 14 t 4 a3 b − ab3 and L{cos bt − cos at} = 2 2 2 2 (s  + a )(s + b )  2 2 a s−b s 1 a sin bt − b sin at ; (b) L−1 = and 2 2 2 2 2 2 2 2 (s + ab(a2 − b2 )  a )(s + b )  (s + a )(s + b ) s cos bt − cos at . L−1 = (s2 + a2 )(s2 + b2 ) a 2 − b2     t k 1 k −1 −1 29. (b) L = 0 sin kτ dτ = (1 − cos kt) and L = k s(s2 + k 2 ) s2 (s2 + k 2 ) t τ 1 0 0 sin kλ dλ dτ = k 2 (kt − sin kt).

27. (a) L{a sin bt − b sin at} =

Exercises 8.3 For each problem, let L{y(t)} = Y (s). Apply the Laplace transform to both sides of the equation, apply the initial conditions, and then solve for Y (S). Find y(t) = L−1 {Y (s)} by taking the inverse Laplace transform. 1. 24Y (s) + s2 Y (s) + 11(sY (s) − y(0)) − sy(0) − y (0). Application of the initial condi−s − 11 tions yields (s2 + 11s + 24)Y (s) + s + 11 = 0 so Y (s) = 2 . Taking the s + 11s + 24 inverse Laplace transform results in y = 15 e−8t (3 − 8e5t ). 3. (s2 + 3s − 10)Y (s) − (s + 3)y(0) − y (0) = 0; (s2 + 3s − 10)Y (s) + s + 2 = 0; −s − 2 Y (s) = 2 ; y = − 17 e−5t (3 + 4e7t ) s + 3s − 10 5. s(s − 1)Y (s) + y(0) − sy(0) − y (0)=1/(s − 1); s((s − 1)Y (s) − 1/(s − 1)) = 0; −s − 2 Y (s) = 2 ; y = tet s + 3s − 10 1 1 7. (s3 + s)Y (s) − (s2 + 1)y(0) − sy (0) + − y (0) = 0; (s3 + s)Y (s) + = 0; 1−s 1−s 1 Y (s) = ; y = 12 (−2 + et + cos t − sin t) s(s3 − s2 + s − 1) 9. s4 y(0) + s3 y (0) + s2 (y (0) + y(0)) − (s5 + s3 )Y (s) + s(y(3) (0) + y (0)) + 1 = 0; 1 (s5 + s3 )Y (s) − 1 = 0; Y (s) = 5 ; y = 12 (2 cos t + t 2 − 2) s + s3

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11. −(s + 2)y(0) + (s + 1)2 Y (s) − 1 ; y = 12 t 2 e−t (s + 1)3

1 1 − y (0) = 0; (s + 1)2 Y (s) − = 0; Y (s) = s+1 s+1

13. s2 (−y(0)) + (s2 − 1)sY (s) − sy (0) + (s2 − 1)2 Y (s) − 2 = 0; Y (s) =

1 1 − − y (0) + y(0) = 0; 1−s s+1

2 ; y = 12 e−t (1 − e2t + t + te2t ) (s2 − 1)2

15. 18s3 y(0) + 9s2 (2y (0) + 3y(0)) − (18s3 + 27s2 + 13s + 2)sY (s) + s(18y (0) + 27y (0) + 13y(0)) + 1 = 0; s(18s3 + 27s2 + 13s + 2)Y (s) − 1 = 0; 1 Y (s) = ; y = 12 − 32 e−2t/3 + 4e−t/2 − 3e−t/3 3 s(18s + 27s2 + 13s + 2) 3 2 ; Y (s) = 3s−3 + Cs−3 es /6 . Because lims→∞ Y (s) = 0, s C = 0 so Y (s) = 3s−3 and y = 32 t 2 .

17. s2 Y (s) − 9Y (s) − 3sY  (s) =

19. Taking the Laplace transform of y + ty − 2y = 0 and applying the initial conditions gives us s2 Y (s) − s − 3Y (s) − sY  (s) = 0, which has solution Y (s) = 2s−3 + 2 s−1 + Cs−3 es 2 . Because lims→∞ Y (s) = 0, C = 0 so Y (s) = 2s−3 + s−1 and hence y = t 2 + 1. 2 2 has solution Y (s) = 2s−3 + Cs−3 es /4 . s lims→∞ Y (s) = 0, C = 0 so Y (s) = 2s−3 and hence y = t 2 .

21. s2 Y (s) − 6Y (s) − 2sY  (s) =

Because

23. Y  (s) + s2 Y (s) = s, −(s2 + 2s)Y  (s) + (s2 + n(n + 1)Y (s) = 1, one √ √ √ 1 −t 1 −t t −t e sin( 3t) 3 + 35 27. y (t) = − 24 24 e cos( 3t) + 24 (−8 + (6 t − 3)e )e 29. y (t) = e−t − 16 te−t + 16 (−3 t 2 + t + 2 t 3 )e−t 27.

28.

29. 1.0

1.0 0.8 0.6 0.4 0.2

1.5 1.0 0.5 1

2

3

4

5

6

7

0.8 0.6 0.4 1

2

3

4

5

6

7

1

2

3

4

5

6

7

Exercises 8.4 Throughout, our notation is that L{ f (t)} = F(S); L−1 {Y (s)} = y(t) and L−1 {X(s)} = x(t). 1. −28e−s s−1 3. (3e−8s − e−4s )s−1 5. − 7.

42e−4s s−1

12e−2s s2 − 1

Answers, Hints, and Solutions to Selected Exercises

14e−πs/3 s2 + 1 es (es − 1 − s) 11. s2 (e2s − 1) 9. −

13.

2es + 3 se2s + ses + s

15. e−πs 17. 100e−2s + e−s  −2πs 19. e 100e3πs/2 +

1 2 s +1



 2 −st 1 e−s e δ(t − 1) dt = 0 1 − e−2s 1 − e−2s 23. −3U(t − π)

21. L{ f (t)} =

25. −3U(t − 4) + 2U(t − 1) 27. −3U(t − 6) + 3U(t − 5) − 4U(t − 3) 29. e3t−6 U(t − 2) 31. cos(2t − 6) U(t − 3) sin2 (t − 5) U(t − 5)

33.

1 2

35.

2(1 − 2e2s − e4s ) 1 2(1 − 2e2s − e−4s ) ∞ −4ks 2(1 − 2e2s − e−4s ) = = k=0 e s2 e4s (1 − e−4s ) s2 e4s 1 − e−4s s2 e4s so L−1



2(1 − 2e2s − e4s ) s2 e4s (1 − e−4s )

 =

∞

L−1

k=0

=2



2(1 − 2e2s − e−4s ) −4ks e s2 e4s



∞

(t − 4k − 4)U(t − 4k − 4) − (2t − 8k − 4) k=0

 U(t − 4k − 2) − (t − 4k)U(t − 4k)

37. cosh(6 − 2t)U(t − 3) 39.

1 s2 (1 + e−4s )

=

1 ∞ −4ks e . Then, s2 k=0 ∞ k=0

41.

L−1

 ∞

k=0

L−1



 ∞ 1 −4ks e = (t − 4k)U(t − 4k). s2 k=0

 1 1 ∞ −ks e = sin(2t − 2k)U(t − k) 2 2 k=0 s +4

43. f (t) = −δ(t − 1/2), 0 ≤ t < 1, f (t) = f (t − 1), t ≥ 1

175

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CH A P T E R 8:

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1 − e−2s . Then, taking the Laplace transform s of both sides of the equation and applying the initial condition gives us sY (s) + 3Y (s) − 1 − e−2s s − e−2s + 1 1= . Solving for Y (s), Y (s) = , and taking the inverse Laplace s s2 + 3s

1 −3t 3 3t 3 2 + e + (e − e )U(1 − t) . transform results in y = 3 e

45. The Laplace transform of f (t) is F(s) =

e−2s (es − 1)2 . Taking the Laplace transs form of both sides of the equation, applying the initial conditions, and solving for e−2s (es − 1)2 Y (s) gives us Y (s) = . Taking the inverse Laplace transform results in s2 (s + 4) ⎧ 1 ⎪ (4t − 1 + e−4t ), 0 ≤ t ≤ 1 ⎪ ⎨ 16 1 −4t y = 16 e (1 − 2e4 + e4t (9 − 4t)), 1 < t ≤ 2 . ⎪ ⎪ ⎩ 1 −4t 4 (e − 1)2 , t > 2 16 e

47. The Laplace transform of f (t) is L{f (t)} =

49. Taking the Laplace transform of both sides of the equation x  + x = 1 + δ(t − π) and applying the initial conditions x(0) = x  (0) = 0 gives us X(s) + s2 X(s) = 10e−πs + 1/s. 1 + 10se−πs 1 10 −πs Solving for X(s) we have X(s) = = + e and taking the s + s3 s(s2 + 1) s2 + 1 inverse Laplace transform results in x = 1 − cos t − 10 sin t U(t − π). 51. Taking the Laplace transform and applying the initial conditions gives us −10 + 13X(s) + s2 X(s) + 4sX(s) + s2 X(s) = 20e−πs/2 . Solving for X(s) and simplifying the 10 + 20e−πs/2 result gives us X(s) = 2 . Taking the inverse Laplace transform yields s + 4s + 13 10 −2t π x = 3 e [sin 3t + 2e cos 3t U(2t − π)]. 53. Taking the Laplace transform of both sides of the equation, applying the initial conditions, and solving for X(s) results in X(s) = 1 + e−s . Taking the inverse Laplace transform yields x = 2 s(s +  4s + 13) 2t 1 −2t 3e − 3 cos 3t − 2 sin 3t, 0 ≤ t < 1 e 39 6e2t − 3 cos 3t + e2 (−3 cos(3t − 3) − 2 sin(3t − 3)) − 2 sin 3t, t ≥ 1. 55. L{ f (t)} =

10e−s (es/2 − 1) so s(1 − e−s ) ∞

X(s) =

10e−s (es/2 − 1) 10e−s (es/2 − 1) −sk = e . s(s + 2)(1 − e−s ) s(s + 2) k=0

Taking the inverse Laplace transform gives us x(t) =

∞

−1



L

k=0

= 5e−2t

10e−s (es/2 − 1) −sk e s(s + 2)



∞   (e2+2k − e2t )U(t − k − 1) + (e2t − e2k+1 )U(t − 1/2 − k) . k=0

Answers, Hints, and Solutions to Selected Exercises

e−2s − e−s + s e−2s − e−s + s ∞ −2ks so = k=0 e 2 −2s s (1 − e ) s2

57. L{ f (t)} =

x(t) =

∞

L−1



k=0

=

e−2s − e−s + s −2ks e s4



∞

1 (t − 2 − 2k)3 U(t − 2 − 2k)+ 6 k=0

 (1 − t + 2k)3 U(t − 1 − 2k) + 3(t − 2k)2 U(t − 2k) .

