Differential Equations

t

= 50(1.395)

32.17

= 2 17 ·

sec,

the required time.. We could have let

e-g t / 50

= 8,

or

50 t=--log8 g ,

(4)

thus reducing equation (1) to 50 1 = -g (-log 8

+8 -

8 - log 8 = 1.643,

1), (5)

in which event we would solve equation (5) for 8, then find t from equation (4). Care must be exercised, however, in solving equation (5), in order to choose the right solution. One might find from Peirce's "Tables" that 8 = 2.598 is a solution of equation (5), but this value of 8, when substituted in (4), would give a negative value for t. Before solving equation (5) it should be noticed that we must have 0 < 8 < 1, from equation (4), since t must be positive. This vigilance was not demanded in

Chapter 3

40

solving equation (3) since there it was only necessary to take

log 8

8 -log 8

0.25 0.24 0.248

-1.386 -1.427 -1.394

1.636 1.667 1.642

Then equation (4) gives, as before, -50( -1.394) t= 32.17

= 2.17 sec.

14. Chemical reactions; first and second order processes. In a chemical reaction the rate of change of a substance may be proportional to the amount of that substance present at a given time. Such a reaction is known as a first order process. If x is the amount of the substance present at time t, the differential equation representing the process is tb;

dt

= kx,

(1)

where k is a constant of proportionality. Suppose, on the other hand, that a molecule of one substance A combines with one molecule of a second substance B so as to form one molecule of C. If a and b are the amounts of A and B respectively at time t = 0, and x is the amount of C at time t, then a - x and b - x are the amounts of A and B present respectively at time t, and z = 0 when t = o. When the rate of change of x is proportional to the product of the amounts of A and B remaining at any given time, the reaction is known as a second order process and is represented by the differential equation dx dt = k(a - x) (b - z).

(2)

Article 15

41

It is the process which is known as second order; the differential equation is of the first order. 1. Suppose that, in the process represented by equation (2), a = 5 and b = 4, and that x = 1 when t = 5 min; find the value of x when t = 10 min. Separating the variables, we have EXAMPLE

dx = ( 1 _ 1 ) dx = k dt. (5 - x)(4 - x) 4 -, x 5 - x

Integration between limits, using the pairs 'of values (0, 0), (1, 5), (z, 10), gives 1

5 -

X

5-

X

log 4 log 4

5

= ktJ o' ] -Xo

(3)

z 10

]

-Xo

=

ktJ o •

(4)

Dividing (4) by (3), we obtain l?g

(~ =:·:)

(4 4)

10

5 '

--~~-=-=2

log 3· 5 or

5- x 5 (16)2 64 4 - x = 4 15 = 45 ' 225 - 45x = 256 - 64x, 19x = 31, x = 1.63.

16. Pressure on a spherical surface; atmospheric and oceanic pressure. Imagine a spherical surface h ft above and parallel to the surface of the earth. Let r be the radius of this sphere; then r = R + h; where R is the radius of the earth, rand R also being measured in feet. At any point of this spherical surface, let p (Ib/ft") be the atmospheric pressure and

42

Chapter 3

(lbfftS) the density of the atmosphere. We wish to find a relation connecting the variables p; p, and h.* If P (lb) is the total pressure on the spherical surface of radius T, due to the weight of the air outside the surface, then

p

dP = 4r(r 2 dp + 2rp dr). Now dP is the (negative) increment in the total pressure on the spherical surface when we increase the radius by an amount dr, and is equal to the negative of the weight of the thin spherical shell of air of inner radius r and thickness dr; hence 4r(r2 dp + 2rp dr) = -4rr2 dr· p, Dividing by 4rr, then substituting r = R + h, dr = dh, we have r dp + 2p dr = - pr dr, dp 2p - = -p . (1) dh R +h In problems on atmospheric pressure where it is sufficiently accurate to regard the surface of the earth as plane, equation (1) may be reduced to a simpler form; letting R become infinite in (1), the relation connecting p, p, and h is simply

P = 411"T 2p,

dp dh

=

-p.

(2) -

If p is oceanic pressure at a depth h below sea level, then h changes sign in equations (1) and (2), and they become respectively (1') and

dp dh =

p,

(2')

where p is now the density of sea water. * For simplicity we idealize the conditions by ignoring motions and variable composition of the air, temperature, latitude and longitude effects, and the variation of gra.vitationa.l force with distance. The last-mentioned varia.tion could readily be included by multiplying p by a suitable factor depending on RIT.

Article 16

43

When we know a relation involving two or three of the variables p, h, p, we may combine it with (1) or (2), eliminate one of the variables, and find a relation between the other two. ExAMPLE

1. Assuming the relation p

= 192p(144 -

O.OOlh),

waere p (lb/ft2) is the pressure and p (lb/ft3) is the density of the atmosphere at a height h (ft) above sea level, find the pressure at a height of 10,000 ft if the pressure at sea level is 2120 Ib/ft 2 • Substituting the value of p from the given relation into equation (2), we have dp = -p dh 192(144 - O.OOlh) Separation of the variables and integration between limits give

1 = l p

10

dp 192 --2120 P p

0.192 log PJ2120

0

,000

-

144 - O.OOlh

= log (144 p

0.192 log 2120

dh

- O.OOlh)

,

]10.000 0 ,

134

= log 144·

I

Using logarithms to the base 10, we get p

IOg10 2120 =

-0.03126 0.192

= -0.1628 = 9.8372 - 10,

p = 2120 X 0.687 = 1460 Ib/ft2•

16. Steady-state heat Bow. Experimental study of the flow of heat has shown that the rate of flow across an area is proportional to the area and to the temperature gradient, i.e., the rate of change of the temperature with respect to distance normal to the area. We now state this law in the form of a differential equation, confining ourselves to the case in which the flow has settled down into a steady state and the rate of flow is therefore independent of time. Furthermore, we consider here only the case in which the temperature depends on

44

Chapter 3

only one space coordinate, x, measured in the direction of flow. * If the flow is radially outward from the axis of 8J cylindrical surface of radius x, all points of the surface will have the same temperature, i.e., the cylinder will be an isothermal surface. If the flow is radially outward from a point, x will be the radius of an isothermal sphere, whereas if the flow takes place in parallel straight lines, the isothermal surfaces will be parallel planes perpendicular to the lines of flow, x measuring the distance of one of these planes from a fixed plane of reference. Let q (cal/see) be the constant rate at which heat is flowing in a body, perpendicularly through an isothermal surface of area A (em"), all points of which have the coordinate x and the temperature u(OC). Then q will be proportional to A du/dx, and we have the differential equation

du

q = -kA dx·

(1)

The constant of proportionality, k (cal/em deg sec), is a property of the material of the body and is called the thermal conductivity. Since the temperature decreases as x increases, the temperature gradient, du/ dx, is negative and the minus sign must be used in equation (1) in order to make q positive. 1. A pipe 10 em in diameter contains steam at 100°0. It is insulated with a coating of magnesia 3 cm thick, for which k = 175 X 10-6 callcm deg sec. If the outer surface of the insulation is kept at 40°C, find the rate of heat loss from a meter length of pipe, and the temperature halfway through the insulation. The thickness of the pipe is not taken into account; it is assumed that the inner surface of the insulation is at 100°C. By letting u represent the temperature on an isothermal cylindrical surface of radius z, within the insulation, then A = 211"xL will be EXAMPLE

* The more general problems of heat flow at a rate varying with the time, and of steady-state flow in which the temperature depends on more than one coordinate, lead to partial differential equations. See Reddick and Miller, Advanced M athematic8 fur Engineers, Chapter VII, Art. 70.

Article 16

45

the area of a portion of the surface of length L. For the rate of heat loss, which is equal to the rate of flow across A, equation (1) gives q

du dx

= -k·21rxL • - .

(2)

Separating the variables and integrating from the inner surface where z = 5, u = 100, to the outer surface where x = 8, U = 40, we have 8 dX = -21rkLl:f:°dU,

q1 IS

q log

x

100

:18

XJIS

=-

q log 1.6

or, when k

]40

21rkLu

100'

= 1201rkL,

(3)

= 175 X 10-6 and L = 100, in cgs units, 2.11r 6.597 q = log 1.6 = 0.4700 = 14.0 cal/sec.

To find the temperature halfway through the insulation, substitute in (2) the value of q given by (3), then integrate from x = 5, U = 100, tc x = 6.5, U = u:

1201rkL = _ 21rkLx du , log 1.6 dx

l U -

100

=-

U,dU = _ 100

60 log 1.6

60 log 1.3 15.74 log 1.6 = - 0.4700

1

65 .

5

=

~

x'

-33.5;

U

= 66.5°0.

PROBLEMS 1. A 2Q-Ib weight moves in a horizontal straight line under the joint action of a constant force of 12 lb, in the direction of motion, and a resisting force whose magnitude in pounds is equal to four times the instantaneous velocity in feet per second. If the weight starts from rest, find its velocity and the distance traveled after l sec. 2. If x dx 2 dy = y(x (],x dy), and y = 0 when x = 2, find by the method of trial and error the value of y when x = V6. ,

+

+

46

Chapter 3

3. A mass of 200 lb falls from rest. If the resistance of the air is proportional to the velocity, and if the limiting velocity is 173 ft/sec, find (a) the velocity at the end of 10 sec; (b) the distance fallen at the end of 10 sec; (c) the time when the velocity is half the limiting velocity; (if) the time when the distance fallen is 173 ft. 4. A ship weighing 8 X 1071b, and starting from rest, reaches a limiting velocity V ft/sec under a force of 2 X 1()5 lb exerted by the propellers. Assuming a resistance to the motion proportional to the square of the velocity, find (a) the time required to attain half the limiting velocity; (b) the time required for the velocity to increase from one-third to two-thirds of the limiting velocity. 6. A body falls from rest in a medium offering resistance proportional to the square of the velocity. The limiting velocity is 80 ft/sec. When the velocity has reached 56 ft/sec, find (a) the time elapsed; (b) the distance traversed.

• ( Note that, smce

d3; dv

dv )

v = dt ' dt = v dx·

6. A charged sphere, of weight w lb, falls from rest under the combined forces of gravity and an electric field, the latter producing an upward force of E lb. If the air resistance is proportional to the speed and the limiting speed is V ft/sec, find the time required to reach p per cent of the limiting speed. Use 9 (ft/sec2) for acceleration due to gravity. 'I. A body falling from rest in a liquid acquires a velocity which approaches 10 ft/sec as a limit. Assuming the resistance of the medium to be proportional to the velocity, and the specific gravity of the body to be 3 times that of the liquid, find (a) the velocity at the end of 1 sec; (b) the time required to traverse the first 6 ft. 8. A body falls from rest in a liquid whose density is l/n that of the body. The liquid offers resistance proportional to the velocity. If the velocity at the end of 1 sec is 20 ft/see and at the end of 4 sec is one-half the limiting velocity, find the limiting velocity and the value of n. . 9. Work Prob. 8 if the resistance is proportional to the square of the velocity. 10. In a certain chemical reaction, the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. If one-quarter the original amount of the substance has been converted when t = 3 min and if an amount A has been converted when t = 6 min, find the original amount of the substance. 11. If radium decomposes at a rate proportional to the amount present, and half the original quantity disappears in n years, (a) what percentage will disappear in n/2 years? (b) in how many years will 25 per cent disappear? . 12. According to Newton's law of cooling, the rate at which heat is lost by a heated body is proportional to the difference in temperature between the

Article 16

47

body and the surrounding medium. If a thermometer is removed from a room in which the temperature is 70°F into the open where the temperature is 30°F, and if its reading is 60°F at the end of ! min, (a) how long after the removal will the reading be 40°F? (b) how long will it take for the reading to drop from 50°F to 35°F? (c) what is the temperature drop during the first 2 min after the removal? 13. In the chemical reaction represented by equation (2), Art. 14, suppose that x = 2 when t = 10 min. Find x at the end of 15 min (a) when a = 6, b = 3; (b) when a = b = 6. 14. In the chemical reaction represented by equation (2) Art. 14, suppose that a = 4, b = 3, and that x = 1. when t = 3. Find the value of t when x = 2. 16. In equation (2), Art. 14, given a = b and x = a/n when t = h; find the value of x when t = 2th [Note: In Probs. 16 to 20, use equation (2), Art. 15.] ft

16. In Ex. 1, Art. 15, at what height will the pressure be half the pressure at sea level? 1'1. Assuming that p is proportional to p, and taking p equal to 14.7 and 12.0 Ib/in. 2 at the surface of the earth and at a height of 1 mile respectively, find the atmospheric pressure at a height of 2.5 miles. 18. If the density of the atmosphere is proportional to the t power of the pressure, and the pressures at heights 0 and 10,000 ft are respectively 14.7 and 10.1Ib/in.2, at what height is the pressure 8Ib/in.2? 19. If p is equal to 14.7 and 12.0 Ib/in. 2 , at the surface of the earth and at a height of 1 mile respectively, and if p = kp%, find the density of the air at & height of 2 miles. 20. How much will the barometer reading decrease in ascending 1000 ft above sea level under the following assumptions? The pressure of the atmosphere is proportional to the density. The barometer reading at sea level is 29.9 in. of mercury. The density of air at sea level is 0.0807Ib/ft3• The specific gravity of mercury is 13.6. The density of water is 62.4Ib/ft3 • 21. If R is the radius of the earth and p is proportional to p, show that the atmospheric pressure (per unit area) at a height R/10 is about 17 per cent less than the result obtained by regarding the surface of the earth as plane. 22. At what distance from the surface of the earth is the total weight of air above equal to the weight of air below? Assume that p is proportional to p and that at the surface their values are 2120 Ib/ft2 and 0.0807 Ib/ft3 respectively.

48

Chapter 3

23. If sea water under a pressure of p Ibjft2 weighs 64(1 + 2 X 10-8p) Ibjft3, find the weight of a cubic foot of water and the pressure at a depth of 2 miles below sea level. (Use equation (2'), Art. 15.) 24. Find the heat loss in calories per day from 20 meters of pipe 30 cm in diameter containing steam at 100°C, if the pipe is covered with a layer of concrete 10 cm thick, and the outer surface of the concrete is kept at 35°C. Find also the temperature halfway through the concrete. (Assume k == 225 X 10-6 cal/em deg sec.) 26. Work Ex. 1, Art. 16, replacing the meter length of pipe by a hollow sphere 1 meter in diameter. 26. A steam pipe has inner and outer radii, fl and f2 respectively, and the temperatures at its inner and outer surfaces are respectively Ul and U2. (a) Derive a formula for the steady-state temperature U at radial distance r(fl < r < f2). (b) Compute the value of u if n = 5 em, f2 = 10 cm, Ul = 100°C, U2 = 80°C, r = 8 om. 27. A hollow spherical shell has inner and outer radii, f1 and ra respectively, and the temperatures of its inner and outer surfaces are respectively Ul and U2. (a) Derive a formula for the steady-state temperature U at radial distance r(fl < r < f2). (b) Compute the value of U if fl = 10 em, f2 = 20 cm, Ul= 100°C, U2 = 64°C, r = 15 em, 28. A cylindrical pipe and a hollow spherical shell both have inside diameters of 3 cm and outside diameters of 5 em. If the thermal conductivity of the material of the pipe is -to that of the shell, find how long the pipe must be in order that the quantities of heat conducted through pipe and shell walls shall be the same under the same inner and same outer temperatures in the respective cases. 29. If a flat furnace wall of thickness tt cm, conductivity kl eal/cm deg sec, and area A cm2 is covered by an insulating material of thickness t2 cm and conductivity k 2 caljcm deg sec, find the heat loss in calories per second through the wall if its inner surface is at UloC and the outer surface of the insulation is at U2°C. 30. A steam pipe of length L cm and radius Xo em is covered by n layers of insulation of conductivities ks, k 2, ••• , k n (cal/cm deg sec) and outer radii Xl, X2, ••• , X n (cm) respectively. If the steam is at Uo°C and the outer surface of the outside layer of insulation is at Un °C, find the heat loss in calories per second through the insulation. 31. A steam pipe of radius Xo is to have two layers of different insulating materials of outer radii Xl, X2 (Xo < Xl < X2). Show that for maximum efficiency of insulation, the better insulator should form the inside or outside layer according as Xl ~ ~, but that if Xl = the two insulating materials are interchangeable. 32. Assuming that the velocity of efflux of water (volume per unit time) through an orifice in the bottom of a tank is proportional to the product of the

-v;;;,

Article 16

49

area of the orifice and the square root of the depth of the water, the differential equation is Adh = -kBV'h. dt '

where h (ft) is the depth of the water and A (ft2) is the area of the water surface at any time t (sec), and B (ft 2) is the area of the orifice. The constant of proportionality, k (ftH/sec), may be determined empirically. Find the time required to empty a cubical tank whose edge is 4 ft. The tank has a hole 2 in. in diameter in the bottom and is originally full of water. (Take k = 4.8.) 33. Suppose that the tank of Frob. 32 is originally empty and that water runs into it at the rate of V ft3/ seC. Show that, if the tank is to fill, it is necessary to have V > 11'/15. Find the time required to fill if V = 211'/15 ft3/ sec. 34. A funnel has the shape of a right circular cone with vertex down and is full of water. If half the volume of water runs out in time t, find the time required to empty. 36. A cylindrical tank 8 ft long and 4 ft in diameter is full of water. Find the time required to empty through a hole 1 in. in diameter in the bottom (a) if the cylinder is standing on end, (b) if the cylinder lies horizontal. (Take k = 4.8.) 36. For the tank of Probe 35 in horizontal position, find (a) the time required to half empty, (b) the depth of water after 30 min. 37. Show that a cylindrical tank full of water will empty faster through a hole of given diameter in the bottom if it stands on end or if it lies horizontal, according as (D/L) S (64/911'2), where D and L are, respectively, the diameter and the length of the tank. 38. A parabolic bowl of equal depth and radius, and a hemispherical bowl of the same radius, are full of water. Find the ratio of the times required to empty through a hole of the same size in the bottom. 39. Ether flows at the rate of 411" ft 3/hr into a bowl which is a paraboloid of revolution 10 ft across the top and 10 ft deep. If the rate of evaporation is proportional to the area A (ft2) of the liquid surface, the constant of proportionality being k = -.h ftjhr, find the time required to fill the bowl. 40. A body with initial velocity u ft/sec downward falls under gravity in a medium offering resistance proportional to the square of the velocity. Show that the distance fallen in t sec is Y=

gV2 log [ cosh gtV + Vu.sinh Vgt] ft,

where V ft/sec is the limiting velocity and g ft/sec 2 is the acceleration due to gravity.

50

Chapter 3

17. Integrable combinations. The differential equations encountered so far in this chapter have been of such a nature that the variables could be separated. We shall see now that a differential equation in which the variables are not separable can be solved if it can be arranged in integrable combinations. The process will be illustrated by some examples. EXAMPLE

1. Solve x dy == (3x 2

-

y) dx.

(1)

This equation cannot be separated into terms containing x only and terms containing y only, but, by transposing the term y dx, we have xdy+ydx=3x2dx.

The right member is now free of y and integrates at once into x 3 • The left member contains both x and y, but it is an integrable combination of x and y; we recognize it as the exact differential of xy, so that it integrates, not term by term, but as a whole, into xy. The general solution of (1) is therefore xy

= x3 + c.

If we make the substitution xy = u in (1), a substitution suggested by the combination x dy y dx, the equation takes the form du = 3x 2 dx ,

+

in which the variables u and x are separated, and integration • gives u = x3 + C, or xy = ~

+ C.

Differential equations integrable by combinations can always be reduced to separable equations in new variables by means of substitutions, but we usually integrate the combinations which are recognizable as exact differentials without substituting new variables.

Article 17

51

7

Some of the frequently occurring integrable combinations and the functions of w~h they are exact differentials are: x dy

+ y dx

= d(xy),

xdy - ydx = x2

d(Y) X

'

ydx - xdy = d(X) y2 Y , 2xy dy - y2 dx = d (y2)

x2

X

2 2xy dx - x dy y2

2 y2

x dy - dx +y x

=

'

= d (x 2)

Y ,

xdy - ydx x2 1+ y

(-1 y)

()2

= d tan -x ,

x

xdy - ydx 2 xdy-ydx= x =d(llo X+Y). 2_ X (y)2 2 gx-y y2 1- -

z EXAMPLE

_~(;J,.

-

2. S o l v e ' x dy - y dx = y dYe

i'--f~)

J't:' h ,,l' VV/tl \i

-::JI

~

(2)

If we multiply both sides by 1/x2 , the left side becomes an integrable combination, but the right side is not integrable. However, multiplying by 1/y2 and changing signs, we have y dx - x dy = _ dy. y2 Y

(3)

Integration now gives x '- = C

y

-log y.

A differential equation such as (3), whose terms are exact differentials or form combinations which are exact differentials,

52

Chapter 3

is called an exact differential equation. The function 1/y2 is called an integrating factor of equation (2), since multiplication by this factor produces the exact equation (3). Rules can be devised for finding integrating factors of certain types of differential equations but the method of finding these factors by inspection will suffice for our needs. EXAMPLE

3. Solve

+ y dy = v'x 2 + y2 dx.

x dx

(4)

Division by the radical on the right gives

+ ydy d + y2 = x,

xdx Vx 2

(5)

an equation whose left member is an integrable combination of the form dU/2~, where = x 2 y2, and. therefore integrates into V u; thus ltVx 2 y2 is an integrating factor of (4). Integration of equation (5) yields v'x 2 y2 = X C,

u

+

+

+

+

the general solution of equation (4). EXAMPLE

4. Solve dy dx

y x

I 2y

- = - --.

Clearing fractions, we have 2xy dy

=

(2y2 - x) dx,

or 2y(x dy - y dx)

=

-x dx.

If, now, we divide by x 2 y, i.e., try l/x 2 y as an integrating factor, we have an integrable combination on the left, but the right side is not integrable; also, a similar situation occurs if we divide by y3. However, division by x 3 gives 2 (y) x dy - y dx X

x

2

= _

dx.

x2

Article 18

53

The left side is now an integrable combination of the form 2u du, where u = y/ x, and the right side is also integrable. Integrating, we have y) 2 1 ( x = x + C, or

y2 = X + Cx 2•

18. Homogeneous equations. A function lex, y) is homogeneous of degree n if the replacement of z by kx and y by ky, where k is any quantity, constant or variable, multiplies the function by k"', that is, if

f(kx, ky) = k"'f(x, y). In particular if n = Qthe function is homogeneous of degree zero and is unchanged by replacing x by kx and y by ky; such a function, if we take k = l/x, can be written as a function of the combination y/x, as follows:

f(x,

y)

= f(kx,

ky)

=f

(1, ~) = F (~).

As illustrations, observe the functions (a) I(x, y) = x 2 - 2xy + y2, (b) I(x, y) = x - 7y

+ 6,

(c) I(x, y) = y log x, (d) I(x, y) = y(log x - log y), xy2 x 2y (e) I(x, y) = 3T _ 411·

+

2kx·ky + k 2y2 = k 2(X2 - 2xy + y2) = k 2/(x, y); hence the function x 2 - 2xy + y2 is homogeneous of second degree. (b) I(kx, ky) = k» - tky + 6. The k cannot be factored out; the function x - 7y + 6 is not homogeneous. If!(x, y) is a polynomial, it is homogeneous only if each of its terms is of the same degree in x and y. (a) !(kx, ky)

= k 2x2 -

54

Chapter 3

(c) j(kx, ley) = ky log kx; the function y log x is not homogeneous. (d) j(kx, ky) = ky(log k» - log ky) = ky log x/y = kj(x, y); hencej(x, y) is homogeneous of first degree. k 3(X2y xy2) . (e) f(kx, ky) = k3(3x3 _ 4y3) = lex, y). The funetion j'(z, y)

+

is homogeneous of degree zero and therefore can be written as a function of y/x; division of both numerator and denominator by x3 gives x2

y + xy 2

3T - 4y3

-YX + (y)2 -x =3

- 4 ~r

The quotient of two functions, both homogeneous of the same degree, is a homogeneous function of degree zero, since, when x is replaced by kx and y by ky, the power of k occurring as factor of the numerator is the same as in the denominator and cancels out. A differential equation dy dx = j(x, y) (1) is called homogeneous if lex, y) is a homogeneous function of degree zero. If the differential equation is written in the form

M(x, y) dx

+ N(x, y) dy = 0,

(2)

it will then be homogeneous if M and N are homogeneous functions of the same degree. If a differential equation is homogeneous it can be solved by either of the following methods: (a) Substitute y = vx. The resulting equation is separable in v and x; solve it and replace v by y / x. (b) Substitute x = vy. The resulting equation is separable in v and y; solve it and replace v by x/y. We have' to prove that the substitution y = vx reduces a homogeneous differential equation to one which is separable in

55

Artic\e 18

v and z, The analogous proof for method (b) is left as an exercise for the student in the first problem of the next group. Since the differential equation is homogeneous, it may be written in the form (3)

Letting y = vx, hence dy = v dx + x dv, equation (3) takes the form v dx + x dv = F(v) dx, or dv dx F(v) - v =

x'

. an equation in which v and x are separated. In the exceptional case where F(v) = v, equation (3) is itself separable: dy/dx= y/x. EXAMPLE

1. Solve

+ (x2 + y2) dy = Let y = vx, dy = v dx + x dv. 2xydx

O.

Method (a). It will be noticed that, since the differential equation is homogeneous of second degree, x 2 will cancel out of each term after the substitution y = vx. In general xfl, will cancel out if n is the degree of homogeneity. Similarly yfl, will cancel out after the substitution x = vy. The substitution and cancellation can be made at the same time. We have

+ (1 + v2)(v dx + x dv) = 0, (v3 + 3v) dx + x(1 + if) dv = O.

2v dx

or

Separating the variables and multiplying by 3 in order to make the second numerator the exact differential of its denominator, we get

+ (3if + 3) dv = O. X v3 + 3v 3 log x + log (VJ + 3v) = log C, 3 dx

Integrating,

Chapter 3

56

Taking antilogarithms, Replacing v by y/ x,

+ 3x2y = C. Letting x = flY, dx = v dy + y dv, then cancelling y2, y3

Method (b). we have

2v(v dy

+ y dv) + (v2 + 1) dy = 0,

or

(3v2

+ 1) dy + 2vydv = O.

Separating the variables and multiplying by 3, we get

3 dy

+

y

Integrating, 310gy

6vdv = 0 3v2 + 1 .

+ log (3v2 + 1) =

10gC.

Taking antilogarithms, Replacing v by x/y,

3x2y + y3

=

C.

Shorter method. We have used this problem as an illustration of the standard methods for solving a homogeneous equation, but there is a shorter way to solve Example 1. Writing the equation in the form x 2 dy + 2xy dx + y2 dy = 0,

we see that the first two terms form an integrable combination, the differential of x 2y, so that integration yields at once y3 C 2Y x + 3" = 3'

or

3x2y + y3 = C.

It is desirable to use the shorter method of integrable combinations whenever possible, provided that too muchtime need not be spent. , in recognizing the integrable combination. EXAMPLE

2. Solve (x

+ v' y2 -

xy) dy - y dx =

o.

57

Article 18

In Example 1 there was little to choose between the methods (a) and (b), although (b) is perhaps a trifle simpler. In this example, however, method (b) is definitely easier, as the student may verify by solving the problem using method (a). There is no general rule for determining beforehand which method is easier, but usually it is better to substitute for the variable whose differential has the simpler coefficient. Here dx has the simpler coefficient and we substitute x = vy. Then, after division by y, the equation becomes (v

+ VI -

v) dy - (v dy

+ y dv) = O.

Separating the variables, we get dy_ Y

Integrating, log y Taking antilogarithms,

dv

VI -

+ 2V1 ye2 v I-v =

-0 v- .

v = log

C:

c.

x/y,

Replacing v by

PROBLEMS 1. Prove that the homogeneous differential equation (1) or (2) of Art. 18 becomes separable upon substituting x = vy, thus verifying method (b). Solve each of the following differential equations. 2. (x - 2y3) dy \ Y dx. 3. dy = 1 +?t. dx x 4. x dy = (x 2 + y2 fl

+ y) dx.

5. dy = e dx 2y - xe'V 6. (x2 + y2) dx = 2xy dYe

7. dy

= '!J.

dx

X

8. dy dx

=

+ y2.

2y x

x2

+ 3x. 2y

58

Chapter 3

-----------------------9.

x:

= y(logy -logx). 2

10. dy = x - y2 - 2y. dx y2- x2_2x

u.

dy dx

12. dy dx

= x - y. x+y

+

3xy

+ y2

3x2

=

o.

+ x~) dy = y. dx 14. 3xy2dy = (x 3 + 2y3) d»,

13. (x

16. 3x dy

= 2y dx - xy cos x dx.

16. y dx + x

1'1.

(log; - 2) dy = O.

(2x tan; +Y)dx = x dy.

18. x dy ydx

+ 3x

- y2 = 3y2 - x 2

.,

o.

+ 3y + 5 = o. dx 3x + lly + 17 3 2y 20. x dy + 2x - x - y3 = o. 19. dy

+

2

y dx

7x

2y 3

-

xy - x3

G

~ xdy - + ;)dY. 22. x(y + V x 2 + y2) dy = y2 d», l)dx + Vx 1(1 - XV y 23. V y2 - 1(1 - YVx 24. v;y dx = (x - y + v;y) dYe 26. x(x2 + y2) dy = y(x + yV'-x + y) da,

21. ydx

2

2

2 -

-

2 -

l)dy

= O.

2-+-y-2

19. Linear equations. The standard form of the linear differential equation of first order is dy dx + Py

= Q,

(1)

where P and Qare functions of z. I t is called linear because it is of the first degree in the dependent variable and its derivative,

59

Article 19

that is, of the first degree in y and dyjdx. Either P or Q may be a constant, which can be regarded as a special case of a function of z, but if their ratio is constant the equation is separable. Writing equation (1) in the form

.

+ Py dx =

dy

Q dx,

(2)

let us try to integrate it by changing the left side into an integrable combination. We multiply equation (2) by an unknown function of x, say R, obtaining R dy

+ yRP dx

= RQ da:

(3)

The right side of (3) is in integrable form, since Q is a known function of x, and R when it is found will be a function of x. Can we determine R so that the left side of (3) will be an integrable combination? The combination R dy

+ y dR

= d(Ry)

(4)

is at once suggested. Comparing the left member of (3) with this exact differential (4), we see that they will be identical provided that (5) dR = RPdx, in which case the left member of (3) will integrate into Ry. Now (5) is a separable equation determining the integrating factor R: dR

R

log R

= Pdx =

R =

f

'

P i1:£,

efPd:x;.

(6) (7)

The constant of integration in (6) is omitted, or taken equal to zero, since the simplest value of R is desired. With the value of R given by (7), the left member of (3) \ntegrates into Ry and the integral of the right member is

60

Chapter 3

expressed by j RQ'il:£

+ C, so that the general solution of the

linear equation (1) is Ry = jRQil:£ where

=

R

+ C,

(8)

e!"Pda:.

Therefore, to solve the linear equation (1), first compute the integrating factor R, then insert it in (8), perform the indicated integration, and simplify the result if possible. EXAMPLE

1. Solve x dy = 2(x 4

+ y) dx..

The equation is linear, but it must be put into standard form (1) before using 'formula (8), thus: : dy 2 3 ---y=2x. dx x

Here P = -2/x, Q = 2x3 • Computing R, we have R = ef -(2/a:) da: = e-2 1o g a:= e10g a:-2 = x-2• Substitution in (8) gives

x- y = jx2

2.2x3

dx

+0

=

x2 + 0,

or EXAMPLE

2. Solve

(x - sin y) dy

+ tan y dx = 0

and find the particular solution satisfying the condition y = 1r /6 when x = 1. At first glance this equation does not seem to be linear, for if we write dy/dx as the first term, the equation takes the form dy dx

+

tan.y x~slny

= 0,

which is not in standard form (1). The equation is not linear with y as dependent variable, that is, it is not linear in y and dyjdx; but,

Article 19

61

if x is taken as the dependent variable and the equation is written in the form dx dy

+ cot y·x = cos y,

we see that it.is linear in x and dx/dy. Hence it can be solved by using formula (8) with x and y interchanged, P and Q now being functions of y, namely, P = cot y, Q = cos y. We have R = e f cot y dy = e10 g sin y = sin y. Then formula (8) gives. siny·x =

fSin

Y oosydy

or, changing 0 to 0/2, 2x sin y

+ C = ~ sin2 y + C,

= sin2 y + C,

which is the general solution of the given equation. To find the required particular solution of the given equation, substitute x = 1, sin y = sin (11"/6) = ~; then 2.(~) =

l

+0,

0= :,

and 8x sin y = 4 sin2 y

+ 3.

Sometimes a differential equation may be solved in more than one way. For instance the above example may be solved also by use of an integrable combination. 2. (Shorter solution.) The differential equation may be written

EXAMPLE

x dy

+ tan y dx

= sin y dy.

Multiplication by the integrating factor cos y produces on the left the integrable combination x cos y dy + sin y dx or d(x sin y); hence x cos y dy

+ sin y dx

=

sin y cos y dy,

x sin y = ~ sin2 y

+ C,

When x = 1, y = 11"/6, ~ = ~ + 0, 0 = :, and the required particular solution is 8x sin y = 4 sin2 y + 3.

62

Chapter 3

20. Substitutions. A differential equation which is not in one of the forms previously discussed may be capable of transformation into one of these forms by substituting one or two new variables. When this is the case the new equation may be solved and the result transformed back in terms of the original variables. We shall now consider a few equations of this kind. (a) Bernoulli's equation. A differential equation of the form dy _+py=Qyn

dx

(1)

'

where P and Q are functions of x, and n is any constant except o or 1, is called a Bernoulli equation. It is assumed that n is neither 0 nor 1, for then the equation would be linear or separable, respectively. , We shall show that the equation (1) can be transformed into a linear equation by substituting a new, dependent variable in place of y; but, in order to see the appropriate substitution, we first multiply equation (1) by u:" so that a function of x only appears on the right, thus: y-n dy

dx

+ Pyl-n = Q.

We notice now that, if the equation is multiplied by 1 - n, the first term becomes the derivative of yl-n: (1 - n)y-"

:= +

Hence the substitution

dz

dx

(1 - n)Pll-"

yl-n

+ (1 -

= (1 -

n)Q.

= z produces the equation n)Pz

= (1 -

n)Q,

which is linear in z and dz/dx, and can be solved in terms of z and x by formula (8) of.Art. 19, after which the replacement of z by yl-n will give the general solution of equation (1).

Article 20

63

However, in practice we do not need to use the letter s, but proceed as follows to solve equation (1). First multiply through by y-n and by the new exponent of y in the second term. Then apply formula (8), Art. 19, .replacing P by the new coefficient in the second term, Q by the new right member, and y by the new power of y in the second term. If, instead of equation (1), we had

dx -dy + Px = Qxfl,,

(2)

where P and Q are functions of y, we would follow the same procedure with x and y interchanged. EXAMPLE

1. Solve dy dx =

X

3

y2

+ xy.