 2 −st 1 es es e f (t) dt = . Then, s(s + 1)X(s) = so 0 s(es + 1) s(es + 1) 1 − e−2s −s −s ∞ 1−e 1−e (−1)k e−sk so X(s) = 2 = 2 −s s (s + 1)(1 + e ) s (s + 1) k=0

59. First, L{f (t)} =

x(t) =

∞

−1

L

k=0

=

∞

  −s k 1−e −sk (−1) 2 e s (s + 1)

  (−1)k (2 − e1+k+t − k + t)U(t − k − 1) + (t − k + 1 + ekt )U(t − k) .

k=0

61. L{f (t)} =

x(t) =

2e−πs/2 2e−πs/2 ∞ −πks/2 so = k=0 e (s2 + 4)(1 − e−πs/2 ) s2 + 4 ∞

−1

L

k=0



2e−πs/2 −πks/2 e (s2 + 4)2



∞

=−

1 [(π − 2t + kπ) cos(2t − kπ) + sin(2t − kπ)] U(t − π/2 − πk/2). 8 k=0

63. L{f (t)} =

 e−s −2ks . Then, = e−s ∞ k=0 e −2s 1−e −s

s(s + 2)Y (s) = e

∞

e−2ks

k=0

Y (s) =

∞ k=0

e−s e−2ks s(s + 2)

 1  1 − e2+4k−2t U(t − 1 − 2k). 2 ∞

y(t) =

k=0

177

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CH A P T E R 8:

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Exercises 8.5 1. 3. 4. 5.

1 3 3t t 2 0 (t − ν)ν dν

=

1 4 12 t

1 −t 5 (2e − 2 cos 2t + sin 2t) t    1 3 2 2 0 (t − ν) sin 4ν dν = 128 3 −8t + 16tν − 8ν + 1 sin(4ν)      t 4 8t 3 − 24t 2 ν + 3t 8ν2 − 1 − 8ν3 + 3ν cos(4ν) 



=

1 3 128 (32t − 12t + 3 sin 4t)

0

1 1 − s 1+s 1 1 9. 2 − 2 s s +1 1 s 10. − s s2 + 1 1 1 1 11. − 2− s−1 s s 7.

13.

e−3s s2

15.

e−3πs/2 (eπs − 1) s2 + 1

17.

1 −3t 54 (2 − 2e

19.

1 −t 10 (2e

− 6t + 9t 2 )

− 2 cos 2t + sin 2t)

21. (1 + cos t)U(t − π) 23. G(s) −

1 G(s) 1 = − 2 ⇒ G(s) = 2 ⇒ g(t) = sin t 2 s s s +1

4 = L{h(t)} − 25. aaaa s+2 H(s) −

s s2 + 1

− L{sin t ∗ h(t)}

4 s 1 = 2 − H(s) s+2 s + 1 s2 + 1 s 2(s − 2) 10 + + s2 + 2 3(s + 2) 3(s2 + 2) √ √ √ 2 2 10 5 h(t) = cos( 2t) − sin( 2t) + e−2t 3 3 3

H(s) =

4 3! sY (s) − y(0) − 4Y (s) + Y (s) = 27. aaaa s (s − 2)4 (s2 − 4s + 4)Y (s) =

6s (s − 2)4

−

Answers, Hints, and Solutions to Selected Exercises

6 12 + 6 (s − 2)5 (s − 2)   1 5 1 4 2t y(t) = t + t e 10 4

Y (s) =

Exercises 8.6 1. sX(s) − 2X(s) + 3Y (s) = 0, 9X(s) + sY (s) + 4Y (s) − 4 = 0; X(s) = − Y (s) =

4(s − 2) ; x = e−7t − e5t , y = 3e−7t + e5t s2 + 2s − 35

3. sX(s) − 5X(s) − 5Y (s) − 2 = 0, Y (s) = −

8 s2 − 2s + 5

4X(s) + sY (s) + 3Y (s) = 0;

12 s2 + 2s − 35

X(s) =

,

2(s + 3) , s2 − 2s + 5

; x = 2et (cos 2t + 2 sin 2t), y = −4et sin 2t

5. sX(s) − 1 = 2X(s) − Y (s) + 3Z(s), sY (s) = 6X(s) − 2Y (s) + 6Z(s), s (Lt [z(t)](s)) = −s − 5 6 −2X(s) + Y (s) − 3Z(s); X(s) = − , Y (s) = , (s + 1)(s + 2) s(s + 2) 2(s − 1)  ; x = e−2t (−3 + 4et ), y = 3e−2t (−1 + e2t ), Z(s) = −  2 s s + 3s + 2 z = e−2t (3 − 4et + e2t ) 1 −s − 4 7. sX(s) = 2X(s) − 3Y (s) + , sY (s) = 4X(s) − 4Y (s); X(s) = − 3 , s−1 s + s2 + 2s − 4 4 Y (s) = 3 ; x = e−2t − 2e−t + et , y = 23 e−2t (et − 1)2 (2 + et ) s + s2 + 2s − 4 1 6 9. sX(s) = 2X(s) − 6Y (s), sY (s) = X(s) − 3Y (s) + , ; X(s) = − s+1 s(s + 1)2 2−s Y (s) = − ; x = e−t (6 − 6et + 6t), y = e−t (2 − 2et + 3t) s(s + 1)2 1 1 11. sX(s) = −2X(s) − 4Y (s) + , sY (s) = 5X(s) + 2Y (s) − ; s+1 s+1 −s − 2 s−3 1   , Y (s) = −   ; x = 34 (2e−t − 2 cos 4t + X(s) = − (s + 1) s2 + 16 (s + 1) s2 + 16 1 (16et − 16 cos 4t − 13 sin 4t) 9 sin 4t), y = 68 1 1 −s2 − s + 5  , Y (s) = 13. sX(s) = X(s) − 5Y (s) + , sY (s) = X(s) − Y (s) 2 ; X(s) = − 2  2 s s s s +4 1 − 2s  ; x = 18 (2 − 10t − 2 cos 2t + 9 sin 2t), y = 18 (4 − 2t − 4 cos 2t + sin 2t) − 2 2 s s +4   1 −t cos t + sin t, 0 ≤ t < π 1 t sin t, 0 ≤ t < π 15. x = ,y= 2 −π cos t, t ≥ π 2 π sin t, t ≥ π 17. x = cos 4t − 12 (U(4t − π) − 2U(8t − π)) sin 4t, y = (U(4t − π) − 2U(8t − π)) cos 4t + 2 sin 4t   2 −st  1 1 − e−s 1 −st 19. L{f (t)} = e dt − e dt = = 0 1 s(1 + e−s ) 1 − e−2s −s ∞ k 1 − e e−ks . Take the Laplace transform of the system and apply the initial k=0 (−1) s

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CH A P T E R 8:

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condition to obtain sX(s) = X(s) + Y (s) + L{f (t)}, sY (s) = −2X(s) − 2Y (s). Solve for X(s) and Y (s) to get ∞

s+2 (s + 2)(es − 1)e−(1+k)s X(s) = L{f (t)} = (−1)k s(s + 1) s2 (s + 1) k=0

and ∞

Y (s) = −

(es − 1)e−(1+k)s 2 L{f (t)} = 2 (−1)k+1 . s(s + 1) s2 (s + 1) k=0

Compute the inverse Laplace transform to obtain x(t) =

∞

k −t

(−1) e



k=0

 −ek+1 + et (3 − 2t + 2k)) U(t − 1 − k) +



  ek + et (2t − 1 − 2k) U(t − k)

and y(t) = 2

∞

(−1)

k=0

k+1

  − t − 2 − k + e1−t+k U(t − 1 − k) + 

  t − 1 − k + ek−t U(t − k) .

 π −st 1 1 − e−πs 1 − e−πs ∞ −2sk e dt = . Then = k=0 e s 1 − e−2πs 0 s(1 − e−2πs ) taking the Laplace transform, applying the initial conditions, and solving for X(s) and Y (s) results in

21. First, L{f (t)} =

X(s) =

∞ s+1 s + 1 1 − e−πs −2sk L{f (t)} = e s s2 + 1 s2 + 1 k=0

and Y (s) =

∞ 2 2 1 − e−πs −2sk e . L{f (t)} = s s2 + 1 s2 + 1 k=0

Taking the inverse Laplace transform gives us x(t) = (sin(t − 2kπ) − cos(t − 2kπ) + 1)U(t − 2kπ) − (cos(t − 2kπ) − sin(t − 2kπ) + 1)U(t − π − 2kπ) and y(t) = −2(cos(t − 2kπ) + 1)U(t − π − 2kπ) + 4 sin2 (t/2 − kπ) U(t − 2kπ). 23. s2 X(s) = 3sX(s) − 2X(s) − sY (s) + Y (s) − 1, sX(s) + sY (s) + 1 = 2X(s) − Y (s); s−2 2 , Y (s) = − ; x(t) = −1 − e2t + 2et , y(t) = −2 + et X(s) = − (s − 2)(s − 1)s (s − 1)s

Answers, Hints, and Solutions to Selected Exercises

25. s2 X(s) + s2 Y (s) = X(s) + Y (s) = −

1 1 , sX(s) − X(s) + sY (s) = 0; X(s) = , 2 s (s − 1)s2

1 ; x(t) = −1 + et − t, y(t) = − 12 t 2 s3

Exercises 8.7 1. Q(t) = 2 − e−t − 2(1 − e1−t )U(t − 1), I(t) = 2e−t − 2(1 − e1−t )δ(t − 1) − 2e1−t U(t − 1)  n −(t−n) )U(t − n) 3. I(t) = 12 ∞ n=0 (−1) (1 − e 5. I(t) = (t − 1 + e−t ) − (t − 1)U(t − 1) + (t − 3 + e−(t−2) )U(t − 2) − (t − 3)U(t − 3) + (t − 5 + e−(t−4) )U(t − 4) − (t − 5)U(t − 5) + · · · 7. I(t) = 100te−3t √  √  √ √ 9. I(t) = 23 3e−t/2 sin 23 t − 23 3e−t/2+π/2 sin 23 (t − π) U(t − π) 11. (a) I(t) = e−4t+4 U(t − 1); (b) I(t) = e−4t + e−4t+4 U(t − 1) 13. I(t) = t − 1 + e−t − (t − 2 + e−t+1 )U(t − 1) 15. I(t) = e2−t U(t − 2) + e6−t U(t − 6) 19. x(t) = 12 ((−t cos t + sin t)U(t) − ((π − t) cos t + sin t)U(t − π)) 21. x(t) = 3 cos 3t − 23 sin 3t 23. x(t) = e−2t (2t + 1) 25. x(t) = 13 e−4t (4e3t − 1) 27. x(t) = e−2t (4 − 4et + 2t + 2tet ) 29. x(t) = 19 e−4t (1 − e3t − 3t + 6te3t ) 1 2 −3t (−12e3 + 18et+2 − 3e3 π 2 + 2π 2 et+2 + 31. x(t) = 36+13π 2 +π 4 [((π − 6) cos πt − e 5πe3t sin πt)U(t − 1)) − e−3t (−12 + 18et − 3π2 + 2π2 et + e3t (π2 − 6) cos πt − 5πe3t sin πt)U(t)]

33. x(t) = − 12 e−2t ((3e4 + e2t − 2te4 )U(t − 2) − (−2 + e2 + e2t + 4t − 2te2 − 2t 2 ) U(t − 1) − 2t 2 U(t)) 1 −2t e (3t sin 3t U(t) − (6π cos 3t − 3t cos 3t + sin 3t)U(t − 2π) + 35. x(t) = 18 (3π cos 3t − 3t cos 3t + sin 3t + 3π sin 3t − 3t sin 3t)U(t − π)) 1 [(3(π2 − 4) cos πt − e−4t (15πe4t sin πt + 16e3t+1 − 4e4 + 3(π4 +17π2 +16) π2 e3t+1 − 4π2 e4 ))U(t − 1) − e−4t (3e4t (π2 − 4) cos πt + 16e3t + π2 e3t − 4π2 −

37. x(t) =

15πe4t sin πt − 4)U(t)]

39. x(t) =

1 1 15 (sin t − 4

41. x(t) =

1 −2t 6 (e sin(3t − 9)U(t − 3) + e2 sin(3t − 3)U(t − 1)) 3e

sin 4t − 15 sin(4 − 4t)U(t − 1))

−5t + 250 cos t − 1250 sin t, bounded 43. x(t) = 200 + 127150 13 e 13 13 −5t − 1250 cos t − 250 sin t, bounded 45. x(t) = 200 + 63650 13 e 3 13

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47. x(t) = −250 + 5350e2t − 200 sin t − 100 cos t, unbounded 49. x(t) = −10000 + 17500et + 2500 cos t + 2500 sin t + (10000 − 10000et−5 + 2500et−5 cos 5 − 2500 cos 5 cos(t − 5) + 2500et−5 sin 5 − 2500 sin 5 cos(t − 5) − 2500 cos 5 sin(t − 5) + 2500 sin 5 sin(t − 5))U(t − 5)  51. x(t) = −5000 1 − 52 et + 12 (cos t − sin t) + 12 (−2 + 2et−1 + et−1 cos 1 − cos t − et−1 sin 1 + sin t)U(t − 1) +

  1 − 32 et−2 + 12 (cos(t − 2) − sin(t − 2)) U(t − 2) + · · ·



53. x(t) = et−2 (100e2 + 200U(t − 2)) so x(5) = e3 (200 + 100e2 ) ≈ 18858 1 −kt (−c + cekt + ck − ck 2 + ck 2 ekt + kx + k 3 x − ckekt cos t + 55. x(t) = k+k 0 0 3e 2 kt ck e sin t)

 6 c0 −a(t−4n) −a(t−4n−1/2) 57. (b) x(t) = n=0 a (1 − e )U(t − 4n)−(1 − e )U(t − 4n − 1/2) ;

 6 1 1 −b(t−4n) −a(t−4n) −b(t−4n) )U(t − 4n) − b−a (e +e )U(t − 4n) (c) y(t) = n=0 c0 b (1 − e

 1 − 6n=0 c0 b1 (1 − e−b(t−4n−1/2) )U(t − 4n − 1/2) − b−a (e−a(t−4n−1/2) +  e−b(t−4n−1/2) )U(t − 4n − 1/2)

√ √ 59. (d) x(t) = cos 5t − cos 5 3t, y(t) = 2(cos 5t + cos 5 3t) √ √ 63. x(t) = 25 (cos t − cos 6t), y(t) = 15 (4 cos t + cos 6t) 65. x(t) = 13 (cos t + 2 cos 2t), y(t) = 23 (cos t − cos 2t)       √ √ 67. x(t) = 15 4 cos 3t + cos √1 t , y(t) = 25 − cos 3t + cos √1 t 2

69. x(t) = + 7 12 cos 2t + 1 4

71. x(t) = 73. 75. 77. 79. 81.

1 7 1 1 18 sin 2t + 12 cos 2t + 18 sin t + 6 4 1 1 9 sin t + 3 cos t − 3 t cos t

2

cos t −

1 6 t cos t,

y(t) =

1 4

1 − 18 sin 2t −

11 1 1 cos t + 18 cos 2t + 12 t sin 2t, y(t) = 79 cos t − 79 cos 2t − 12 t sin 2t     √ √ θ1 (t) = − 18 3 sin √2 t + 18 sin 2t, θ2 (t) = − 14 3 sin √2 t − 14 sin 2t 3 3     θ1 (t) = 12 cos √2 t + 12 cos 2t, θ2 (t) = cos √2 t − cos 2t 3 3     √ √ θ1 (t) = 14 3 sin √2 t + 14 sin 2t, θ2 (t) = 12 3 sin √2 t − 12 sin 2t 3 3     θ1 (t) = − 14 cos √2 t + 14 cos 2t, θ2 (t) = − 12 cos √2 t − 12 cos 2t 3 3     √ √ x(t) = − cos √1m t 2k + kω2 , y(t) = cos √1m t 2k + kω2 7 18

√ 83. x(t) has frequency ω1 , where ω1 2 = 9k/(2m) − k 17/(2m); y(t) has frequency ω2 , √ where ω2 2 = 9k/(2m) + k 17/(2m)

Answers, Hints, and Solutions to Selected Exercises

√ √ 292175000 −100(5+2 6)t 292175000 − 7387691623 √ e−100(5+2 6)t + 159587072641 159587072641 e 1276696581128 6 √ √ √ √ 584350000 e400 6t−100(5+2 6)t + 7387691623 √ e400 6t−100(5+2 6)t − 159587072641 cos 377t − 1276696581128 6 204799950 584350000 204799950 159587072641 sin 377t; (c) I(t) = − 159587072641 cos 377t − 159587072641 sin 377t

Q(t) =

87. (a)

√ √ √ √ 91. x(t) = − 13 3 sin 3t, y(t) = 13 3 sin 3t

    √ 93. θ1 (t) = 18 4 cos 2t + 4 cos √2 t + sin 2t − 3 sin √2 t , θ2 (t) = − cos 2t + 3 3     √ 2 1 3 2 cos √ t − 4 sin 2t − 4 sin √ t 3

3

97. x1 (t) = −2 sin t, x2 (t) = sin t, x3 (t) = 2 sin t

Chapter 8 Review Exercises 1.

s−1 s2

3.

1 − e−5 s s

5. 2 s−3 + 5 s−1 7. (s − 2)−2 9. 6 (s − 1)−4 s2 − 9 11.  2 s2 + 9 13.

s+5 s2 + 10s + 34

15. e−3/2 sπ 17. −2

−3 e−7 s + 2 e−4 s s

e5−s s−5   −2 e s 2 s2 + 2 s + 1 21. 2 s3 s 1 1 −st dt = (s − 1)e + 1 23. L{f (t)} = (1 − t)e 1 − e−s 0 s2 (es − 1)

19. −42

1 1 2e2 (2ses + π) π −st cos t e dt = 2 1 − e−2s 0 (4s2 + π2 )(e2s − 1)    t sin τ 1 1 dτ = tan−1 27. L 0 τ s s     1 − e−t 1 29. L = ln 1 + t s 25. L{f (t)} =

183

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31.

L−1

37. 39.

41. 43.