Written in the standard form (1), the equation is dy dx - xy

= X 3y2.

Multiplying by y-2 and by the new exponent of y in the second term, which is -1. we have

_y-2: +

xy-l

= -x3.

Since the first term is the derivative of y-l, the equation is linear with y-l as dependent variable. Using formula (8) of Art. 19, we have R = efz dz = ez 2 / 2 , e'"'/2y-l

=

f

e;"'/2( -x3 dx)

+ C = -2

t;

e""/2(x dx)

Integration (Peirce, 402), with x replaced by x2 /2, gives e;"'/2y-l

=

_2e"'/2 ( ; -

1) +

Multiplying by e-z'/2 y and rearranging, we have (2 -

x2 + Ce-z 2 / 2 )y = 1.

C.

+ C.

64

Chapter 3

(b) Equations reducible to homogeneous. A differential equation of the form

where the a's and b's are constants, can be reduced to a homogeneous equation by making a substitution for both variables, z and y. The method would not be used in the special cases where the constants have values that render the equation solvable by one of the previous methods; for instance, When a2 = bi = 0, or al = b2 = 0, or al/b l = a2/b2 = aa/ ba, the equation is separable. When a2 = b, ~ 0, the equation is solvable by means of the integrable combination, x dy + y dx = d(xy). When as = ba = 0, the equation is homogeneous. When al = 0, or b2 = 0, the equation is linear. We shall employ a transformation which moves the origin to the point (h, k), then choose the point (h, k) so that the resulting equation is homogeneous. Let x = X + h, y = Y + k (making dx = dX, dy = dY), and substitute in (3): (alX

+ a2 Y + Uth + a2k + as) dX + (blX + b2Y + blh + b2k + ba) dY =

0.

Now choose hand k so that

+ a2k + CIa = 0, blh + b2k + ba = o.

alh The result is

(4)

a homogeneous equation which we can solve in terms of X and Y, and then go back to the original variables by means of the substitution X = x - h, Y = y - k.

Article 20

65

In practice, however, it is not necessary actually to change to the new variables X, Y, and afterwards change X to x - h and Y to y - k. The same effect is produced if, after dropping aa and b3 from equation (3), we solve the resulting equation and then change x to z - hand y to y - k, hand k having been determined previously by solving equations (4). The method can be stated as follows: 1. Solve equations (4) for hand k. 2. Drop a3, b3 from (3) and solve the resulting equation. 3. In the result of step 2 change x to x - h, y to Y - k, EXAMPLE

2. Solve (x

+ y + 1) dx + (6x + lOy + 14) dy

Solving

=

O.

+ k + 1 = 0, 3h + 5k + 7 = 0, h

we obtain h = 1, k = -2. We now solve (x + y) dx

+ (6x + lOy) dy

= O.

The substitution x = vy gives (v + 1) (v dy

+ y dv) + (6v + 10) dy =

O.

Separating the variables, dy

+

y

(v + 1) dv = O. (v + 5)(v + 2)

Resolving the second fraction into partial fractions,

[4 11 + + dy

3(v

5) - 3(v

1] +

2) dv = O.

~

Multiplying by 3 and integrating, 3 log y

+ 4 log (v + 5) -

log (v

+ 2) =

Taking antilogarithms, y3(v + 5)4

= O(v + 2).

log C.

66

Chapter 3

Replacing v by a] y,

+ 5y)4 = C(x + 2y). Finally we change x to x - 1, y to y + 2, and obtain (x + 5y + 9)4 = C(x + 2y + 3), (x

which is the required solution.

There is one exceptional case where the equations (4) cannot be solved for hand k, This occurs when in equation (3) btx

+ b2y

= m(at x

+ a2Y),

where m is a constant. It happens, however, that this exceptional case is more easily treated than the regular one, for if in equation (3) we substitute atX + a2Y = v, btx + b2y = mv, and the value of either dx or dy from the differential relation at dx + a2 dy = dv, we obtain a separable equation. EXAMPLE

3. Solve dy x+y+4 -=) . . dx 2x+2y-l

+ y = v, dx = dv - dy, we have (2v - 1) dy = (v + 4)(dv - dy), (3v + 3) dy = (v + 4) dv,

Substituting x or

which is separable: v+4

(

3)

3 dy = v + 1 dv = 1 + v + 1 dv. Integrating, 3y

Replacing v by x

+ C = v + 3 log (v + 1).

+ y, x - 2y

..

+ 3 log (x + y + 1) = C.

(c) Substitution suggested by the form of equation. Sometimes a substitution of variables, suggested by the form of a differ-

Article 20

67

entia! equation, will reduce the equation to a solvable form in the new variables. EXAMPLE

4. Solve

~ = sin (x + y). A little reflection discloses that this equation' does not belong to one of the types previously considered, nor does it help to expand the right member and write dy dx

.

.

= SIn x cos y + cos x sin y;

but the form of the equation suggests letting :t

+ y == v,

dy

= dv -

dz.

The result of this substitution is

dv -dx -

1

. = sIn v,

an equation separable in the variables v, z.

We have

dv = da; 1 sm e

+

.

a-

Integration (Peirce, 294) gives

Of,

replacing

f)

C-tan(1r _ 42' by x + y,

~=C-

tan

(~ _

x -;

y}

The above example suggests the more general case: A differential equation of the form dy - =f(ax

dx

+ by)

(6)

68

Chapter 3

is reducible to a separable equation by the substitution ax + by = v. The proof is immediate, for, using this substitution together with the value of dy obtained therefrom, equation (6) becomes

(dV

1 b dx - a

) = f(v) ,

or dv

bf(v)

+a

= dx '

in which the variables are separated. 21. The equivalence of solutions. Suppose that we have obtained by different methods two general solutions of the same differential equation and we wish to show that they are equivalent. One way is to show that by changing the C of one solution into the proper function of the C of the other, the solutions become identical. For instance, if the two solutions are

x

+ Cxy + y

= 0,

C'(x

+ y)

- xy

= 0,

we show that they are equivalent by letting C = -l/C', whereupon the first reduces to the second. On the other hand, if the functional relation between the C's is not obvious, we may show that the solutions are equivalent without finding this relation. Let u(x, y)

= C,

vex, y) = C', -

(1)

I

be two general solutions of the same differential equation; then they must give the same value for dy/ da, The values of dy/dx obtained by differentiating these implicit functions (1) are respectively dy dx

-=

iJu iJx iJu ' iJy

--

dy -= dx

iJv iJx iJv' iJy

--

69

Article 21

and the condition that these two values of dy/ ax are identical is

au av avau_ o ax . ay - ax · ay = , or

au ax av ax

au

-ay

J=

av

== o.

(2)

-ay

The determinant J is called the Jacobian of the functions u, v with respect to x, y. The two solutions (1) may be shown to be equivalent by showing that their Jacobian vanishes identically. This means that there is a functional relation * between C and C', but it does not tell what the relation is; the search for the relation is sometimes an interesting exercise. 1. Show that the two following solutions of differential equation (2), Art. 10, are equivalent: EXAMPLE

arc tan x

+ arc tan y =

+Y

x

C,

1- xy

= C'•

(3)

Here we can see the functional relation between C and C', namely, C = arc tan C'; for, if we substitute C = arc tan C' in the first equation and take the tangent of both sides, we get the second equation. But, if this relation were not apparent, we could, without finding it, show the solutions equivalent as follows: Let u = arc tan x

+ arc tan y,

-

V -

x+y • , 1- xy

then, forming the Jacobian (2), we have 1

J=

1

= (1 -

1 xy)2

- (1 -

1 xy)2

=0•

* For a proof that the vanishing of the Jacobian is a necessary and sufficient condition for functional dependence, see Goursat-Hedrick, Mathematical Analysis, Vol. I, p. 52.

70

Chapter 3

Hence the solutions (3) are equivalent; they are general solutions of the same differential equation. The vanishing of the Jacobian of u, v does not prove that the solutions u = 0, v = 0' are correct, but it shows that if one is correct the other is also; if both solutions were incorrect and the Jacobian vanished, they would still be equivalent solutions of some differential equation.

We now consider an example in which the functional relation between the C's is not so obvious. EXAMPLE

2. Solve the differential equation (x 2 + y2) dx + x2 dy = 0

(4)

by substituting y = vx, then by substituting x = vy, and prove the solutions equivalent. Also find the functional relation between the

0'8. First 8olution. Letting y = vx in (4), we have (1 + if) dx

-; +

dv 1 + v + if = 0,

2

2v+ 1

dx

log x

+ v dx + x dv = 0,

+ va arc tan

log x +

va

2 2y+ x va arc tan v'3x

= 0, = C.

(5)

= v:y in (4), we have (r + 1) (v dy + y dv) + if dy = 0,

Second solution.

Letting x

dy

+

y dy y logy

(v2 + 1) dv = v(v2 + V + 1)

° ,

+ (1 _ 2 +1 )dv = v+l v

v

0'

2 2v+ 1 va arc tan va = 0 ', 2 2x + Y va arc tan yay C',

+ log v -

log x -

=

I

(6)

71

Article 21

We have to prove that the two solutions (5) and (~) are equivalent. LettingJ and g stand for the left members of (5) and (6) respectively, we shall prove that these two solutions are equivalent by showing that the Jacobian of J, g with respect to a, y is identically zero. We find iJx = x(x2 iJJ iJy

+ y2 + xy + y2)

x2

iJJ

iJg = iJx' iJg -_.

x x2+xy+y2

- iJy'

hence aJ

aJ

ax

ay

ag

ag

=0,

J=

-ax

ay

since the two rows of the determinant are identical. To find the functional relation between C and C', subtract equation (6) from equation (5): arc tan

2y+x

20

2x+y

0x + arc tan 0y

=

,

(C - 0 ).

Taking the tangent of both sides of (7), we find

0(2r + xy + 2x2 + xy) 3xy -

(2y

= tan

+ x) (2x + y)

0

(0 _ C')

2

'

or

hence

o

2

_

/is

(C - C') = arc tan (-v 3)

and the required functional relation is

o = 0' _

2'Jr

30

•

=-

'Jr

3'

(7)

Chapter 3

72

r

PROBLEMS Solve the following differential equations. dy 1. dx = eZ y.

+

2. x dy dx

+ 3y = x 2 + 1.

3. dy = xy - 6. dx 2x 2 4. 2x 3 dy 3(x 2y 1) dx = o. 6. (1 x 2 ) dy = (x 3 - xy x) th.

+

+

+

+

+

6. dy = xy 1. dx 1 - x2 7. x dy - 2y dx = 2y4 dy.

8. x ( : +

y) = x - y.

9. 2x dy = (y + 2x2log x) th. dy y 10. x =-. dx x dy x 11. y =-. dx Y

+ +

12. 3x dy dx

+ y + x 2y4 =

O.

13. (x - 1) dy = [y + (x - 2)e:l:] dx.

+

14. dy = 6x y - 12. dx 6x - y - 12

16.

dy

d; -

(x

+ y) =

1.

16. dy _ 1 = 1. dx x+y 17. dy dx

=

18. x 3 dy

y 19. x

=

+

x y - 1. x-y-l

+ 3x2 ~ = 2 cos Y dYe dy (2 - x2y2)_. dx

Article 21

+ y + 1 = o. dx + 3y - 1 21. x 2 dy + y = x 2 + 2xy.

20. 2 dy

+

73 x 2x

dx

22. cos

X: -

y sin x

+ y2 =

O.

23. (2x + y2)y dy = dx. 24. (x + 2y) d:x + (y - 1) dy =

o.

Solve each of the following differential equations by two different methods. 26. dy =

Y

x

dx

+ 2y

26. x dy = y(xy - 1) d», 3

27. dy = 4x y • d:x x4 + y2 28. x dy

ydx 29. dx dy

+1

+~ =

30. Y dx

xdy

x

'3 y2. :.

y

= xy3 + y2 - 1.

31. Find the solution of (4 - x 2 + y2) dx

+ 4y dy =

0

for which y = 1 when x = 2. 32. If

: + 2x2 + 3 =

2y,

and y = 2 when x = 0, find the value of y when x = 1. 33. Find the curve, passing through the point (n/6, - i), whose slope at any point is given by the function (y + sin3 x) / (sin x cos x). 34. Obtain a solution of the differential equation dy

- - y cot x dx

. 2 = sin x

such that y will vanish when x = 1r /2. Find the maximum and mi~imum values of y, and sketch the curve representing the solution from x = 0 to x = 21r.

74

Chapter 3

35. Solve

x2 : :

2xy = 3y n,

-

where n is a constant. For what values of n is the solution valid? Find the solution for any exceptional values of n. 36. Solve dy - = cos (5x - 3y). dx 37. Solve dy dx

=

(x

+ 1)2 + (4y + 1)2 + &y + 1.

38. Solve x dy

dx

+ y log ~ = x

y(l

+ x sin z)

by using each of the following substitutions: (a) Let y/x = v, obtaining an integrable combination in e and e. (b) Let log (y/x) = v, obtaining an equation linear in v and dv/dz. 39. Show that the differential equation x dy

= J(xy)

ydx

is reducible to an equation separable in v and x by the substitution xy = v, hence solve (a) y(1

+ x2y2) dx =

x dy;

40. Show that the differential equation dy

dx

+ ~y + as) btx + btU + bs

= f(a1x

is reducible to a homogeneous equation by the substitution x = X y

= Y + k, if hand k are chosen so that

a1h + a2k +

Solve ()

~=

a dx

(Z+4 )2. 4x-4' y

(b) dy = dx

as

=

0,

(x + + 1)3. 11 x +y

+ h,

Article 22

75

41. The general form of Riccati'8 equation is dy dx = P

+ Qy + Ry2,

where P, Q, and R are functions of x. Solve the following special cases: (a) P = x3 , Q = 2/x, R = liz. Let y = x2v. (b) P = sec2x, Q = tan x, R = -1. Let y = tan x (I/v).

+

42. Show that for the differential equation

dy

dx

+: = I y

(a) the substitution y = vx leads to the general solution

V3

-log (x2 - xy 2 (b) the substitution x

va

+ y2) + arctan 2y-x va3x = el;

= vy leads to the general solution

-log(x2 2

-

xy + y2) - arc tan

2x-y

V

3y

= C2.

Show that the two above solutions are equivalent and find the relation connecting eland O2 • 43. Solve the differential equation dy dx = cosh (!x (a) by means of the substitution je

+ y)

+ y = v, obtaining the solution

tanh(~ +~) = vatan(~ x + C} (b) by means of the substitution !x

+ y = log v, obtaining the solution

26~/21+. + 1 = v'3tan(~ x + C2} Show that the two above solutions are equivalent and find the relation connecting 01 and O2•

22. Chemical solutions. We shall consider now some physical problems leading to differential equations of the first order which are not senarable. ..

76

Chapter 3

1. A tank contains 100 gal of brine in which 50 lb of salt are dissolved. Suppose that brine containing 2lb/gal of salt runs into the tank at the rate of 3 gal/min and that the mixture, kept uniform by stirring, runs out of the tank at the rate of 2 gal/min. Find the amount of salt in the tank at the end of 30 min. Let z be the number of pounds of salt in the tank at the end of t minutes, then dxldt (lb/min) will be the rate at which the amount of salt in the tank is increasing. Now the rate of increase equals the tate of inflow minus the rate of outflow. The rate of inflow of salt is 2(lb/gal) .3(gal/min) = 6(lb/min). EXAMPLE

Since the brine runs into the tank 1 gal/min faster than it runs out, the amount of brine in the tank at the end of t min is 100 + t gal, and the concentration of salt present at that time is xl (100 + t) lb/gal, Hence the rate of outflow of salt is x l00 + t (lb/gal) ·2(gal/min)

=

1~~ t (lb/min).

Therefore the differential equation is dx 2x dt = 6 - 100 + t

When the last term is transposed to the left side, the difierential equation takes the standard linear form dx dt

2

(1)

+ 100 + t . x = 6,

*

which we solve by the method of Art. 19:

R

= ~2 dt/(lOO+t) =

(100

+ t)2X =

e2 log

f

(lOO+t)

6(100

= (100

+ t)2 dt + C,

(100 + t)2X = 2(100 + t)3 +

* The substitution x

+ t)2,

c.

(2)

= c(l00 + t), where c is the concentration of salt present

at time t, would reduce equation (1) to an equation separable in the variables c and t: (See Probe 1, Art. 24.)

Article 22

77

The constant of integration is determined from the condition that x = 50 when t = 0, so that C = 1002.50 - 2.1003 = -1002.150.

Hence (100

+ t)2X = 2(100 + t)3 -

1002 ·150,

or x = 2(100

+ t)

150 - (

t

)2'

(3)

1 + 100

which gives the amount of salt in the tank at any time t; for t min we have 150 X]t=30 = 260 - 1.69 = 171 lb.

=

30 (4)

Instead of adding and evaluating a constant of integration, we could have integrated equation (1) between limits, writing, instead of equation (2), t=30

(100 + t)2X];:~

= 2(100 + t)3]:O.

a:=50

Then 1302x - 1002.50 = 2(13OS - 1003 ) , =

x

3

2.130

2

150.100 = 260 - 150 = 171lb 1302 1.69' -

Variations of Example 1. In certain cases, problems of this type

lead to separable differential equations. For instance, suppose that in the example just solved the brine is running out at the same rate that it is running in, namely, 3 gal/min. The amount of brine in the tank remains constant, 100 gal, and the differential equation is dx

=6_

dt

3x .

100

Separation of variables gives dx

3dt

200 - x = 100'

78

Chapter 3

and the solution proceeds as follows:

-3t]30

z

log (200 - X)]50 = 100 0 ' log 200 - x x

200 - x 150 = -0.9,

= 150e-o.9 = 150 X 0.4066, = 200 - 61 = 1391b,

the amount of salt in the tank at the end of 30 min. Again, suppose that the conditions of the original problem are the same except that pure water is running into the tank. There is now no inflow of salt, and the differential equation is

dx dt

-= -

2x 100

+t

Separating the variables, and solving, we have

dx

2dt

-:= -

+ t' log x]:O = - 2 log (100 + t)]:o, 100

X

x

log 50

=

130

-2 log 100 x

=

=

50 1.69

-2 log 1.3

=

1

= log 1.69·

29.6 lb,

which in this case is the amount of salt in the tank at the end of 30 min.

is. Electric circuits.

In the subsequent differential equations for electric circuits the following symbols will be used: R (ohms) = resistance; L (henries) = inductance; C (farads) = capacitance; e (volts) = electromotive force (emf), g (coulombs) = quantity of electricity, or charge on a condenser, and i (amperes) = current, at time t (sec). The capital letters are constants; the small letters are, in general, variables.

Article 23

79

Under certain conditions the circuit equation is a linear differential equation of first order, a special case of which reduces to a separable equation. For a simple circuit containing a resistance and an inductance in series with a source of emf *

L di dt

+ R·'t = e.

(1)

In general, when e is a function of t, the equation is linear, but when e is a constant the equation is separable. For a simple circuit containing a resistance and a capacitance in series with an emf (2) or upon differentiating, since i = dq/dt,

R di dt

+ .!. i (;

= de. dt

(3)

In general, when the right member of (2) or (3) is a function of t, the equation is linear, but it becomes separable if the right member is constant. If R, (;, and e are given, q may be determined from (2), then i can be obtained from the relation, i = dq/dt; or i may be determined directly from equation (3). The general solutions of equations (1), (2), and (3) can be written down in a form involving an integral, but, unless e is constant, the integrations cannot be carried out until the form of e as a function of t is given. For example, to obtain the general solution of (1), where e = jet), we have di R. 1 (4) t). dt + L 't = L i C

* The differential equation for a circuit containing resistance, inductance, and capacitance will be found in Art. 37: For a derivation of these differential equations, see Bedell and Crehore, Alternati~ Currents, or Reddick and Miller Advanced Mathematics!(f/' Engineers, Arts. 7 and 14.

80

Chapter 3

The integrating factor

* is =

E(R/Llt. i

Ef(R/L) dt

lf

=

E(R/L)t;

iR/Llc'!(t)

hence

dt + K,

where K is a constant of integration, and the general solution is i

=

l

E-(R/Llt

f

E(R/Ll'!(t)

dt

+ KE-(R/Llt.

(5)

The integration cannot be carried out until the form of jet) is specified. Suppose that the emf is a simple harmonic function of the time, e = j( t) = E sin wt. Here E is the amplitude or maximum value of the emf, and w (rad/sec) is the angular velocity, or 211" times the frequency in cycles per second, so that when t is given in seconds, the angle wt is in radians. Equation (5) now becomes (Peirce, 414)

i=

l

E-(R/Llt

[(~):+

IJJ2 E(R/L)t

(~ sin IJJt -

IJJ

cos IJJt)]

+K E-(R/L)t, or i

= R 2 +EL 2IJJ2 (R sin IJJt -

1M cos

IJJt)

+ KE-(R/Llt.

(6)

The value of K is determined when a corresponding pair of values (i, t) is known; thus if i = 0 when t ,= 0, we have K = ELw/(R2 + L2(2), and equation (6) reduces to

.

t

= R 2 +EL2w2 ('R·SIn wt - L w COS wt) + R2 ELw + L2W2

-(R/L)t ( ) E '. 7

Hence, for a simple circuit containing a resistance R and an inductance L, in series with an impressed emf of the form e = E sin wt, equation (7) gives the current i at any time t after the introduction of the emf under the condition that i = 0 when t = o. This condition means that the circuit was * We write now E instead of e for the constant 2.718· .. to distinguish it from the electromotive force e.

81

Article 23

idle, i.e., no current was flowing, just before the emf was introduced, so that at the time when the emf is applied the current starts at zero. There can be no sudden jump in the current when the emf is introduced, for that would make dildt infinite and equation (1) would no longer be valid. The last term of (7) is called the transient term, since it dies out with increasing t; theoretically it becomes zero only when t is infinite, but practically it is negligible for t fairly small. The remaining term on the right is called the steady-state term, since it gives the value of the current when a steady state has been reached, i.e., after the transient has died out. We now use the method of Example 3, Art. 2, to reduce the equation of the steady-state current to another form:

i = R2 : L2W2 (R sin wt - L» cos wi)

=

VR2

E

+ L2w2

[

VR2

R

+ L2CJJ2

sin CJJt _

VR2

or .=

't

V R2 E+ L2

w2

. ( wt - t an -

SIn

Lw

+ L2w2

lLw) R .

cos wt]

'

(8)

Equation (8) shows that the maximum value, or amplitude, of the steady-state current is E IVR2 + L2W2, which is less than EIR, thevalue it would have if there were no inductance. The expression V R2 + L2W2 is called the impedance, and represents apparent resistance. The steady-state current is alternately positive and negative and oscillates between +EI v'R 2 + L2W2 and - E Iv'R2 + L2W2; it is a simple harmonic function of the time, having the same frequency as the impressed emf. Consider now the case where e = E, a constant. Equation (1) reduces to di E - Ri , dt = L

82

C1tapter 3

a separable equation whose solution proceeds as follows: di 1

E - Ri

L dt,

=

=-

log (E - Ri) E - Ri =

R

Lt

+ log K,

KE-(R/L)t,

1 (E - K E- (R/L)t) ,

"

(9)

~=R

which is the general solution of equation (1) for a constant emf. The condition i = 0 when t = 0 would give K = E and reduce the equation to the form

i = ~ (1 -

(10)

E-(R/L)l).

As t increases, the exponential term becomes negligible and the current takes on its Ohm's law value, E/ R. 1. A simple electric circuit contains a resistance of 10 ohms and an inductance of 4 henries in series with an impressed emf of 100 sin 200t volts. If the current i = 0 when t = 0, find (a) the current when t = 0.01 sec; (b) the ratio of the steady-state current when t = 4 sec, to its maximum value. (a) We could use equation (7) as a formula and substitute the given values of the letters, but instead let us start with the diiierential equation and integrate between limits: ExAMPLE

4 di dt

+ 10i = 100 sin 200t)

: + 2.5i = t.= ~.Ol ; .5ti]'='

t=O

E°J)2f,i

1°.

= 25

i-O

25 sin 2OOt, 01

°

;

.5t

sin 200t dt,.

= 2.52 ~ 2002 ;.5'(2.5 sin 200t EO.025i =

1 8

E2.5t

200 COS 2OOt)]~·Ol,

(.!..~ sin 200t _ cos

200t)]0.01

°

Article 24

83

(where 1/8 is written instead of 1/8.00125),

(-!.

i = 1 sin 2 - cos 2 + 8 80 .

~ =

i

E-0 0025)

1(0.9093 ) 8 80 + 0.4161 + 0..9753

' ,

= ~ (1.4~) = 0.175 amp.

(b) From equation (8) the ratio of the steady-state current, when t = 4 sec, to its maximum value is sin (800 - tan-I 80). Expressing the angle in degrees, we have

= 89.28°, 57.29578° = 45,836.62°,

tan-1 80 800 rad

= 800 X

800 - tan-I 80 rad

= 45,747.34°.

Reducing this angle by 127 X 360° = 45,720°, we obtain for the required ratio sin 27.34° = 0.459. Without using degrees, the computation could have been made as follows: tan-l 80 = 1.5583, 800 - tan-ISO

= 798.4417.

Reducing this angle by 25471" = 797.9645, we obtain for the required ratio sin (0.4772) = 0.459.

24. Rope wound on a cylinder. A rope is wound on a rough circular cylinder whose axis is horizontal. Let p. represent the coefficient of friction, a (ft) the radius of the cylinder, T (lb) the tension in the rope at any point P, p (lbjft) the linear density of the rope, and (J the angle POA. (See Fig. 3.) Let Q be a point on the rope somewhat to the left of P, ~(J the angle POQ, ~8 the length of arc PQ, and M the midpoint of PQ. Suppose that the rope is on the point of slipping from

84

Chapter 3

right to left; then the tension at Q will be T + aT, a quantity larger by aT than the tension at P. Let us resolve along the tangent line at M the forces acting on the small portion of the rope, PQ. The tension T + aT, N acting tangentially to the cylinder at Q, yields toward the left along the tangent line a force (T + aT) cos (afJ/~). We equate this to the ............... sum of all forces toward the right T.... along the tangent line. Notice that BI----~-+--1 the frictional force along the tangent line is toward the right, since it opposes the motion, and that it is equal to p, times the resultant along FIG. 3 the normal MO of the tension T, the tension T + aT, and the weight pas of PQ. The resulting equation is (T

fJ) + aT) cos 2afJ = T cos 2afJ + pas cos (a fJ + 2 + II- [

T sin

~IJ + (T + 11T) sin ~IJ + pl1s sin (IJ + ~IJ)

l

(1)

°

.

If we let afJ ~ 0, and hence as ~ and aT ~ 0, in order to obtain the tension T at P, we obtain T = T; but if, after cancelling the equal terms, T cos (afJ /2) we divide equation (1) through by AfJ and then take limits as afJ ~ 0, we obtain the differential equation which determines T as a function of fJ: AT afJ as ( afJ cos 2 = p afJ cos fJ

+ 2AfJ) + • AfJ

II.

r:

2T

+2 AT

SIn-

2

afJ 2

+ p afJ as. (fJ + afJ) SIn 2'

Article 24

85

.

or, smce

.

~8

and

-=a ~()

~()

sIn· 2 L1m A9-+0

~()

=1

J

2

:~ = pa cos 0 + JL[T + pa sin 0]. The linear differential equation

:~ -

JLT

=

pa(cos 0

+ JL sin 0)

(2)

therefore states the law connecting the tension T at any point P with the angle corresponding to this point. The general solution of equation (2) may be found by the method of Art. 19; it is T

=1:

JL2 [(1 - JL2) sin 0 - 2JL cos 0]

+ Ce",

(3)

where C is an arbitrary constant. (See Problem 14 at the end of this artiele.) Two special cases occur: when T is so large compared to p that p can be taken equal to zero, and when friction is negligible so that p can be taken equal to zero. These cases are represented respectively by the following differential equations and general solutions: dT d() dT

d()

= pT,

T

=

= pa cos (),

T =

Ce"; pa

sin ()

(4)

+ C.

(5)

1. On a rough circular cylinder with horizontal axis and a diameter of 8 ft and in a plane perpendicular to the axis lies a chain with one end at the level of the axis of the cylinder. If the coefficient of friction between the chain and cylinder is !, how far below the axis must the other end of the chain hang down so that the chain is on the point of slipping? EXAMPLE

86

Chapter 3

Referring to Fig. 3, suppose that the right end of the chain is at A and that the left end hangs down a distance L below B. Equation (2) is the differential equation for this problem. Assuming that the general sol ution (3) has been obtained, the constant C is determined from the condition, T = 0 when fJ = 0, as follows: l

o=

4p

9[0 -:J + 0,

1 + 25 C Also T

=

pL when fJ

=

lOOp . 6 34 5

= 60P .. 17

= 1r, so that we have from

_ lOOp [16. _ ~ ] pL - 34 25 sm 1r 5 eos e L =

~ (1 + e3r / 5 )

=

(3):

311'/6 + 60p 17 e ,

~ (1 + 6.586)

= 26.8 ft.

The same result is obtained by substituting II = g, a = 4, in the formula found for L in Problem 15 of the following group. EXAMPLE 2. A rope weighing ~ lb/ft is wrapped once around a smooth horizontal cylinder 6 ft in diameter. If the rope is cut at its lowest point so that the ends hang down, find the maximum tension in the rope. Assuming that the cylinder is so smooth that friction is negligible, equation (5) is applicable:

T

=

~·3 · sin 8

+ 0.

The length of rope hanging down on each side is equal to a quarter of a circumference or i ·61r ft and its weight is (611"/4) ~ = :1r lb, so that C is determined by the condition, T = : 1r when fJ = 0 or 11"; hence C = :1r, and 3. 3 T = 2 SIn 8 4 11".

+

The maximum tension occurs at the top where 8 = 1r/2; hence

3 T max = 2

3

+4

11"

=

3.86 lb.

87

Article 24

PROBLEMS 1. In Ex. 1, Art. 22, what is the concentration of salt in the tank at the end of 30 min [obtained from equation (4)]? Obtain the same result by solving the separable equation mentioned in the footnote to equation (1). 2. A tank contains 100 gal of fresh water. Brine containing lib/gal of salt runs into the tank at the rate of 2 gal/min, and the mixture, kept uniform by stirring, runs out at the rate of 1 gal/min. Find (a) the amount of salt present when the tank contains 125 gal of brine; (b) the concentration of salt in the tank at the end of 1 hr. 3. A tank contains 100 gal of brine holding 25 lb of salt in solution. Three gallons of brine, each containing 1 lb of dissolved salt, run into the tank per minute, and the mixture, kept uniform by stirring, runs out of the tank at the rate of 1 gal/min. Find (a) the amount of salt in the tank at the end of 40 min; (b) the time when the concentration of salt in the tank is i lb/gal. 4. (a) Brine, containing 1 lb of salt per gallon, runs into a 2Oo-gal tank initially full of brine containing 3lb of salt per gallon, at the rate of 4 gal/min. If the mixture runs out at the same rate, when will the concentration of salt in the tank reach 1.01Ib/gal? (b) Work Probe 4(a) if the mixture runs out at the rate of 5 gal/min. 5. A tank contains 100 gal of saturated brine (3 lb/gal of salt). Brine containing 2lb/gal of salt runs into the tank at the rate of 5 gal/min and the mixture runs out at the rate of 4 gal/min. Find the minimum content of salt in the tank and the time required to reach the minimum. 6. A tank contains V gal of fresh water. Brine containing c lb/gal of salt runs into' the tank, and the mixture runs out, the ratio of the rate of inflow to the rate of outflow being r( <1). Find the maximum amount of salt in the tank and the corresponding volume of brine in the tank. 7. A resistance of 3 ohms and an inductance of 1 henry are connected in series with an emf of 10 sin lOt volts. If the current is zero when t = 0, find (a) the current when t = 0.1 sec; (b) the current when t = 3 sec; (c) the ratio of the numerical value of the steady-state current, when t = 3 sec, to its maximum value. S. A resistance of 20 ohms and an inductance of 2 henries are connected in series with an emf of e volts. If the current is zero when t = 0, find the current at the end of 0.01 sec if (a) e = 100;

(b) e

= 100 sin 150t.

9. Find the general solution of equation (2), Art. 23, if e = /(t). From this result obtain the general solution when (a) e = E, a constant; (b) e ::w: E sin ",t.

88

Chapter 3

10. Obtain the general solution of equation (3), Art. 23, for an emf of E sin wt volts, (a) by substituting e = E sin ClJt in equation (3) and integrating; (b) by differentiating the result of Prob. 9(b). 11. An electromotive force is introduced into a circuit containing in series a resistance of 10 ohms and an uncharged condenser of capacity 5 X 10-4 4il.rad. Find the current and the charge on the condenser when t = 0.01 sec, Ia) if e = 100 volts; (b) if e = 100 sin 120rt volts. 12. An emf of 100 volts is introduced into a circuit containing in series a resistance of 10 ohms and an uncharged condenser of capacity 5 X 10-4 farad. When a steady state has been reached, the emf is removed from the circuit. Find the current and the charge on the condenser 0.01 sec after the removal of the emf. 13. An emf of 100 sin 1201rt volts is introduced into a circuit containing in series a. resistance of 100 ohms and a condenser of capacity 5 X 10-4 farad. There is already a charge on the condenser at the time t = 0, when the emf is introduced,. such that the current at that time is 1 amp (positive). Find the current 0.1 sec later. 14. Obtain the general solution (3) of the differential equation (2), Art. 24. 16. In the problem leading to differential equation (2), Art. 24, assume that the right end of the rope is at the level of the axis of the cylinder and that the other end hangs down a distance L below the axis; find L, in terms of p, and a, when the rope is on the point of slipping. 16. In the problem leading to differential equation (2), Art. 24, assume that the right end of the rope is at the top of the cylinder and that the other end is at the level of the axis. (a) Find the equation which determines the coefficient of friction if the rope is on the point of slipping. (b) Solve the equation of part (a) by the method of trial and error, obtaining the value of p, to three figures. 17. A rough circular cylinder with horizontal axis has a piece of flexible cable lying across the uppermost quarter of its circular cross section. Find the force which, applied tangentially at one end of the cable, would just cause it to slip. Radius of cylinder = 2 ft. Weight of cable = lIb/ft. Coefficient of friction = i. 18. A flexible cable weighing Ilb/ft hangs a distance of L ft below the horizontal axis of a drum of radius 6 in. and reaches three-quarters of the distance around the-upper half of a circular cross section. Assuming that the cable is on the point of slipping and that the coefficient of friction is i, find L. 19. A piece of cable weighing 1.5 lb/ft rests along the circumference of a smooth cylinder of radius 4 ft. If the maximum tension in the cable is 3 lb, what is the length of the cable? 20. A ship, pulling with a force of 10,500 lb, is held by 2.5 turns of rope wound around a post. If the coefficient of friction is 0.4, with what force

Article 25

89

must a man pull at the other end of the rope in order to hold the ship? How many turns of rope would be necessary if the man pulls with a force of 40 lb? 21. On a smooth cylinder rests a 4o-lb weight attached to a chain which passes over the cylinder and hangs down 2 ft below the horizontal diameter of the cylinder. A 2o-lb weight is attached to the lower end of the chain. The diameter of the cylinder is 12 ft and the weight of the chain is 5Ib/ft. Find the position in which the 40-lb weight rests in equilibrium. 22. A flexible cable weighing p lb/ft hangs a distance of 6 in. below the horizontal axis of a drum 5 ft in diameter and reaches around the upper half of a circular cross section for a distance equal to one-third the circumference of the drum. Assuming that the cable is on the point of slipping, obtain a formula for finding the coefficient of friction, IJ., and show that p, = 0.44 satisfies it closely. 23. Assuming p, = 0.7324 in Prob. 16, find the angle BOP, where B is the lower end of the rope, 0 is the center of the cylinder, and P is the point on the rope where the tension is a maximum. 24. A trough is 2 ft long and of width 1/(10 - e) ft at a distance of x ft from its flat base. Ether flows into the trough, originally empty, at the rate of 10/(10 + t) ft3fhr, where t (hr) is the time after the flow starts. If the rate of evaporation is proportional to the area A (ft 2) of the liquid surface, the constant of proportionality being k = -to ft/hr, find the depth of the liquid in the trough after 2 hr. How much greater would the depth have been if none of the liquid had evaporated? 25. Tests, taken after an overhaul of a plastic molding machine producing 100 articles Per second, showed approximately the following rate of production of defective articles: (1) 20 per min due to uncontrollable causes, (2) 1 per hr increase in defectives Per 104 additional articles, due to steady wear of machine parts, (3)2 Per hr increase in defectives per 103 additional defective articles. Find the total number of defectives produced during 10 hr after an overhaul.

25. Curves determined from geometric properties.

It

often happens that a family of curves may be characterized by a geometric property stated in terms of the coordinates of a point on one of the curves and the first derivative of one of the coordinates with respect to the other. Such a statement will be a differential equation of first order whose general solution represents the family of curves, each of which possesses the given property. Then, if an additional condition is given, such as she requirement that the curve shall pass through a given point, this condition may be used to determine the arbitrary

90

Chapter 3

constant in the general solution. Using this value for the arbitrary constant, there is obtained the particular solution which represents a curve possessing the given geometric property and satisfying the required condition. The geometric property may be stated either in rectangular coordinates or in polar coordinates. If x, yare the rectangular coordinates of a point P on a curve (x = abscissa, Y = ordinate), dy/dx represents the slope of the curve at that point, that is, the tangent of the angle T

0"-----1--'-----FIG. 4

FIG.

5

measured counterclockwise from the positive direction of the z-axis around to the tangent line at P. Denoting this angle by O! (Fig. 4), dy (1) ax = tan O!. If p, (J are the polar coordinates of a point P on a curve (p = radius vector, (J = vectorial angle), dp/ d(J does not represent an important geometric concept such as that expressed by equation (1) in rectangular coordinates. The combination p d(J/ dp, however, is shown in calculus to represent the tangent of the angle QPT (Fig. 5), measured counterclockwise from the radius vector OP (extended) around to the tangent line at P. Denoting this angle by 1/1, d(J p dp

= tan 1/1.

(2)

Article 25

91

By means of relations (1) and (2) we can set up differential . equations in rectangular and polar coordinates which define systems of curves having certain geometric properties; sometimes we shall bring in other formulas from calculus, such as those for area under a curve, distance along a curve, area of a surface of revolution, etc. We shall make a distinction between the terms "length" and "distance." The length of a line segment or arc will mean its actual length, a positive' number of linear units, irrespective of the direction in which it is measured, whereas distance, if positive in one direction, is negative in the opposite direction. When dealing with lengths, the terms "tangent" and "normal" will mean the length of the segment of tangent line and normal line respectively, between a point P on a curve and the z-axis, 1. At every point of a curve the projection of the normal on the z-axis has the length k. Find the family of curves possessing this property. The length of DB in Fig. 4 is y tan a = y (dyldx) if a is an acute angle. When a is obtuse, dyldx stands for a negative number,'and the length of DB is -y (dyldx). The differential equation for the required family of curves is therefore EXAMPLE

dy ±ydx = k,

Integration gives

y2 = ±2kx + 0,

•

a family of parabolas each of which possesses the required property. The axis of x is the axis of all the parabolas; they open toward the right or left according as the + or - sign is used, and shift to right or left with varying C. 2. Find the family of curves having the property that the tangent is of constant length k, and determine the particular curve of the family which passes through the point (0, k). The length of AD in Fig. 4 is ± y cot a = ±y (dx/dy), the + or sign being used according as a is acute or obtuse. Then, the EXAMPLE

92

Chapter 3

Vy2 + A[)2, the differential equa-

length of the tangent AP being tion of the family of curves is

(:r

~y2 + 11 Hence

11 [ 1 +

(:YJ2

dX)2= ( dy

k y2

= k.

(3)

= Tc2,

_1

,

Vk2 _ y2 dx = ±

y

dYe

(4)

Integration (Peirce, 130) gives

±x + C = v'k2 - y2 - k cosh-1 k. Y

(5)

If 'U = k when x = 0, then C = 0, and the particular curve passing through (0, k) is

±x = Vk 2

-

y2 -

k eosh'"! k. y

(6)

The curve (6) is a tractrix (Fig. 6), the + or - sign on the left being used respectively for the left and right branches of the curve; these y A

k

o

Q

FIG. 6 signs denote respectively positive and. negative values for the slope dy/dx, as shown by equation (4). Equation (5) includes all the curves obtained by shifting the tractrix of Fig. 6 to the right and left; furthermore, it includes the reflection of these curves in the z-axis, since equation (5) is unchanged when y is changed to -yo The complete family of tractrices repre-

Article 25

93

sented by equation (5) is symmetrical to the x-axis, and each curve of the family has all its tangents of length k. EXAMPLE 3. Find the curves for which the length of any arc equals the area under the arc and above the x-axis. . The statement that length equals ,area means that the number of inches of arc equals the number of square inches of area. If the integral representing the arc length from any fixed abscissa a to a variable abscissa x is equated to the integral representing the corresponding area under this arc and above the x-axis, we have

i

Z

~1+

(:r

dx = [Ydx.

(y

> 0.)

(7)

Differentiating with respect to x,

~l+(:r =Y. or

dy ( dx

)2= y

2

-

(8)

1,

whence dy

±v'2

y - 1

Integrating,

+ C (y ¢ cosh (x + C).

± cosh-1 Y = or y =

X

(9)

=dx.

1),

(10) (11)

y Equation (11) represents a family of eatenariee obtained by shifting the catenary, y = cosh x, to the right and left (Fig. 7). Each catenary has the property 1 that the length of any arc equals the area between the arc and the x-axis. Direct substitution of (11) into (7) verifies that -----:!~---% o the curves (11) satisfy the required conFIG. 7 dition. It should be noticed that equation (10) does not follow from (9) if y = 1, so that this value of y should be tested to see if it satisfies equation (7). It is seen that

Chapter 3

94

y = 1 does satisfy (7) and hence also possesses the required property; this line is the envelope of the curves (11). The complete solution of the problem is therefore the family of curves (11) together wiUl their envelope, y = 1. 4. At every point P of a curve the radius vector and the tangent at P intersect at the same angle, cot-1 k. Find the curves having this property. Referring to Fig. 5 we see that t/; = cot-1 k or 180 0 - cot-1 k, so that tan 1/1 = ± 11k; then from equation (2) the differential equation of the required family of curves is EXAMPLE

pdO

1

dp = ± k' or dp

-

Integration gives log

p

p

=

± kdfJ.

= ± kfJ + log C,

or p = Ce±k8,

two singly infinite families of equiangular spirals, so named because of the equal angles between radius vector and tangent at all points of the curves. PROBLEMS 1. At any point P of a curve, the projection of the normal on the z-axis and the abscissa of P are equal in length. Find the curve if it passes through the point (2, 3). 2. Find the curve through the point (0, 2) such that the projection of the tangent on the z-axis is always of length 2. 3. Find the curve having both the following properties: (1) Its ordinate equals the logarithm of its slope. (2) It crosses the line x = 1 at an angle of 60°. 4. Find the curve through the point (1, 2) whose normal at any point (except where x = 0) is bisected by the y-axis. 6. Find the family of curves having the following property: The perpendicular from the origin to the tangent line .and the abscissa of the point of tangency are of equal length. 6. Find the curve through the point (2, 1) such that the x-intercept of the tangent is twice the ordinate of the point of tangency. Draw the curve.

Article 25

95

7. Find the curves for which each normal and its x-intercept have the same length. 8. Find the curve through the point (4, 2) such that the line through each point of the curve and the origin bisects (a) the angle between the corresponding ordinate and tangent; (b) the angle between the corresponding ordinate and normal. 9. Work Prob. 8, using polar coordinates, if the curve passes through the point (2, 60°). 10. The normal at a point P of a curve meets the x-axis at Q. Find the equation of the curve if it passes through (0,3), and if the locus of the midpoint of PQ is y2 = x. 11. The tractrix (Fig. 6) may be thought of as the path-of a heavy particle P being dragged along a rough horizontal plane by a string PQ of length k. The end of the string, Q, is pulled along the z-axis, the initial positions of P and Q being at A and 0 respectively. If k = 10 in., how far have P and Q traveled from their initial positions when P is 5 in. from the x-axis? 12. If the length of the tangent to a tractrix is 2 ft, find the length of.the line joining the two points on the curve where the slope is ± 1. 13. (a) Show that the family of catenaries obtained in Ex. 3, Art. 25, may also be derived from the following property: If at any point on a curve (y > 0) an ordinate and a tangent are drawn, the perpendicular from the foot of the ordinate to the tangent is of unit length. (b) Obtain the equation of the family of curves characterized by the above property if the perpendicular is of length k, instead of unit length. State the corresponding characteristic property involving arc length and area under the arc. 14. Find the curve for which the length of any arc equals half the area under the arc and above the x-axis, if the curve has slope 1 when x = o. 16. An arc of a curve extends from a point P (a, b) to a point Q upward and to the right of P. If for all positions of Q the area under the arc and above the x-axis is k times the vertical distance of Q above P, find the equation of the curve. 16. An arc of a curve is drawn upward and to the right, starting at the point (a,O). The area between the arc and the x-axis is equal to the nth power (n > 1) of the ordinate which forms the right boundary of the area. Find the equation of the curve. Is the restriction n > 1 necessary? 17. On a curve which passes through the origin, take an arc from (0, 0) to any point (x, y) of the curve. The area between this arc and the x-axis, when rotated about the x-axis, produces a volume V~. The area between the same arc and the y-axis, when rotated about the y-axis, produces a volume V". Find the curves characterized by the property: V ~ = V". 18. A curve has the property that any arc of it, when rotated about the x-axis, produces an area equal to the difference of the ordinates at the end-

96

Chapter 3

points of the arc. Show that the curve is one of the branches of a tractrix whose tangent is of length 1/2r. 19. Find the curves for which any arc, rotated about the x-axis, produces a surface of revolution whose area is l/n the volume enclosed by it. 20. Find the curve characterized by the following property: An arc of the curve in the first quadrant, from (0, k) to any point (x, y), when rotated about the x-axis, generates a surface of revolution whose area is equal to the area of the cylinder formed by rotating the straight line from (0, y) to (x, y) about the x-axis. Draw the curve. 21. At any point P of a curve the normal bisects the angle between the ordinate and the radius vector (or the radius vector extended through P). Find the curves characterized by this property, (a) using rectangular coordinates, (b) using polar coordinates. 22. Find the family of curves for which the vectorial angle is n times the angle between radius vector and tangent. 23. Find the family of curves for which the radius vector through any point trisects the angle between the corresponding tangent and ordinate. 24. A reflector is to be constructed so that light emanating from a point source will be reflected in a beam of Parallel rays. Taking the origin as the point source and the x-axis as the direction of the reflected beam, the reflector . will be a surface of revolution generated by rotating a certain curve about the x-axis. Find the equation of the curve, (a) using rectangular coordinates, (b) using polar coordinates. Note the physical law that the angle of incidence (the angle between the incident ray and the normal to the surface) equals the angle of reflection (the angle between the reflected ray and the normal to the surface). 26. Find the equation of an axial section of the convex surface of a lens, whose concave surface is spherical, if a ray of light parallel to the axis of the lens is focused at the center of the sphere, (a) using rectangular coordinates, (b) using polar coordinates. Take the origin at the focus, and the x-axis along the axis of the lens. Employ the physical law, sin a/sin {j = k, where k is the constant index of refraction, a is the angle of incidence, and {j is the angle of refraction. (c) What is the equation of the axial section if the index of refraction is 1.5 and the distance along the axis of the lens, from the convex surface to the point where the light is focused, is 3 ft?

26. Orthogonal trajectories. Suppose that a singly infinite family of curves is given. Consider a second family composed of all the curves which intersect all the curves of the given family at right angles. When two families of curves are so related each family is called the orthogonal trajectories of the other.

97

Article 26

As an illustration, the equation x 2 + y2 = k represents a singly infinite family of concentric circles. Here k is a parameter, constant for any particular circle but varying from one circle to another. The orthogonal trajectories of this family of circles are represented by y = ex, the family of straight lines through the common center of the circles. The circles are also the orthogonal trajectories of the family of.straight lines. The relation is geometrically obvious in this case, but in general we need an analytical method for finding orthogonal trajectories. If we differentiate the equation of the given family of curves and eliminate the parameter, we obtain the differential equation of the given family-an equation which gives the slope dyjdx of anyone of the curves in terms of the coordinates x, y of a point on the curve. Now the slope of the orthogonal trajectory at the point (z, y) must be the negative reciprocal of the slope of the given curve in order for the condition of perpendicularity to hold. The differential equation of the orthogonal trajectories is therefore obtained by writing dyjdx equal to the negative reciprocal of the value it has in the differential equation of the given family. Integration of the differential equation so obtained then yields the equation of the orthogonal trajectories. Let us apply this method to the family of concentric circles used in the above illustration. 1. Find the orthogonal trajectories of the family of circles concentric about the origin. The equation of the given family of circles is EXAMPLE

x2 + y2 = k, Differentiation with respect to x gives 2x

(1)

dy + 2y= 0, dx

or dy x - = - -. dx y

(2)

In this case the parameter k was eliminated automatically by differentiation; equation (2) is the differential equation of the original family

98

Chapter 3

of circles (1). The differential equation of the orthogonal trajectories is therefore y dy - -- , dx x or dy dx -=-. (3) y x Integrating (3), we have log y = log x

+ log e,

or '

y = ex,

(4)

the equation of the orthogonal trajectories.

When the parameter does not disappear automatically by differentiation, it must be eliminated by combining the original equation with the one resulting from differentiation, in order to obtain the differential equation of the original family. 2. Find the orthogonal trajectories of the family of circles with centers on the z-axis and passing through the origin. The equation of the given family is EXAMPLE

x2

+ y2 = ka,

(5)

Differentiating, dy

2x

+ 2y dx = k.

(6)

The elimination of k between (5) and (6) gives 2x2

or

dy + 2xy= x 2 + y2, dx dy y2 _ x 2 2xy , dx =

(7)

the differential equation of the original family (5). The differential equation of the orthogonal trajectories is therefore dy 2xy dX=X 2 _ y 2

'

(8)

Article 26

99

which can be written in the following form involving an integrable combination, 2 2xy dx - x dy + d _ 0 Y-

2

Y

Integration then yields x

•

.' I' ,

2

- +Y = Y

C,

or (9)

the equation of the orthogonal trajectories, a family of circles also passing through the origin, but with centers on the y-axis.

If the equation of a singly infinite family of curves is given in terms of the polar coordinates p, 8, and a parameter k, we differentiate the equation and eliminate the parameter, thus obtaining the differential equation of the given family in polar coordinates. This equation enables us to express the value of p doldp, that is, by equation (2), Art. 25, the value of tan l/I, for anyone of the curves in terms of the coordinates p, 0 of a point' on the curve. Now the l/I of the orthogonal trajectory at the point (p, 8) differs by 90 0 from the l/I of the given curve, hence the tan l/I of the orthogonal trajectory equals the - cot 1/1 or -l/tan l/I of the given curve. The differential equation of the orthogonal trajectories is therefore obtained by writing p doldp equal to the negative reciprocal of the value it has in the differential equation of the given family. Integration of the differential equation so obtained then yields the equation of the orthogonal trajectories. EXAMPLE 3. Work Example 2, using polar coordinates. In polar coordinates the. equation of the given family of circles is p =

k cos o.

(10)

Differentiating with respect to 0, dp dO

•

= -k SIn O.

(11)

100

Chapter 3

The parameter k may be eliminated by dividing (10) by (11): pd8 = - cot 0, dp

-

(12)

the differential equation of the given family (10). To obtain the differential equation of the orthogonal trajectories, we write p dO/dp equal to the negative reciprocal of its value in (12): pdO -dp = tan 0,

or dp

-p = cot 0 dJJ.

(13)

Integration then yields log p = log sin 8 + log 0, or p =

C sin 0,

the equation of the orthogonal trajectories. EXAMPLE

4. Find the orthogonal trajectories of the family of

curves. p = sin 0

+ k,

The differential equation of the given family is dp dB = cos 8,

or p dO dp

p

-=--.

cos 0

Hence the differential equation of the orthogonal trajectories is p

dJJ

cos 0

-=---

dp

p

,

or dp - 2" = sec8d8. p

101

Article 26

Integration then yields 1

- = log (sec 8 + tan 8) + C, p

the equation of the orthogonal trajectories.. PROBLEMS

Find the orthogonal trajectories of the following families of curves. 1. The family of curves of Ex. 1, Art. 25. 2. The family of curves of Ex. 4, Art. 25. 3. (x - 1)2 + y2 + lex = o. .4. x 2 = y2 + ky3. 6. eZ cos y = k. 6. sinh y = k sec x. 7. y2 = 4k(x + k). 8. x 2 + 3y2 = kyo 9. y = x tan t (y k). 10. iJ = sin x sinh y + k, 11. y = log tan (x + sin x: k). 12. y = k sin 2x. 13. p = k sin 28. 14. e2P = k cot 8. 16. p2 = k(p sin 9 - 1).

+

+

16.

(p +;) sin

/I = k,

Find the orthogonal trajectory, through the point specified, of each of the following families of curves. 17. y2 = kx, (-2,3). 18. y2 = x2 + ky, (1, -2). 19. p = k sin (8/4), (i, 180°). 20. p = k(l + sin 8), (2,7r/6). 21. Find the curve through the point (7r/4,0) belonging to the family sin x

+ sinh y =

k cosh 'Y

and the curve through the same point belonging to the orthogonal family; verify that the two curves intersect at right angles. . 22. Find the orthogonal trajectories of ax 2 + y2 = lex, where a is a fixed constant and k is a parameter.

102'

Chapter 3

23. Find the orthogonal trajectories of cos y - a cosh x

= k

sinh x,

where a is a fixed constant and k is a parameter. 24. Find the orthogonal trajectories of x2 + ay2 = b (a) if a is a parameter and b is a fixed constant; (b) if b is a parameter and

a is a fixed constant. 26. Find the orthogonal trajectories of p

= k tan (8 + a)

(a) if a is a fixed constant and k is a parameter; (b) if k is a fixed constant

and a is a parameter. 26. Find the orthogonal trajectories of pm

= k sec nO,

where m and n are fixed eonstants and k is a parameter. 27. Show that the family of tractrices represented by equation (5), Art. 25, with k = r, are orthogonal trajectories of the family of circles in Prob. 12, Art. 8. 28. Consider a plane sheet of conducting material into which a current is fed at a point A (1,0) and out of which it flows at a point B (-1,0). It can be shown that the streamlines are represented by the family of circles passing through the points A and B. Find the orthogonal trajectories, i.e., the equipotentials. Draw both families of curves. Show that the equipotentials can be characterized by the property that at any point P on an equipotential the ratio PA/PB is constant. 29. In Prob. 28 the polar equation of the streamlines, taking the origin at B and the initial line along BA, is p = 2 cos

0 + k sin O.

Find the polar equation of the equipotentials. 30. Given the fixed points A (1,0) and B (-1,0). A curve has the property that for any point P (x, y) on it, the difference between the angles P AB and P BA is constant. Show that the curves characterized by this property are a family of hyperbolas, and find their orthogonal trajectories. Show that the orthogonal trajecteries can be characterized by the property that the product P A .PB is constant and that they are therefore Cassinian ovals.

eluzp;te4 I/LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS

27. The 'linear equation. In Art. 19 of Chapter 3 we discussed the linear equation of first order. We shall now deal with linear equations of nth order, the word "linear" signifying that the equation is of first degree in the dependent variable and its derivatives. It is assumed in this chapter that the coefficients of the dependent variable and of its derivatives are constants; in the next chapter linear equations with variable coefficients will appear. The type-form of linear equation of nth order with constant coefficients is dny dn-1y dy ao dxn + al dxn-1 +...+ an-l dx + anY = R, where the a's are constants, ao '¢ 0, n is a positive integer, and the right member R is a function of x (or a constant). 28. The differential operator. Throughout this chapter we shall use symbolic methods in which the symbol D, called

the differential operator, stands for ~ and symbolizes the operation of taking the derivative with respect to x of what follows it. Thus the equation ~ x 2

= 2x would be written, in operator

notation, Dx2 = 2x. Similarly D2 stands for ::2' and D" for tr -da":'. thus D2X2 = 2, and Dneaz = aneaz· In operator notation the equation of the preceding article is aoDny + a1Dn-l y +...+ an-1Dy + anY = R, or (aoDn + a1Dn-l +...+ an_1D + an)y = R. 103

104

Chapter 4

The expression in parentheses, a polynomial in D, may be denoted as a function of D by the symbol f(D). The typeform linear differential equation of nth order will then be written f(D)y = R, where f(D) = aoD" + aID,,-t +. ·. + an-ID + an is a polynomial difierential operator which, operating on y, produces R. We shall now examine an important property of the operator f(D). Take, for example, the case where 11, = 2, a.o = 1, at = -1, a2 = -2, and let the operator be applied to the function x 2 + x; that is, consider the expression

(D2 - D - 2) (x 2

+ z).

(1)

This means: Take the second derivative of x 2 + X, minus the first derivative of x 2 + x, minus twice x 2 + z. Thus

(D2 - D - 2) (x 2

+ x)

= 2 - (2x + 1) - 2(x 2

+ x) = 1 - 4x - 2x 2•

Now, if D2 - D - 2 is regarded as a quadratic expression in the letter D, without regard to the meaning of D, the expression can be factored by algebra into CD - 2) (D + 1). Let us inquire whether, when D is regarded as a differential operator, the operator (D - 2) (D + 1) has the same effect on x 2 x as the operator D2 - D - 2. To evaluate the expression

+

(D - 2) (D

+ 1) (x2 + z),

(2)

we must first operate on x 2 + x with the operator D operate on the result with D - 2: (D - 2) (D

+ 1) (x + x) 2

=

(D - 2) (x 2

+ 1, then

+ 3x + 1) = 1-4x-2x2 ,

which is the same value as that obtained for the expression (1).

Article 28

105

Finally, we may factor D2 - D - 2 in the other order and obtain (3) (D + 1) (D - 2) (x 2 + x). Evaluating this expression, we have (D + 1) (D - 2) (x 2 + x)=(D + 1) (1 - 2x 2)=1 - 4x - 2x 2• We see that the expressions (1), (2), and (3) are equal. This illustrates an important property of an operator which is a polynomial in D with constant coefficients. D may be regarded as playing a dual role. The polynomial feD) may be factored with D playing its algebraic role, then, with the factors taken in any order, D may assume its role as operator. This may not be true, however, if the coefficients of f(D) are not all constants, as the following example shows. We have (D

+ x)Dy = (D2 + xD)y,

but D(D + x)y

= D2y + Dxy = D2y + xDy + yDx

=

(D2 + xD

so that (D

+

l)y;

+ x)Dy ~ D(D + x)y.

An operator, however, even with variable coefficients, is commutative with a constant; for example, (D2

+ xD + 3)2y = 2(D2 + xD + 3)y.

Consider next the effect of operating with f(D) on the product of an exponential function (fJZ and y, a function of x. We have Deazy = ea:&Dy aeazy = eaz(D + a)y, D2 eazy = ea:&(D2 + aD)y aeaz(D a)y = eaz(D a)2y ,

+ +

and, in general,

+

Dneazy = eaz(D

+ a)n y,

+

(4)

a formula which may be proved by mathematical induction [Problem 1, Art. 29(b)]. Hence, substituting in f(D)eazy = (aoDn + a1Dn-l + an-1D + an)eaZy

+...

106

Chapter 4

the values of the derivatives of eazy as given by (4), we have

+ e)" + al (D + a)n-l +...+ an-l (D + a) + an]y,

f(D)eazy = eaz[ao(D

or (5)

This formula states that when the operator feD) operates on a product eazy, the exponential e" may be shifted from right to left across the operator provided that the D in the operator is changed to D + a. We may refer to this process as the exponential shift. For example, (D2 - 4D + 4)e2Zx3 = (D - 2)2e2zx3 = e2zD2x3 = 6xe2z. Reversing formula (5) and changing D to D - a, we have eazf(D)y = feD - a)eazy; (6) that is, we may shift the exponential eaz from left to right across the operator feD) provided that the D in the operator is changed to D - a. This process may be called the reverse exponential shift. For example, e-Z(D2 - 2D + 5)y = e-Z[(D - 1)2 + 4]y = (D2 + 4)e-Zy.

We shall find considerable use for formulas (5) and (6) in the work which follows. 29. The linear equation with R = o. Before developing methods for solving the general linear differential equation f(D)y

=

(ao Dn + a1Dn-l

+...+ O-n-lD + an)y = R,

(1)

we consider first the case where R = 0, which is a step toward the solution of the more general case; that is, we shall now be concerned with solving the equation f(D)y = (aoDn

+ a1Dn-l +...+ an-1D + an)y = O.

(2)

Take first the case where all the a's, except ao, are zero: (3)

107

Article 29

Integrating n times in succession, we obtain

=

Dn-1y Dn-2 y

= CIX + C2, 2

Dn-3 y •

•

•

= Cl (n =

~

+ C2X + Ca, •

X n -3

_ 2)!

+ C2 (n _

n-l

y

CI

• Xn - 2

Dy

=

CI,

X Cl (n _

I)!

+"C2

3)!

n-2 x_ (n 2)!

+... + Cn-2X + Cn-h +...+ Cn_1X + c,..

We shall find it more convenient to write this value of y in a different form by letting Cn = Ch Cn-l = C2, ••• , C2 = (n - 2)! Cn- h Cl = (n - I)! Cn; hence the general solution of the differential equation (3) is y = C1 + C2x + Cax 2 +...+ Cnxn-1 • (4) We are now almost ready to solve the differential equation (2). We need to notice one rather obvious fact, however. Suppose that Yl and Y2 are two particular values of y that satisfy (2) ; that is, j(D)Yl = 0 andf(D)Y2 = O. It follows that the sum of Yl and Y2 will satisfy (2), since j(D)(Yl + Y2) =j(D)Yl + f(D)Y2 = O. Furthermore, if YI, Y2, ... , Yn are particular solutions, then Yl + Y2 +... + Yn is a solution. (a) Solution oj (2) when the roots of the polynomial equation feD) = 0 ate real and distinCt. Suppose that the polynomial

equation aoDn

+ a1Dn-1 + .. ·+ an-ID + an =

0

(5)

has all its n roots real and distinct: rl, r2, .. " rn • When we say that rl is a root of equation (5) we are not saying that a value of the operator D is rIo Equation (5) is not a differential equation; it is an algebraic equation of the nth degree in the letter D. Here D is playing its algebraic role as explained in Art. 28. When we have factored the polynomial in (5), we

108

Chapter 4

shall rewrite equation (2) in factored form; it will be a differential equation in which D is an operator. There need be no confusion regarding the dual role which D plays, and we shall not dwell further on this point. Since D - TI, D - T2' ••• , D - Tn are factors of the polynomial in (5), the differential equation (2) may be written (D -

TI)

(D -

T2) •••

(D -

Tn-I)

(D -

Tn)Y

N ow the equation (6) is evidently satisfied if (D SInce (D - TI) (D - T2) ••• (D - Tn_I)O = 0. But, if (D - Tn)Y = 0, then dy dx = TnY, •

dy Y

log y

= O. Tn)Y

(6)

= 0,

= r.n, dx

= TnX + log Cn, Y=

Cn({",z.

(7)

Hence (7) is a particular solution of (6). However, we saw in Art. 28 that the factors in (6) may be taken in any order, so that anyone of them could be placed last, just before the y. Therefore (7) is a particular solution for each integral value of the subscripts from 1 to n and, by the principle enunciated earlier in this article, Y

=

C1ef'J.Z

+ C2e +.··+ Cn({ftZ f'2Z

(8)

is a solution of (6) and hence of (2). Furthermore, it is the general solution since the C's are all essential-that is, the solution (8) cannot be expressed with fewer than n arbitrary constants. The problem of solving the differential equation (2) is merely a problem in algebra-finding the roots Tt, T2, ••• , r« of the polynomial equation feD) = 0, then, when these roots

Article 29

109

are distinct, substituting in (8) to obtain the general solution of the differential equation. Four methods of finding the roots of a polynomial equation are illustrated in the four following examples. EXAMPLE

1. Solve d,2y

2 dx 2

dy

+ dx -

y = O.

In operator notation, the differential equation is

+D -

(2D2

l)y

=

o.

By inspection, the roots of the quadratic equation 2D 2 + D - 1 = (2D - I)(D + 1) = are ~ and -1. Hence the general solution of the differential equation is

°

y EXAMPLE

= Ole OOz + 02e-z •

2. Solve (D4

+ 2D 3 -

D2

-

2D)y = O.

Here we factor by grouping terms: D 4 + 2D 3 - D 2 - 2D = D3(D + 2) - D(D (D3

D)(D

-

+ 2) = 1)(D + 1)(D + 2) = o.

+ 2) = D(D -

Roots: 0, 1, -1, -2. General solution: y EXAMPLE

= 0 1 + C2ez + C3e- z + C4e- 2z •

3. Solve (D 3

-

8D

+ 3)y =

O.

Here we notice that the cubic D3 - 8D + 3 will vanish when D = -3, so that D + 3 is a factor; we may find the other factor by synthetic division: 1 o -8 3 -3 9 -3 1

-3

1

o

The other factor is D 2 - 3D + 1 and the other roots are those of the quadratic equation D 2 - 3D + 1 = 0, namely, (3 ± VS)/2.

110

Chapter 4

General solution:

y = Cl e EXAMPLE

3z

+ C2e (1 / 2)(3 + V5)z + Cae (1/2) (3-V5)z. '.

4. Solve

(D

4

, - 6D

2

+ l)y = O.

In this case the quartic can be factored by arranging it as the difference of two squares: D4

6D 2 + 1

-

= (D 2 -

1)2 -

4D 2

=

(D 2 Roots:

2±VB = 2

1±

0,

-

2D - 1)(D 2

+ 2D -

-2± VB \. = -1 2

1) = O.

± 0.

General solution: y

= Cl e (l + V2)Z + C2e (1- v'2)z + C3e <- 1+ v'2)z + C4e <- 1- v'2)z .

(b) Solution of (2) when the roots of the polynomial equation f(D) = 0 are real but not all distinct. Suppose that the polynomial equation aoDn

+ alDn-l +... + an_1D + an

= 0

has all its n roots equal to r, The general solution of the differential equation (2) cannot then be found from the formula (8), for this formula would now yield y = Clerz + C2erz + + Cne rz = (C 1 + C2 +...+ Cn)erZ, and, since Cl + C2 + stant C, the result is

+Cn can be replaced by a single con-

which is only a particular solution. Since D - r is an n-fold factor of f(D), equation (2) may be written To solve this equation, multiply by e-rz : e-rZ(D - r)ny = O.

Article 29

111

Now, performing the reverse exponential shift [formula (6), Art. 28], we get which is the same as equation (3) with y replaced by e-ny• Hence, from (4), the solution is e-rzy = C1 + C2 x + Cax 2 +...+ Cnxn-1 , or y = (C1 + C2 x + Cax 2 +.. ·+ Cnxn-l)erz. (9) }If the polynomial equation f(D) = 0 has some single roots and other multiple roots, the solution of the differential equation f(D)y = 0 is obtained by adding the terms corresponding to the single roots and the terms corresponding to the multiple roots. EXAMPLE

5. Solve

The roots of J(D) = 0 are 0, 0, 1, 1, 1, - 2, 3. The part of the solution corresponding to the double root 0 is, by (9), (C1 + C2x )eO or C1 + C2x; the part corresponding to the triple root 1 is (Ca + C4x + C5~)eZ; and the parts corresponding to the single roots - 2, 3 are respectively C6e-2z and C~eaz Hence the general solution is <,

y

= C1 + C2x + tc,

+ C4x + C5X2)~ + C6e-2z + C7eaz. PROBLEMS

1. Prove formula (4) of Art. 28 by mathematical induction. 2. Find the results of the indicated operations:

+ 3D - 1)x3eZ, (DS + 4D2 + 4D)x (D" - 8D3 + 24D2 - 32D)e

(a) (DS - 3D2

(b) (c) (<1) (D - a)naz •

2e-2Z ,

2Z

sin 2x,

3. Show that the general solution of Ex. 4, Art. 29, may be written in the form y = Aez sinh + a) + Be-z sinh (j).

(V2x


112

Chapter 4

4. Write the general solution of the differential equation

d2y = k2y d,x2

in terms of exponential functions and in terms of hyperbolic functions. Solve the following differential equations. 6. (D2 - 5D - 6)y = o. 6. (D2 6D - 5)y = O.

+

d4y

7. dx 4

-

d2y 8 dx 2 + 16y

=

o.

+

8. (2D3 SaD2 + 2a2D)y = O. 9. (2D3 - D2 - 2D + l)y = O. 10. (D" - 7D2 + l)y = o. 5

3

11. d y = 2 d y . dx 5 dx 3 12. (D4 - 6D2 8D - 3)y = O. 13. [abD2 - (a2 b2)D ab]y = O. 14. (10D4 - 9D3 - 23D2 4)y = O. 16. (36D4 - 37D2 4D 5)y = O.

+ + + + + +

16. Given the differential equation

d4y d3y dx4+ dx3 = 0 subject to the conditions y

= 0, dyjdx = 0, when x = 0,

y = 1, dyjdx = 1, when x = 1.