2s 2 (s + 1)2

 = −2 sin t

 2s2 − 7s + 20 = 2 − et sin 3t s(s2 − 2s + 10)   14 −1 L − 2s = −14U(t − 2) se  6−s  3e L−1 = −3e6t U(t − 1) 6−s      −18 −18 ∞ −3ks −1 −1 e L =L = ∞ k=0 18 sin(3k − t) (s2 + 1)(1 − e−3s ) s2 + 1 k=0 U(t − 3k)   s cos φ − ω sin φ −1 L = cos(ωt + φ) s2 + ω 2

33. L−1 35.

 −

2 3 (cos t − cos 2t)

45. y = e2t (cos t − 2 sin t) 47. y = e−2t (1 + e2 U(t − 1))

  2 es − 1 1 1 −st −st dt = te dt + (2 − t)e = 1 1 − e−2s 0 s2 (es + 1) 1 1 − e−s ∞ 1 − e−s k −sk . Then, taking the Laplace transform = k=0 (−1) e −s 2 1+e s s2 of both sides of the differential equation, applying the initial conditions, and solving for Y (s) gives us

49. First, L{ f (t)} =

∞ es − 1 1 − e−s (s + 4)(s + 8)Y (s) = 2 s = (−1)k e−sk s (e + 1) s2 k=0

Y (s) =

1 − e−s

∞

s2 (s + 4)(s + 8)

k=0

(−1)k e−sk .

Taking the inverse Laplace transform gives us  y(t) = L =

−1

   ∞ ∞ 1 1 k −sk −1 k −sk−s (−1) e (−1) e −L s2 (s + 4)(s + 8) s2 (s + 4)(s + 8)

1 −8t e 256

∞

(−1)k

k=0



k=0

 e8+8k − 4e4(t+1+k) + e8t (11 + 8k − 8t) U(t − k − 1)    − e8k − 4e4(t+k) + e8t (3 + 8k − 8t) U(t − k) .



51. x(t) = e1−2t e(−e2 + et )U(t − 2) + (−e + et )U(t − 1) 53. g(t) = 1 + sin t − cos t

k=0

Answers, Hints, and Solutions to Selected Exercises

 57. x(t) =

1 15 (cos(t/2) − cos 2t − 30 sin(t/2)), 1 15 (cos(t/2) − 31 sin(t/2)), t ≥ π

 59. Q(t) =

0≤t<π

2 11 250 sin 50t, 0 ≤ t < 2 11 500 ((−1 + cos 200) cos 100t + sin 200

sin 200t), t ≥ 2

  61. x(t) = 100e2t−2 100e2 + U(t − 1) 65. x = 12 (t + cos t sin t), y = 18 (−2t + sin 2t)   67. x = 4e−t (et − e2 (−1 + t))U(t − 2) + (1 − et + t)U(t) ,   y = e−t (−2et + te2 )U(t − 2) + (−2 + 2et − t)U(t)  √   √ √ 69. x(t) = 25 2 2 sin( 2t) + 3 sin √1 t ,  √ 3  √ √ y(t) = 15 −2 2 sin( 2t) + 4 3 sin √1 t 3 √ √ 1 71. x(t) = 2 (cos t − cos 3t + t sin t), y(t) = 12 (− cos t + cos 3t + t sin t) 73. x(t) =

1 8

   √ −3 sin 2t − 3 sin √2 t , y(t) = 3

1 4

   √ 3 sin 2t − 3 sin √2 t 3

Differential Equations at Work A. The Tautochrone  dx =

1.

 dx =

k2 − 1 dy k 2 y2 k2



− 1 · 2k 2 sin θ cos θ dθ =

k 2 (1 − sin2 θ)

k 2 sin2 θ k 2 sin2 θ cos θ = · 2k 2 sin θ cos θ dθ = 2k 2 cos2 θ dθ. sin θ

· 2k 2 sin θ cos θ dθ

  Then, x(θ) = 12 2k 2 θ + k 2 sin 2θ + C1 so x(0) = 0 means that C1 = 0 so that   x(θ) = 12 2k 2 θ + k 2 sin 2θ . 2. y(θ) = k 2 sin2 θ = 12 k 2 (1 − cos 2θ). 3. Increasing the value of k increases the length of the curve.

k51

y 1.0 0.8 0.6 0.4 0.2 1

2

3

4

5

6

x

185

186

CH A P T E R 8:

Introduction to the Laplace Transform

y

k52

4 3 2 1 5

10

15

y

20

25

x

k53

8 6 4 2 10

20

30

40

50

x

4. The time is independent of the choice of y (that is, the choice of θ). Therefore,  time = 0

y

φ(z) dz = √ y−z

 0

y

ky−1/2 dz = −2k √ y−z



y−z y

y = −2k · −1 = 2k. 0

B. Vibration Absorbers 1. First, we apply the Laplace transform to the system m1 s (X1 (s) − x1 (0)) − x1 (0) + (k1 + k2 )X1 (s) − k2 X2 (s) =

F0 ω s2 + ω 2

m2 s (X2 (s) − x2 (0)) − x2 (0) − k2 X1 (s) + k2 X2 (s) = 0 and then rearrange the equations and apply the initial conditions to obtain (m1 s2 + k2 + k2 )X1 (s) − k2 X2 (s) =

F0 ω 2 s + ω2

−k2 X1 (s) + (m2 s2 + k2 )X2 (s) = 0. We eliminate X2 (s) from the system by multiplying the first equation by m2 s2 + k2 and the second equation by k2 and then adding:     F ω   0 m2 s2 + k2 m1 s2 + k1 + k2 − k22 X1 (s) = m2 s2 + k2 2 s + ω2    F ω  0 m1 m2 s4 + (m2 k1 + m2 k2 + m1 k2 )s2 + k1 k2 X1 (s) = m2 s2 + k2 2 . s + ω2 √ 2. The amplitude is zero when ω/ k2 /m2 = 1.

Answers, Hints, and Solutions to Selected Exercises

6 4 2

0.5

1.0

1.5

2.0

22 24

C. Airplane Wing √

√     2 3√ 2 3√ 1. x1 (t) = √ sin 2c t , x2 (t) = − √ sin 2c t , x3 (t) = 2 2 6 c √ 3 c  √ 2 3√ EI − √ sin 2c t , c = ; period: p(c) = 2π/(3 2c/2), 2 m1 3 6 c p(0.0001) ≈ 296.19, p(0.01) ≈ 29.62, p(0.1) ≈ 9.37, p(1) ≈ 2.96, p(2) ≈ 2.09

D. Free Vibration of a Three-Story Building 1. Notice that ⎛ m1 ⎝0 0

⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ 0 k1 + k2 −k2 0 x1 x1 0 ⎠ ⎝x2 ⎠ + ⎝ −k2 k2 + k3 −k3 ⎠ ⎝x2 ⎠ = m3 x3 0 −k3 k3 x3 ⎞ ⎛ ⎞ ⎛ ⎞ ⎛   m1 x1 (k1 + k2 )x1 − k2 x2 m1 x1 + (k1 + k2 )x1 − k2 x2 ⎝m2 x  ⎠ + ⎝−k2 x1 + (k2 + k3 )x2 − k3 x3 ⎠ = ⎝m2 x  − k2 x1 + (k2 + k3 )x2 − k3 x3 ⎠. 2 2 m3 x3 −k3 x2 + k3 x3 m3 x3 − k3 x2 + k3 x3 0 m2 0

2-4. y

x 0.5

2

4

6

8

10

12

14

t 20.5 21.0 21.5

20.5

2

4

6

8

10

12

14

t

20.5 21.0 21.5

y

x

50

100

150

200

t 20.5 21.0 21.5

2

4

6

8

10

12

14

z 2

1.5 1.0 0.5

0.5

20.5

z 1.5 1.0 0.5

1.5 1.0 0.5

1 50

100

150

200

t

50 21 22

100

150

200

t

t

187

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CH A P T E R 8:

Introduction to the Laplace Transform

5.

7. Increasing the number of stories increases the number of equations in the system of differential equations. A five-story building leads to a system of five secondorder differential equations; a fifty-story building leads to a system of fifty second-order differential equations.

CHAPTER 9

Eigenvalue Problems and Fourier Series

Exercises 9.1 3. y(x) = c1 x + c2 , y (x) = c1 , y (0) = c1 = 0; y (x) = 0, y (1) = 0  = 1; no solution 7. y(x) = c1 cos 3x + c2 sin 3x, y(π) = c1 = 0; y (0) = 3c2 = 0, c2 = 0; y(x) = 0

y(x) = c2 sin 3x ;

y (x) = 3c2 cos 3x;

11. y(x) = c1 cos 2πx + c2 sin 2πx, y(1) = c1 = 2π; y(0) = 2π; y (x) = −2πc1 sin 2πx + 2πc2 cos 2 πx, y (0) = 2πc2 , y(0) + y (0) = 2π + 2πc2 = 0, c2 = −1 ; y(x) = 2π cos 2πx − sin 2πx 17. Case I: λ = 0, y(x) = c1 x + c2 , y(0) = c2 = 0, y (x) = c1 , y (1) = c1 = 0; λ = 0 is not an eigenvalue. Case II: λ = −k 2 < 0, y(x) = c1 ekx + c2 e−kx , y(0) = c1 + c2 = 0   or c2 = −c1 , y (x) = c1 kekx − c2 ke−kx , y (1) = c1 k ek + e−k = 0, c1 = 0, c2 = 0, no negative eigenvalues. Case III: λ = k 2 > 0, y(x) = c1 cos kx + c2 sin kx, y(0) = c1 = 0, y (x) = c2 k cos kx, y (1) = c2 k cos k = 0, k = (2n − 1) π/2, n = 1, 2, . . . ; yn (x) = sin [(2n − 1)xπ/2] , n = 1, 2, . . . 21. Case I: λ = 1/2, y(x) = e−x/2 (c1 + c2 x), y(0) = c1 = 0, y(x) = c2 xe−x/2 , y (x) =   c2 e−x/2 − 1/2xe−x/2 , y (1) = c2 e−1/2 = 0, c2 = 0, λ = 1/2 is not an eigenvalue.   Case II: 4 − 8λ = − k 2 < 0 (λ > 1/2), y(x) = e−x/2 c1 ekx + c2 e−kx , y(0) = c1 +   c2 = 0 or c2 = − c1 , y (1) = c1 e−1/2 ek (k − 1/2) + e−k (k + 1/2) = 0, c1 = 0, c2 = 0, no eigenvalues λ, λ > 1/2. Case III: 4 − 8λ = k 2 > 0 (λ < 1/2), y(x) = e−x/2 (c1 cos kx + c2 sin kx), y(0) = c1 = 0, y(x) = c2 e−x/2 sin kx, y (x) = c2 e−x/2 × (k cos kx − 1/2 sin kx), y (1)=c2 e−1/2 (k cos k − 1/2 sin k), tan k = 2k, yn (x) = e−x/2 × sin kx where k satisfies tan k = 2k. 1 27. d/dx(dy/dx)+ λy = 0, s(x) = 1. If m  = n, sin[(2m − 1)πx/2]sin[(2n − 1)πx/2]dx = 0