Obtain a formula for Y]~=-l in terms of e, and use this formula, with e = 2.718, to compute the value of Y]~=-l. ~

(c) Solution oj (2) when the po'(ynomial equation f(D) = 0 has complex roots. Suppose that the polynomial equation,

aoDn

+ atDn-1 + ... + an-ID + an = 0,

where the a's are real constants, has a complex root. We know that such roots must occur in conjugate pairs: if 2 + 3i is a root, 2 - 3i must also be a root, and, in general, if a + ~i is a root, a - ~i must also be a root, a and ~ being real constants. -

Article 29

113

Suppose first that n = 2 and the equation has a pair of complex roots: rl = a + (ji, T2 = a - (3i. Then formula (8) would give for the solution of (2), y = C1e(a+tH)~ + C2e(a-{Ji)z. This is a correct solution but it is not customary to leave it in this form with complex exponents. We transform it as follows. Factoring the right member, we have y = eaz(CleiPz + C2e- ilk) For the exponentials eif~z and e-i /3z we may substitute the values obtained from Euler'8 relation, ei8

by first putting

(J

= cos 8 + i sin 8,

= (jx, then (J =

-(jx, namely,

eiPz = cos ~x + i sin ~x, e- i/3z = cos ~x - i sin ~x. This gives y

= eaz(Cl cos ~x + iC l sin ~x + C2 cos ~x

- iC 2 sin ~x),

or, changing C 1 + C2 to A, iC l - iC2 to B, which is merely a change of notation on the two arbitrary constants, we have y

= eaz(A cos px + B sin px).

(10)

In Chapter 1, Art. 2, we saw that the expression in parentheses may be written in either of two equivalent forms, K sin (px + a) or K cos (px + E); hence alternative forms of (10) are (10') y = Ke" sin (px + a), y

== Ke" cos (px

+ E).

(10")

Formula (10) gives the general solution of (2) whenf(D) = 0 is a quadratic equation with a pair of complex roots. If feD) = 0 is a quartic equation having, besides a ± Pi, another distinct pair of complex roots, 'Y ± ai, the general solution of (2) is obtained by adding to the right member of (10) the

114

Chapter 4

expression e'YZ (E cos ax + F sin ax) corresponding to the pair 'Y ± ai. If f(D) = 0 has both complex and real roots, the solution of (2) is obtained by adding the terms corresponding to the complex roots and the terms corresponding to the real roots. Finally, if the polynomial equation f(D) = 0 possesses a double pair of complex roots, a ± Bi, a ± ~i, the corresponding part of the solution of (2) is obtained by writing the expression for a single pair, namely, eClZ(A cos ~x + B sin ~x), with A replaced by C1 + C2 x and B replaced by C3 + C4 x, a process analogous to that used for double real roots. For a triple pair of complex roots A would be replaced by C1 + C2X + C3 X 2 , and B by C4 + C5 x + C6 X 2 , etc. EXAMPLE

6. Solve (D3

+ D + 10)y = o.

Since the cubic D 3 + D + 10 vanishes when D = - 2, we may find the other roots of the cubic equation D 3 + D + 10 = 0 by synthetic division: 1 0 1 10 -2 4 -10 -2

I

1

-2

5

o

The other roots are those of the quadratic equation D 2 - 2D + 5 = 0, namely, (2 ± V 4 - 20)/2 = 1 ± 2i. Hence ex = 1, (3 = 2, in (10), and the general solution of the differential equation is y = EXAMPLE

C1e- 2z

+e

7. Solve

Z(C

2

cos 2x

+ C3 sin 2x).

+ 2D 2 + l)y = O. The roots of D 4 + 2D 2 + 1 = (D 2 + 1)2 = 0 are ±i, ±i. Here the pair ±i, with ex = 0, {3 = 1, is repeated; hence we use (10) with A replaced by C1 + C2 x and B replaced by C3 + C4 x. (D 4

General solution:

Article 29

115

PROBLEMS Solve the following differential equations. d,2y

+ k 2y = o. 2. (D2 - 2D + 4)y = O. 3. (36D2 + 36D + 13)y = iP8 dB 4. - 2 - 4 - + 138 = O. dt dt 1. dx2

O.

d2x

dx + 2a - + k2x = 0, where 0 < a < k. dt dt 6. (4D4 + 5D2 - 9D)y = O.

5.

-2

d6y

7. dx6

-

Y = O.

8. (4D3 - 107D - 222)y = O. 9. (D6 - 3D2 + 2)y = O. 10. 11. 12. 13. ·14. 15. 16.

d48

648 = o. dt d48 -4 + 648 = 9· dt (D6 - 12D2 - 16)y = O. (D4 - D2 + 2D + 2)y = O. (D6 + 12D4 + 48D2 + 64)y = O. (D5 + 2D3 + 10D2 + D + 10)y = O. (D5 - 4D4 + 2D3 - 8D2 + D - 4)y = O.

-4 -

17. According to the preceding theory, if J(D)y == 0 is a fourth order differential equation with a repeated pair of complex roots, a ± ~i, a ± ~i, the general solution is y = ea.1:[(Cl

+ C2X) sin ~x + (Ca + C..x) cos ~x].

Are the following also correct forms for the general solution:

=

(a)

y

(b)

y =

+ Fx)e«~(A sin px + B cos px), ea.1:[A sin (~x + E) + Bz sin (~x + F)], (E

where A, B, E, and F are arbitrary constants?

116

Chapter 4

30. Rectilinear motion. Before developing the theory for solving the linear equation f(D)y = R with R ~ 0, we shall consider some applications of the equation f(D)y = 0 to problems in rectilinear motion. These problems will involve differential equations of second order, that is, the case where feD) is a quadratic in D; certain cases which could be treated by means of first order equations have already been studied in Chapter 3. (a) A ttractive force proportional to displacement; resistance negligible. Consider a particle moving without resistance along a straight line, which we may take as the z-axia, under the action of an attractive force located at the origin O. Suppose that the force is proportional to the displacement z of the particle from 0 at time t, x being positive to the right and negative to the left of O. Since force is proportional to acceleration (cf. Art. 12), the acceleration d2x/dt2 will be proportional to the displacement x. If the particle is to the right of 0, the displacement is positive and the force (acting toward the left) and acceleration are negative; if the particle is to the left of 0, the displacement is negative and the force and.acceleration are positive; hence the acceleration and displacement are oppositely signed and the differential equation of the motion is

d2x 2x, dt2 = -k

(1)

where k 2 is used for convenience, instead of k, as the constant {)f proportionality. Equation (1), written in operator notation, with D = d/dt,

.

IS

(1') Its general solution may be written down by the method of Art. 29(c) and is the same, as that of Problem 1, Art. 29(c), with x and t taking the places of y and x, namely,

x = C1 sin kt

+ C2 cos kt.

(2).

Article 30

117

A form equivalent to (2) is (Art. 2) z

=A

sin (kt

+ a),

(3)

where a

= tan"?

C2 •

C1

Differentiation of (2) and (3) gives two forms for the velocity v = dx/dt: (4) v = kC 1 cos kt - kC 2 sin kt,

v = kA cos (kt

+ a).

(5)

The motion defined by (1), (2), or (3) is called simple harmonic motion. Equation (3) shows that the particle vibrates between the extreme positions x = ±A. A is called the amplitude of the motion. The motion is periodic. If, at any time t, the displacement and (signed) velocity of the particle are noted, then after a certain time T, called the period, the particle will again have the same displacement and velocity. To find T we observe from equations (3) and (5) that x and v will both return to their original values if the angle Jet + a increases by 21r, that is, if t increases by 21r/ k; hence 21r T =-. (6) k

Any integral multiple of 21r/k would also be a period, but by the period T we mean the smallest period. In particular, T is the time of a complete vibration from one extreme position to the other and back again. (b) Repulsive force proportional to displacement; resistance negligible. In the case of a repulsive force directed away from 0, the force and acceleration have the same sign as the displacement; the differential equation of the motion is then d2x 2x. dt2 = k

(7)

118

Chapter 4

The general solution of (7) is [Problem 4, Art. 29(b)]

+ C2e-kt, = C1 sinh kt + C2 cosh let. x

or x

=

Clek t

(8)

(9)

The analogy between the solutions of equations (1) and (7) may be exhibited by considering two particles, each starting from rest at a distance a from 0, the one under the action of an attractive force and the other under the action of a repulsive force emanating from O. We assume that the initial accelerations are numerically equal; that is, k will be the same in both equations. The initial conditions, x = a and v = 0 when t = 0, serve to determine the arbitrary constants in the solutions in both cases. ATTRACTION 2

dx 2x dt2 = -k X =

REPULSION 2

dx dt

-2=

C1 sin kt

+ C2 cos kt

X

= C~ sinh kt

(z = a, t

v=

so: cosh kt + ak sinh kt

o=

~

(v = 0, t = 0)

a =

o=

= 0, t

(v

= 0)

+ C; cosh kt

C;

a = C2 (x = a, t = 0) v = kC 1 cos kt - ak sin kt C1

k 2x

=

0)

x = a cosh kt

x = a cos kt

The displacement is a circular function of the time in the case of the attractive force, and the corresponding hyperbolic function of the time in the case of the repulsive force. This x, t relation is called the equation of motion. EXAMPLE

1. In the case of the two particles just discussed,' let

h and t2 be respectively the times required to travel the first a/2 units of distance; find the ratio tt/t2 • When each particle has traveled a/2 units, starting at x = a, the first will be at x = a/2 and the second at x =;: 3a/2; hence kt1 cos-1 0.5 kt 2 = cosh-1 1.5 '

t1 t2

1.0472 = 0.9625 = 1.09.

119

Article 30

2. A particle moves with simple harmonic motion in a straight line. When t = 0, the acceleration is 9 ft/sec 2 , the velocity is 4.5 It/sec, and the displacement x = -4 ft. Find (a) the displacement when the time equals half a period; (b) the first time when the displacement is zero and when it is 1 ft. The differential equation is EXAMPLE

d2x dt2-

= -k2x.

Since d2x/dt2 = 9 when x = -4, k

-- .a2·

Then x and v [equations (2) and (4)], together with the conditions for determining the arbitrary constants, are x = 0 1 sin ~ t

v

= -101 cos

+ O2 cos it

~t - ~02 sin ~t

(t

= 0, x = - 4),

(t

= 0, v = 4.5).

Substitution of the conditions in these equations gives C2 CI = 3; hence the relation between displacement and time is x

=

3 sin ~t - 4 cos ~t.

=

-4,

(10)

The amplitude of the motion is A = V 3 2 + (- 4)2 = 5 ft, and the period is T = 211"/(~) = 411"/3 sec. (a) To find the displacement when the time equals half a period) substitute t = 211"/3 in equation (10): X]t=21r/3 =

3 sin 11" - 4 cos 11" = 4 ft.

(b) To find the time when x

= 0 or 1 it is more convenient to use

the alternate form for x [equation (3)], x

=

5 sin (~t

+ a),

(11)

where a = tan- 1 - -j-. But care must be exercised in choosing the appropriate value for a, which could be an angle in either the second or fourth quadrant if there were no restrictions. Here, however, x is given negative when t = 0, hence sin a is negative and a must be in the fourth quadrant. We write a = -tan-1 t = -0.9273

Chapter 4

120

radians, then equation (11) gives, for the first positive value of t which makes x = 0, ~t - 0.9273 = 0,

t

= 0.618 sec.

When x = 1, we solve for the smallest positive t which satisfies 1

= 5 sin

~t - 0.9273

~t

(~t - 0.9273).

= sin-1 0.2 = 0.2013, = 1.1286,

t = 0.752 sec. 3. A cubical block of wood 6 in. on an edge and weighing 4 lb floats in water (62.4 Ib/ft3 ) . If it is depressed slightly and released, find the period of vibration, assuming that resistance is negligible and that the top remains horizontal. If the initial depression below the floating position is 2 in., find the equation of motion and the distance of the top of the block above water at the end of t of a period. EXAMPLE

Q

p

~

a

FIG. 8 In Fig. 8 the first position of the cube is the floating or equilibrium position. The plane PQ, which coincides with the plane of the water surface when the cube is floating at rest, may be called the equilibrium plane. Take the origin in the water surface and let x, measured in feet, positive downward and negative upward, represent the displacement of the equilibrium plane PQ from the water surface at any time t (sec). In the second position in the figure, PQ is x feet below the water surface. •

Article 30

121

In the equilibrium position the weight of the cube, 4 lb, is exactly balanced by the buoying force of the water, which, according to the Principle 0/ Archimedes, is numerically equal to the weight of the water displaced. The volume of water displaced is PBX (~)2 ft 3 and its weight is PB X (~)2 X 62.4 = 15.6 PB lb. Hence

15.6 PB

=

4,

P

B

1

12 .

.

= 3.9 ft = 3.9 In. = 3.08 m.,

and the cube floats with 3.08 in. submerged, or AP = 2.92 in. above water. In the second position, where PQ is x feet below the water surface, the resultant force, that is, the excess of the total buoying force over the weight of the cube, is numerically equal to the weight of the volume of water having cross section PQ and height z. This force, acting upward, is negative, and is therefore equal to -62.4(~)2X = -15.6x lb. If PQ is x feet above the water surface, x is negative, and -15.6x represents the resultant positive force acting downward, that is, the excess of the weight of the cube over the total buoying force. Hence, for any position of PQ, the resultant force tending to restore the cube to its equilibrium position is -15.6x; equating this to the symbol for pounds force (Art. 12), we have 4 d2x

-2

g dt

= -15.6x,

or ~x

dt2 = -3.9gx.

(12)

The acceleration is negatively proportional to the displacement, and the cube, when depressed slightly and released, will execute a simple harmonic motion, the plane PQ vibrating up and down through its equilibrium position x = O. To find the period it is not necessary to solve equation (12). Equation (6) gives at once, with g = 32.17 ft/sec 2 , 211" T = ~ ~ = 0.561 sec. v 3.9g Now suppose that the initial depression below the equilibrium position is 2 in. = ~ ft. In order to find the equation of motion we must

122

Chapter 4

solve equation (12). Writing the general values of x and v [equations (2) and (4)], together with the initial conditions, we have x

+ C2 cos V3.9g t (t = 0, x = ~), vagg(Cl cos V3.9g t - C2 sin V3.9g t) (t = 0, v = 0).

= C1 sin vagg t

v=

Substituting the conditions, we find C2 = ~ and C1 = 0; hence the equation of motion is x = ~ cos V3.9g t. At the end of ~ of a period, t = ; T = t(1r/V3.9g), and 1

X]t=5T/8

51r

= 6 cos "4 = -

V2

12 ft

.

= -1.41 In.

Hence the equilibrium plane PQ is 1.41 in. above the water surface, and the top of the block is 1.414 + AP = 1.414 + 2.923 = 4.34 in. above water at the end of ~ of a period.

In all the problems of the following group resistance is assumed to be negligible. They are all problems in rectilinear motion except those concerned with the motion of a pendulum, but here the angular displacement (J takes the place of linea.r displacement x. Whenever the numerical value of g is needed] the value 32.17 ft/sec 2 should be used. PROBLEMS 1. A particle P starts from rest at x = a in accordance with the law = - k 2x. At the same time a second particle Q starts from rest at x = a in accordance with the law d2x/dt2 = k 2x. Find (a) the distance traveled by P while Q moves its first a/3 units of distance; (b) the distance traveled by Q while P moves its first a/3 units of distance. . 2. For the motion described in Ex. 2 of Art. 30, find (a) the relation between x and v; (b) the values of x and v when t equals a quarter-period; (c) the time when the particle first reaches its extreme left position. 3. A particle moves on a straight line under the action of a repulsive force, according to the law d2x/dt2 = x/4. If it is projected from the point x = 0 with a certain initial velocity, how much time will elapse until its velocity is double the initial velocity?

d2x/dt2

Article 30

123

4. In Ex. 3 of Art. 30, assume that the block of wood is depressed until its upper face lies in the water surface and is then released. . At the end of 2 sec, (a) how much of the block is submerged? (b) how fast is it moving and in which direction? 6. A cylinder! ft in diameter and weighing lOr lb floats half submerged in water. It is pressed down sq that its top is in the water surface; then it is released, and vibrates with axi~ vertical. Find (a) the period of vibration; (b) the location of the midpoint of the cylinder after 1 sec. 6. A cylinder weighing 40 lb floats in water with its axis vertical. It is pressed down slightly and released, whereupon it vibrates with a period of 1.8 sec. Find the diameter of the cylinder. 7. A cubical block of wood 28 in. on an edge floats in water. It is pressed down slightly and released, whereupon it vibrates with a period of 1.35 sec, its top remaining horizontal. Find the specific gravity of the wood. 8. A spring, fixed at its upper end, supports a weight of 10 lb at its lower end, which stretches the spring 6 in. If the weight is drawn down 3 in. below its equilibrium position and released, find the period of Vibration and the equation of motion of the weight. Assume that Hoqke'8 law holds, i.e., force is proportional to stretch. 9. A rubber band of natural length AB is suspended vertically from A and a weight is attached to it at B. The weight stretches it 18 in. If the weight is drawn up to the point B and then projected downward with a velocity of 5.67 (= Vg) ft/sec, find the greatest extension of the band. 10. The resultant attraction of a spherical mass on a particle within the mass is directed toward the center of the sphere and is proportional to the distance from the center. Suppose a straight tube bored through the center of the earth and a particle weighing w lb dropped into the tube. If the radius of the earth is 3960 miles, find how long it will take the particle (a) to pass through the tube; (b) to drop halfway to the center. 11. A rod of length L (ft) is pivoted at 'one end and has a weight w (lb) attached to the other end, forming a simple pendulum. If 8 is the angular displacement of the rod from the vertical at time t (sec), show that

d28

- 2 = dt

g

-sinO.

L

If 0 is so small that sin 0 may be replaced by 0, the equation takes the form of equation (1), Art. 30. In this case, with a equal to the amplitude, or maximum angular displacement, find the period of vibration and the equation of motion when the initial conditions are 8 = a, w = } = 0, when

t

=

o.

~

"'~

124

Chapter 4

12. Find the equation of motion of a pendulum of length L (ft) if the amplitude of the vibration is small and the pendulum is started from its lowest position by giving it an angular velocity Wo rad/sec. 13. A seconds pendulum is one which swings through its complete arc in one second, that is, one whose period is two seconds. (a) Find the length of a seconds pendulum with small amplitude. (b) If the pendulum of Probe 12 is a seconds pendulum with amplitude 4°, find woo 14. A pendulum swings through an arc of 6°. How much of its period is consumed in traveling from its extreme position halfway to its lowest position?

(c) Attractive force proportional to displacement; resistance proportional to velocity. The vibratory motion with constant amplitude, which we have just studied, resulted from a force proportional to the displacement, acting without resistance. If friction or other resisting forces are taken into account, the particle may execute damped vibrations in which the oscillations continually decrease in size, or, if the resistance is sufficiently large, the particle may gradually come to rest at its equilibrium position without vibrating. 'We consider here only a resistance which is proportional to the velocity. Suppose that a particle moving along the z-axis suffers such a resistance. If the particle is moving toward the right, the velocity dx/dt is positive, and -K dx/dt represents the negative resisting force, K being a positive constant. If the particle is moving toward the left, the velocity dx/dt is negative, and -K dx/dt represents the positive resisting force. If, therefore, for either direction of motion, we add the resisting force -K dx/dt to the force -k 2 x, proportional to the displacement, we obtain the resultant force acting on the particle. Equating this resultant force, measured in pounds, to the symbol for pounds force, (w/g)(d 2x/dt2 ) , where w (lb) is the weight of the particle, we have the differential equation \ w d2x 2 dx - -2= -kx-K-· g dt dt

Now multiply by g/w, write for convenience k 2g/ w = b2, Kg/w = 2a, and use the operator notation D = d/dt, thus

125

Article 30

obtaining the standard form of the differential equation of motion, (13) (D 2 + 2aD + b2)x = O. The roots of the quadratic D2 + 2aD + b2 = 0 are -a ± Va 2 - b2 , so that the form of the general solution depends on the relative values of a and b. If a < b, that is, if the resistance is sufficiently small, the roots will be complex, and the solution will contain trigonometric functions and represent vibratory motion. In this case we write the roots in the form -a ± Vb 2 - a2 i, and the general solution is z

= e-at(Cl

sin

Vb 2

-

a2 t

+O

2

cos Vb 2

or

z

= Ae-at sin (Vb 2

-

a2 t

+ a).

-

a2 t), (14) (15)

If a were equal to zero, that is, if there were no resistance, we would have the case of simple harmonic motion discussed in Art. 30(a), where the z, t relation, if plotted on rectangular t, x axes, would represent a sine curve oscillating between the straight lines z = ±A. Here the x, t relation represents a distorted sine curve oscillating between the curves x = ±Ae-at. The oscillations are damped down and gradually die out with increasing t; equation (14), or (15), represents damped oscillatory motion. The factor e-at is called the damping factor (d.f.). It should be remembered, however, that the x, t curve is not the path of the particle. The particle oscillates back and forth through the origin in a straight line, the x-axis; the x, t curve pulls this motion out, so to speak, in the t direction. We call T = 2rI ~ the period, where ~ = V b2 - a2 • If 'a = 0, T reduces to the period for simple harmonic motion, T = 2rlb. The period represents the time interval between two successive passages of the particle through the origin in the same direction. If a > b, both roots -a ± v' a2 - b2 are real and negative. Writing rl = -a + Va 2 - b2 and T2 = -a - V a2 - b2,

126

Chapter 4

we obtain the general solution of the differential equation (13): x

= Cte1'

+ C2({2',.

T2

< Tl < O.

(16)

Then, differentiating equation (16), (17) ,

Suppose that at time t = 0 the particle is started at the point x = Xo > 0 with initial velocity v = Vo. Substitution of these conditions in equations (16) and (17) determines the constants C1 and C2 : Xo

= C1 + C2

Vo = TIC I

+ T2C2

C =

}

2

-Vo

+ TlXo.

Tl -

By inserting these values of C l and C2 in equation (16), comes x =

1 rl -

T2

[(vo -

r 1t

T2x O)e

-

(vo -

TlxO)e~].

T2

~t

be(18)

Since Tl and T2 are both negative, equation (18) shows that the particle will settle down to x = 0 as t increases indefinitely; the way it moves, however, depends on the value of the initial velocity vo. Since Tl - T2 is positive, x will be positive provided (19) For t > 0 the exponential in (19) is greater than 1; also for Xo > 0, Vo - T2Xo > Vo - TIXO. Therefore if Vo > T2Xo the inequality (19) is satisfied and the particle remains to the right of x = 0; for Vo > 0 the particle starts at x = Xo, moves toward. the right, then reverses its direction of motion, and approaches x = 0; for T2XO < Vo < 0 the particle starts toward the left and approaches x = o. If Vo < T2XO the inequality (19) is reversed for sufficiently large t; the particle starts toward the left at x = xo, passes through x = 0, reverses its direction of motion, and approaches x = o.

Article 30

127

4. A particle starts from rest at time t = 0 (sec) with a displacement x = 5 (ft) to the right of the origin, and moves along the x-axis according to the law EXAMPLE

d 2x dt 2

dx

+ dt + 1.25x =

O.

Find (a) the time required for the damping factor to decrease 50 per cent, (b) the percentage decrease in the damping factor after one period, (c) the location of the particle after one period. The differential equation is given in standard form [equation (13)], with a = ~, b2 = 1.25, a < b. (a) The d.f, is e-at = e-t / 2 • Its original value, for t = 0, is 1; it reduces to ~ when

e- t / 2

-

-

!.

2'

t

- - = -log 2, 2

t

= 2 log 2 = 1.39 sec.

(b) The period is 21r/Vb 2

-

a2

= 21r sec. At this time the d.f.

has the value e--r = 0.0432. Hence the d.f. has decreased 95.7 per cent. (c) Only in part (c) do we need to use the initial conditions and solve the differential equation

+ D + 1.25)x = O. The roots of the quadratic D 2 + D + 1.25 = 0 are (D 2

(20) - ~ ± i; hence

the general solution of (20) is

x = e-t / 2 (C1 sin t + C2 cos t).

(21)

Differentiation of (21) gives

I

=

e-t / 2

[ (C1 -

~2) cos t- (~1 + C2 )

sin

t].

(22)

Substituting the initial conditions x = 5, v = 0, t = 0 in (21) and (22), we obtain

128

Chapter 4

Hence the equation of motion is

x = e-t /2 (g sin t + 5 cos t). Then

x] t=2r = e

-11"

(5)

= 0.216 ft,

and the particle is located 0.216 ft to the right of the origin. EXAMPLE

5. A particle moves along the z-axis in accordance with

the law

d2x dt2

dx

+ 10 dt + 16x = O.

From a point 1 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 9 ft/sec. Find (a) the time when the particle passes through the origin; (b) the numerically greatest negative displacement; (c) the maximum (positive) velocity. The differential equation is in standard form (13), with a = 5, b = 4, a > b, Tl = - 2, r2 = - 8. Since Xo = 1 and Vo = - 9, the condition Vo < T2XO is satisfied and we have the type of motion in which the particle passes toward the left through the origin, reverses its direction of motion, and approaches x = o. For x and v [equations (16) and (17)] we have 2t

+ C2e -8t,

x =

C1e -

v =

-2C1e-

2t -

8C2e -8t.

Substitution of the initial conditions x = 1, v = -9, t = 0, gives 1 -9

= C1 + C2 } = -2C1 - 8C2

Hence

+ ~e-8t,

(23)

2t !.e2 8 e-8t 3 -3 .

(24)

x = - ~e-2t V --

7 C2 -- 6·

(a) When x = 0, equation (23) gives

e6t = 7,

t = ~ log 7 = 0.324 sec.

(b) The numerically greatest negative displacement occurs when v = 0, that is, from equation (24), when e6t = 28, from which e-2t =

Article 30 28-~, e-8t

129

= 28-%.

Substituting these values in equation (23),

we obtain x = ~( -28

+ 7)28-% =

-3.5(28)-% = -0.0412 ft.

(e) When the velocity is a maximum,

dv

2 3

- = - - e-2t dt

+ -224 e-8t = 3

0

.

Then e6t = 112, from which e-2t = 112-~\ e-8t = 112-%. stituting these values in equation (24), we find Vmax

= ~ (112 - 28)112-%

= 28(112)-%

Sub-

= 0.0519 It/see.

EXAMPLE 6. A spring, fixed at its upper end, supports a weight of 10 lb at its lower end, which stretches the spring 6 in. If the weight is drawn down 3 in. below its equilibrium position and released, find the period of vibration and the equation of motion of the weight, assuming a resistance in pounds numerically equal to ~ the speed in feet per second. (Problem 8, Art. 30(b), where the resistance is negligible, is a special case of this problem.) We first determine the spring constant. Since a force of 10 lb stretches the spring 6 in. or ~ ft, we have 10 = e( ~), and the spring constant is e = 20 lb/ft. In solving a problem of this type, it is important to specify and keep clearly in mind the direction chosen as positive, and the origin from which the displacement is measured. We shall take the downward direction as positive and the origin 0 at the equilibrium position of the weight, that is, the position of the weight when it is hanging at rest. Let x (ft) be the displacement of the weight from its equilibrium position at any time t (sec); then the elastic force tending to restore the weight to its equilibrium position is - 20x lb, the minus sign denoting that this force and the displacement are opposite in sign. We also have a resisting force whose sign is opposite that of the velocity dx/dt, namely, - ~ dx/dt lb, so that the resultant force tending to restore the weight to its equilibrium position is - 20x ~ dx/dt lb. The force acting on the moving 1Q-Ib weight is also (10/g)(d2x/dt2 ) lb. Equating these two expressions for pounds force, we obtain the differential equation 10 d2x

- -2=

g dt

1 dx -20x---

2 dt '

130

Chapter 4

or, in the standard form of equation (13), (D 2

+ 2aD + b2)x = 0,

where a (j

g

= 40 = 0.804,

= Vb 2

-

a2 = V64.34 - .65 = 7.98.

The period is therefore 211"/(3 = 0.787 sec. The general solution is x

=

«r«, sin (3t + C2 cos (jt).

(25)

Differentiating (25), we get

v = e-at[(pC 1

-

aC2 ) cos (3t - (aC.

+ (3C2 ) sin (3t].

(26)

Substitution of the initial conditions x = ~, v = 0 t = 0, in (25) and (26) gives

C2

1

a C1 = 4(3

=-

4'

=

0.0252.

Hence the equation of motion is x = e-O.804t(0.0252 sin 7.98t

+ 0.250 cos 7.98t).

PROBLEMS 1. For the motion represented by equation (13), Art. 30(c) , where a < b, show that if v is the velocity at any time, the velocity one period later is e-aTv, where T is the period. 2. Discuss the motion of the particle in Art. 30(c) for the case where a = b, with initial conditions x = xo > 0, v = Vo, when t = O. 3. For each of the two cases in Art. 30(c) where a > b and the particle reverses its direction of motion, obtain a formula for the time elapsed until the reversal takes place. 4. The motion of a particle is represented by equation (13), Art. 30(c) , with a > b and initial conditions x = Xo > 0, v = 0, when t = 0; at what time will the particle be moving fastest? 5. Discuss the motion of a particle along the x-axis under the action of a repulsive force proportional to displacement and a resistance proportional to velocity, .assuming as intial conditions x = Xo > 0, v = Vo when t = O.

Article 30

131

6. A particle executes damped vibrations of period 1.35 sec. If the damping factor decreases by 25 per cent in 16.3 sec, find the differential equation, with numerical coefficients, representing the motion. 7. A weight hung on a spring and vibrating in air with negligible damping has a period of 1 sec. It is set vibrating with a practically weightless damping vane attached to it, causing a resistance proportional to the speed, and the period is found to be -1.5 sec. Find the damping factor and write down the differential equation, with numerical coefficients, corresponding to the damped vibrations. 8. A weight suspended from a spring executes damped vibrations of period 2 sec. If the damping factor decreases by 90 per cent in 10 sec, find the acceleration of the weight when it is 3 in. below its equilibrium position and is moving upward with a speed of 2 It/sec. 9. A weight of 4 lb is hung on a spring causing an elongation of 2 in. It is set vibrating and its period is 1r /6 sec. Find the time required for the damping factor to decrease 75 per cent if resistance is proportional to velocity. 10. In Ex. 3, Art. 3O(b), suppose that the water offers a resistance (lb) numerically equal to K times the velocity (ft/sec), thus increasing the period of vibration by 0.04 sec. Find the value of K and the time required for the damping factor to decrease 50 per cent.. 11. A particle moves along the z-axis in accordance with the law d 2x

dt2

dx

+ 10 dt + 9x =

O.

From a point 2 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 20 ft/sec. Find (a) the time when the particle reaches its leftmost position; (b) the distance traveled and the velocity at the end of 1 sec. 12. If the particle of Prob. 11 is projected toward the right at the rate of 20 ft/sec, find (a) the time when it reaches its rightmost position; (b) the distance traveled and the velocity at the end of 1 sec. 13. If the particle of Prob. 11 is projected toward the left at the rate of 10 ft/sec, find the distance traveled and the velocity at the end of 1 sec. 14. A particle moves along the z-axis in accordance with the law d2x

dx -+ 6dt2 dt -16x =

0

.

From a point 2 ft to the right of the origin the particle at time t = 0 (sec) is projected toward the left at the rate of 10 ft/sec. Find (a) the time when the particle reaches its leftmost position; (b) the distance traveled and the velocity at the end of 1 sec.

132

Chapter 4

16. A lo-lb weight is suspended by a spring which is stretched 2 in. by the weight. Assume a resistance whose magnitude (lb) is 40/v'Y times the speed (ft/sec) at any instant. If the weight is drawn down 3 in. below its equilibrium position and released, find (a) the displacement of the weight from its equilibrium position after 1/(2v'Y)sec; (b) the time required to reach the equilibrium position. 16. A lo-lb weight hanging on a spring stretches it 2 ft. If the weight is drawn down 6 in. below its equilibrium position and released, find (a) the highest point reached by the weight; (b) the location of the weight after 1/v'Y sec. Assume a resistance whose magnitude (lb) is 10/0 times the speed (ft/sec). 1'1. A 10-lb weight stretches a certain spring 1 ft. The weight is drawn down 6 in. below its equilibrium position and then released. Assuming a resistance whose magnitude (lb) is equal to the speed (ft/sec) at any point, find (a) the period; (b) the percentage decrease in the damping factor when the weight is at its highest point; (c) the distance (in.) of the weight from its starting position when t = 0.6 sec. 18. A weight of 16lb hanging on a spring, which it stretches 6 in., is given a downward velocity of 10 ft/sec. If the resistance (lb) of the medium is numerically equal to the velocity (ft/sec), is the weight above or below the starting point at the end of 0.5 sec, and how much? 19. A simple pendulum makes small oscillations with a period of 2 sec in a resisting medium. Assume an angular retardation whose magnitude (rad/ sec2) is 1/25 the magnitude of the angular velocity (rad/seo). If the pendulum is released from rest at an angular displacement of 10 , find the displacement at the end of 10 periods. 20. A 6-lb weight of specific gravity 3 stretches a spring 4 in. when immersed in water. If the period of vibration in water is g sec, by what percentage will the damping factor decrease in 1 sec? Assume resistance proportional to velocity. 21. A weight having specific gravity p is immersed in water and supported by a spring which it stretches a in. It is set vibrating and the resistance of the water to the motion of the weight is proportional to the velocity.' If at the end of r periods the damping factor is l/n its initial value, obtain a formula for the period T in terms of p, a, n, r, and g. Use the formula to compute the value of T if p = 3, a = 8 in., n = 6, r = 1, 9 = 32.17 ft/sec 2•

31. The linear equation with R ~ 0; simplest case. We consider now the simplest case of the differential equation f(D)y = (ao D n + a1Dn-l +...+ an-1D + an)y = R, with R ~ 0, namely, the case in which an and all the other a's

133

Article 32

except one, sayan _ r , are equal to zero; the differential equation then has the form dry an - r dxr

= R,

where R is a function of z, or dry dxr

= F(x).

(1)

The solution of the differential equation is effected merely by integrating r times, adding each time a constant of integration. EXAMPLE

1. Solve

d3 y

dx 3 = 6x

+ 2 cos 2x.

Integrating three times, adding each time a constant of integration, we have

tPy dx 2 =

3

X

2

+ SIn• 2x + Ch

dy 3 cos 2x dx = x 2 x4

y="4-

sin 2x 4

+ CIX + C2,

2 CIX +2+C2X+C3.

General solution: 4y

=

x4

-

sin 2x

+ C1X 2 + C2x + C3 •

32. Deflection of beams. As an application of equation (1), Art. 31, we consider some problems on the deflection of beams. Fig. 9 shows a piece of a bent beam. I t can be regarded as made up of fibers such as F'F, all originally of length 8 (ft). The surface PQ, containing fibers whose lengths are unaltered when the beam is bent, is called the neutral surface. The curve of one of these fibers B'B is called the elastic curve of the beam. Fibers below the neutral surface are stretched and those above are compressed when the beam is bent. Suppose that the

Chapter 4

134

fiber F'F at a distance z below the neutral surface has been stretched an amount e by the force S dA, where S (lb/It") is the stress per unit cross-section area and dA (ft 2 ) is the cross-section area of the fiber. Also let R (ft) be the length of BC, the

SdA

FIG.

9

radius of curvature at B of the fiber B'B, and let QQ' be perpendicular to B'B and BC. Now, by Hooke's law, the stress S per unit area is proportional to the stretch e/8 per unit length in the fiber F'F; that is e S = E-

(1)

8 '

where the constant of proportionality E (lb/ft 2 ) is Young's modulus, or the" modulus of elasticity. Furthermore, from the figure we have the proportion (2)

Then, from (1) and (2),

Ez

S =-. R The moment of the force S dA about the axis QQ' is

E zS dA = - Z2 dA. R

Article 32

135

Integrating this over the cross section of the beam we get the bending moment M:

But

jz2dA = I, where I (ft4) is the moment of inertia of the cross-section area of the beam with respect to the axis QQ'; hence

El M=-·

R

(3)

Now take the z-axis horizontal through some point of the fiber B'B chosen as the origin and the y-axis positive upward. The formula for radius of curvature is R = (1 + y'2)3/2/y", but for small bending y' is small and v" may be neglected in comparison with unity, so that a close approximation to the radius of curvature is R = 1/y". By use of this approximation, (3) becomes (4) M = Ely", the expression for the bending moment at any section of the beam. Equating this expression to the algebraic sum of the moments, with respect to the axis QQ' of the section, of all forces on one side of the section which tend to bend the beam about this axis, we obtain the differential equation of the elastic curve of the beam. When the moments of all these forces are expressible in terms of x the differential equation is of the form (1), Art. 31, with r = 2. 1. Find the elastic curve and the maximum deflection of a beam L ft long resting on supports at the ends and slightly bent under a uniform load of q lb/ft, Take the origin 0 at the left support and the y-axis positive upward (Fig. 10). Then take a section of the beam through a point P at a distance x from 0 and equate the expression for the bending moment at P, Ely" [equation (4)], to the sum of the moments of all the forces EXAMPLE

136

Chapter 4

on one side of P, say to the left, which tend to bend the beam at P. Since y" is positive on a curve when the curve is concave toward the positive direction of the y-axis, we write a moment positive when it tends to produce concavity in the direction of the positive y-axis. Since v" is negative on a curve when the curve is concave toward the negative y-axis, we write a moment negative when it tends to produce concavity in the direction of the negative y-axis. The load qL is equally distributed on the supports, causing an upward force of qL/2 at the left support and a moment qLx/2 with respect to P. The load y

qL

qL

T

2'

k----+------:::JIIr--,% p

FIG. 10

qx on OP acting downward at the middle point of OP produces a moment - qx2/2 with respect to P. The differential equation of the elastic curve of the beam is therefore

Ely" = qLx _ qx2. 2 2

(5)

In this equation the units of the various quantities involved are as follows: E (lbjft2 ) , I (ft 4) , y" (ft- l ) , q (lb/ft), L (ft), x (ft). The differential equation (5) is solved merely by integrating twice and determining the two constants of integration from two conditions. Integrating once, we get

Ely' = The condition y'

q~x2 - q~ + (JI'

(6)

= 0, x = L /2, gives

qL3 qL3 o = 16 - 48

+ 01 ;

01

qL3

= -. 24

Article 32

137

Inserting the value of C1 in (6) and integrating again, we find Ely =

qLx3

12 -

qx4

qL3x

24 -

24 + C2 •

The condition y = 0, x = 0, gives C2 = of the beam is

o.

(7)

Hence the elastic curve (8)

There were just enough independent conditions to determine the two arbitrary constants. It may be noticed that a third known condition y = 0, x = L, is not independent; it satisfies (8) automatically, or, if substituted in (7), yields again C2 = 0. If we had not seen at the beginning that the reaction at each support is qL/2, we could have called one reaction R and the other qL - R; then all three conditions would have been needed to determine R, Cit and C2 • It may be noticed further that if we had taken moments to the right of P, instead of to the left, it would have been more complicated but the same differential equation would have been obtained; for in that case

EI "

= qL(L - x) _ qeL - x )2

Y

2

2'

which reduces to equation (5). The maximum deflection of the beam is the value of -y when x = L/2, that is, from equation (8), -ylz:-L/2

q

(L

4

L

4

L

4

= 24EI 16 - 4" + 2

)

5qL

4

= 384EI ft.

EXAMPLE 2. Solve Example 1 if the beam, instead of being simply supported at its ends, is embedded horizontally in masonry at its ends. The differential equation is the same as in Example 1, except that there is an additional unknown moment at each end of the beam, exerted by the masonry, which keeps the beam horizontal at the ends (Fig. 11). Since this moment tends to produce concavity downward it is negative; we denote it by -M. The differential equation is

Ely" =

q~

-q:; -

M.

138

Chapter 4

Integrating once,

Ely' =

qLx2

4

qx3

- {\ -

Mx

+0

The condition y' = 0, x = 0, gives 0 1 = 0. y' = 0, x = L/2, gives

1•

Then the condition

qL3 qL3 ' ML qL2 M=-· 0= 16 - 48 - 2; 12 Integrating again, qLx3 qx4 qL2x2 Ely = 12 - 24 - 24 + O2 • y

FIG.

11

The condition y = 0, x = 0, gives O2 = 0. Hence the equation of the elastic curve of the beam is

-q

y

= 24EI (x4

-

2Lx3

+ L 2x2),

and the maximum deflection is 4

qL -Y]x=L/2 = 24El

(116 - 14 + 41)

4

qL = 384El ft.

Two other conditions y = 0, x = Land y' = 0, x = L are not independent of the ones used and are satisfied automatically. 3. In Example 2, if there is an additional concentrated load of F lb at the midpoint of the beam, find the elastic curve of the left half of the beam and the maximum deflection. EXAMPLE

Article 32

139

The effect of the load F at the midpoint is to increase the reaction at each support to (qL F) /2. But there is another important difference between this example and the preceding one. In Example 2 the point P can be taken anywhere along the beam and the moments to the left of P are given by the same expression; the differential equation obtained holds for the whole beam. In this example, however, if P is moved toward the right across the midpoint of the beam, an additional moment to the left of P is introduced, due to the concentrated load, so that the differential equation of the right half of the elastic curve is different from that of the left half; the two halves of the elastic curve have different equations. This may seem strange, since the elastic curve "looks like" a curve with a single equation, but if the student will draw for illustration the curve y = - x3 from -1 to 0 and the curve y = x 3 from 0 to 1, the result will be a curve whose two halves have different equations. Taking P to the left of the midpoint of the beam and taking moments to the left of P, the differential equation representing the left half of the elastic curve is

+

Ely" = (qL ~ F)x _ q;2 _ M. Following exactly the same steps and using the same conditions as in Example 2 (the student should do this), the elastic curve of the left half of the beam is found to be y

=

-q 4 (x 24EI

-

2Lx3

F + L 2x2) + 4 (4x3 - 3Lx2 ) 8EI·

Notice that the two conditions y = 0, x = Land y' = 0, x = L, which were satisfied automatically in Example 2, are not satisfied here; these conditions are true only for the right half of the beam. For the maximum deflection we have 3

-y]Z==L/2

(1 3)

4

+

3

qL4 FL qL 2FL = 384EI - 48EI 2 - 4 = 384E' ft. PROBLEMS

1. A beam of length 20 ft is simply supported at the ends and carries a weight of 240 lb at its midpoint. Taking the origin at the left end and neglectingthe weight of the beam, find (a) the equation of the elastic curve of

140

Chapter 4

-

the left half of the beam; (b) the equation of the elastic curve of the right half of the beam; (c) the maximum deflection. 2. If the beam of Prob. 1 carries in addition a uniform load of 30 lb/ft, find the maximum deflection. 3. A cantilever beam (one end free and the other fixed horizontal) of length L ft weighs q lb/ft and carries a load of W lb at its free end. (a) Taking the origin at the free end, find the equation of the elastic curve of the beam. (b) What load, distributed uniformly along the beam, would produce the same maximum deflection as the load W at the free end? 4. A cantilever beam of length L ft weighs q lb/ft and carries a load of P lb at its midpoint. Find the maximum deflection of the beam. 6. A beam L ft long, carrying a uniform load of w lbjft, is fixed horizontally at one end and is simply supported at the other. Find (a) the deflection of the midpoint of the beam; (b) the maximum deflection of the beam. 6. A beam 10 ft long, fixed at one end and simply supported at the other, carries a uniform load. Find the distance from the supported end to the point where the maximum deflection occurs. 7. A beam 6 ft long is simply supported at its ends. Two concentrated loads, each equal to P lb, are supported at the points of trisection of the beam. Neglecting the weight of the beam, find the points at which the deflection has half its maximum value. 8. A cantilever beam of length L ft, vertical dimension h ft, and modulus of elasticity E Ib/ft2 carries a load of material whose density is w Ib/ft3 and whose depth at distance x ft from the free end is kx2 ft. Neglecting the weight of the beam, find its maximum deflection. 9. A beam of length 2L ft carrying a uniform load of q lb/ft is supported at its ends and at its middle point. (a) Taking the origin at the middle point, find the elastic curve of the right half of the beam. (b) Show that the maximum deflection is qL4 (39 + 55y'33) 216EI ' and that this is 42 per cent of the value it would have if the beam were cut in two at its middle point. 10. A simply supported beam L ft long carries a concentrated load of W lb at a distance of eft from the left end. (a) Taking the origin at the left end, find the equation of the elastic curve of the portion of the beam to the left of Wand of the portion of the beam to the right of W. (b) If c < L/2, show that the distance from the left end to the point of maximum deflection is greater than c and less than L/2, and find the maximum deflection.

Article 33

141

33. The linear equation with R ~ o. We now rewrite equations (1) and (2) of Art. 29: f(D)y = R, (1) f(D)y = 0, (2) where f(D) = .aoDn + a1Dn-l + ... + an-iD + an. Equation (2) has been solved in Art. 29; we now seek the general solution of equation (1), in which R is a function of x. The general solution of equation (2) is a value of y containing n essential arbitrary constants, n being the order of the differential equation; we call this the complementary function (c.f.) of equation (1) and denote it by Yc. Thenf(D) operating on Yc produces 0: (3) f(D)yc = o. Now suppose that in some way we are able to find a particular value of y, free of arbitrary constants, which satisfies equation (1); we call this a particular integral (p.i.) of equation (1) and denote it by yp. Then f(D) operating on YP produces R: f(D)yp = R.

(4)

We have seen in Arts. 2 and 11 that a particular integral of a differential equation can be obtained from the general solution by assigning particular values to the arbitrary constants which enter into the general solution; but here we are going to find a particular integral before the general solution is known-the finding of a particular integral is now to be a step in the process of finding the general solution. Adding equations (3) and (4), we find

+ yp) = = Yc + YP

f(D)(yc

hence y

R; (5)

satisfies the differential equation (1), and since it contains n essential arbitrary constants it is the required general solution.

142

Chapter 4

Thus the general solution of (1) consists of two parts, the e.f. and a p.i, The operator j(D) , operating on the c.f., produces nothing-the c.f. merely acts as a carrier of the required arbitrary constants. The p.i., however, when operated on by jeD), yields the right member R. EXAMPLE

1. Solve

+ 2D + l)y = 4ex • by solving (D2 + 2D + l)y = 0, or (D 2

The c.f. is found = 0; it is

Yc = (0 1

(D

+ 1)2y

+ 02 x)e- x.

A p.i., found by inspection, is YP = e", since (D 2

+ 2D + l)e = e" + 2ez + e = 4ex• Z

X

Therefore the general solution of the given equation is y

=

(0 1

+ 02 x)e- + e', X

This example illustrates the plan of solving an equation of form (1). We have but to find the c.f. by the method of Art. 29, then find a p.i., and add them together. However, we cannot rely on inspection to reveal a p.i. except in very simple cases. We shall explain in the next two articles two methods for finding a p.i. The first method, undetermined coejficients, will apply when R is of certain special form; the second, variation oj parameters, is general and can be used theoretically for any form of R, although it is impracticable to carry it through for complicated forms of R. 34. Undetermined coefficients. We shall find it convenient in this article to call the part of a term which is multiplied by a constant coefficient the variable part (v.p.) of the term. Thus the v.p, of 7xe2z is xe2z , the v.p, of 6 cos 2x is cos 2x, and the v.p. of 3x' is x. If a term is a constant, say 5, it can be thought of as 5xo, where 5 is the constant coefficient and Xo or 1 is the v.p. Now a function may be of such kind that successive differentiation of it beyond a certain point ceases to yield terms with new variable parts.

143

Article 34

For example, take the function x 2e3z and its two successive derivatives:

x 2e3z ,

3x 2 e3z

+ 2xe

3z

,

9x2e3z

+ 12xe + 2e3z. 3z

Further differentiation would yield only terms with variable parts, x 2e3z , xe3z , e3z • As another example, take the function 3x2 + sin 2x and its two successive derivatives:

3x2

+ sin 2x,

6x + 2 cos 2x,

6 - 4 sin 2x.

These and further derivatives contain only terms with v.p.'s x 2 , X, 1, sin 2x, cos 2x. As an example of a function which does not possess this property, take tan z and its successive derivatives: tan z, sec2 x, 2 sec2 x tan x, Further differentiation continues to produce functions with new variable parts. I t can be shown * that functions possessing the above property are expressible as sums of terms of the form Cxpeaz cos ~x or CxPelXZ sin ~x, where p is a positive integrer or zero, a and {3 are any real constants (including zero), and C is any constant. Thus the function x 2e 3z in the first example above is obtained from the first form by putting C = 1, p = 2, a = 3, P = 0; and the function 3x2 + sin 2x in the second example is obtained by summing the first form with C = 3, p = 2, a = 0, {3 = 0, and the second form with C = 1, p = 0, a = 0, ~ = 2. Suppose now that R is a function possessing the above property; that is, R is a function such that successive differentiation of it beyond a certain point ceases to yield terms with new variable parts. Otherwise stated, R is a function expressible as sums of terms of the form CxPe lXZ cos ~x or CxPe lXZ sin Bx, where p is a positive integer or zero, a and {3 are any real eon* See Reddick and Miller, Advanced Mathematics for Engineers, Art. 11, for proof of this and the rule following.

Chapter 4

144

stants (including zero), and C is any constant; then it is shown (loc. cit.) that the following rule will yield a particular integral of equation (1), Art. 33. Write the variable parts of the terms in the right member Rand the variable parts of any other terms obtainable by succes8ively differentiating R. Arrange the v.p.'s so found in groups such that all v.p.'s obtainable from a single term of R appear in only one group. Any group consisting of v.p.'s none of which is a v.p, of a term of the complementary function Ye is left intact, but if any member of a group is a v.p. of a term of Yo all the members of this group are to be multiplied by the lowest positive integral power of x that will make them all different from the o.p, of any term in Ye. Now multiply each member of all the groups by a general constant (undetermined coefficient) and take the sum of the expressions 80 obtained as Yp. Finally, operate on YP with feD) and equate the complete coejJicients of the various v.p.'s to the coefficients in the corresponding terms of R in order to evaluate the undetermined coefficients in Yp. 1. Solve

EXAMPLE

(D2 - D)y = 3x2

-

4x

+ 5 + 2~ + sin z.

(1)

We give all the work necessary, followed by an explanation of the steps taken in finding yp. Roots of f(D) = (D 2 .

yp=Ax3+ Bx2+ -1)

D2 yp=

-3A=3

A=-l

Ye = C1 + C2,ez • Cx +Eze"

0, 1.

+Fsin x+G cos x (2)

6Ax+2B+E~e:i:+2Ee:i:-Fsinx-Gcosx (4)

6A-2B=-412B-a=5 B=-l 0=-7

YP

= 0:'

D) = D(D - 1)

3Ax2+2Bx+ C+Exe:i:+ Ee:i:;-G sin x+F cos x (3)

Dyp= 1)

-

= - x3

-

,;2 -

7x

2E-E=2 E=2

+ 2x~ -

-F+G=l, -F-G=O F=-!, G=!

~ sin x

+ YP' is x 3 + (C2 + 2x)e

+ ~ cos x,

The general solution, y = Ye

y = C1

-

7x - x 2

-

Z

-

~ sin x

+ ~ cos z.

Article 34

145

The v.p.'s of 3x 2 and its two successive derivatives are x 2 , a, 1. Corresponding to the term - 4x we would get x, 1, and corresponding to the term 5 merely 1, but these are already included in the group r, x, 1, so that this group of v.p.'s corresponds to the terms 3x2 4x + 5 of R. Now one of these v.p.'s, namely 1, is the v.p. of 0 1, a term of the c.f.; hence we multiply each member of the group x 2, x, 1, by x, obtaining x 3 , x 2 , x, none of which is a v.p. of a term of Ye. By multiplying by A, B, 0 respectively we get the first three terms of YP' namely, Ax3 + Bx2 + Ox. The only v.p. obtainable from 2eZ is ~, which itself forms the second-group. Since f? is the v.p, of the term 02ez in Ye, we multiply it by x and prefix a general constant E to obtain the fourth term, Exez , of Yp. From sin x the only new v.p. obtainable by successive differentiation is cos z, so that the third group consists of sin z, cos z. Since neither of these is a v.p. of a term of Ye, we multiply by F and G respectively, obtaining, for the last two terms of YP' F sin x G cos a; Now write down YP' and under it write Dyp and D2yp, arranging terms with the same v.p.'s in the same column. Since YP satisfies equation (1), the combination D 2yp - Dyp must be identical with the right member of (1). To produce this combination we multiply equation (4) by 1, equation (3) by -1, and add, placing the multipliers 1 and -1 at the left of the respective equations. We do not use a; multiplier- (other than 0) for equation (2) since y, unaffected by a D, does not appear in the left member of (1). By equating the resulting coefficients of the various v.p.'s column by column to the coefficients of the corresponding terms in the right member of (1), we obtain just enough equations to solve for the values of the undetermined coefficients; any additional equations, although superfluous, must not be contradictory. In this example there are six equations, obtained from the 2nd, 3rd, 4th, 6th, 7th, and 8th columns, for determining the six undetermined coefficients, A, B, 0, E, F, G. Note that the first column yields nothing, and the fifth gives E - E = 0, or merely 0 = 0; if the fifth column had yielded a contradictory re- ' sult, a mistake would have been indicated.

+

I

A short cut. In the case where R is the product of an exponential function e" and some other function of x, it is usually shorter to multiply the equation through by e-az and then

146

Chapter 4

make use of the reverse exponential shift [equation (6), Art. 28] in finding the p.i. As an illustration we shall work the following example first by the regular method just explained, then by use of the short cut. : :, ~__ EXAMPLE

1 (D l'if)-)

2. Solve

First solution. The roots of the quadratic D 2 (D - 1)2 = 0 are 1, 1; hence Yc

=

(C1

-

2D +/1 =

+ C2x)e

te •

Writing the v.p.'s of 2xete and of the terms obtainable by differentiation, we have the group ze", e". Multiply the members of this group by x 2 in order to obtain the group x 3ete, x 2eX , both of whose members are different from the v.p, of any term in Yc. Now multiply these members by A, B, respectively, to obtain yp. 1)

-2)

Dyp = 1)

Ax3e~

D2yp = Ax3e x

+ (3A + B)x2e~ + (6A + B)x2e~ + (6A + 4B)xe~ + 2Be~ 6A+4B-4B=2 A -- .!3

2B = 0 B=O

(Note that the first two columns give 0 = 0.) General. solution: Second solution. The c.f., as before, is

Multiplying the differential equation through bye-X, we have e-X(D - 1)21/ = 2x.

Article 34

147

Applying the reverse exponential shift, D 2e- Z y = 2x.

A p.i. of this equation is obtained by integrating twice, omitting the constants of integration:

General solution: y

= (C1 + C2x

+ ~x3)ez.

If R contains in addition other terms not of the above type, it is often convenient to divide R into parts, then find a p.i, corresponding to a part of the above type by means of the short cut, using the regular method to obtain a p.i. corresponding to the other parts. The sum of the separate p.i.'s will be the complete Yp. For, suppose that the differential equation is

and we find a p.i. Ypl corresponding to R 1 and a p.i. spending to R 2 , that is f(D)YPl

Yp2

corre-

= Rl,

It follows that so that is a p.i. of the original differential equation. As an illustration, take Example 2 with the terms 3x 2 added to the right member: EXAMPLE

3. Solve

(D2

As before,

-

2D

+ l)y =

2xeZ

+ 3x 2 + 4.

+4

148

Chapter 4

Using the short cut to obtain the p.i. Ypl corresponding to 2xez , we have (second solution of Example 2) _

1 3

Z

Ypl - aX e .

To obtain Yp2, that is, a p.i. of the equation (D 2

-

2D

+ I)y =

3x2

+ 4,

we have I) yp2

-2)

+

= Ax2

Bx

+C

2Ax+B

DYp2 =

I) D2Y p2 =

2A

A=3 Yp2

2A-2B+C=4 6 -24+C=4

-4A +B=O B = 12

= 3x2 + 12x + 22

C == 22

General solution: Y = (C1 EXAMPLE

+ C2x + ~x3)ez + 3x2 + 12x + 22.

4. Solve (D 2

The roots of D 2

2D

-

Yc

2D

-

+ 2)y = e

+2 =

Z

cos z,

0 are 1 ± i; hence the c.f. is

= eZ(Cl sin X

+ C2 cos z).

.

Multiplying the differential equation by e-Z gives

e-Z[(D - 1)2 + I]y == cos z, or, applying the reverse exponential shift, (D 2

+ I)e-Zy =

cos z.

We now find a p.i. of this equation, that is, a particular value of e-Zy, by the method of undetermined coefficients. The group of v.p.'s obtainable from cos x is cos x, sin z. But, if we were solving this equation for e-:l:y, the c.f. would be C1 sin x + C2 cos z. Hence

Article 34

149

the members of the above group must be multiplied by x in order to make them different from the v.p.'s of terms in this c.f. We have 1) e-Zyp

=

+ Bz sin x

Ax cos x

D2e - zyp = -Ax cos x e -ZYp YP

+ A cos x + B sin x Bx sin x + 2B cos x - 2A sin x

Bx cos x - Ax sin x

De -zYP = 1)

= Ix sin x = Ixez sin x

2B

=1

B=I

1-

2A. = 0

A=O

General solution: y

= (01 sin x

+ O2 COS X + ~x sin x)f?

If this example had been done by solving the original equation directly by the method of undetermined coefficients, the work would have been considerably longer, since a YP of the form, Axe~ cos x Bae" sin x, would have had to be differentiated twice.

+

PROBLEMS

Solvethe following differential equations. 1. (D2 + 2D + l)y = e-2z • 2. (D2 - l)y = x2 2x. 3. (D2 - 5D - 6)y = 2e3z - ae-2:1:. 4. (D2 - 5D 6)y = 2e3z - 3e-2z • 6. (D3 + D2)y = x2 - X + 2. 6. (2 + D + D2 - 4D3)y = 1 + 2x2• 7. (D2 - 3D 2)y = x 2 - 2ez • 8. (D2 - D - 12)y = e9z - 2e-z + 1. 9. (D3 - 3D - 2)y = 3eZ e-Z - 4. 10. (D2 - 2D - l)y = 3 sin x. 11. (D2 + 4)y = 2 sin x + cos 2x. 12. (D3 - D)y = 2 sin x 4 cos x 6x.

+

+

+

+

+

+

13. (2D2+5D-3)y=W+l+V;=; 14. 2 illy

dx2

+ dy + 8x -

16. d2x2 + x dt

dx =

4t sin t.

4 = e:e/ 2 + 17 cos 2x.

150

16.

Chapter 4 d2

dx~

= 4y

+ 12x2e- 2z •

1'1. (DS + D2 - D - l)y = xeZ

+ e-

sin x.

Z

18. d4y _ d8y = 120(x3 + eSZ). dx4 dx 3

+ +

19. d2y2 - 2 dy 2y = 10 cos x cosh x. dx dx 20. (3D4 8D3 6D2)y = (XS - 6x 2

+

+ 12x -

24)e-~c.

d2y dy . 21. - 2 - 2 - - 8y = 5e- t SIn 2t. dt dt 22. (DS D 2)y = e-Z cos x. 23. (DS + 3D2 - 4)y = 9xe-2z + 4x.

+ + (D3 - D2 + 4D -

(V + +

24. 4)y = 4 sin2 3x/ 2). 26. (D2 4D 404)y = e-Z(sin 20x 40 cos 2Ox). z 26. (DS - 3D 2)y = (1& 6)e ge- 2z • 2'1. (4D2 - l)y = e-z/ 2[x - cos (x/2)]

+

+ +

+

28. Determine x so that it will vanish when t = 0 and when t = log 2 and satisfy the differential equation

d2x dt2

-

dx 3 dt

+ 2x = e3t •

Show that x will vanish for no other value of t. 29. Find a solution of the differential equation

d2y dy 4 dx2 = 4 - x dx

o.

if y = 1/4 and dy/dx = 0 when x = 30. Find a solution of the differential equation

d8y _ 2 d2y dx

3

dx

2

+ dy =

2(eZ

+ x)

dx

if yand its first two derivatives vanish when x =

o.

31. Solve

Q d22 + 106Q = 100' sin SOOt dt if Q = 0, dQ/dt = 0 when t = 0; find the value of dQ/ dt when t = 1/100 (i.e., when 500t = 5 rad).

Article 35

151

35. Variation of parameters. There was developed in the preceding article the method of undetermined coefficients which enables us to find a p.i. of the differential equationj(D)y = R, where R is of a certain special form. It happens that R is of this special form in most of the linear differential equations which arise in practice, but it is well to have a method for finding a p.i. in case an equation is encountered in which the method of undetermined coefficients does not apply. Such is the method of variation of parameters, due to Lagrange. The method is general and can be used theoretically for any form of R, and even for a linear differential equation with variable coefficients * provided the complementary function is known, although it is impracticable to carry it through except for fairly simple forms of R. The difficulty in carrying _it through increases not only with the complexity of R, but also with the complexity of the complementary function. Even for Problem 22, Art. 34, where the complementary function is rather simple, the method is hardly practicable; in Problem 27 of the same article, however, although R is more complicated, the e.f. is simpler, and, as we shall see in Example 2, this problem can be solved by the method of variation of parameters with little more work than by the method of undetermined coefficients. The chief value of the method from the practical point of view is in solving an equation like that of the following Example 1, where R is a function whose successive derivatives continue to yield terms with different variable parts, so that the method of undetermined coefficients would involve a YP having an infinite number of terms. The method of variation of parameters is powerful for use in theoretical considerations and is justly described by mathematicians as "elegant." We shall explain the theory for the differential equation of second order with constant coefficients, fCD)y = CaoD 2 + aiD + a2)y = R, (1) then indicate how it applies to equations of higher order. * For example, Prob. 12 at the end of this article.

Chapter 4

152

First find the c.f. by the familiar method of Art. 29. We use, for convenience, A and B instead of the arbitrary constants C, and O2 , and write the c.f. in the form

Yc = Au + Bo, where u and v are certain known functions of x. Our problem is now to find a p.i. of equation (1), that is, a function such thatf(D) operating on it produces R. We know that feD) operating on the function Au + Bv produces 0 if A and B are constants; it is reasonable to surmise that if A and B were functions of x (which can be called parameters), it might be possible to determine -them (vary the parameters, so to speak) so that j'(D) operating on Au + Bv would produce R instead of o. Accordingly we write y

= Au + Bv

(2)

where now A and B are undetermined functions of z. If we can determine A and B (free of arbitrary constants) so that (2) satisfies (1), the corresponding value of y will be the YP that we are seeking. We need two conditions on A and B in order to determine them. We may choose the first condition in some way that will simplify the problem. But the second condition is forced on us; it is that y = Au + Bv must satisfy equation (1) after the first condition has been imposed. The form in which these two conditions appear is made clear by the following compact arrangement of the problem. Subscripts denote differentiation with respect to z.

= Au + Bv Dy = Au~ + BVt + A1u D2 y = AU2 + BV2 + Atut y

+ B1v = 0 + BtVI = R/ao

Arrange the four terms obtained by differentiating Au + Bv so that the first group of two terms contains A and B, and the

Article 35

153

second group contains Al and B 1 • Now draw a vertical line separating the two groups and set the second group equal to zero. This is the first condition, mentioned above, which we arbitrarily impose on the two functions A and B, namely, that the sum of their derivatives multiplied by U and v respectively shall vanish. Next differentiate the part of Dy which remains on the left of the vertical line, obtaining four more terms for D2 y arranged as before. If now we imagine that the values of y, Dy and D2 y are substituted in (1), the parts to the left of the vertical line yield nothing, since these would be the values of y and its derivatives if A and B were constant. There remains onlyao(A1UI + B1VI) which must equal R if the second condition mentioned above is satisfied. To say that y = Au + Bv must satisfy (1) when A1u + B1v = 0 is the same as to say that A1UI + B1VI = R/ao, which is the second condition that the functions A and B must satisfy. In order to determine A and B, the equations

+ B1v = 0, A1uI + B1VI = R/ao, A1u

must be solved for Al and B 1 ; these expressions are then integrated (omitting constants of integration) to determine A and B. Substitution of these values of A and B in (2) yields a p.i. YP of equation (1). The general solution of (1) is then

y

=

Yc

+ Yp·

If, when Al and B 1 were integrated to obtain A and B, con.. stants of integration had been added and the resulting values of A and B substituted in (2), the general solution of (1) would have been obtained. Since Yc must be found at the beginning, however, it is expedient to use the method of variation of parameters merely to find yp, then add Yc and Yp to obtain the general solution.

154

Chapter 4

EXAMPLE

1. Solve (D 2

The c.f. is Yc = Ct sin x

+ l)y = tan »:

+ C2 cos a:

Asinx+Bcosx

y=

Acosx-Bsinx+

Dy=

D 2 y ,= -Asinx-Bcosx+ A

B

=

-cosx

= sin x

At sin x

At cos x - B1 sin x = tan x At

- log (sec x

+ tan x)

+ B I cos X = 0

= sin x

sin2 x Bt = cos x

= cos x

- sec x

+ tan x) log (sec x + tan x)] cos z

Yp = -cos x log (sec x

Y = CI sin x

+ [C2 -

In solving the two equations to the right of the vertical line for Al and B, we multiplied the first of these equations by sin x, the second by cos z, then added to obtain At; we then multiplied the first equation by cos x, the second by sin z, and subtracted to obtain B t • The expressions for Al and B I were then integrated (omitting constants of integration) to find A and B. When these values of A and B were substituted in the equation y = A sin x + B cos x, the terms -cos x sin x and sin x cos x cancelled and the expression for 'UP was obtained. Finally the general solution was written down by adding 'Ie andyI'.

In studying this method the student may have been curious about the statement that we may choose a condition on A and B that will simplify the problem. In the above example we arbitrarily imposed on A and B the condition that Al sin x + B, cos X = O. Was it necessary to impose this particular condition? Suppose we had written some other constant or a function of x instead of 0, would the result have been the same? The answer is that this particular condition was not necessary; it was convenient-it simplified the problem. If the student has sufficient curiosity, he should try working this example by

Article 35

155

imposing the condition A 1 sin x + B 1 cos X = »; the final result will be the same but the work will be longer. We shall now solve Problem 27, Art. 34, by the method of variation of parameters, omitting explanations. EXAMPLE

2. Solve (4D 2

l)y = e-.,/2 (x - cos ;)-

-

The c.f. is Yc = C1ez / 2

+ C2e-

Z

/

2

•

y= AeZ/ 2+ B e-z / 2 Dy=

A 1ez / 2 + B 1e- z / 2 =0

~AeZ/2- ~Be-z/2+

D2y= lAez/2+ lBe-Z / 2+ ~Alez/2- ~Ble--z/2 = le-z/ 2[x-cos (x/2)]

z /2

y=C1e

+

x 1 x) e" C - -- -x+-2 . -+2

('

2

X

4

8

5

SIn

2 5

COS -

Z/2'

2

The method of variation of parameters is applicable theoretically to differential equations of any order, but practically it is seldom used for equations of higher order than the second. The amount of work necessary to carry it through increases rapidly with the order of the equation, owing to the necessity of solving a system of simultaneous equations for At, BIt ... , the number of equations in the system being equal to the order of the differential equation.

156

Chapter 4

In order to illustrate the method for a third order equation, we give the following example, although it could be solved much more readily by the method of undetermined coefficients. Note that here we need three conditions for determining At, BI, CI, so we arbitrarily set the tails of Dy and D2y equal to zero for the first two conditions. ExAMPLE

j(D)

=

3. Solve

(D - l)(D - 2)(D - 3)

+ Ce3a: Dy = A~ + 2Be2a: + 3Ce3z + D2y = A~ + 4Be2a: + 9Ce3z + D 3y = A~ + 8Be2z + 27Ce3a: + y

= Aea: +

Be2a:

+ B le2z + Cle3a: = 0 AI~ + 2B 1e2z + 3Cle3a: = 0 Al~ + 4B1e2z + 9Cle3z = e-z Alez

Solving the system of three simultaneous equations to the right of the vertical line for At, Bt, 0 11 we obtain I.

A --

-4: e-2z

- - e-3z B1 -

B =

~e-3a:

4 a: C1 -- .le2

C --

le-4a: -8

Al

=

~e-2a:

4: YP -- -I.e-a:

+ I.e-:l: 3

I.e-a: -- _ 8

Y -- C1ea: + C/2 e2z + C-'3 e3Z -

~-a: :Il4~

Le-a: 24 •

PROBLEMS Solve the following differential equations, working 5, 9, and 10 by method of Art. 34 and also by the method of Art. 35. 1. (D2 l)y = esc z. 2. (D2 4)y = cot 2z. 3. (9D2 l)y = sec (zj3). 4:. (D2 l)y = tan2 z. 6. (D2 - l)y = sin 2 x.

+ + + +

th~

Article 36 6. 7. 8. 9. 10.

157

(4D2 + 4D + 5)y = 4e- z / 2 sec x + 4ez / 2• (D2 - l)y = x2e z2 / 2• (D2 - 2D + 2)y = 3x + eZ tan x. (D2 + l)y = x cos x. (D3 - 7D - 6)y = 26e-2z cos x.

11. (a) Solve by the method of variation of parameters:

(D2 + 3D

+ 2)y =

sin ete•

(b) Show that; although successive differentiation of the function sin ~ continues to yield terms with different variable parts, nevertheless the differ-

ential equation may be solved without using the method of variation of para-

meters. 12. Solve by the method of variation of parameters: (D

+ P)y = Q,

where P and Q are functions of x, obtaining the solution (8), Art. 19.

36. Forced vibrations. In Example 6, Art. 30. we saw that the differential equation 10 d2x

- - 2= g dt

1 dx

-20x--2 dt

(1)

represents the vibrations of a 10-lb weight, hung on a spring with spring constant 20, in a medium offering resistance numerically equal to half the velocity. If resistance is negligible, the last term is absent and the differential equation 10 d 2x

--2 =

g dt

(2)

-20x

represents the special case of simple harmonic motion in a nonresisting medium [Problem 8, Art. 30(b)]. These differential equations, of standard form. (D2

+ 2aD + b2)x = 0, (D2 + b2 )x = 0,

(a

< b),

158

Chapter 4

respectively, represent free or natural vibrations, that is, vibrations due to the inherent forces in the system. The system involved in (1) consists of the spring and the weight, together with the resisting medium in which they are immersed; in (2) the system is merely that of the spring and the weight. Suppose now that there is applied to the weight, vibrating in accordance with equation (1), a periodic force A sin wt, external to the system, of amplitude A and period 211"/w. The differential equation of the motion will then be 10 d 2x

g dt2

1 dx

= - 20x - 2

dt

.