(m + n − 1) sin[(m − n)π] + n sin[(m + n − 1)π] + m sin[(m + n)π] =0 2(m − n)(m + n − 1)π

189

190

CH A P T E R 9:

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30.





d  d   (x) Pn (x) + 1 − x 2 Pn (x) Pm (x) + n(n + 1)Pn (x)Pm (x) = 0, 1 − x 2 Pm dx dx m(m + 1)Pm (x)Pn (x) = 0 Subtract and integrate by parts where 1 −1

1 



1 d  2  2   1 − x 2 Pn (x)Pm (x)dx 1 − x Pn (x) Pm (x)dx = 1−x Pn (x)Pm (x) − −1 dx −1

=0

and 1 −1

1 



1 d  2  2   (x)Pn (x)dx. 1 − x Pm (x) Pn (x)dx = 1−x Pm (x)Pn (x) − 1 − x 2 Pm −1 dx −1

=0

sin 0 cos 0 0 1 36. (a) = = 0, nontrivial solutions exist sin π cos π 0 −1 sin 0 cos 0 0 1 (c) −5 = = 0, nontrivial solutions exist e sin 5 π e−5 cos 5 π 0 −e−5 41. yp (x) = A sin 2x + B cos 2x, yp (x) = −4A sin x − 4B cos x, −4A sin x − 4B cos x = sin 2x, A = −1/3, B = 0, y(x) = c1 cos x + c2 sin x − 13 sin 2x, y(0) = c1 = 0, y(π) = 0 for any c2 ; y(x) = c2 sin x − 13 sin 2x. 45. (a) y(x) = C1 x + C2 , y(0) = C2 = 0, y(x) = C1 x, y(1) + y (1) = 2C1 = 0, C1 = 0. (b) λ = −k 2 < 0, y(x) = C1 ekx + C2 e−kx , y(0) = C1 + C2 = 0, C2 = −C1 , y(1) + y (1) = C1 [(1 + k)ek + (k − 1)e−k ] > 0 unless C1 = 0. Therefore, C2 = 0. (c) λ = k 2 > 0, y(x) = C1 cos kx + C2 sin kx, y(0) = C1 = 0, y(1) + y (1) = C2 [sin k + k cos k] = 0, k = −tan k. (d) Graph y = x and y = −tan x. Approximate the points of intersection. λ1 ≈ (2.02876)2 ≈ 4.11586, λ2 ≈ (4.91318)2 ≈ 24.1393, λ3 ≈ (7.97867)2 ≈ 63.6591. FindRoot[k == -Tan[k], {k, 2}]

Plot[{k, -Tan[k]}, {k, 0, 10}]

{k -> 2.02876}

30

FindRoot[k == -Tan[k], {k, 5}] 20

{k -> 4.91318}

10

FindRoot[k == -Tan[k], {k, 8}] 2 ⫺10 ⫺20 ⫺30

4

6

8

10

{k -> 7.97867}

Answers, Hints, and Solutions to Selected Exercises

Exercises 9.2 1 3. a0 = 2

x3 x dx = 2 3

1 =

2

0

1 an = 2

0

2 3

2nπx cos nπx − 2 sin nπx + n2 π2 x 2 sin nπx x cos nπxdx = 2 n3 π 3

1

2

0

0

4nπ cos nπ 4 = 2 2 (−1)n , n ≥ 1 2 2 n π n π ∞  4(−1)n 1 1 4 4 f (x) = + cos nπx = − 2 cos πx + 2 2 cos 2πx + · · · 3 3 π n2 π 2 2 π =

n=1

 x3 3  1

1  4Cos[(2k – 1)πx] – + 2 (2k – 1)2 π2 n

x2 dx

f[x_, n_] : =

k=1

Plot[f[x, 10], {x, 0, 1}] 1

x2 dx

0

0.8

1 3 

0.6

x2 Cos[nπx]dx 0.4

2nπxCos[nπx] – 2Sin[nπx] + n2 π2 x2 Sin[nπx]  1

n3 π 3

0.2

x2 Cos[nπx]dx

0.2

0.4

0.6

0.8

0

2nπCos[nπ] – 2Sin[nπ] + n2 π2 Sin[nπ] n3 π 3

2 7. a0 = 4

2

2 xdx + 4

0

2 an = 4

2

4 dx = 2; 2

nπx 2 x cos dx + 4 4

0

4 cos 2

nπx 2  nπx 4 2  nπx nπx 2 + dx = 2 2 4 cos + nπx sin sin 4 4 4 0 nπ 4 2 n π

2  nπ nπ = 2 2 −4 + 4 cos + nπ sin , n ≥ 1; 2 2 n π f (x) = 1 +

∞  n=1

2  nπ nπ nπx −4 + 4 cos + nπ sin cos 2 2 4 n2 π 2

1

191

192

CH A P T E R 9:

Eigenvalue Problems and Fourier Series 1 13.

bn = 2

2 cos nπx − n2 π2 x 2 cos nπx + 2nπx sin nπx x sin nπxdx = 2 n3 π 3

1

2

0

−2 + 2 cos nπ − n2 π2 cos nπ + 2nπ sin nπ n3 π 3

 2 1 4 n − − 3 3 , n ≥ 1; = 2(−1) n3 π3 nπ n π

0



=2

  ∞  1 2 4 n − f (x) = 2(−1) − 3 3 sin nπx n3 π3 nπ n π n=1

 =2

17.

2 bn = 4

     1 −4 1 1 −4 + + sin πx + 2 − sin 2πx + 2 sin 3πx + · · · π 2π π3 33 π3 3π

2

nπx 2 x sin dx + 4 4

0

4 sin

nπx dx 4

2

−2  nπx nπx 2 2  nπx 4 nπx cos − 4 sin − cos 4 4 0 nπ 4 2 n2 π 2 −2  nπ nπ 2  nπ = 2 2 2nπ cos − 4 sin + cos − cos nπ 2 2 nπ 2 n π −2   nπ nπ = 2 2 nπ cos − cos nπ − 4 sin , n ≥ 1; 2 2 n π =

f (x) =

∞  −2   nπ nπ nπx nπ cos − cos nπ − 4 sin sin 2 2 4 n2 π 2 n=1





 nπx  dx 4   nπx    2 nπx Cos nπx 4 -4 Sin 4 n2 π 2   nπx  2 Sin dx 4 4   2 Cos nπx 4 nπ 2 4

x Sin

Simplify

-

 2  4  nπx   2 2 x Sin Sin n dx + − 4 0 4 4 2

     4 Cos nπ nπ Cos 3nπ -4 Sin nπ 4 4 4 n2 π 2

 2 2 4 0  1 2

x Sin

 nπx  4

dx +

 4  nπx  2 Sin dx 4 2 4

   4 Cos nπ 2 - 4 Cos[nπ] nπ nπ

Answers, Hints, and Solutions to Selected Exercises

 2  nπx  2 x Sin dx 4 0 4  nπ     2 2nπ Cos nπ 2 -4 Sin 2 n2 π 2  4  nπx  2 Sin dx 4 2 4     1 4 Cos nπ 2 - 4 Cos[nπ] 2 nπ nπ

 nπ     2 2nπ Cos nπ 2 -4 Sin 2 n2 π 2      4 Cos nπ nπ Cos 3nπ – 4 Sin 4 4 b[n_] : = – n2 π 2

Plot[f[x, 20], {x, 0, 4}]

1.5

f[x_, k_] : =

k 

b[n]Sin

n=1

 nπx  4

2

1 0.5

1

23. cos3 x =

3 4

2

3

4

cos x + 14 cos 3x

Exercises 9.3 3. f is odd; a0 = 0; an = 0, n ≥ 1; π      π 1 2 −x n2 x 2 − 6 cos nx 3 n2 x 2 − 2 sin nx 3 x sin nxdx = + bn = π π n3 n4 0 −π  2 2  2 n π − 6 cos nπ =− , n ≥ 1; n3     ∞   12 2π2 3 n 2 2 f (x) = − (−1) sin nx = 2π − 12 sin x + − π sin 2x n 2 n3 n=1  2  2π 4 + − sin 3x + · · · 3 9 Clear[f] f[x_] := x3 p = Pi; 1/p Integrate[f[x] Sin[n Pi x/p], x]  1 x(-6+n2 x2 )Cos[nx] + π n3  3(-2+n2 x2 )Sin[nx] n4

g[x_, k_] := Sum[b[n] Sin[n Pi x/p], {n, 1, k}] g[x, 3] 1 2(-6+π2 ) Sin[x]+ 6-4π2 Sin[2 x]+ 4 2  2 -6+9π Sin[3 x] 27 p1 = Plot[g[x, 10], {x, –p, p}]

193

194

CH A P T E R 9:

Eigenvalue Problems and Fourier Series

b[n_] = 1/p Integrate[f[x] Sin[n Pi x/p], {x, –p, p}]    2π -6+n2 π2 Cos[nπ] 1 + π n3  6(-2+n2 π2 ) Sin[nπ] n4

30 20 10 ⫺3

⫺2

⫺1

2

1

3

⫺10 ⫺20 ⫺30

⎛ 7. a0 =

1⎝ 1

0

1 xdx +

−1

1 x cos nxdx +

−1

=

dx ⎠ =

0

1 n2 π 2

x2 2

−1

0

1 1 + [x]10 = − + 1 = ; 2 2 1

cos nπx x sin nπx cos nxdx = + nπ n2 π 2

1 x sin nπxdx +

0

(1 − cos nπ) =

0 bn =

0

0 an =

⎞

1 n2 π 2

0

sin nπx + nπ −1

1 0

(1 − (−1)n ), n ≥ 1;