+ A SIn wt,

of standard form (D2

+ 2aD + b

2)x

=

C sin wt,

(a

< b).

(3)

Neglecting the resistance due to the medium, the standard form of the ditierential equation is (D2

+b

2)x

= C sin wt.

(4)

Equations (3) and (4) represent forced vibrations respectively with and without resistance proportional to velocity and due to the medium. The solutions of these equations consist of a complementary function Xc plus a particular integral Xp • The form of the solution of equation (4) will depend on whether w = b or ClJ ¢ b, that is, whether the period 211"/w of the impressed external force is or is not equal to the period 2r/b of the natural vibrations. If w = b, we have the case of resonance, where the vibrations get larger and larger. This condition would cause dangerous stresses in some vibration problems; on the other hand a resonant condition is desirable in certain acoustical and radio-circuit problems. EXAMPLE

1. A 12-lb weight hangs at rest on a spring which is

stretched 4 in. by the weight. The upper end of the spring is given the motion y = sin .y2gt, where y (ft) is the displacement, measured positive upward, of the upper end of the spring from its original posi-

Article 36

159

tion at time t (sec). Find (a) the equation of motion of the weight, (b) the position of the weight 7r/ v'3Y sec after the motion starts. Since a force of 12 lb stretches the spring ~ ft, 12 = c- ~, and the spring constant is c = 36 lb1ft. Take the origin at the equilibrium position of the weight and let x (ft), measured positive upward, represent the displacement of the weight from its equilibrium position at time t (sec). Now when the upper end of the spring is raised a distance y the weight will follow a (smaller) distance z. Thus there is an additional stretch of y - x in the spring, producing an upward force of 36(y' - x) lb. The differential equation is therefore 12 fPx

- -2 = 36(y -

z).

9 dt

By substituting y = sin ~t, multiplying by 9112, and using operator notation, the differential equation takes the form (D 2

+ 3g)x = 3g sin v'2Yt.

This equation is of standard form (4) with w ~ b. The eomplementary function is Xc

= C1 sin v'3Yt + C2 cos v'3Yt.

For the particular integral we have 3g) Xp

1)

=

A sin V2!jt

+

B cos V2!jt

D2X p = - 2gA sin V2!jt - 2gB cos V2!jt x p = 3 sin V2!jt

gA = 3g gB A=3

=0

B=O

Hence the general solution is x = C1 sin

The condition x yields

v'3Yt + C2 cos V3gt + 3 sin ~t.

= 0, t = 0,

gives C2

= 0.

Then differentiation

160

Chapter 4

and the condition dx/dt = 0, t = 0, gives

The equation of motion is therefore x

When t

= - V6 sin ~t + 3 sin v'2Yt.

= 7r/~,

Z]'=r/V3U

we find

= -V6 sin 11" + 3 sin ~ = 0 + 3 sin 146° 58' = 3(0.5451) = 1.64.

Hence, when t = 7r/v'3g sec the weight is 1.64 ft above the equilibrium. position.

37. Electric circuits. In Art. 23 electric circuits leading to linear differential equations of first order were discussed. Using the same notation as in that article, the differential equation for a circuit containing resistance, inductance, and capacitance, is * i = Dq.

(1)

Here D = djdt and, as in Art. 23, the capital letters L, R, and C represent constants of the 'circuit whereas the small letters are, in general, variables. 1. An inductance of 1 henry and a capacitance of 10-6 farad are connected in series with an emf e = 100 sin 500t volts. If the charge and current are both initially zero, find (a) the charge and current at time t sec, (b) the value of the current when t = 0.001 sec, (c) the maximum value of the current. Substituting L = 1, R = 0, C = 10-6 and e = 100 sin 500t in (1)1 we have the differential equation EXAMPLE

(D2

+ 106 )q =

100 sin 5OOt.

* For references to the derivation of this equation see footnote, Art. 23.

<,

Article 37

161

(a) The solution-finding the complementary function and par-

ticular integral, and evaluating the constants of integration-proceeds as follows: qc = 0 1 sin l000t + 02 cos l000t 6 10 ) B cos 500t A sin 500t + 1)

D2qp = - 5002 A sin 500t - 5002 B cos 500t (106

5002)A = 100 (106

-

-

1 A = 7500

q

5OO2)B = 0

B = 0 .

= qc + qp = 0 1 sin l000t + 02 cos 1000t + 7:00 sin 500t q

i

i == 0,

= t

Dq =

= 0,

t = 0;

O2 = 0

= lOOOCl cos lOOOt + I~ cos 500t 0 = 100001

0; 1

(

1

+ 15 '

.

.

01

1

= -

15000 ,

)

q = 15 000 2 sin 500t - sin l000t .

,

i =

I~ (cos 500t -

cos lOOOt).

(b) When t = 0.001 sec,

i],=o.OOl =

I~ (cos 0.5 -

cos I)

=

h

(0.8776 - 0.5403)

0.3373 15 = 0.0225 amp.

(c) The maximum (and the minimum) value of the current occurs

when . 500. 00 1000. D t = - 15 SIn 5 t + 15 SIn loo0t = 0, or, if 500t == T, when 2 sin 2T - sin T = 4 sin T cos T - sin T = sin T( 4 cos T - 1) = O.

162

Chapter 4

For sin T = 0:

For cos T

T = 0,

i = O.

T = 7r,

i

= l:

= ~ (cos 7r -

cos 27r)

15

1 cos 2T = 2 cos2 T - 1 = - - 1 8

=-

.!. 15

7

= - 8-;

Hence the maximum value of the current is 3/40 amp. The minimum (maximum absolute value) is attained when T = 7r; the current varies between -2/15 and 3/40 amp. PROBLEMS 1. In Ex. 1, Art. 36, how far will the weight rise before starting downward? 2. Solve Ex. 1, Art. 36, if the upper end of the spring is given the motion y = sin .y3gt. 3. An 8-lb weight is hanging at rest on a spring which is stretched 1 ft by the weight. The upper end of the spring is given the motion y = sin 2..y'gt, where y (ft) is the displacement, measured positive upward, of the upper end of the spring from its original position at time t (sec). Find (a) the position of the weight 7r/(4..y'g) sec after the motion starts; (b) the extreme positions attained by the weight. 4. If the upper end of the spring in Prob. 3 is given the motion y= sin ..y'gt, find (a) the position of the weight when t = 7r/(3yg)sec; (b) the time when the weight first passes through the equilibrium position. 6. A lO-lb weight is hanging at rest on a spring which is stretched 4 in. by the weight. If the upper end of the spring is given the motion y = sin y'2gt, find the position of the weight when t = 7r/..y'g sec. 6. Solve Probe 5 if the upper end of the spring is given the motion y = sin .y3gt. 'I. Solve Prob. 5 if the resistance (lb) of the medium in which the weight vibrates is equal to 20/..y'g times the speed (ft/sec). 8. Draw a graph showing i from t = 0 to t = 7r/250 for Ex. 1, Art. 37. 9. In the circuit of Ex. 1, Art. 37, if e = 100 sin 1000t volts, find the current when t = 0.01 sec. 10. An inductance of 1 henry is connected in series with a capacity of 10-4 farad and an emf e = 100 sin 50t volts. If q = i = 0 for t = 0, find (a) the current when t = 0.02 sec; (b) the maximum value of the current.

163

Article 37

11. Solve Prob. 10 if e = 100 volts. 12. A circuit consists of an impedance coil of inductance L and negligible resistance connected in series with a condenser of capacity C and an emf e = 120 sin 250t volts. Assume i = 0, q = 0, when t = O. If L = 2 henries and a = 2 X 10-6 farad, find (a) the current when t = 0.01 sec; (b) the values between which the current varies. 13. A condenser of capacity 2 X 10-4 farad, having a charge of 0.05 coulomb, is placed in series with a coil of inductance 2 henries and a resistance of 28 ohms, and discharges by sending a current through the circuit. The current being initially zero, find the current and the charge on the condenser when t = 0.1 sec. 14. An inductance of 1 henry, a resistance of 400 ohms, and a condenser of capacity 1.6 X 10-5 farad are connected in series with an emf of 40 cos 250t volts. If the charge and current are both zero when t = 0, find the current when t = 10-3 sec. 16. An inductance of 1 henry, a resistance of 100 ohms, and a capacitance of 10-4 farad are connected in series with an emf of 100 volts. If no charge is present and no current is flowing at time t = 0, find the maximum value of the current. 16. An inductance of 2 henries, a resistance of 100 ohms, and a capacitance of 2 X 10-4 farad are connected in series with an emf e = 100 sin 50t volts. Initially the current is zero and the charge is 0.05 coulomb. Find the charge and current when t = 0.02 sec. . 17. A particle slides freely in a tube which rotates in a vertical plane about its midpoint with constant angular velocity w. If x is the distance of the particle from the midpoint of the tube at time t, and if the tube is horizontal when t = 0, show that the motion 'of the particle along the tube is given by

d2x

dt2

-

w2x = -g sin wt.

Solve this equation if x = Xo, dx/dt = Vo, when t = O. For what values of Xo and Vo is the motion simple harmonic? 18. (a) A cantilever beam of length L in. and weighing w lb/in. is subjected to a horizontal compressive force of P lb applied at the free end. Taking the origin at the free end and the y-axis positive upward, show that the differential equation of the elastic curve of the beam and the maximum deflection are, respectively,

( + P) W;I (1 - ~ D2

Maximum de1lection =

wx

2

EI Y = - 2EI' sec 9 + 9 tan 9) •

9=

~~L. EI

Chapter 4

164

(b) Find the maximum deflection of a wrought iron cantilever beam 2 in. by 4 in. by 12 ft with E = 15 X 1()6 lb/in." and weighing 490 Ib/ft3, if the 2-in. side is horizontal and the compressive force is P = 6000 lb. 19. (a) A cantilever beam of length L in. and weighing w lb/in. is subjected to a horizontal tensile force of P lb applied at the free end. Taking the origin at the free end and the y-axis positive upward, show that the differential equation of the elastic curve of the beam and the maximum deflection are, respectively,

( Maximum deflection

D2 _

.!-.-) y = _ wx EI 2EI'

= w;.I ( 1 + ~ - sech (J -

2

(J tanh (J) ,

(J

=

~P L. EI

(b) Find the maximum deflection of a wrought iron cantilever beam 2 in. by 4 in. by 12 ft with E = 15 X 106 Ib/in,2 and weighing 490 Ib/ft3 if the 4-in. side is horizontal and the tensile force is P = 6000 lb. 20. If, in Probs. 18(a) and 19(a), a weight of W lb is added at the free end, and the weight of the beam may be neglected, show that the formulas for maximum deflection are, respectively,

Compressive force P:

Maximum deflection _

Tensile force P:

Maximum deflection

=

~L

(ta:

(J -

1),

~L ( 1 - ~ (J}

e~5. SOME SPECIAL HIGHER ORDER EQUAliONS

38. Equations reducible to linear with constant coefficients. In this chapter some of the common types of differential equations of order higher than the first will be discussed, beginning with a special kind of linear equation of nth order, dny dy n-l d",-ly aox'" dx'" + alx dX"'-l +...+ an-IX dx + anY = R. (1) Here the a's are constants, R is a function of X (or a constant) and n is an integer greater than 1; if n were equal to 1 the equation would be a special case of the linear equation of first order treated in Art. 19. It will be noticed that this equation differs from that studied in Chapter 4 by having variable coefficients; the exponent of x, however, in each coefficient must be the same as the order of the derivative which it multiplies. Such an equation can be solved by reducing it to the type treated in Chapter 4, that is, by reducing it to a linear equation with constant coefficients. First write the equation in operator notation: (aox"'n'"

+alx",-lD",-l + ... + an-2x2D2 + a"'_lxD + an)y = R. (2)

It will now be shown that a change of independent variable from X to z by means of the transformation x = e' will reduce equation (1) or (2) to a linear differential equation with constant coefficients. Use will be made of two operators-that of equation (2) indicating differentiation with respect to z, D = d/dx, and a script P) indicating differentiation with respect to z, §l) = djdz. 165

166

Chapter 5

We shall express xD, x 2D2, etc., in terms of ~.

x = eZ ,

or z = log x. dy dy dz dy 1 xDy = x dx = x dz dx = x dz x = ~y. Hence

xDy =

~y,

(3)

and

xD =~.

(4)

Operating on equation (3) with D, XD2 y + Dy = D~y. Transposing the term Dy, multiplying by x, and making use of (4), Hence (5) and

x 2D2 =

~(~

- 1).

(6)

Operating on equation (5) with D, x 2D3y + 2xD2y = D~(!?) - l)y. Transposing the term 2xD2y, multiplying by z, and making use of (4) and (6), x3D3y = xD~(~ - l)y - 2x 2D2 y = [~2(~

- 1) - 2~(~ - l)Jy = ~(~ - 1)(~ - 2)y.

Hence

x3D3y =

~(~

-

1)(~

x3D3 =

!!)(~

- 1) (~ - 2).

- 2)y,

(7)

and

(8)

Looking at (4), (6), and (8), we have a strong suspicion that

xnDn = ~(~- 1)(~- 2)·· .(~- n + 1). (9) This formula may be established by mathematical induction. The solution of equation (1) or (2) now proceeds AS follows: (aoxnDn + alxn-1Dn-l +... +an_2x2D2 + an_lxD + an)y = R ...

Article 38 .

ON

ON

THE LEFT, SUBSTITUTE

xD = x 2D 2

167

~

x=tr

= ~(~ -

x3D3 =

THE RIGHT J SUBSTITUTE

log x = z

1)

~(fJ) -

1)(~

- 2) •

•

Solve the resulting equation, by the method of Chapter 4, for y in terms of z, Then substitute z = log x, or eZ = x, to obtain y in terms of x. EXAMPLE

1. Solve 3

X

d3y 2 d2y dy 2 dx3 + X dx 2 - 4x dx = 3x

+ log x + 2.

We write the steps in the solution, following the preceding theory without further explanation. (x3D3 + x 2D2 - 4xD)y = 3x 2

+ log x + 2

+ fJ)(~ - 1) - 4~]y 3e2 z + z + 2 3fJ)y = ~(fJ) + 1)(~ - ~)y = 3e +z + 2 Yc = ci + C2e-a + c3e3a 2a YP = Ae + B Z2 + Oz +2Bz + C

[fJ)(fJ) - 1)(fJ) - 2) (fJ)3 - 2!t)2 -

z::

2z

+2B

SA-SA-6A=3

A = -~

-6B

B 2z YP -- _ !'e 2

Y

=

Cl

2+ C3 x 3 +C ~

-4B - 30 = 2

0=-:

= -~ _

_ Cl + C2e-z + Cae3z -

Y-

=1 l 6 Z2

1 _2z 2e- -

1 __2 2x- -

'!'z 9

1 _.2 6Z- -

4

"fZ

112 41 og x og x - "9"

"6

168

Chapter 5

A generalization of equation (1) is obtained by replacing the x's in the coefficients of the derivatives by a + bz, The transformation a + bx = eZ ~~en reduces the equation to an equation with constant coefficients. If the preceding theory is worked out for this transformation, we have the following process for solving the differential equation. [ao(a + bx)nDn +...+ an_2(a + bX)2D2 + an-lea + bx)D + an]y = R. ON

ON

THE LEFT, SUBSTITUTE

+ bx)D = bPJ (a + bX)2D2 = b2P)(~ (a + bX)3D3 = b3P)(PJ -

THE RIGHT, SUBSTITUTE

(a

a+bx=e' 1) 1)(~

- 2)

x=-b log (a + bx) = z

Solve the resulting equation for y in terms of z, then substitute z = log (a + bx) to obtain y in terms of z, EXAMPLE

2. Solve [(3x

+ 1)2D2 + (3x + I)D -

Solution: [9~(~ - 1)

+ 3~ -

3]y

= (~- 1)(3~+ l)y = = C1e z + C2e- z / 3

-1) ~yp = ~2yp =

+ Aez Azez + 2Aez Azez

6A-2A=1

A -=

1 4

-B =-1

B=1

+ C2e- z /3 + lze z + 1 C l(3x + 1) + C2(3x + 1)-1/3 + ~(3x + 1) log (3x + 1) + 1 Y

y

e- - 1

+B

-2)

3)

9x.

3]y = 3(e' - 1)

(3!J)2 - 2~ - l)y Yc

=

=

Cle z

169

Article 38 PROBLEMS

1. If D = d/dx, g) = d/dz, x x"D"

S. If D

=

eZ, prove by mathematical induction that

=

g)(g) - 1) (g) - 2) ... (g) - n

=

d/dx, g) = d/dz, a + bx = eZ, show that

(a

+ bx)D = b~,

(a

+ bX)2D2 =

+ 1).

b2g)(~ - 1).

-

Solve the following differential equations.

d3y 3. x dx3 3

4. x 2

-

Z+

dy 2x dx + 2y 2y

+ x log x.

x:

=

4x3•

=

5. x 2d2y _ 2x dy + 2y = 4x2 + sin log z. dx2 dx d2y 6. x2 - 2 dx

-

dy dx

X -

-

3y

8x3 log x

=

+ 6.

7. (x3D3 + ~2D2 + 8xD

+ 2)y = x2 + ax -

8. (x4D4 - l1x 2D2)y

+ log x.

= X

4.

d3y d2y 4 9. x dx3 + 2x3d;2 + 2xy - 10 = O. 10. (xSD4

+ 6D)y = 11 log x. x

3

11. x 2 d y + dy = 7 log5 X dx3 dx

d2y 12. 4x2 dx2 + y

=

x

_ J: 3v x - 2x3.

13. (4z8D3 - 8x 2D2 - xD

14. (x2D2 + 4xD

+ '#..

+ l)y = x + log x + log2 z.

+ 2)y = x + sin x.

(Cj. Prob. 11, Art. 35.)

d2y 15. (2x - 3)2 dx2 = 2x - y.

16. (1 + 3.1:)3:; - 18(1 + 3.1:) :

= 81(1 + x).

17. [(2x - 1)3D3 + (2x - I)D - 2]y = 4x.

Chapter 5

170 18. Find the solution of d2y

x2 -

tb;2

dy

-

x- = dx

ax - x2

if y = 3 and dy/dx = 2 when x = 1. 19. A steam pipe has inner and outer radii rl and r2 respectively, and the temperatures at its inner and outer surfaces are, respectively, Ul and U2. Show that the temperature U at radial distance r(rl < r < r2) is given by the differential equation

Solve this equation under the conditions U = Ul when r = rl and u = U2 when r = r2. (Cf. Probe 26, Art. 16.) 20. A hollowspherical shell has inner and outer radii T1 and r2 respectively, and the temperatures at its inner and outer surfaces are, respectively, Ul and U2. Show that the temperature U at radial distance r(rl < r < r2) is given by the differential equation

d2u du r-+2-=O. dr

dr

Solve this equation under the conditions u = Ul when r = rl and when r = r2. (Cf. Prob. 27, Art. 16.)

U

= U2

39. Dependent variable absent.

Consider a differential equation involving an independent variable x and a dependent variable y, but one in which y is absent except in the derivatives. If in such an equation we substitute for dyjdx a single letter p, * then substitute for the higher derivatives the corresponding expressions in terms of p and x, the result is an equation of lower order in p and x. If the latter equation can be solved in terms of p and x, we then substitute dyjdx for p and try to solve the resulting equation for the relation between y and x. The substitutions are dy dx

= p,

* H dy/ dx also is absent, but two or more higher derivatives are present, we set the derivative of lowest order present equal to p and make corresponding substitutions for the higher derivatives.

171

Article 40 EXAMPLE

1. Solve d 2y

x dx2-2

dy

3

ax=x

+x.

Solution:

x

dp 3 +x, dx-2p=x dp 2 - -p dx x

-

= x 2 + 1.

The integrating factor of this equation of first order, linear in p and dp/dx, is ef<-2/z) dz = e-2 1og z = 1/x2• Hence

1 = /1-

-2 . p

x

x

2

(~

1

+ 1) ax = x - - + 0 11 X

40. Independent variable absent. If, in a differential equation involving the independent variable x and the dependent variable y, x is absent except in the derivatives, we substitute for dyjib; a single letter p, * then substitute for the higher derivatives the corresponding expressions in terms of p and y. The result is an equation of lower order in p and y; if it can be solved in terms of p and y, we then substitute dyjib; for p and try to solve the resulting equation for the relation between y and x. The substitutions are dy

dx

* See footnote,

=

Art. 39.

p,

d2y

dp dy

dp

- 2- - - - p dx - dy dx dy ,

172

Chapter 5

If the dependent and independent variables are both absent, as in Problem 12 of the following group, then either the method of Art. 39 or that of Art. 40 may be used, but the former is usually simpler. EXAMPLE

1. Solve lPy dx2

=

dy. 2y dx'

and find the value of y when x = -rjl2 under (a) the conditions y = 1, dyjdx = 2 when x = 0; (b) the conditions y = 0, dyjdx = -1 when x = O. Substituting p for dyjdx and p dp/dy for d2yjdx2 in the differential equation (1)

it becomes dp p dy

=

2yp.

Dividing by p [p = 0 would give the trivial solution, y = 0, of equation (1) which would not satisfy the given conditions], dp

= 2ydy.

Integrating, p =

'If + 0 1 . , - J

(2)

In problems of this kind the constants of integration should be evaluated as soon as they appear. . (a) Under the first set of conditions we have, from equation (2), . 2 = 1 C1, C1 = 1, so that

+

dy y2

+1=

The next integral is tan-1 y = x

dx.

+ C2 ,

and, under the given conditions, tan-II = 0 tan -1 y =

+ -4 , 'lr

X

+ C2, C2 =

'lr/4; hence

• Article 40

173

which is the solution of equation (1) satisfying the first set of conditions. For the value of iJ when x = -7('/12, we have 7(' Y]Z=-1I" /12 = tan 6 = 0.577. (b) Under the second set of conditions we have from equation (2), -1 = 0 01, 0 1 = -1, so that

+

dy

.

y2 - 1 = dx.

The next integral is

-tanh-1 Y = x

+ O2 ,

and, under the given conditions, - tanh " ! 0 = 0 hence

tanh'"!

y =

+ O2 ,

O2

=

0;

-x,

y = -tanh z, which is the solution of equation (1) satisfying the second set of conditions. For the value of y when x = -7('/12, we have y1';=-11" /12 = tanh 1~ = tanh 0.2618 = 0.256. PROBLEMS Solve the following differential equations.

1. x dl-y = dy _ X(dy )2. dx 2 dx dx 2. xy" + (x2 - l)y' + 1 = x 2• 3. xy" '= y' y'3 (cf. Prob. 11, Art. 8). 4. x2y" + y'2 = 3xy'. d4y d3y 6. X dx4 + dx 3 = 6.

-+

6. (y

d2y + 1) dx - 2 =

(dY)2 •

2 -

dx

7. 2yy" - s" = o. 8. 2yy" + v" = o. 9. 2yy" - y'2 = 4y2. 10. 3 ('::.)'

~ 2:~ = o.

174

Chapter 5

11. Show that the differential equation of Probe 7 can be changed into the differential equation of Probe 8 by changing y to y3. 12. Solve y" y'2 4 = 0 by (a) the method of Art. 39; (b) the method of Art. 40. 13. Solve Prob. 18 of Art. 38 by the method of Art. 39. 14. Find a curve which satisfies the differential equation

+

+

2y'2 + y2

w" = and has slope

va at the point (0, 1).

Find a solution of each of the following differential equations satisfYing the given conditions.

15 f1ly == • fb;2

~ y2

(: =

2

1,11= 1, when x = 1)= 1, when x == 0).

16. (1 - x 2 )y" == xy'

(y' == 2, y

17 2 d2y = ell

(y' == -1, y = 0, when x

•

fb;2

=

0).

18. (1 - x) (y" -.y') + y' = 0 (y' == -1, y = 0, when x == 0). d2y dY , 19. dx 2 = sec2 y tan y ( dx == -1, x = log 2, when y == 4 .

1(')

d2

SO. cos" Y ~ = sin y

(: =

0,11 = 0, when x =

o)

Find the particular value of the variable specified for each of the following differential equations with given conditions.

dy 31r) • " ( Y = 1, dx == 2, when x == 4" · Find YJz=-tr/O

(11 = 8, :

= 4,

(11 = 0, :

= -1, when x

(: =

1, x

(11 = 2, : +

26. Given v" yy' = 0, with y values of y and of y' when x = -1.

= -

when x

= 1)=

o

Findll].-40

0). Find 11].-.

~ , when 11 = 0)-

Find x]

.

= 1, when x = 1)- Findll] ... =

-1 and y' '

= f when x == 0; find the

Article 41

175

41. The general linear equation of second order. differential equation d2y

dy

11 dx2 + 12 dx + faY

=

The

/4,

where the/'s are functions of a, may, by dividing by /1, be written in the form v" + Py' + Qy = R, (1) where primes denote differentiation with respect to », and P, Q, and R are functions of z. In an attempt to solve equation (1) suppose that we try the effect of substituting y = UV, where u and v are unknown functions of x. Then y' = uv' + u'v, y" = uo" + 2u'v' u"v, and equation (1) becomes

+

or

us" + 2u'v' + u"v + P(uv'

+ u'v) + Quv = R, uv" + (2u' + Pu)v' + (u" + Pu' + Qu)v = R.

(2)

One way to simplify equation (2) is to choose u so that u" + Pu' Qu = 0, thus causing the terms in v to disappear from equation (2). This means that y = u satisfies equation (1) when R = 0; that is, y = u is a particular solution of the equation . y" + Py' + Qy = O. (3)

+

If then we have given, or can find in some manner, a value u of y which satisfies equation (3), the substitution of y = uv in (1) reduces it to

dv' dx

+ (2U' + p) v' = R .

(4)

u u This equation is linear in v' and dv'/ dx, and may be solved in terms of v' and z by the method ofArt. 19. (If R = 0 the equation is separable in v' and z.) Then, writing v' = dv/dx, a second integration yields v. Having u and v, we multiply them together to obtain the general solution, y = uv, of equation (1). Quite often a particular solution of (3) may be determined by inspection. For example, y = x is a particular solution if p = - Qx, and y = eZ is a particular solution if 1 + P + Q = o.

176

Chapter 5

EXAMPLE

1. Solve (1 - XJ)y" - xy'

+y =

3XJ.

By inspection y = x is a solution of (1 - x 2)y" - xy' hence we substitute y = xv,

+ v,

y' = xv'

y" = xv"

+ 2v',

+ v) + xv

= 3x2 •

+y =

0;

in' the given equation, obtaining

(1 -

x 2)(xv"

+ 2v')

- x{xv'

The two terms containing v cancel, as they should, and the equation reduces to or dv' dx

+

(2

x) , x - 1 - XJ v

3x

= 1 - x2 •

By the method of Art. 19, the integrating factor is =

ef[(2/Z)-(z/1-z2)] dz

e2 log z+~ log

(1-z2)

+ c;. =

-(2

= x2 y 1 _ x 2 •

Hence 2

2

x V 1 - x v'

=

f-

3

/3x dx v 1- x 2

.:»

v --- dx -

+ C~,

c;.

2 x

x

V1 -

C~Yl - x2

X

X

v= - - x-

y

1 - x2

---1+--n-~=== 2 2

2

and the general solution,

+ x 2 )V

y = xv,

x2

'

+ C2 ,

is

= 2 - x 2 + C1 V 1 - x 2 + C2x.

42. The inverse square law. If a particle moves from rest under the action of a force varying inversely as the square of the distance z from a point 0, the differential equation of motion, taking the z-axis as the line of motion, is d 2x k2 dt2 = =F x 2

,

(1)

Article 42

177

the negative or positive sign being used according as the force is attractive or repulsive. To solve these equations we use the method of Art. 40, ietting dx/dt = v (velocity) and d2x/dt2 = v dv/dx; then equations (1) become (2)

Assuming that the particle starts from rest at x = a, the initial conditions are v = 0, x = a, when t = o. We continue the solution of equations (2) in parallel columns: FOR A REPULSIVE FORCE

FOR AN ATTRACTIVE FORCE

k2

".

= - - +01 Z

-

2

" = 0, z = a; v=

-VIDe~!z

:-

v

(;

k'\J2

Let e

f

t =

t

')n.

dz

ooe' 8 d8

,

~k'\J2~ (cos 8 sin 8 + 8) + C2

= 0, z = at 8 = 0;

Z a

Vz-a=V~sinh8

= V~ sin 8

~f

(3)

Let z = a cosh2 8

dz = -2a cos 8 sin 8 d8

t = To

= V2k~!a !z

t-! ~fV;dz - k"';2 Vz - a

V;dz v' a - z

= acos2 8

Va - z

01 = a

:=k~~Z

_k~~a z z 1

t = -

!a

k2

C2

=0

t

=!k "';"2 ~ f2a cosh2 8d8

t t

= 2a cosh 8 sinh 8 d8

i~

(cosh 8 sinh 8

= 0, z = a,

+ 8) + C,

8 = 0; C2 = 0

178

Chapter 5

y

108

o FIG. 12

T sin

(J

=

(1)

W8,

T cos (J = H.

(2)

Take the y-axis vertical through A and the z-axis horizontal at a distance a (ft) below Aj a value for a will be chosen later which will make the equation of the curve come out in simplest form. Dividing (1) by (2) and writing dyjdx for tan (J, we have the differential equation of the curve, dy

W8

(3)

dx = H·

Equation (3) contains three variables x, y,

8.

The

8

may be

179

Article 43

eliminated by differentiating with respect to x and replacing 'dsjdx by VI + (dyjdx)2:

~ = ;; ~1 + (~t --,

(4)

Equation (4) can be solved by the method of Art. 39, letting dyjdx = p, d2yjdx2 = dpjdx; then

= w da:

dp

VI + p2 Integrating, sinh"? p = Since p

= dyjdx = 0 when x =

H

;;x + 0

0, 0 1 = 0 and

dy . h W dx = sin H

Integrating again, Y=

H W

1•

W

cosh H X

X•

+O

2•

But Y = a when x = o. In order to make O2 a = Hjw, and the equation of the curve is y

= 0, we choose

x = a cosh-, a

(5)

the standard equation of a catenary; x and yare the coordinates of any point P, referred to the origin 0 which is at a distance a = Hjw ft below A. The length of the arc AP is 8

=

is ~ + (~r is~ + dx =

1

=.

l

z 0

x a

cosh - dx

1

= a sinh -ax .

sinh" : dx (6)

Chapter 5

180

Suppose (Fig. 13) that a cable of length S ft is suspended from two points at the same level and L ft apart and that it dips d ft. Then, from equations (5) and (6),

d=

Y]z=L/2 -

a

=

a

(COSh ~ -

1),

S = 2a sinh L . 2a Y

(7)

(8)

L 2'

T

'Y

o FIG.

The tension at P is T

=H

sec

(J

= H ~1 +

13

(Zr =

H cosh : = wy.

(9)

1. A wire, fastened at the same level to two poles 120ft apart, dips 30 ft. Find (a) the length of the wire; (b) the tension at the lowest point and at a point of support if the weight of the wire is 0.10 lb/ft. Substituting d = 30, L = 120, in equation (7), EXAMPLE

30

= a

(Cosh ~ - 1),

from which the value of a is to be found. It is simpler to let 6O/a = A, then solve for A the equation

Article 43

181

Using the method of trial and error, we find, from Peirce's "Tables," X

x -+1 2

cosh x

1.0 . 0.9 0.930

1.500 1.450 1.465

1.543 1.433 1.465

Hence A = 0.930, a = 60/0.930. (a) Substituting in equation (8), the length of the wire is 120 .

S

128.4

= 0.930 SInh 0.930 = 0.930 = 138 ft.

(b) The tension at the lowest point is

H = aw = 6/0.930 = 6.45 lb. The tension at a point of support is, from equation (9), T = w(d + a) = wd + H = 3 + 6.45 = 9.45 lb. •

PROBLEMS Solve the following differential equations. 1. x2y" - (x2 + 2x)y' (x + 2)y = 2x 4• 2. xy" - y' + (1 - x)y = o. 3. xy" - (2 + x)y' + 2y = 2x 3ez • 4. (2x2 + 3x + l)y" + 2xy' - 2y = O. 6. xy" (2x + l)y' + (x l)y = (2x + l)ez • 6. Solve the differential equation xy" + (2 - x)y' - y = 2 cos x, given that y = l/x is a particular solution of xy" + (2 - x)y' - y = o.

+

+

+

7. In the motion discussed in Art. 42, assuming that k is the same for both attraction and repulsion, i.e., that the two forces have the same magnitude at a given distance from 0, find the ratio of the times required to travel the first a/2 units of distance in the two respective cases. 8. In the motion discussed in Art. 42, assuming that k = 8 ft%/sec and a = 8 ft, find the distances traveled during the first second in the two respect. rve cases.

182

Chapter 5"

9. A particle is repelled from 0 by a force obeying the inverse square law. If the particle starts 1 ft from 0 with an acceleration of 2 ft/sec 2, find the time required to travel 3 ft. 10. Find the number of hours required for a particle to fall from the distance of the moon to the surface of the earth. Assume that the radius of the earth is R = 4000 miles and that the distance from the center of the earth to the moon is 60R = 240,000 miles. 11. Suppose that the earth's attractive force should suddenly become repulsive, remaining numerically the same. How many hours would be required for a particle to shoot out from the earth's surface to the distance of the moon? Assume distances as in Prob. 10. 12. Find the velocity with which a particle would strike the earth's surface if it started from rest at a very great (practically infinite) distance and moved subject only to the earth's attraction. Take R (radius of earth) = 3960miles. (This is the "velocity of escape," i.e., the velocity with which a particle would have to be shot from the surface of the earth in order not to return.) 13. In Prob. 12 find the time consumed by the particle in traveling the last R miles. 14. Suppose that a particle, starting from rest at a distance R = 3960miles from the surface of the earth, falls into a straight tube bored through the center of the earth. If the only force acting is that of the earth's gravitation, find the time required to reach the center. (Cf. Prob. 10, Art. 30(b).) 15. A chain has its ends fastened at the same level to two posts 60 ft apart. (a) Find the dip in the chain if it is 70 ft long. (b) How long is the chain if the dip is 5 ft? (c) Find the tension at the lowest point and at a point of suspension of the chain of part (b) if the weight is 1.5Ib/ft. 16. A cable 60 ft long hangs from two supports at the same level and dips 6 ft. Find the distance between supports. 17. A flexible cable of length b ft is hung over two smooth pegs at the same level L ft apart. Show that there are two, one, or no positions in which the . equiilibnum . aceordimg as L b > oable WI·11 hang In <: e. 18. If the length of the radius of curvature at any point P of a curve is proportional to the length of the normal drawn from P to the x-axis, 1 [

+

~)'r ky~l + (~)2. =

dx 2

Find and identify the family of curves possessing this property in the four cases: (a) k = 1; (b) k = -1; (c) k = 2; (d) k = -2. Explain the significance of positive and negative values for k,

Article 43

183

19. Solve the differential equation formed by setting the expression for radius of curvature equal to a constant r, and thus obtain the general equation of a circle of radius r. 20. At any point of a curve the product of the lengths of the ordinate and radius of curvature is equal to the length of the normal. Find the equation of the curve if it passes through the origin with slope 1 and is concave upward. What is the length of the radius of curvature at the points x = 0, -r/4, -r/2? Sketch the curve for -3r/4 < x < r/4. 21. Find the curve satisfYing the conditions of Prob. 20 except that it is concave downward. Show that this curve is symmetrical with respect to the origin to the curve in Prob. 20. 22. Find and identify the family of curves such that the length of an arc of any curve of the family is proportional to the difference of the slopes at the ends of the arc. 23. A curve passes through the point P (1, 0) with slope 0 and has the property that the slope at any point is twice the length of the arc from P to that point. Find the ordinate and the slope of the curve at the point whose abscissa is 2. 24. The length of an arc of a curve is equal to the distance intercepted on the y-axis between the tangents at the ends of the arc. Find the equation of the curve having this property and passing through the point (1, 0) with slope

o.