−x cos nπx sin nπx sin nπxdx = + 2 2 nπ n π

0

− cos nπx + nπ −1

1 1 (1 − 2 cos nπ) = (1 − 2(1−)n ), n ≥ 1; nπ nπ  ∞  1  1  1  n n+1 (−1) (−1) f (x) = + 1− cos nπx + 2 + 1 sin nπx 4 nπ n2 π 2 =

n=1

=

1 2 3 1 + 2 cos πx + sin πx − sin 2πx + · · · 4 π π 2π

p = 1; a[0] = 1/p (Integrate[x, {x, –1, 0}]+ Integrate[1{x, 0, 1}]) 1 2

b[n_] = 1/p (Integrate[x Sin[nPix/p], {x, –1, 0}]+ Integrate[x Sin[nPix/p], {x, 0, 1}]) 1 2Cos[nπ] Sin[nπ] + 2 2 nπ nπ n π

1 0

Answers, Hints, and Solutions to Selected Exercises

a[n_] = 1/p (Integrate[x Cos[n Pi x/p], {x, –1, 0}]+ Integrate[1 Cos[n Pi x/p], {x, 0, 1}]) 1 Cos[nπ] - 2 2 2 2 n π n π Integrate[x Cos[n Pi x/p ], x] Cos[nπx] x Sin[nπx] + nπ n2 π 2 Integrate[Cos[n Pi x/p], x] Sin[nπx] nπ p1 = Plot[g[x, 20], {x, –1, 1}]

Integrate[x Sin[n Pi x/p], x] x Cos[nπx] Sin[nπx] + nπ n2 π 2 Integrate[Sin[n Pi x/p], x]

-

Cos[nπx] nπ g[x_, k_] : = 1/4+ Sum [1/(nˆ2 Piˆ2(1–( – 1)ˆn) Cos[nPix]+ 1/(n * Pi)(1 – 2( – 1)ˆn)Sin[nPix], {n, 1, k}] g[x, 3] -

1 2Cos[πx] 2Cos[3πx] + + + 4 π2 9π2 3Sin[πx] Sin[2πx] Sin[3πx] + π 2π π 1 0.5 ⫺1

⫺0.5

0.5

1

⫺0.5 ⫺1

13. Neither; f (−x) = (−x)2 − (−x) = x 2 + x  = −f (x) or f (x)  x2 , x ≥ 0 17. Odd; f (x) = , f (−x) = −f (x) −x 2 , x < 0 2 + (−1/2) 3 1/2 + (−1) 1 = ; (c) =− ; 2 4 2 4

2  π π 2 2 πx π2 x3 x(π − x)dx = = − , 41. a0 = π π 2 3 0 3 37. (a) 0;

(b)

(d)

2 + (−1) 1 = ; 2 2

0

an =

2 π 

=

π x(π − x) cos nxdx 0

2 π



π   −2 − n2 πx + n2 x 2 sin nx (π − 2x) cos nx − n2 n3

−2 = 2 (1 + cos nπ), n ≥ 1; n

0

(e)

2 + (−1) 1 = 2 2

195

196

CH A P T E R 9:

Eigenvalue Problems and Fourier Series

f (x) =

∞ ∞  π2  2  π2  4 − 2 1 + (−1)n cos nx = − cos 2 nx + + 6 6 n (2n)2 n=1

n=1

1 = 1 − cos 2x − cos 4x − · · · 4

p = Pi;

f[x_, k_] : =

a[0] = 2/p Integrate[x(Pi – x),

1/2 a[0] + Sum [a[n] Cos[nx], {n, 1, k}] a[n_] =

{x, 0, p}[ π2 3

2/p Integrate [x(Pi – x) cos[n Pix/p], {x, 0, p}[ [nπ] 2 Sin[nπ] + 2 - nπ2 - π Cos 2 3 n n 

2/p Integrate [x(Pi – x), x]  2 3 x 2 πx 2 - 3

π p1 = Plot[f[x, 6], {x, 0, Pi}]

π

2.5

2/p Integrate [x(Pi – x) Cos[n Pi x/p], x]   1 (π-2x)Cos[nx] 2 π n2 (-2-n2 πx+n2 x2 ) Sin[nx] n3

2 1.5



1 0.5 0.5

45. bn =

4 n3 π

1

1.5

2

2.5

3

[1 − cos nπ], n ≥ 1;

∞ ∞    4  8 n sin nx = 1 − (−1) sin(2n − 1)x 3 n π (2n − 1)3 π n=1

n=1

=

8 8 8 sin x + 3 sin 3x + 3 sin 5x + · · · π 3 π 5 π

p = Pi;

f[x_, k_] := Sum [b[n]Sin[nPix/p],

2/p Integrate[x(Pi – x) Sin[nPix/p], x]    -2-n2 πx+n2 x2 Cos[n x] 1 2 + π n3  (π-2x) Sin[n x] n2

{n, 1, k}] p1 = Plot[f[x, 6], {x, 0, p}] p1 = Plot[f[x, 6], {x, 0, p}];

Answers, Hints, and Solutions to Selected Exercises

b[n_] =

p2 = Plot[x, (Pi – x), {x, 0, Pi}];

2/p Integrate[x(Pi – x)Sin[n Pi x/p],

Show[p1, p2]

{x, 0, p}[  [nπ] π Sin[nπ] 2 2 n3 - 2 Cos n3 n2

2.5 2

π

1.5 1 0.5 0.5

49. a0 =

π2 3

, an = −

1

1.5

2

2.5

1 , n ≥ 1; bn = 0, n ≥ 1; n2

∞

f (x) =

π2  1 π2 1 1 − cos 2nx = − cos 2x − cos 4x − cos 6x + · · · 2 6 6 4 9 n n=1

p = Pi;

b[n_] = 2/p Integrate[f[x]Sin[2n Pi x/p],

f[x_] := x(Pi – x) a[0] = 2/p Integrate[f[x], {x, 0, p}] π2 3 a[n_] = 2/p Integrate[f[x]Cos[2n Pi x/p], {x, 0, p}]  π + -nπ Cos[2nπ]+Sin[2nπ] 2 - 4n 2 3 4n

{x, 0, p}] 1 + -Cos[2nπ]-nπSin[2nπ] 2 - 4n 3 3 4n 

π Simplify[b[n]] Sin[nπ](-nπ Cos[nπ]+Sin[nπ]) n3 π

π Simplify[s] Cos[nπ](-nπ Cos[nπ]+Sin[nπ]) n3 π

61. (a) an = 0, n ≥ 0; bn = (c) a0 = n≥1

1 π

π

x 4 dx =

0

1 π

2π4 5 ,

π

x 3 sin nxdx =

0

an =

1 π

π

  −2 −6+n2 π2 3 n

x 4 cos nxdx =

0

cos nπ, n ≥ 1

  8 −6+n2 π2 4 n

Exercises 9.4

1 2 3. p(x) = e2x , s(x) = e2x , 0 e2x (e−x sin nπx) dx = 12 , cn =  1 2x 2 0 e f (x)(e−x sin nπx)dx; (a) cn =

2nπ (1 − e cos nπ) n2 π2 +1

(b) cn =

2 n2 π2 +1

=

[1 − e(−1)n ] ;  + e1/2 sin nπ 2

2nπ n2 π2 +1

 nπ − e1/2 nπ cos nπ 2

cos nπ, n ≥ 1; bn = 0,

3

197

198

CH A P T E R 9:

Eigenvalue Problems and Fourier Series

c[n_] =

f[x_, k_] := Sum[c[n]Exp[–x]Sin[nPix],

2Integrate[

{n, 1, k}]

Exp[2x]Exp[–x]Sin[nPix],

Plot[f[x, 20], {x, 0, 1}]

{x, 0, 1}] nπ + 2 1+n2 π2  E(-nπCos[nπ]+Sin[nπ]) 1+n2 π2 Integrate[ Exp[2x](Exp[–x]Sin[nPix]) ˆ 2, 

1.1

0.2

{x, 0, 1}] 1 Sin[2n π] 2 4n π c[n_] =

0.4

0.6

0.8

1

0.8

1

0.9

0.8

2Integrate[

f[x_, k_] :=

Exp[2x]Exp[–x]Sin[nPix],

Sum[c[n]Exp[–x]Sin[nPix],

{x, 0, 1/2}] nπ 2 1+n2 π2  nπ     √  E nπ Cos nπ 2 -Sin 2 1+n2 π2 

{n, 1, k}] Plot[f[x, 20], {x, 0, 1}] 1.2 1 0.8 0.6 0.4 0.2 0.2

√

0.4

0.6

√

7. r = −1± 1−4(1−λ) = −1± 2−3+4λ ; Case I: λ = 3/4, y(x) = e−x/2 (c1 + c2 x), 2 y(0) = c1 = 0, y(1) = e−1/2 c2 = 0, c2 = 0, λ = 3/4 is not an eigenvalue; 2 Case II: −3 +4λ = k 2 > 0 yields y(x) = 0; Case III: −3 + 4λ = −k < 0, kx −1/2 c sin k = 0, y(x) = e−x/2 c1 cos kx 2 2 + c2 sin 2 , y(0) = c1 = 0, y(1) = e 2 = nπ, k = 2nπ, λn = 3−n4 π , n = 1, 2, . . . , yn (x) = e−x/2 sin kx 2, n= 1 x d x x x −x/2 1, 2, . . . ; dx (e y) + e (1 − λ)y = 0, s(x) = e ; cn = 2 0 e g(x)(e sin nπx)dx k 2

2 2

Answers, Hints, and Solutions to Selected Exercises

Chapter 9 Review Exercises 1 5. bn =

nπx x sin dx + 2

2

0

(2 − x) sin 1

nπx −2  nπ nπ dx = 2 2 −nπ cos − 2 sin 2 2 2 n π

2  nπ nπ 4 8 nπ − 2 sin − 2 2 sin nπ = 2 2 sin , n≥1 − 2 2 nπ cos 2 2 2 n π n π n π 1 9. a0 =

2 xdx +

0

1 an =

(2 − x)dx = 1; 1

nπx x cos dx + 2

0

2 (2 − x) cos 1

nπx 4  nπ dx = 2 2 −1 − cos nπ + 2 cos ,n ≥ 1 2 2 n π

p = 2;

f[x_, k_] :=

a[0] =

1/2a[0]+

2/p

Sum[a[n]Cos[nPix/p],

(Integrate[x, {x, 0, 1}]+ Integrate[2 – x, {x, 1, 2}]) 1 a[n_] = 2/p (Integrate[xCos[nPix/p], {x, 0, 1}]+ Integrate[ (2 – x)Cos[nPix/p], {x, 1, 2}]) 4Cos[nπ] 4 + - 2 2n π  n2 π2  nπ  2 2Cos nπ 2 -nπSin 2 + 2 2  nπ    nπn π 2 2Cos 2 +nπSin 2 n2 π 2

{n, 1, k}] approx[x_] = f[x, 20] 1 4Cos[πx] 2 π2 4Cos[3 πx] 4Cos[5 πx] 9 π2 25 π2 4Cos[7 πx] 4Cos[9 πx] 49 π2 81 π2 Plot[approx[x], {x, 0, 2}, PlotRange → All] 1 0.8 0.6 0.4 0.2

0.5

1

1.5

2

199

200

CH A P T E R 9:

Eigenvalue Problems and Fourier Series

 13. a0 =

1 π

0

−π



an =

1 π

bn =

1 π

0

π

xdx + dx 0

 = 1 − π2 ; π



x cos nxdx + cos nxdx = n−1 2 π (cos nπ − 1); −π 0   0 π x sin nxdx + sin nxdx = −1 nπ [(π + 1) cos nπ − 1] , n ≥ 1 −π

0

21. (a) sin x = b1 sin x + b3 sin 3x, b1 = 34 , b3 = − 14 , bn = 0, n  = 1, 3; 3

(b) cos3 x = a1 cos x + a3 cos 3x, a1 = 34 , a3 = 14 , an = 0, n  = 1, 3

CHAPTER 10

Partial Differential Equations

Exercises 10.1 3. X  Y − yXY  = 0, X  = kX, X(x) = C1 ek x , Y  /Y = k/y, Y (y) = eC2 yk = C3 yk , u(x, y) = C1 C3 ek x yk = Cek x yk 7. uy = f (y), u(x, y) = F(y) + G(x) where F  (y) = f (y); F and G are arbitrary functions. 13. ux = ex sin y, uxx = ex sin y, uy = ex cos y, uyy = −ex sin y 19. ux = 2 cos 2x cos 2t, uxx = −4 sin 2x cos 2t, ut = −2 sin 2x sin 2t, utt = −4 sin 2x cos 2t 25. ux = 10 − 4e−16t sin 4x, uxx = −16e−16t cos 4x, ut = −16e−16t cos 4x 31. ux = 4 cos ct cos 4x, uxx = −16 cos ct sin 4x, ut = −c sin ct sin 4x, utt = −c2 cos ct sin 4x, 4c2 = 16, c = ±2 u[x_, t_]:= Cos[ct]Sin[4x] D[u[x,t],x]

4 D[u[x,t],{t,2}] -4 c2 Cos[ct]Sin[4x] Solve[-16 == -4c2,c]

4 Cos[ct]Cos[4x] D[u[x,t],{x,2}]

{{c → –2}, {c → 2}}

-16 Cos[ct]Sin[4x] 4 D[u[x,t],t] -4 cSin[ct]Sin[4x]

35. Yes, ut (x, t) = −2 sin 2x sin 2t, ut (x, 0) = −2 sin 2x sin 0 = 0

Exercises 10.2 3. bn = 2

1 0

x(1 − x) sin nπxdx =

f[x_]:= x(1 - x) p = 1;

∞  −4 2 2 (cos nπ − 1); u(x, t) = bn sin nπx e−n π t/2 3 3 n π n=1

b[n_]= 2/p Integrate[f[x]Sin[n Pi x/p],{x,0, 1}]   2 2 Cos [nπ] Sin [nπ] 2 − − n3 π 3 n3 π 3 n2 π 2

201

202

CH A P T E R 10:

Partial Differential Equations

9. S(x) = 20 − 10x, u(x, t) = S(x) + 2 bn = 1

1 0

=

∞  n=1

bn sin nπx e−n

2 π2 t



−20 (20 − 10x) sin nπxdx = 2 2 ((x − 2)nπ cos nπx − sin nπx) n π

1 0

20 (cos nπ − 2) nπ

2 Integrate[(10x – 20)Sin[n Pi x], x]

b[n_] =

1 – 2 2 n π (20( – 2nπ Cos[nπx] + nπx Cos[nπx] – Sin [nπx]))

2Integrate[(10x – 20)Sin[n Pi x], {x, 0, 1}]   20 10( – nπ Cos [nπ] – Sin [nπ]) 2 – – nπ n2 π 2

13. S  = 0, S(0) = T0 , S  (1) = 0, S(x) = c1 x + c2 , S(0) = c2 = T0 , S  (x) = c1 , S  (1) = c1 = 0, S(x) = T0 Dsolve[ {s’’[ x ] == 0, s[0] == T0, s’[ 1 ] == 0}, s[x], x] {{s[x] → T0}}

19. u(x, 0) = A0 +

∞  n=1

an cos nx = 4 sin2 x = 4





1 2 (1 − cos 2x)

= 2 − 2 cos 2x, A0 = 2,

a2 = −2, an = 0, n  = 2, u(x, t) = 2 − 2 cos 2x e−4t 23. (This problem has homogeneous BCs.) u(x, t) = X(x)T (t), X  + λX = 0, X(−π) = X(π), X  (−π) = X  (π), (This eigenvalue problem was solved in Section 9.1 Ex. 35) X0 (x) = 1, Xn (x) = an cos nx + bn sin nx, n ≥ 1, u(x, t) = ∞ ∞   1 1 2 a0 + (an cos nx + bn sin nx)e−n t , u(x, 0) = a0 + (an cos nx + bn sin nx) = 2 2 n=1 n=1 f (x), all coefficients are Fourier series coefficients for f (x) with p = π. π 1 1 π 2 π 29. f is even; A0 = a0 = |x|dx = xdx = π, an = π1 |x| cos nxdx = 2 π −π π0 −π ∞  2 π 2 −4 x cos nxdx = 2 (cos nπ − 1), bn = 0, n ≥ 1; u(x, t) = π/2 + (2n−1)2 π π0 n π n=1 cos(2n − 1)xe−(2n−1) t . 2

Answers, Hints, and Solutions to Selected Exercises

p = Pi;

b[n_] =

a[0] = 2/Pi Integrate[x, {x, 0, Pi}] π a[n_] =

1/Pi (Integrate[–x Sin[n Pi x/p], {x, –Pi, 0}+ Integrate[x Sin[n Pi x/p], {x, 0, Pi}]

2/Pi Integrate[x Cos[n Pi x/p], {x, 0, Pi}]   1 Cos [nπ] + nπSin [n π] 2 – 2+ n n2 π

nπCos [nπ] – Sin [nπ] –nπCos [nπ] + Sin [nπ] + n2 n2 π Plot[f[x, 10], {x, –p, p}]

f[x_, k_] :=

3

1/2a[0] +

2.5

Sum[a[n]Cos[n Pi x/p] + b[n]Sin[n Pi x/p], 2

{n, 1, k}]

1.5 1 0.5

35. S(x) = 50 + 50x, 1 

bn = 2a x 1 − x 3 − (50 + 50x) sin nπxdx

23

22

21

1

2

3

0



∞  2 −24 − 50n4 π4 + 4 6 − 3n2 π2 + 25n4 π4 2 2 = cos nπ; u(x, t) = bn sin nπx e−n π t n5 π 5 n=1

p = 1; b[n_] =

u[x_,t_,k_] := Sum[b[n]Sin[n Pi x/p] Exp[ – n∧ 2Pi∧ 2t], {n, 1, k}]+(50 + 50x)

2/p Integrate[

 2 –

(x(1 – x∧ 3)–(50 + 50x)(Sin[n Pi x/p],

uTap = Table[u[x, t, 12], {t, 0, 1, 0.1}];

{x, 0, 1}]

Plot[Evaluate[uTab],{x, 0, 1}],

24 + 50n4 π4 + n5 π 5

(24 – 12n2 π2 + 100n4 π4 )Cos[nπ] – n5 π 5  ( – 24 + 53n2 π2 )Sin[nπ] n4 π 4 Simplify[s]

plotRange → All] 100 80 60 40 20

1 5 n π5 (2( – 24 – 50n4 π4 + 4(6 – 3n2 π2 + 25n4 π4 ) cos [nπ] + 220 (24nπ – 53n3 π3 ) sin[nπ]))

0.2

0.4

0.6

0.8

1

203

204

CH A P T E R 10:

Partial Differential Equations

39. u(x, t) = X(x)T (t), X  + λX = 0, X(0) = 0, −X  (1) = X(1); λ = 0 is not an eigenvalue; no negative eigenvalues; λ = k 2 > 0 : X(x) = c1 cos kx + c2 sin kx, X(0) = c1 = 0, X  (x) = kc2 cos kx, kc2 cos k = sin k or k = − tan k; eigenvalues are λn = kn2 ∞  2 where kn satisfies kn = − tan kn , n ≥ 1; u(x, t) = cn sin kn x e−kn t , ∞ 

u(x, 0) =

cn sin kn x = f (x), cn

1

n=1 1

sin2 kn xdx = f (x) sin kn xdx when m = n, 0    1 1 2 1 1 1 1 1 1 sin kn xdx = (1 − cos 2kn x)dx = x− sin 2kn x = 1− sin 2kn , 20 2 2kn 2 2kn 0 0       1 1 1 1 1 1 sin 2kn = 2 sin kn cos kn = 1− 1− 1 − (−kn cos kn ) cos kn = 2 2kn 2 2kn 2 kn 1  1 