25. Solve the differential equation * x 2 y"

+ xy' + (x

2

-

i)Y = O.

Use a modification of the method of Art. 41, choosing u so that the coefficient of v' in equation (2) vanishes. 26. Solve Probs. 7, 10, 11, and 12 under the assumption that the earth attracts according to the inverse cube law, but that the acceleration due to gravity at the surface of the earth remains numerically the same (32.17 ft/ see"). 27. A particle moves under the action of a force which varies inversely as the nth power of the distance from a point O. If x is the distance from 0 at time t, and if the particle starts from rest at x = a, show that the t, x relations are FOR AN ATTRACTIVE FORCE

1'"

t=

1

~la

k \)""2

:I:

t = -1~11a k

2

1:

dx

v'a1- n _ Vx

n

(n

a1-

n

(n> 1)

dx 1- n -

• This is Bessel's equation (1), Art. 52, for n =

< 1)

x 1-

I.

184

Chapter 5 FOR A REPULSIVE FORCE

I#nj%Vx

t=-

k

t

2

a

1~lj% ~

= -

k

2

a

dx 1- n -

/ V a1-

a1-

n

x 1-

n

dx n -

(n

< I)

(n> 1)

28. Using the formulas of Probe 27, with equal k's, find the ratio of the times required to travel a units of distance starting from rest at x = a, iii the two cases of an attractive and of a repulsive force varying inversely as the square root of the distance from a point O. 29. Subject to an attractive force which varies inversely as the i power of the distance from a point 0, a particle starts from rest 16 ft from 0 with an acceleration numerically equal to 4 ft/sec. Find the time required for the particle to reach O. 30. A flexible chain L ft long is hung over the upper end of a smooth inclined plane which makes an angle 8 with the horizontal. The end of the chain which hangs down is a ft below the end which rests on the plane. Show that the time required for the chain to slide off is

t

=

J L cosh- ~ (sec) "Jg(1 + sin 8) a' 1

where g = 32.17 ft/sec 2• Assume a space of at least L ft below the upper end of the plane. 31. Using 8 = 0° and 90° respectively in the formula of Prob. 30, find the time required for a chain 13 ft long, with a = 5 ft, to slide (a) off a smooth -, horizontal table; (b) off a smooth peg. 32. If "upper" is replaced by "lower" in Probe 30, obtain a formula for the time required for the chain to slide off the plane. 33. If the lower end of the chain in Probe 32 is initially at the lower end of the plane, show that the time required for the chain to slide off is

t=

"JJg(1 - L.SIn 8) cosh"! esc 8j

also show that, as 8 approaches 1r /2, t approaches the time required for a freely falling body to fall L ft. Assume that the plane is at least L ft long. 34. A particle Q starts from the origin and moves uniformly along the x-axis. It is pursued by a particle P which starts at the same time from a point on the y-axis at a distance a from the origin. The velocity of Q is k times that of P. Find the curve of pur8uit, i.e., the path of P, (a) i(k;= 1; (b) if k = 1. Show that capture will occur if k < 1 after a time ak/(I - k2)r sec, where a is measured in feet and r (ft/sec) is the rate of Q.

e~6 SIMULTANEOUS EQUATIONS

44.. System of two first order equations. The simplest system of simultaneous differential equations contains two equations of first order in three variables-one independent variable and two dependent variables. Taking x as independent variable and y and z as dependent variables, we now consider a system of two first order equations which can be written in the form dy = M dx

'

(1)

dz = N dx

'

where M and N are functions of x, y, and z, or in the form (2)

where P, Q, and R are functions of x, y, and z, Any or all of the functions M, N, P, Q, and R may actually contain less than three variables; they may even all be constant. Solving such a system as (1) or (2) will be understood to mean finding two relations, free of derivatives, which together involve ,the three variables and two arbitrary constants, and which satisfy the equations. We take first an example of the simplest case where the capital letters are all constants, then an example in which M and N are not constant, but one at least of equations (1) contains only two variables. 185

186 EXAMPLE

Chapter 6

1. Solve dy _ 1

dx

,

-

dz = 2.

(3)

dx

Here each of the two equations may be integrated immediately, giving the solution y = x + Ch Z

EXAMPLE

= 2x

(4)

+ C2 •

2. Solve

dz

dy

- = 2x, dx

-=

y

dx

+z

(5)

x

Integration of the first equation gives y = x2

+ C1•

(6)

Substituting the value of y from (6) into the second of equations (5) and rearranging the terms, we find dz dx

-

-

z x

- =

x

C1 +_. x

This is a linear equation which may be solved by the method of Art. 19. It may be solved also by use of an integrable combination, as follows: _x _dz_ - _z_d_x _ 2 + 2 dx,

(1 0.)

X

X

(7)

Equations (6) and (7) constitute the required solution of the system (5).

Geometric interpretation. The term space will be used in this chapter to denote three-dimensional space in which a point is z. determined by its three rectangular coordinates, z, y, A single equation (not a differential equation) in the three variables represents a surface. The surface is a plane if the

anti

Article 44

"187

equation is of first degree. If the equation contains only two variables it represents a cylindrical surface with elements parallel to the axis denoted by the missing variable; if the equation is linear in the two variables the surface is a plane parallel to the axis denoted by the missing variable. Two simultaneous equations in three variables represent the intersection of the two surfaces represented by the single equations, i.e., a curve in space; if the two surfaces are planes, the curve is a straight line. In Example 1, the first of equations (4), for a fixed C 1 , represents a plane parallel to the z-axis; the second, for a fixed C2 , represents a plane parallel to the y-axis; hence the two equations taken simultaneously, for fixed 0 1 and O2 , represent the straight line intersection of the two planes. As 0 1 and C2 independently take on an infinity of values we have 00 1 parallel planes intersecting another 00 1 parallel planes in a family of 00 2 parallel straight lines, one through each point in space. This doubly infinite family of straight lines is the geometric picture of the solution of the system of differential equations (3). In Example 2, for a fixed C1 , the parabolic cylinder (6), with elements parallel to the z-axis, intersects the singly infinite family of parabolic' cylinders (7), with elements parallel to the y-axis, in a family of 00 1 curves. Thus, for each value of C1 we have 00 1 curves of intersection; for C1 arbitrary, there are 00 2 curves of intersection forming the geometric picture of the solution of the system of differential equations (5). What does the system of differential equations tell us about the direction of the curves representing the solution? Let us look at the equations of the system written in the form

dx

P

dy

=Q=

dz

R·

(2)

The functions P (z, y, z), Q (z, y, z), and R (z, v, z) determine at any point (z, y, z) of space three numbers, and these three numbers, regarded as direction components, fix a direction at

Chapter 6

188

that point.' Thus if P = z + y, Q = -2y, R = x + y + z, then, at the point A (3, 2, 1), P = 5, Q = -4, R = 6. If from A we travel 5 units in the positive direction of the z-axis, then 4 units in the negative direction of the y-axi~, then 6 units in the positive direction of the z-axis, we arrive at a point B, and the line joining A to B fixes a direction at the point A; we denote this direction by the notation [5, -4, 6]. Furthermore, the three differentials dx, dy, and dz are, at the point (z, y, z), the direction components of the tangent line to the curve through this point and satisfying the differential equations. Relations (2) state that dx, dy, and dz are proportional to P, Q, and R, and hence that the directions determined by these two sets of direction components are the same. Therefore the system of simultaneous differential equations (2) defines a family of curves, one passing through each point of space in a direction [P, Q, R]. In Example 1 the differential equations can be written in the form

dx 1

dy

== -

1

dz

= -. 2

The solution (4)

represents a family of parallel straight lines in the direction [1, 1, 2]. If y = -2 and z = 3 when x = 1, then C1 = -3, C2 = 1, and the particular solution through the point (1, -2,3) is

y

=x

- 3,

z = 2x

+ 1.

(8)

This particular solution could be written down directly from the differential equations, since the equations of a line through (1, - 2, 3) in the direction [1, 1, 2] are x-I

1 the equivalent of (8).

-

y+2 z-3 12'

(8')

Article 44

189

In Example 2 let us find the particular solution through the point (1, 2, 3) and its direction at this point. Substituting x = 1, y = 2, Z = 3 in equations (6) and (7),

01

= 1,

and the particular solution is

y = x2

+ 1,

Z

= x2

+ 3x -

1.

(9)

Writing the differential equations in the form

dx dy dz -X=2x2 = y + z ' we have for the direction of the curve (9) at the point (1, 2, 3), [P, Q, R]

= [1, 2, 5].

Systems in which both equations (1) contain all three variables. Certain systems in which both equations (1) contain all three variables may be solved by proceeding as in the following example which we shall solve by two different methods. EXAMPLE

3. Solve dz dx = x

+ y,

dy dx = x

+ z,

(10)

Ftrst solution. By using a system of multipliers l,m,n, which may be either constant or variable, any of the fractions in equations (2) may be set equal to

1dx + m dy + n dz lP + mQ + nR·

w.· . ' ( ). riting equations 10 In

form (2), we have dx

- 1

dy dz x+z x+y

.

Then dx

- - dy-dz 1 z-y

dy+dz+2dx y+z+2x+2

190

Chapter 6

Integrating the first two fractions, then the first and third, we obtain log (y - z) = - x

+ log 0h (11)

log (y

+ z + 2x + 2) = x + log O2 , Y + z + 2x + 2 = 02ez.

(12)

By adding and subtracting (11) and (12),

= Aez z = Aez

y

+ Be?" -

x-I,

(13)

Be-z - x-I.

(14)

Equations (11) and (12), or equations (13) and (14), constitute the solution of the system (10). Second solution. Subtracting the first of equations (10) from the second, we have d

dx (y - z) = z - y,

(D

+ 1)(y -

z)

= 0, (15)

Substituting the value of z from equation (15) into the second of equations (10), we obtain dy dx - Y

=x

-

0 -z

Ie ,

a linear equation, whose solution is

«» =

f

(xe-" - C1e- 2") ax + C2 ,

= e- -x Z

(

1)

+ c21 e-2z + O2,

or (16) Equations (15) and (16) constitute the solution of the system (10); if the value of y from (16) is substituted in (15), the equivalent solution consisting of equations (13) and (14) is obtained.

191

Article 44

PROBLEMS 1. Find the equation of the cylinder formed by the straight lines which satisfy the equations dy dz = 3 - = -1, dx ' dx and (a) pass through the circle y2 + Z2 = 1, x = 0; (b) pass through the parabola 3y2 = x, Z = o. 2. Find the particular solution of

dz

Y

-=--1 d» x '

through the point (1,2,3), and its direction at this point. 3. Find a curve through the point (1, 1, 1) satisfying the equations dy

- = -2xz

dx

'

dz

dx = 2xy.

4. Find the two most general functions of x such that the derivative of each one equals the other one. Solve the following systems of differential equations.

dy y - --=0 dt x '

5.

dx dy =-l+y l+x

6. 7.

dx t - - - = O. dt x

dy dx

+ z = sin. x,

dz

dx

+ y = cos x.

8. In a certain type of chemical reaction, if a substance A forms an intermediate substance B, which in turn changes into a third substance C, the respective concentrations, x, y, e, of the three substances obey the relations dx -+ax=O, dt

x

+y +z =

c,

where a, b, and c are constants. Solve for x, y, and e, using the conditions y = z = 0 when t = 0, (a) if a ¢ b; (b) if a = b.

Chapter 6

192

9. There .are two tanks, each of 100 gal capacity, the first being full of brine holding 50lb of salt in solution and the second being full of water. If water runs into the first tank at the rate of 3 gal/min and the mixture, kept thoroughly stirred, passes out at the same rate through the second tank, when will the first tank contain twice as much salt as the second, and how much salt will have passed through the second tank at that time? 10. There are two tanks, each of 100 gal capacity, the first being full of brine holding 50 lb of salt in solution and the second being full of water. If the brine runs out of the first tank into the second at the rate of 3 gal/min and the mixture, kept thoroughly stirred, runs at the same rate out of the second tank into the first, when will the first tank contain twice as much salt as the second? 11. Tank A initially contains 100 gal of brine in which 20 lb of salt are dissolved. Two gallons of fresh water enter A per minute and the mixture, assumed uniform, passes at the same rate from A into a second tank B initially containing 50 gal of fresh water. The resulting mixture, also kept uniform, leaves B at the rate of 1 gal/min. Find the amount of salt in tank B at the end of 1 hr.

45. Systems of two linear equations. A differential equation is linear if it is of first degree in the dependent variables and their derivatives. All the differential equations encountered so far in this chapter, except those of Problems 5 and 6, are linear but of the simplest type, each being of first order and containing only one derivative. We now develop a method for solving a system of two linear equations of any order, with constant coefficients, involving one independent and two dependent variables. The method may be used also for systems of three or more equations, and for systems of equations with variable coefficients, of the type discussed in Art. 38, which can be reduced to linear equations with constant coefficients. Two examples follow which illustrate the method. EXAMPLE

1. Solve dy

dz

-+-=4y+l dx dx '

Article 45

193

Using operator notation, we place all terms containing the dependent variables y and z on the left, those containing y in the first column and those containing z in the second:

+ Da = 1, 3)y + z = x 2 •

(D - 4)y

(1)

(D -

(2)

It can be shown * that for a system of differential equations written in this form the number of arbitrary constants appearing in the gen. eral solution is equal to the exponent of the highest power of D in the expansion of the determinant of the (operator) coefficients of the dependent variables. The highest power of D in the expansion of the determinant D-4 D D - 3 1

is D2 ; hence two arbitrary constants will appear in the solution of the system of equations (1) and (2). If more than two arbitrary constants appear in the process of solution, the extra ones must be evaluated, or expressed in terms of only two. Three methods of procedure are possible. We may eliminate z from equations (1) and (2), solve the resulting equation for y, then substitute this value of y in (2) to obtain z, If the value of y, containing two arbitrary constants, were substituted in (1) to obtain z, another constant of integration would appear and would have to be evaluated by substituting the values of y and z in (2); hence it is better to obtain z from equation (2). Another method, obviously more complicated in this case, is first to eliminate y from equations (1) and (2), solve the resulting equation for z, then substitute the value foundfor z in (1) or (2) to obtain y; this method introduces a third arbitrary constant which must be evaluated. A third method would be to eliminate z from equations (1) and (2), then solve the resulting equation, obtaining y in terms of x and two arbitrary constants, say C1 and C2 • Next eliminate y from equations (1) and (2), solve the resulting equation and obtain z in terms of x and two arbitrary constants, say C~ and C;. Finally C~ and C; would be evaluated or expressed in terms of C1 and C2 by substituting the values of y and z in (1) or (2). • See Forsythe's A Treatise on Differential Equations, fifth edition, Art. 171.

~apter 6

194

We proceed to solve Example 1 by each of the three methods. First solution. Multiplying equation (2) by D, (D 2

3D)y

-

+ Dz = 2x.

(3)

Subtracting (1) from (3), (D2 - 4D + 4)y = (D - 2)2 y= 2x - 1.

(4)

Solving (4) by the method of Art. 34, we find

=

Yc

(C1

+ C2x )e2x •

YP

= Ax + B

4) -4)

=

A

D2 yp =

o

Dyp 1)

4A = 2

A Y

=

- .l. -2

(C1

4B - 4A = -1

B

_1 4

+ C2x )e2x + ~x + ~.

(V"/ ,.

(5)

Differentiating (5), Dy

= (2C 1 + C2

+ 2C2x)e2z + ~.

From (5) and (6), (D - 3)y

"

= (-C 1 + C2

-

.' _2

C2x)e- x

Then. from (2),

-

3

Vx -

1 4-

!

= (C1 -

Z

(6)

C2

(7) /'

+ C2x )e2x + '~2 + ~x + ~. ,

(8)

Equations (5) and (8) comprise the general solution of the system of equations (1) and (2). Second solution. Multiplying equation (1) by D - 3 and (2) by o - 4, we have

+

(D - 3)(D - 4)y (D2 - 3D)z = -3, (D - 3)(D - 4)y + (D - 4)z = 2x - 4x 2•

(9) (10)

Subtracting (10) from (9), (D2

-

4D

+ 4)z =

CD - 2)2Z

= 4x2 -

2x - 3.

(11)

Article 45

195

The solution of (11) proceeds as follows: Zc =

(a~

+ a~x)e2:ll.

4)

zp=Ax2+ Bx+ a

-4) 1)

4A = 4

De;

=

D2Zp

=

4B - 8A = -2

2Ax+ B 2A 40 - 4B

+ 2A =

B=~

A=1 z

=

(a~

-3

0=1

+ a;x)e2:ll + x2 + ~x + 1.

(12)

. Substituting the value of z from (12) into (2), we obtain the linear equation (D - 3)y = - (C~ + C;x)e211: - ~x - l, whose solution is found by the method of Art. 19:

e-&y = -

f

[(0;.

+ C;x)e-" + (~x + ~)e--3Z] dx + 0;,

3 e-311: X x .= a;e- - C;e- ( -x - 1) - 2 9 (-3x - 1)

y = (C~ To evaluate

+ 12 + a;,

+ C; + a;x)e211: + ~x + l + c;e3x.

a; we multiply (12) by D and (13) by D Dz

=

e-311:

(2~

(13) 4, giving

+ a; + 2a;x)e 211: + 2x + i,

(D - 4)y = (-2C~ - a; - 2a;x)e211: - 2x - ~ - a;e 311:,

which, substituted in (1), yield

1 - a;e3x = l. Hence

a; = 0, and equation (13) becomes y = (~

+ a; + a;x)e2x + ~x + l.

(14)

Equations (14) and (12) comprise the general solution of (1) and (2). If = O2 and a~ = 0 1 - O2 , equations (14) and (12) reduce to (5) and (8).

a;

196

Chapter 6

First eliminating z and then eliminating y from equations (1) and (2), and solving the resulting equations, as above, Third solution.

y

z=

= (C1 + C2x)f!'z +

+ l, (C~ + C;x)e2z + x 2 + ~x + l. ~x

(5)

(12)

In order to find the relations connecting the constants C1, C2 , O~, C;, substitute (5) and (12) in (2), making use of the value of (D - 3)y given by (7): (-C 1 + O2 - 02X)f!'z - ~x -l + (C~ + C;x)e2Z+x2 + ~x + l = x 2, (0 1 - O2 + 02x)e2z = (O~ + C;x)e2Z, O~

= 01 -

C2 ,

C;

= O2 •

Hence (12) reduces to

z = (0 1

-

C2 + 02x)e2z

+ x2 + Ix + ~;

(8)

equations (5) and (8), as before, comprise the general solution of the given system of equations. EXAMPLE

2. Solve

d2x + 2 (dY _ a) = dt dt

0

2

'

~: -2(: - b) = O. First write the equations in the form

+ 2Dy = 2a,

(15)

2Dx - D 2 y = 2b,

(16)

D 2x

where D now denotes d/dt. The determinant of the coefficients of the dependent variables x and y yields D 4 as the highest power of D; hence the general solution will contain four arbitrary constants. Adding D X (15) and 2 X (16), we get (D3

Then Xc

+ 4D)x =

4b.

= 0 1 + O2 sin 2t + 0 3 cos 2t, X

=

C1

Xp

= bt,

+ C2 sin 2t + 0 3 cos 2t + bt.

(17)

197

Article 45

-

Substituting in (15) the value of D 2x = -4C2 sin 2t - 4C3 cos 2t, we find Dy = 2C2 sin 2t + 2C3 cos 2t + a. Integration of (18) gives

(18)

cos 2t + C3 sin 2t + at + C4 • (19) Equations (17) and (19) contain the proper number of arbitrary constants and constitute the general solution of the given system. This solution may also be written in the form y

= -C2

x y

where A, B, C, and

= A cos (2t + a) = A sin (2t + a)

a

+ bt + B, + at + C,

are arbitrary constants. PROBLEMS

Solve the following systems of differential equations. dy 1. - - y - z = 2 cos 2x, dx 2. dy dx

+ dz

3. :

+ 2x = t -

dx

dz - - 3y dx

+ e = o.

= cos x - 3y

dy = 11 sin x 'dx

y,

dx

4. dt - Y = cos 2t,

::

+: = 1 -

+y -

z.

y.

dx dy . dt - dt - x - y = SIn 2t

+ cos 2t.

dy dy dz 6. - = e~ - y - z, 2 - + - = cos x - z. dx dx dx dy liz . dy 6. - - - + y + z = 4 SIn x, 2 - - y + 3z = 2 cos z; dx dx . dx dz . dy dz dy 7. - + - + y = 2 SIn 2x, - + - + 2y + 2z = COS 2x. dx dx dx dx

+

. t, 8• -dx = y - x 3 SIn dt dy d 2y 9. dx 2 - dx = 1 - s, 2y

10. d _ dy dx 2 dx

+-

dx dy = 2y - 3x dt dt dy dz X dx + dx = 4e- + y.

+ z = 10e-2~

'

dy dx

+ dz

dx

_ y

+ 2et +· 6 SIn t -

= 1 + 7e- 2x

.

2 cos t.

198

Chapter 6

dx 11. t dt

dy t dt + x

+ y = t + x,

d2y 12. t2 dt2

dy

+ t dt + x = 1 + log t,

= t + y. dx

t dt

dy

+ t dt

- x - y

= log t.

13. Show that the system of differential equations

+ (D + l)z = x, l)y + [(n 2 - n + I)D + n]z = Dy

(D -

eZ ,

where D = djdx, has none, one, or two arbitrary constants in its general solution when n = 0, 1, and t respectively. Find the general solution in each of these cases. 14. Find a solution of the system

(7D2

+ 23)x -

8y

=

0,

(3D2 - 13)x + 2(D2 + 5)y = 0,

where D = djdt, subject to the conditions: x = 0, y = 0, Dx = 1, and Dy= 3, when t = 0. Find the values of x and y when t = 1. 16. Solve the system d2x dy --8-+9x=at+h dt2 dt ' d2y -2

dx +8-+9y=bt+k. dt dt

16. Solve the three simultaneous equations dx

- =x+y dt '

dy - =z-x dt '

dz

-=y+z. dt

46. Motion of a projectile. The theory of the motion of a projectile in the atmosphere is quite difficult, owing partly to complications connected with the law for variation of resistance with velocity. We shall treat here only the case in which resistance is assumed to be proportional to velocity, and the path is considered to be a plane curve traced out by the center of gravity P (z, y) of the projectile. If t is the time, d3;/dt and dy/dt are the components of velocity in the x- and y-directions, and d2 x/ dt2 and d2y/dt 2 represent the corresponding components of acceleration. Let x and y be measured in feet and t in seconds; also let w (lb) be the weight of the projectile and g (ft/see"), as usual, the gravity

199

Article 46

constant. We form a system of two simultaneous differential equations by equating (w/ g) (d2 x/ dt2 ) and (w/ g) (d2y/ dt2 ) respectively to the components of force in the x- and y-directions:

wd

2x

g dt

2

= -k dx dt ' .

wd

2y

g dt

= -w

2

-k dy.

(1)

dt

Here each of the equations contains only two variables and hence can be solved separately. The solution will consist of two equations giving the parametric equations of the path, x and y in terms of t. Four arbitrary constants will appear in the course of solution, for whose determination four conditions are necessary. We shall solve equations (1) under the assumption that the projectile. starts at the origin with an initial velocity Vo (ft/sec) in a direction making an angle a with the horizontal; that is, the four conditions are x = 0, y = 0, dx/dt = Vo cos a, dy/dt = Vo sin a, when t = O. After writing equations (1) in operator notation, with D = d/ dt, the solution proceeds as follows:

(D + :OD)Y = 2

x

= C1 + C2e - (k

y

g /w ) t

Dz = - kg C2e-(kg /w)t W

.

o = C1 + C2

= Cs + C4e-

Dy

=

-

o=

Os

4

w

wt

(kg /w) t -

kg C e -

-

-g

k

(k g /w) t -

+ C4

Vo cos a = - kg C2

· kg O4 - w Voslna=--

C1 = -C2 =

C3 = -C4 =

W

W

kg

x

Y=

vocosa

k

W

= -w Vo cos a (1 kg

~ ( Vo sin a + ~) (1 -

w k

-

w (voSina+ w) kg k

e-(kg/w)t)

e-(kg/

1D

lt) -

~ t.

(2)

200

Chapter 6

Equations (2) are the parametric equations of the path and locate the position of the projectile at any time t. The a, y, equation of the path could be obtained by eliminating t between the two equations. PROBLEMS

1. Assuming that there is no resistance, i.e., k == 0 in the example of Art. 46, find (a) the x, y equation of the path; (b) the maximum height of the projectile and the maximum range. 2. (a) Find the maximum height of the projectile in the example of Art. 46. (b) Show that the result found in (a) approaches the value found for the maximum height of the projectile in Prob. 1(b) as k --. o. 3. The small oscillations of a certain system with two degrees of freedom are given by the equations

dh

d2x

-+3x-2y=0 lit2 '

d2y

-+--3x+5y=0. lit2 dt2

If x = 0, y = 0, £lx/dt = 1, liy/lit = 3, when t = 0, find (a) the values of x and y when t = 1; (b) the maximum and minimum values of x and y. ' 4. A particle of unit mass moves in accordance with the law

d2x lit2

=

li2y lit2

y,

- .'

= -x.

If it starts from rest at the point ( -1, 0), find the parametric equations of its path. Plot the path up to the point where it crosses the x-axis, and find the coordinates of the maximum point on this portion of the path curve. 5. The primary of a transformer has inductance L1 henries and resistance R1ohms, the secondary has inductance L 2 henries and resistance R2 ohms, and the mutual inductance is M henries, where L1L2 > M2. The free oscillations in the two circuits are given by

L1lii1- + M lii -

2

lit

. lit

+ R1't1. - 0

'

M liil lit

+ L2 lii + R21.2. = 0, lit 2

where il and i2 (amp) are the respective currents in the primary and secondary at time t (sec); show that i1 and ~ approach zero as t becomes infinite. Suppose that the values of the circuit constants are L1 = 3, L2 = 6, R1 = 7, R2 = 10, M = 4; also that il = 2, i 2 = 3, when t = O. (a) Show that ~"-2 continually decreases but that it increases to a maximum and then decreases indefinitely as t becomes infinite. (b) Find the maximum value of il

Article 47

201

and the time required for it to reach the maximum. (c) Find the values of il and i 2 when t = 0.01 sec. 6. If a particle of unit mass at P (x, y) is attracted toward the origin 0 by a force F"the x- and y-components of the force are -F cos 0 and -F sin fJ, where 0 is the vectorial angle of P. The differential equations of motion are then . Fy d2y

-2= - r

dt

where r is the radius vector of P . ....(a) Multiply the second equation by x, the first by y, and subtract; then, making use of the formula for differential sector area, Art. 3, show that the areal velocity (time derivative of sector area) is constant. , (b) Multiply the first equation by~2 dx, the second by 2 dy, and add, then letting r = 1/R, show that

d2R d0

2

R_

+ -

L

h2R2'

'where h is twice the areal velocity. (c) Solve the differentialequation in (b) if F = kR2 (inverse square law), subject to the conditions R = l/ro, dR/dO = 0, when 8 = 0, and show that the path is a conic. 7. In Thomson's experimental determination of the ratio m/e of the mass to '9te charge of an elec.tron, in which the electrons were subjected to an electric field of intensity E and a magnetic field of intensity H, the equations

d2x dy m-+He-=Ee 2

'dt

dJ,

d 2y dt

dx dt

m--He-=O 2

,

were employed." If x = y = dx/dt = dy/dt = 0 for t path is a cycloid whose parametric equations are'

x = Em(l _ cos He H2e m

t),

y

'

= 0,

show that the

= Em (He t _ si~ He H2e m

m

t).

47. The roots of unity. The n roots of the equation x", = 1,

(1)

where n is a positive integer, are called the nth roots of unity. De Moivte's theorem gives the n values of x which satisfy equation (1) in the form 21rk

x = cos'n * See Phil.

21rk + .•SInn ' 1,

(k

Mag., Vol. 48 (1899), p. 547.

= 0, 1'" 2 ... n - 1),

(2)

202

Chapter 6

where i = v=I. These values of x are the nth roots of unity. They may be represented graphically by n points equally spaced around a unit circle centered at the origin, the first point being at x = 1, corresponding to k = O. If r denotes the next of these points in the counterclockwise direction, corresponding to k = 1, 21r .. 211" ( ) r=cos-+'tsln3

n

n'

and we may call r the basic nth root of unity; all the nth roots of unity are obtained by raising r to the powers 0, 1, 2, .'., n - 1. Thus instead of (2) we may write 'r" tIV

= 1

r r2

" "

r ...

,

rn-

l • ,

(rn

=

1).

(4)

> 1),

(5)

The relation 1

+ r + r 2 +...+ r n - l

= 0,

(n

follows from the symmetrical arrangement of the points. Furthermore 1 +rk

+r2h

+.. ·+r(n-l)k = 0,

and ",nk

(k = 1,2,3,·· ',n -1),

= 1.

(6) (7)

When n = 2, r = -1; the square roots of unity, 1, r, are 1, -1.

When n = 3, r = -~ + iy'3/2 and r 2 = -~ - iy'3/2. It is customary to denote r by the Greek letter '" when n = 3. Thus", = -~ +iv'3/2, (JJ2 = -~ - iv'3/2, and the cube roots of unity are 1, "', ",2. When n = 4, r = i; the fourth roots of unity, 1, r, r 2 , r 3 , are 1, i, -1, -i. 48. Cyclic systems of differential equations. By a cyclic interchange of the letters Xl, X2, X3, " ' , Xn-l, Xn, we mean a change of Xl to X2, X2 to X3, ••• , Xn-l to X n, Xn to Xl' We call a system of differential equations cyclic if it is unchanged by a cyclic interchange of the dependent variables; for example, the

Article 48

203

following system in which the x's are the dependent variables and D = d/dt: DXl DX2

= =

X2,

Xg,

(1) DXn_l = X n , DXn

= Xl.

Certain cyclic systems of n equations in the dependent variables Xl) X2, ••• , X n , may be solved by changing to a new set of dependent variables, Ul, U2, " ' , Un, connected with the old set by means of. the substitutions

= Xl + rk-lx2 + (rk-l )2Xg +...+ (rk-l )n-lxn, (2) where k = 1, 2, 3, ... , n, and r is the basic nth root of unity. Uk

The original dependent variables Xl, X2, •• " X n and the independent variable t are assumed to be real. Usually we wish to express the x's in a form involving t and n arbitrary constants but involving no fixed complex constants, that is, to obtain a solution in real form, although complex quantities are used to effect the solution. . . EXAMPLE

1. Solve the system Dx=ax+by+ez,

+ bz + ex, Dz = az + bx + cy,

Dy = ay

where D = d/dt, and a, b, e are real constants. In the substitutions (2) let n = 3, r = w, and replace Xt, x, y, z, and Uh U2, Ua by u, v, w. Then

(3)

X2, Xg

+ y + z, = z + wy + w2z,

by

u = x v

(4)

w=x+w2y+wz.

The last term of the third equation is w4z, but it is written tJJa = 1.

wZ

since

204

Chapter 6

Adding equations (4), we find

u+v+w=3x, since 1 + w + w2 = 0 [equation (5), Art. 47]. Multiplying equations (4) by 1, w2 , w respectively and adding, then by 1, w, w2 and adding, we find similarly

+ w2v + ww = 3y, u + wv + w2w = 3z, u

so that the solution of equations (4) for x, y, z gives

+ v + w), ~ (u + w2v + ww), ~ (u + wV + w 2w).

x = ~ (u y = Z

=

(5)

Now, adding equations (3), we obtain Du = (a -Ir b + c)u, whose solution is

u

= 3CleCa+b+c),.

(6)

Multiplying equations (3) by 1, w, w2 respectively and adding, we find Dv = (a + bw2 + cw)v, whose solution is (7) Finally, multiplying equations (3) by 1, w2 , w respectively and adding, we have Dw = (a + bw + c (2 )w, 'W

= 3CseCa+oo,+cw2)t.

(8)

Substituting (6), (7), and (8) in (5) gives

x

= Cte(a+b+c),t + C2eCa+oo,2+cw)t + C se(a+bw+cw2)t,

+ w2C2eCa+bw2+cw)t + wCse(a+bw+cw2)t, z = CleCa+b+c)t + wC2e(a+bw2+cw)t + w2Cse(a+bw+cw!)t. y = CleCa+b+cH

(9)

Equations (9), where the C's are arbitrary constants, represent the general solution of the system (3), but the solution is not in real form

Article 48

205

since it involves the fixed complex constant w. This solution, however, may be transformed into the following form (see Problem 1 at the end of this chapter):

x

= Ge(a+b+c)t

+ KJ't sin (J.Lt + a),

(pt + a - 2;), z = Ce(a+b+o)' + Kt!" sin (pt + a - ~), y

= Ce(a+b+o)I + KeA• sin

(10)

where b

c

2

2'

X=a----

and C, K, and a are arbitrary constants. Equations (10) represent the general solution of the system (3) in real form.; for any real values of the arbitrary constants G, K, and a, the values of z, y, and z are real. EXAMPLE 2. Solve the system (1). This system may be solved by use of the substitutions (2), but here it is easy to eliminate all but one of the dependent variables, say xl, then solve the resulting equation for Xl and find the other x's by successive differentiation. Substituting the value of X2 from the first equation into the second, then the value of Xa from the second equation into the third, ... , then the value of X n from the (n - l)th equation into the nth, we have or (D" - I)XI = 0, from which we write down the value of Xl, then obtain DXl = X2, ••• , Xn from DXn-l = Xn• The general solution of system (1) is therefore Xl

=

Gle

t

X2 = Glet •

X2

+ G 2ert + G aer2t + + c.s">, + rG2ert + r2Gaer2t + + rn-lGnern-lt, •

•

•

•

•

•

•

•

from

(11) •

•

•

•

206

Chapter 6

in which the derivative of each equation gives the following one and the derivative of the last gives the first. As in the case of equations (9), the equations (11) represent a solution in imaginary form for n > 2, but in any particular case the solution can be exhibited in real form, for example in the following special case of (1).