1 2 f (x) sin kn xdx 1 + cos2 kn , c2n 1 + cos2 kn = f (x) sin kn xdx, cn = 1 + cos 2k n 2 0 0 n=1

0

Exercises 10.3 3. ut (x, 0) = n ≥ 0,

∞ 

bn nπ sin nπx = sin 3πx, 3b3 π = 1, b3 = 1/(3π), bn = 0, n  = 3, an = 0,

n=1

u(x, t) = 1/(3π) sin 3πt sin 3πx ∞  7. u(x, t) = (an cos nt + bn sin nt) sin nx; a2 = 1, an = 0, n  = 2, bn = 0, n ≥ 1; n=1

u(x, t) = cos 2t sin 2x 13. ux = ur ; uxx = urr ; uxy = urr + urs ; uyx − uxx = urs , urs = 0, u(r, s) = F(r) + G(s), u(x, y) = F(x + y) + G(y) 17. B2 − 4AC = (−4)2 − 4(1)(2) = 8 > 0, hyperbolic ∞  27. p = 1, u(x, 0) = an sin nπx = sin πx, a1 = 1, an = 0, n ≥ 2, bn = 0, n ≥ 1, n=1

u(x, t) = cos πt sin nx, u(x, t) = 12 [sin π(x + t) + sin π(x − t)] = 12 [sin πx cos πt + sin πt cos πx + sin πx cos πt − sin πt cos πx] = sin πx cos πt

Exercises 10.4 

 nπ(1 − x) nπx nπy + Bn sinh sin , u(1, y) = 2 2 2 n=1 ∞ ∞   nπx nπy nπ nπy An sinh sin = 0, An = 0, n ≥ 1, u(0, y) = Bn sinh sin = y, 2 2 2 2 n=1 n=1 2 1 nπy −4 cos nπ Bn = y sin dy = sinh(nπ/2) 0 2 nπ sinh(nπ/2)

3. u(x, y) =

∞ 

An sinh

1/Sinh[n Pi/2] Integrate[y Sin[n Pi y/2], {y, 0, 2}]   2Csch nπ 2 (2nπCos[nπ] – 2Sin[nπ]) – n2 π 2

Answers, Hints, and Solutions to Selected Exercises

7. u(x, y) =

∞ 

(An cosh nπy + Bn sinh nπy) sin nπx, An = 0, n ≥ 1, u(x, 2) =

n=1

∞ 

Bn sinh

n=1

2nπ sin nπx = sin πx, B1 =1/ sinh 2π, Bn = 0, n ≥ 2, u(x, y) = sinh πy sin πx/ sinh 2π 2 1 2T0 T0 sin nπydy = (1 − cos nπ) 10 nπ

13. Bn =

2 Integrate[T0 Sin[n Pi y], {y, 0, 1}]   T0 T0 Cos[nπ] 2 – nπ nπ

17. bn =

2 2 nπ 4 x sin dx = (−1)n+1 20 2 nπ

23. Bmn =

2



− 4mnπ cos mπ + mnπ(1 + π) cos(m − n)π − 2mnπ2 cos nπ+  mnπ(1 + π) cos(m + n)π √ 27. u(x, y, t) = 1/(4λ1,2 ) sin λ1,2 t sin x sin 2y, λ1,2 = c 5   ∞  nπx nπ(a − x) nπy  33. (c) w(x, y) = An sinh + B˜ n sinh sin , B˜ n = b b b n=1 b b 2 nπy 2 nπy h(y) sin dy,  An = k(y) sin dy nπa nπa b sinh b 0 b b sinh b 0 b

  π 2cschn 2 + −2 + n2 (π − 1)π cos nπ 2 x(1 − x) sin nxdx = 37. bn = , dn = π sinh n 0 n3 π π 1 2 −2(−1 + cos nπ) 2 sin nxdx = sin nπydy = cschnπ, cn = π sinh nπ 0 nπ π sinh nπ2 0 ∞  −20(−1 + cos nπ) csch nπ2 , n ≥ 1, u(x, y) = (bn sinh ny sin nx + cn sinh nπx sin n 2 nπ n=1 πy + dn sinh nπ(π − x) sin nπy) m 2 n2 π 3

1  2 −12csch2nπ cos nπ x 1 − x 2 sinnπxdx = , bn = 0, cn = 0, sinh 2nπ 0 n3 π 3 2 1 −4 cos nπ dn = y sin(nπy/2)dy = csch(nπ/2), u(x, y) = sinh (nπ/2) 0 nπ ∞  (an sinh nπy sin nπx + dn sinh nπ(1 − x)/2 sin(nπy/2))

41. an =

n=1

Clear[a] a[n_] = 1/Sinh[2 n Pi]2 Integrate[x(1 – x∧ 2)Sin[n Pi x], {x, 0, 1}] 1 (2Csch[2nπ] n4 π 4 ( – 6nπ Cos[nπ] + 6Sin [nπ] – 2n2 π2 Sin [nπ]))

205

206

CH A P T E R 10:

Partial Differential Equations

Clear[d] d[n_] = 1/Sinh[n Pi/2] Integrate[y Sin[n Pi y/2], {y, 0, 2}]   2Csch nπ 2 (2nπCos [nπ] – 2Sin [nπ]) – n2 π 2

45. u(x, y) =

1 1 sinh y sin x − sinh 2y sin 2x sinh π 2 sinh 2π

Exercises 10.5 3. u(1, θ) = A0 +

∞ 

(An cos nθ + Bn sin nθ) =

n=1

1 2

+ 12 cos 2θ, A0 = 12 , A2 = 12 , An = 0,

n  = 0, 2, Bn = 0, n ≥ 1, u(r, θ) = 12 + 12 r 2 cos 2θ π 7. Bn = π2 0 sin nθdθ = −2 nπ (1 − cos nπ), n ≥ 1; An = 0, n ≥ 0, u(r, θ) = ∞  2n−1 − 4r /((2n − 1)π) sin(2n − 1)θ n=1

2/(Pi) Integrate[ – Sin[n t], {t, 0, Pi}]  Cos[nπ] 2 –1 + n n π

13. u(r, 0) = J0 (α1 r), 17. 1/αn

1 0

∞ 

An J0 (αn r) = 0, An = 0, n ≥ 1,

n=1 B1 =

ut (r, 0) =

∞ 

cαn Bn J0 (αn r) =

n=1

1/(cα1 ), Bn = 0, n ≥ 2, u(r, t) = 1/(cα1 ) sin(cα1 t)J0 (α1 r)

αn rJ0 (αn r)dr = 1/αn

1 0

d/dr[rJ1 (αn r)]dr = 1/αn [rJ1 (αn r)]10 = 1/αn J1 (αn )

  2 J1 (αn ) J2 (αn ) − ; Bn = 0, n ≥ 1 21. An = αn [J1 (αn ]2 α2n  23. Note that r = x 2 + y2 and tan θ = y/x (or θ = arctan(y/x)); rx = 12 (x 2 +  y2 )−1/2 (2x) = x/ x 2 + y2 = r cos θ/r = cos θ, ry = y/r = r sin θ/r = sin θ, θx = x 1 y −y −r cos θ − cos θ 1 (− ) = = = , θ = = y r 1 + (y/x)2 x 2 x 2 + y2 r2 1 + (y/x)2 x 2 x r sin θ sin θ sin θ = = , ux = ur rx + uθ θx = cos θ ur − uθ , uy = ur ry + r r x 2 + y2 r2 cos θ 2 sin θ cos θ sin2 θ sin2 θ uθ θy = sin θur + uθ, uxx = sin2 θurr + urθ + 2 uθθ + ur + r r r r 2 sin θ cos θ cos2 θ cos2 θ 2 sin θ cos θ uθ , uyy = sin2 θurr + urθ + 2 uθθ + ur + r r r r 2 sin θ cos θ uθ r 4 1 1 4 29. cos 2φ = 2 cos2 φ − 1; P2 (cos φ) − ; u(r, φ) = r 2 P2 (cos φ) − 3 3 3 3

Answers, Hints, and Solutions to Selected Exercises

33. (a) A0 =

π 2 1 2π θ cos θdθ −π

= −2, An =

π 2 1 π θ cos θ cos nθdθ −π

=

 4π − 4n4 π cosnπ

, π (n − 1)3 (n + 1)3 Bn = 0, n ≥ 1

 π 2 8n − 8n3 cosnπ 1 (b) A0 = 0, An = 0, Bn= π θ sin θ cos nθ dθ = , n≥1 (n − 1)3 (n + 1)3 −π 

  1 π 2 1 π 2 θ 1 + θ 2 dθ = π2 5 + 3π2 15, An = θ 1 + θ 2 cos nθdθ = (c) A0 = 2π −π π −π



4 −12 + n2 1 + 2π2 cos nπ , Bn = 0, n ≥ 1 n4

Chapter 10 Review Exercises 3. (a) uxx = 0, uyy = 0;(b) uxx = 9e3x cos3y, uyy = −9e3x cos 3y; 2 y2 − x 2 2 x 2 − y2 (c) uxx = 2 , uyy = 2 x 2 + y2 x 2 + y2 7. u(0, θ) = A0 ; u at the center of the disk equals the average value of f around the boundary of the disk. 11. rR + R + λ2 rR = 0, R(a) = 0, Z  − λ2 Z = 0, Z(0) = 0, R(r) = c1 J0 (λr) + c2 Y0 (λr), c2 = 0 (bounded), J0 (λa) = 0, λn = αn /a, αn -nth zero of J0 . ∞  Z(z) = c3 cosh λz + c4 sinh λz, c3 = 0, u(r, z) = an sin λn zJ0 (λn r), a 2T0 rJ0 (λn r)dr an = sinh bλn 22 J12 (aλn ) 0

n=1

207