EXAMPLE,3. Write in real form the solution of the system

DX2 =

Xs, .

Dxs

X4,

The first of equations (11) for n Xl

=

(12)

= 4, r == i, is

= Glet + G2eit + Gse-t + C4e- it•

By means of Euler's relation we may replace 02eit + C4e-it by K sin (t ex) as in Art. 29(c). Also let 0 1 == A, 0 3 = B. We thus write Xl in real form. and obtain the other x's by successive differentiation:

+

+ Be-t + K sin (t + a), t t X2 = Ae - Be- + K cos (t + ex), t t Xs == Ae + Be- - K sin (t + ex), t t X4 = Ae - Be- - K cos (t + ex). Xl =

Aet

(13)

Equations (13), in which A, B, K, and a are arbitrary constants, form the required solution of the system (12).

49. A special case of Einstein's equations. Einstein's law of gravitation is expressed by a system of ten partial differential equations of second order. It has been shown by Kasner * * "Solutions of the Einstein equations involving functions of only one variable," by Edward Kasner, Transactions oj the American Mathematical Society, Vol. 27 (1925).

207

Article 49

that in a special case these equations reduce to the following system of three ordinary differential equations of first order:

Dz = yz - x2 , Dy = z:c - y2, Dz

= xy -

(1)

Z2,

where D = d/dt. We notice that this system of equations is cyclic in :1;, y, z, but not linear as in the case of the system in Example 1, Art. 48; however, the method used in that example will be followed in solving system (1). Letting

+ =x+

u =x 1)

W

we find uv

= x2 +

vw = wu

+ z, 6J'Y + CJ)2Z, y

= X + ClJ2y +

+ ClJ2,,2 x 2 + y2 + Z2 ClJy2

= 3:2 + (JJ2y2 +

C1JZ2 -

(2)

ClJZ,

ClJ2:tJJJ -

yz -

~,

xy -

yz -

zx,

611:Y -

yz -

6)2ZX•

(3)

Now, adding equations (1), multiplying them by 1, "', 6)2 respectively and adding, then multiplying by 1, lA>2, W respectively and adding, we reduce system (1) to the system

Du =-vw, Dv = -uv, Dw

(4)

=-wu.

From the last two of these equations dv/v = dw/w, hence v = k2w, where k 2 is an arbitrary constant written in this form for convenience, and the system (4) is equivalent to

Du

= -k 2w2 ,

Dw =- wu, v

= k 2w.

(5)

208

Chapter 6

Eliminating u from the first two of equations (5), we get -k 2w2 = -WD2W + (DW)2

w2

or

,

WD2W - (DW)2 = k 2w4.

Solving this equation by the method of Art. 40, letting Dw = p, D2W = p dp/dw, we have wp dp _ p2 = k2w4 dw '

then, as in Art. 20(a) , 2p dp _ 2 p2 = 2k2u1 dw w '

where the constant of integration is taken as -c 2 ; if the + sign is used the resulting solution will involve hyperbolic instead of circular functions. Solving for p( = dw/dt), separating the variables, and integrating again, we find

1

kw

c

c

- sec" -

c

= ± (t

w = k sec c(t

+ b) '

+ b).

The last two of equations (5) then give v and u: v u

= ck sec c(t + b), = -c tan c(t + b).

Article 49

209

The relations (5) of Art. 48 yield the values of x, y, z: k2

x =

=

y

+ 1 c sec c(t + b)

3k

CJ)2k 2

+ CJ) c sec c(t + b) -

3k

CJ)k 2

Z =

c

- 3 tan c(t

+ CJ)2 csec c(t + b) -

c

3 tan c(t C

3k

+ b),

3 tanc(t

+ b),

(6)

+ b).

Equations (6), in which k, b, and c are arbitrary constants, represent the general solution of the system (1). It may be noticed that if we write k

al =

2

+1

CJ)

a2 =

3k'

2k2

+ CJ)

3k

,

'

then at, a2, and aa satisfy the relations

+ a2 + aa = 0, ala2 + a2aa + aaal = k +1 ala2aa = 27k3 ; al

~,

6

hence at, a2,

aa are the roots of the cubic a3 -

+

k6 1 la = 0• 3 27k3

(7)

The solution (6) of the system (1) may therefore be written in the form z, y, z

=

aiC sec c(.t

+ b)

c

- 3 tan c(t

+ b),

(i = 1,2,3), (8)

where the a's are the roots of the cubic (7). This is the solution obtained by another method in the paper by Kasner previously cited. An integral of a system of differential equations expressed as a relation free of derivatives, involving some or all of the dependent variables and an arbitrary constant, is of importance

210

Chapter 6

in the discussion of certain physical problems. From the third of equations (5) we see that

+ wy + Z = x + w y + wZ 2

x

W

2

C

(9)

is an integral of the system (1). PROBLEMS

1. Transform equations (9), Art. 48, into equations (10), making use of Euler's relation, et.fJ = cos 8 + i sin 8. 2. Obtain the solution of the system dx dt

= Y+~

dy dt

= z + x - y,

dz dt

=

in the real form

x

-

x,

+ y -~,

+ Be-2t , Ae t + 06-2t , Ae t - (B + O)e-2t ,

x = Ae t y = Z =

(a) by putting a = -1, b = c

=

1 in equations (9), Art. 48; (b) by putting

a = -1, b = e = 1, hence ~ = -2, p, = 0 in equations (10), Art. 48. Solve in real form the following systems, where D = d/&:

3. Dx = ay, Dy

=

D~

= ax.

(lB,

4. Dx=Dy=Dz=x+y+~. 6. Dz + Dy = z, Dy+D~=x,

Ds + Dz = y. 6. Dx = I (y + z + t), Dy = ! (z + x + t),

+ y + t). 7. tDx = y + s, tDy = z + x, tDz = x + y. D~ =

i

(x

"

211

Article 49

8. t Dx1 = ax2, t DX2 = axa, tDxa = ax4, tDX4 = axl. 9. Show that the substitutions

u=3:+Y+s, v = xy + yz + zx,

w = X'gZ, reduce·system (1) of Art. 49 to the system

Du+u2 = 3v, De = 0,

Dw+3uw = v2• 10. Obtain the integrals

z-y ---=C z-x

and

xy+yz+zx=C

of system (1), Art. 49. 11. Prove that the three integrals of system (1), Art. 49, given in equation (9) and in Prob. 10, namely, J(x, y, s)

=

+ ",y2 + ",2s = C1, X + '" y + "'z

g(x, y, z)

=

z-y = C2, z-x

x

h(x, y, s) = xy + yz + zx = Ca, are not independent by showing that the Jacobian of J, g, h with respect to x, y, z vanishes. (Cj. Art. 21.) 12. Letting x/z = a, y/z = {j, the first two integrals in Prob. 11 may be written

0(0:, (j)

= 1-{j = I-a

C2 •

Show that F and G are not independent and find the relation connecting them.

e~7 SERIES SOLUTIONS 60. Power series. There are many simple-looking differential equations, for example, d2 y/ dx 2 + xy = 0, for which the methods discussed up to this point are inadequate to yield a general solution. When such an equation is encountered it may be possible to solve it by use of an infinite series. We shall consider in this chapter only certain =Ie equations of the form (1)

where the f's are functions of x and primes denote differentiation with respect to ». Assume that a value of y, expressible in the form of a power series in x, satisfies equation (1), and write down the expressions for y and its first two derivatives: (2) y = ao + alX + a2 x2 + aax3 +...+ anxn +...,

+ 2a2x + 3aax2 +...+ (n + 1)an+lXn + , y" = 2a2 + 3 ·2aax + 4·3a4x2 + ... + (n + 2)(n + l)an+2xn +....

y' = al

(3)

(4)

Not all differential equations of form (1) can be solved by the methods of yf - y = 0 (cf. Probe 26, this chapter. For example, the equation 4x 3y" + Art. 8) has a general solution which is not expressible by series of type (2), Art. 50, or of type (1), Art. 51. In this case the assumption of a solution of type (2), Art. 50, leads merely to the trivial solution y = 0, and the assumption of a solution of type (1), Art. 51, leads to a contradiction ao = o. Furthermore, in other cases these assumptions might lead to series which do not converge except possibly for x = o. It is understood that a solution expressed in series form is valid only for values of x for which the series is convergent. A fuller discussionmay be found in Ince's Ordinary Differential Equations, Chapter VII, or in Cohen's· Differential Equations, Chapter IX. 212 III

6r-

Article 50

213

Now substitute the values of y, y', and y" from (2), (3), and (4) in equation (1), then equate to zero the complete coefficient of each power of x in order that (1) shall be satisfied identically. It may happen that two of the a's thus remain arbitrary but that the others are determined or expressible in terms of the two arbitrary ones. In this case equation (2) expresses y as a power series in x involving the two a's as arbitrary constants. If the series converges, equation (2) furnishes the general solution of (1), valid for values of z for which the series is convergent. Some of the following examples illustrating the series method could be solved without the use of series; in such cases it is interesting to notice that the results obtained by the use of series are equivalent to those obtainable by previous methods. EXAMPLE

1. Solve (1

+ x 2)y" + xy' -

y

=

o.

(5)

Substitute the values of y, y', and y" from (2), (3), and (4) in (5) and arrange the result in tabular form. At the left of the table are the terms of equation (5), at the top the various powers of x which occur, with their coefficients in the body of the table.

-y xy' y" x 2y"

con.

x

x2

Xs

· ..

-ao

-al al

-a2

-as

· ..

2a2

3as

3·2aa

4·3a4

5· 4a6

~

3·2as

~

·.. ·.. ·..

·..

xn

-an nan

(n

+ 2) (n + 1)an+2 n(n - l)a n

· ..

· .. ·.. · ..

The sum of each column of coefficients must vanish. The first two columns give 2a2 - ao = 0,

= 0,

ao arbitrary,

a3

=

0,

al

arbitrary.

214

Chapter 7

The next two columns give

4·3a4

+ 3a2 = 0,

5·4as

+ 8as =

0,

as

= 0,

and so we see that all a's with odd subscripts vanish, except at which is arbitrary, and that each a with even subscript can be expressed in terms of ao which is arbitrary. Perhaps the best way to compute the a's is to obtain a general formula by equating to zero the column of coefficients of x": (n

+ 2)(n + l)an + 2 + (n2 an + 2

Then, letting n

= 0,

=-

l)an

= 0,

n-l n

+ 2 an·

1, 2, 3, 4, "., we find

n=O, 11,

= 1,

as = 0, 1

n = 2,

a, --

n = 3,

as = -faa = 0,

- .l.,., 4:Ul2 --

- a0,

-

2·4

1·3

n=4,

a6

•

= -la, = 2.4.6 ao,

•

Substituting these values of the a's in equation (2) and remember.. ing that at is arbitrary, we have

Y = ao ( 1 + -1 x 2 - - 1 x4 + 1·3 x 6 - ... ) 2 2·4 2·4·6

+ alX,

(6)

which, with ao and al arbitrary constants, is the general solution of equation (5). We might recognize the series in parentheses as the expansion of (1 X2)~ and write the solution (6) in the finite form

+

y = aoV

I + x2 + atx.

(7)

215'

Article 50

However, if the series in parentheses were not recognized we could obtain the finite form of solution as follows. Putting ao = 0, at = 1 in (6) yields the simple particular solution y = x. We now apply the method of Art. 41 to find the general solution of (5) when a particular solution is known. By substituting y

=

y' = xv' + v,

xv,

y"

= xv" + 2v',

in equation (5); it is reduced to

~ +G+l~X2)dx=O. Two integrations give log v'

+ 2 log x + ~ log (1 + x 2 ) = log Ct,

or

dv dx

-=

C1 x 2V l

+ x2

,

then

hence, multiplying by x, we obtain y

= A VI + x2 + Bx,

where A and B are arbitrary constants, which is form (7). Notice that the series solution (6) is valid only for -1 < x < 1, whereas there is no such restriction on x in solution (7). The solution may be verified by eliminating the arbitrary constants from the solution in finite form and thus obtaining the original differential equation (Problem 27, Art. 8). EXAMPLE

2. Given the differential equations

x2 y" - 2xy' 4x 2y"

+ 2y = 0,

+ 4:vy' -

y = 0,

(a) (b)

obtain the general solution of (a) by the power series method and show that this method is inadequate to yield the general solution of (b).

216

Chapter 7

Substitution of the values of y, v', and y" from equations (2), (3), and (4) in equation (a) results in the following table:

2y -2xy' x 2y"

con.

z

x2

2ao

2al - 2al

2a2 -2·2a2 2a2

XS

2aa -2·3aa 3·2as

·.. · ..

·..

·..

xn

·..

2an -2nan. n(n - l)a n

·.. ·.. · ..

Equating to zero the sum of each column of coefficients, we see that all the a's are zero except al and a2 which are arbitrary since the columns in which they occur vanish identically. This could also be seen by summing the coefficients of z": (n2

-

3n

+ 2)an = (n -

l)(n - 2)an

= OJ

hence an = 0 unless n = 1 or 2, al and a2 being arbitrary. Equation (2) then reduces to

the general solution of equation (a) in finite form. Applying the same method to equation (b) we have the table:

-y 4xy' 4x 2y"

con.

x

-ao

-al 4al

x2

x3

-tl2

-aa

4·2~

4·3aa 4·3·2aa

4·2~

·..

xn

· ..

·.. ·.. ·..

-an 4nan 4n(n - l)an

·.. ·.. ·..

Equating to zero the sum of each column we find that all the a's are zero; equation (2) gives only the trivial solution y = O. There is no power series in x which satisfies equation (b). The power series method is inadequate for the solution of (b) but we shall obtain the solution by a more general series method in the next article. Both equations (a) and (b) can be solved by the method of Art. 38.

217

Article 50 EXAMPLE 3. Solve by the power series method

xy" - (x

+ 2)y' + 2y =

and express the solution in finite form. Substituting the values of y, y', and (8), we have the following table: con.

2y -2y' -xy' xy"

2ao -2al

x2

x

0,

(8)

v" from

(2), (3), and (4) in

2~

2al -2· 2a2 -al

-2·3as

2~

3·2as

2as -2·4a4 -3as 4·3a4

-2~

xn

·..

2an -2(n l)an+l -nan (n l)nan+l

·.. ·.. ·.. ·..

·..

x3

·.. ·.. ·.. ·..

+ +

Equating to zero the sum. of each column, the first column give~_ al = llo, with ao arbitrary; the next gives - 2Cl2 + al = 0, a2 = al/2~ the x2 column vanishes identically, making aa arbitrary; the remaining columns determine a4, a5, ..• in terms of aa. The general formula, obtained by summing the coefficients in the x" column, is (n

+ 1) (n -

2)an +l

=

(n - 2)an ,

or

an an+l = n

(n ~ 2),

+ 1'

from which we find

n = 0, n = 1, n = 2,

aa arbitrary,

n = 3, n

= 4,

•

•

•

•

•

•

Chapter 7

218

Substitution of these values of the a's in (2) gives y=ao (

1+x+2

X2)

4

5

X X ) +a3 x3+-+-+ ... 4 4·5 ' (

(9)

the general solution of equation (8) in series form, ao and a3 being arbitrary constants. In order to express the solution in finite form we might use the method of Example 1, but it is simpler to notice that the second part of the solution may be written 3 ) .3!a3 (x3!+4!+5!+··· x4 x5

,

and that the series in parentheses is now the series for first three terms, 1 + x + x 2 /2. Hence Y = ao

(1 +

x

~

minus its

+ ~) + 3la3 [e'" - (1 + x + ~)],

or, letting ao - 3 tas = A, 3 !a3 Y= A (1

= B,

+ x + ~) + s«,

(10)

where A and B are arbitrary constants. Equations (9) and (10) are equivalent, the latter being the solution of (8) in finite form. This example could have been solved without using series, by the method of Art. 41. Verification of solution (10) is found in Problem 28 of Art. 8. EXAMPLE 4. Solve

v" + xy = o.

Substituting the values of y, y', and (11), we have the following table: con.

xy y"

x2

x

ao 2112

3· 2a3

al 4·3lZ4

x3

tl2

5·4a6

(11)

v" from (2), (3), and (4) in

·.. ·.. ·..

·..

xn

(n

+ 2) (n +

an-l 1)an+2

·.. · ..

Article 50

219

Equating to zero the sum. of each column, we see that a2 = 0; ao and al are arbitrary; aa, a6, ll9, .•. are expressible in terms of llo; a4, a7, alO, ... are expressible in terms of al; as, as, au, ... are expressible in terms of a2 and are all zero. The general formula is

an+~

=-

(n

+ 1)(n + 2) •

Then, letting n = 1, 2, 3, ... , we have

ao aa = - 2.3' aa

4ao

a6=--=5·6 6!' ll9=

1·4· 7ao 9!.' alO =

- -8·9 = •

--=

•

•

101

9·10

•

•

2X4 al ( x - 4!

+

= 0,

as

= 0,

' au

= 0,

•

•

2· 5a l

6·7 7! ' 2·5· 8a l a7 ---=

xa 1.4x6 1.4.7x9 , ) y=ao ( 1- 3 1+ 6! 91 +...

as

•

•

•

•

+

2.5x7 2.5.8x10 7! 10 !

)

+... ,

(12)

the general solution of equation (11), where ao and al represent arbitrary constants. Here we leave the solution expressed in terms of two infinite series which do not represent expansions of any elementary functions so far encountered. The solution may be expressed in terms of Bessel functions. * PROBLEMS Solve the following differential equations by the power series method; also, in case the result involves an infinite series, express the solution in finite form if possible. In the first six problems verify the result by obtaining the original differential equation from the general solution as in Chapter 2. 1. x(x 2)y" - 2(x l)y' 2y = O. 2. x(x + l)y" (x - l)y' - y == O.

+

* See Reddick

+

+

+

and Miller's Advanced Mathematics for Engineers, Chapter VI. Following the theory of this chapter it may be shown that (12) is equivalent to y = Ax~J-~(ix%) + BX~J~(ix%).

220 3. 4. 5. 6. 7. 8. 9. 10. 11.

Chapter 7

+ +

(1 - X 2)y" - xy' 4y = O. (1 - 4x 2)y" - 4xy' 4y = O. xy" - (2x2 3)y' + 4xy = o. xy" - y' 4x 3y = O. y" - xy' - y = O. (1 - x 2)y" - 2xy' + 2y = O. y" - xy' 2y = O. xy" - 2y' + xy = O. (Cf. Prob. 29, Art. 8.) 2(x2 + 8)y" + 2xy' + (x 2)y = O.

+ +

+

+

12. (a) Show that the power series method gives only a particular integral of the differential equation xy" + 2y' - xy = OJ find the particular integral and express it in finite form. (b) By making use of the particular integral found in (a), obtain the general solution of the differential equation. (c) Eliminate the arbitrary constants from the general solution found in (b), thus obtaining the original differential equation. (Cf. Prob. 30, Art. 8.)

51. The series of Frobenius. In Art. 50 we found that the differential equation (b) of Example 2 could not be solved by the method of that article since it has no solution in the form of a power series in z. We shall solve this differential equation as the next illustrative Example 1 by employing the more general Frobenius series. The series of Frobenius is obtained from the power series (2) of Art. 50 by multiplying it by xc: y

=

XC(ao

+ alX + a2x2 +... + a",x'" +. ·.), ao ~ O.

(1)

Here c is a constant to be determined, as well as the a's, by substituting the series in the equation to be solved and equating coefficients as in Art. 50. For zero or positive integral values of c, (1) is merely a power series, but for negative or nonintegral positive values of c the series represents a new type. The Frobenius series (1) is therefore a generalization of a power series, including it as a special case. The assumption ao ~ 0 means that XC is the lowest power of x appearing in the series. The method of solving a differential equation by use of series (1) is similar to that of Art. 50. Assume that a value of y

Article 51

221

expressible in the form of series (1) satisfies the differential equation of form (I), Art. 50, and write down the expressions for y and its first two derivatives:

+ a1XC+1 + a2XC+2 + aax + + ..., = Cac¢-l + (c + l)alx + (c + 2)a2XC+1 +. ·.,

y = aoXC

y'

C

C

3

(2) (3)

v" =c(c-l)ao1f-2+(c+ l)catxc-l+ (c+2) (c+ 1)a2xc+. .•• (4) Substitute the values of v, y', and v" from (2), (3), and (4) in the differential equation, form a table like those in Art. 50, starting with the lowest power of x which occurs after the substitutions are made, then equate to zero the complete coefficient of each power of x. Suppose that two values of c are thus determined. If, for one value of e, we are able to determine all the a's in a manner involving only one of them which is arbitrary, we substitute these values in (2) and obtain a value of y, say Yh which is a particular solution involving one arbitrary constant. * If, for the other value of c, we can determine the set of a's, which we now call a"s, involving one of them which is arbitrary, we have another particular solution, say Y2. The sum

involving two arbitrary constants, is the general solution of the differential equation. It may happen that for one value of c the a's are determined in terms of two of them which are arbitrary; the corresponding value of Y, involving two arbitrary constants, then represents the general solution. Solutions in series are valid for values of x for which the series converge. There follow some illustrative examples, the first being the differential equation (b) of Example 2, Art. 50. * It is understood here as in Art. 50 that the series represents a solution only for values of x for which it converges.

222

Chapter 7

EXAMPLE

1. Solve 4x2y"

+ 4xy' -

y = O.

(5)

Substitute the values of y, y', and y" from (2), (3), and (4) in (5) and arrange the result in tabular form, starting with the lowest power of x present after the substitutions, namely XC:

-y 4xy' 4ry"

x c+2

xc+1

XC

-ao 4cao

-a2 -al 4(c+l)al 4(c+2)~ 4c(c-l)ao 4(c+l)cal 4(c+2) (c+l)~

·..

x c+r

·..

·.. ·.. ·..

-a r 4(c+r)ar 4(c+r) (c+r-l)a r

· .. ·.. ·..

Equating to zero the sum. of the coefficients of xc, we find c = ~ or - ~,

(4c2 - l)ao = 0,

ao arbitrary.

The set of a's corresponding to each of the two values of e will now be determined. We obtain a general formula by equating to zero the sum of the coefficients of xc+~, first for c = ~, then for c = - ~. For c = ~: [4{~

+ r)2 -

I]«,

=

(4r

+ 4r)a, = 4r(r + l)a, = OJ

hence a, = 0 unless r = 0, with ao arbitrary. For c = - ~: [4( - ~

+ r)2 -

l]a~

= (4r2

-

4r)a:.

=

4r(r - 1)~

= OJ

hence a:. = 0 unless r = 0, 1, with a~ and a~ arbitrary. Then equation (2) gives, for c = ~ and c = - ~ respectively, Yl

=

aox~,

Y2

=

~-1/2

+ a~x~.

The first of these results is superfluous since the second contains two arbitrary constants and represents the general solution. However, we would get the same general solution by setting y = Yl + Y2 and replacing ao + a~ by a new arbitrary constant. The general solution of equation (5) is therefore

Y = Ax~

+ Bx-~,

where' A and B are arbitrary constants. This solution could be found without the use of series, by the method of Art. 38.

Article 51

223

As a second example we solve the differential equation of Problem 12, Art. 50, by use of a Frobenius series. EXAMPLE

2. Solve xy"

+ 2y' -

xy = 0,

(6)

and express the result in finite form. Substituting the values of y, y', and y" from (2), (3), and (4) in (6), we have the following table: xC-I

x c+ 1

XC

-xy 2y' 2cao 2(c+l)at xy" c(c-l)ao (C+l)cal

XC

-ao 2(c+2)~

(c+2)

(c+l)~

...

+2

X

... ... ...

-al 2(c+3)aa (c+3) (c+2)a,J

C

+T

-ar-l 2(c+r+l)ar+l (c+r+l) (c+r)ar+l

·.. ·.. ·.. ·..

Adding the first column and equating the sum to zero, we have

c(c + l)ao = O. Hence c = 0 with ao arbitrary, or c = -1 with ao arbitrary (= ~). The second column gives (c + l)(c

+ 2)at = o.

Hence if c = 0, at = 0, and if c = -1, at is arbitrary (= a~). The general column gives FOR

FORC=O (r

+ 1) (r + 2)ar+t = ao ao a2- · !, 2 3

3

ar-I,

r(r

C

= -1

+ l)a~+l ,

=

a:.-h ,

, ao ao a ---2 - 1.2 - 2!'

,

,

, at al a3- - -! , - • 2 3

,

3

,

ao a4 - 3.4 - 4!' ,

a2

,

a3

,

a3

as

= 5.6 =

0,

ao a6 - 6.7 - 7!' a4

•

•

,

at

a ---5 -

4.5 - 51'

,

,

a4 ao a ---6 - 5.6 - 61' ,

224

Chapter 7

It happens here that for o = -1 all the a's are determined in terms of two of them, ~ and a~, which are arbitrary; hence this value of e yields the general solution: 3

5

2

4

6

I x x x ) ( x x x ) . y=a~ ( - + - + - + - + ... +a~ 1+-+-+-+··· . (7) x 2! 4! 6! 3! 5! 7!

The second series of (7) would give the particular solution corresponding to c = o. T~ express (7) in finite form, we make use of the expansions: x3 x5 x1 sinhx=x+3t+51+71+ •.. , x2 x4 x6 coshx=I+-+-+-+··· 21 4! 61 •

Then (7) reduces to ,coshx + ,sinhx y = ao al · x x

(8)

A trick method. If we notice that

D(xy) = xy' where D

+ y,

D 2 (xy)

= xy" + 2y',

= d/dx, then equation (6) can be written (D 2

-

l)xy

=

o.

Hence we have at once by the method of Art. 29, xy = Clf?

+ C2 e- z ,

or

(9) which is the general solution of equation (6) and is equivalent to (8). EXAMPLE

3. Show that the differential equation xy" - xy'

+y =

0

(10)

has only one particular solution of type (2). Make use of this particular solution in finding the general solution.

Article 51

225

Substitute the values of y, and form the table: xc- l

y', and v" from (2), (3), and (4) in (10)

·..

xc+l

XC

·..

xc+r

y

ao

al -(c + l)al 2) (c 1)t12

-xy' -000 xy" c(c - l)ao (c + l)cal (c +

+

-

·.. ar · .. ·.. -(c +r)ar ·.. ·.. (c +r + 1) (c +r)ar+l ·..

The first column gives c(c - l)ao = 0, from which c = 0 or e = 1 with ao arbitrary. The second column gives (c + 1)001 = (c - l)ao which cannot be satisfied by c = 0, since ao ~ 0; but for c = 1, at = o. Furthermore all the successive a's are 0 when c = 1, as can be seen from the formula (r + 2)(r

+ l)ar+1 = rare

Therefore the only solution of type (2) is for c = 1, ao arbitrary, a1 = a2 = ... = 0, namely, the particular solution y

= aox.

(11)

To find the general solution we use the method of Art. 41 with y = x as a simple particular solution obtained from (11) by putting ao = 1. Substituting y = xv, y' = xv' + v, y" = xv" 2v', in equation (10), we obtain

+

+ 2xv' - x2v' = 0, dv' + 2 - x dx = O.

x 2v" v'

Integrating,

x

log v' + 2 log x - x dv

Olf?

;k=7· Then

= log 0 1,

226

Chapter 7

Substituting for eZ its power series expansion,

v = 01

f 0 + ~ + ;1 + :1 + :~ + ...) ax +

= 01 (

I -

X

x

2

O2

3

x ) + log x + 2! + 2.3! + ~4 t + ... + O2 , x

and, since y = xv, the general solution of equation (10) is

x3 x4 ) y=OI ( -1+xlogx+2t+2.3t+3-;4t+ ... X2

+02 X•

52. Bessel functions of zero or positive integral order. Bessel functions are named after the German mathematician and astronomer, Friedrich Wilhelm Bessel, who was director of the observatory at Konigsberg. He obtained them in solving a differential equation connected with a problem in planetary motion, but they occur in various other problems in applied mathematics. Bessel functions are particular solutions of the differential equation x 2 y"

+ xy' + (x 2 -

n 2 )'11

= O.

(1)

The number n may have any real or complex value, but we limit this discussion * to the case where n is zero or a positive integer: n = 0, 1, 2, 3, . . .. With this restriction on n let us make use of a Frobenius series to find a particular solution of equation (1). Substituting in (1) the values of y, y', and y" from equations (2), (3), and (4) of Art. 51, we form the following table of coefficients: ZC

-n2y x2y X'/I' x 271"

XC+ 2

zC+l

-n2ao

- n2al

-n2~

(c+l)al c(c-l)ao (C+l)cal

(c+2)~

ao cao

* See footnote to Ex. Bessel functions.

(c+2)

(c+l)~

x C+ 3

·..

- n2aa al (c+3)aa (c+3) (c+2)aa

·.. ·.. ·.. ·..

4, Art. 50, for reference to

8.

ze+ r

-n2ar ar-2 (c+r)a,. (c+r) (c+r-l)a,.

·..

·.. ·.. ·.....

more extensive discussion of

Article 52

227

Setting the sum of the first column equal to zero, we find (c - n 2 )ao = 0, which with ao arbitrary is satisfied if c = n. We now obtain the particular solution of equation (1) corresponding to c = n. The second column gives al = 0, then the fourth column gives aa = 0, ete.; all a's with odd subscripts vanish. We compute the a's with even subscripts from the formula obtained by equating to zero the sum of the coefficients in the x c +r column with c = n: [(n + r)2 - n 2]a r + ar-2 = 0, ar - 2 . a - r r(2n + r) Hence 2

a2 = - 2 (2n a2

4(2n •

a2k

•

-

+ 4)

ao

2·4(2n

•

=

+ 2) , + 2)(2n + 4) ,

•

(-I)k

+

•

ao

2·4 ... 2k(2n 2)(2n + 4) ... (2n . 2 .4 . . . 2k = 2k (1 .2 .'. . k) = 2k k r

Now since

+ 2k) .

and (2n+2) (2n+4) ... (2n+2k) -

2k(n+l)(n+2) .•. (n+k),

we may write a2k

=

(

-1

)

ao

k

22kk !(n

+ 1) (n + 2)

... (n

+ k) .

Multiplying both numerator and denominator by 2n n !, 2n n!ao (-1) 2n +2k k !(n k)! ' k

a2k

=

+

or, setting 2nn !
,

a2k

= (-l)k

2"+2kk~ + k) '"

228

Chapter 7

In this formula let k = 0, 1, 2, 3, ... and substitute the resulting a's, together with c = n, in the Frobenius series (2) of Art. 51 (remembering that all a's with odd subscripts vanish); then , [ xn xn+2 y = ao 2nn I - 211.+21 !(n \

+ 1) I +

x n+4 ] 211.+42 !(n + 2)! - · .. •

The series in brackets is known as the Bessel function J ~ (e) of order n(n = 0, 1, 2, 3, ... ), which, written in:E notation, is In(x) =

~

f::t

+2k (-I)k 2 n+2kk I(n + k)!· X

n

(2)

We therefore have the following particular solution of the 'differential equation (1): (3)

where ~ is an arbitrary constant and In(x) is the Bessel function of zero or positive integral order given by (2). '. The power series (2) is convergent for any finite value of z, and the derivative or integral of In(x) may be found by differentiating or integrating the corresponding series term 'by term. Writing out the expansion of Jo(x) and J 1 (x ) from (2) we have Jo(x)

J 1 (X)

=

=

x2

1 - 22

x4

x6

+ 24(21)2 2 +... 2k X + (-I)k 22k(k!)2 +.·.,

x z3 2 - ~2!

6(3!)2

x5

+ 2 52!3!

x7 2 73 !4I

(4)

+.. ·

+ (-I)k 22k+

X 2k+1 1kl(k

+ I)! +....

(5)

We notice that .. the derivative of the second term in Jo (x) is the negative of the first term of J 1 (x ) , the derivative of the third term in J 0 (x) is the negative of the second term in J 1 (z),

-

Article 52

229

~ etc. In general the derivative of the term containing ~k+2 in Jo(x) is

+

d X2k+2 k (2k 2)X2k+1 - ( l)k+l _ -( -1) 22k+2(k 1) 12 = dx 22k+2(k 1)!2 X 2k+1 - (-I)k 22k+1k !(k

+

+

+ 1) 1'

which is the negative of the term containing ~k+l in J 1 (x) ; also the derivative of the first term of Jo(x) is zero. Hence '(6)

PROBLEMS

1. Solve the differential equation x 2y"

+ 4xy' + 2y =

0

(a). by use of a Frobenius series; (b) by the method of Art. 38, obtaining the

same result by both methods. 2. Solve the differential equation xy"

+ (2 -

x )y' - y

=

0

(a) by use of a Frobenius series; (b) by reducing the differential equation to

(D2 - D)xy = 0,

obtaining the same result by both methods. Solve each of the four following equations by use of a Frobenius series and express the result in finite form if possible. 3. 4. 6. 6.

+

4xy" 2y' - y = o. 4x(1 - x)y" 2(1 - 2x)y' + y 2x(1 + x)y" + y' - 4y = o. 2xy" + 3y' + xy = O. .

+

=

o.

7. Show that the method of Art. 51 Yields only a particular solution of xy" Find the general solution.

+ (2 -

x)y' - 2y

=

o.

230

Chapter 7

8. Show that the method of Art. 51 yields only the particular solution y = ao(1 + x) for the differential equation x 2y" + (x + 1)y' - y = O. Obtain the general solution by using a series in descending powersof a, y

I

aQXn

+ a1xn-1 + ~xn-2 + "',

and express the result in finite form. 9. Compute to four decimal places the values of J o(3) and J1(2). 10. Prove that

f

~ [xJ,(x)] = xJo(x),

xJo(z) dx

= x.Tt(x).

11. (a) Find a particular solution of the differential equation

xy" - y'

+ xy =

0

and express the result in terms of a Bessel function. (b) Prove that the value for y found in (a) satisfies the differential equation 12. Find a solution of

4x2y"

+ (x + 1)y =

0

such that y = 1 when z = 4. 13. (a) Find a solution of x 2y" - xy'

+ (x2 + l)y =

0

such that y = -6 when x = 3. (b) Find the area under the curve,represented by the solution found in (a) from x = 0 to x = 2. 14. (a) Find a particular solution of the differential equation d2y

y

dx

x

-= 2

and express the result in terms of a Bessel function. (b) Prove that the value for y found in (a) satisfies the differential equation. 15. Solve the differential equation of Prob. 7 by writing it in operator ' notation and factoring the operator.

/

/

ANSWERS Art. 3

+ +

+

+

+

1. (a) y = sin x C; (b) y = x 4 C1X 2 CcaX Ca. 2. y - x = 0(1 xy). 3. (a) 5; (b) 1; (c) 3. 4. (ax 5. Cl(f;2 - C1) = '(C2 X)2. 9. 1. 10. 0.533.

+

+ b)(ay + b) =

C.

Art. 8 xy' = x + 2y. 6. (1 + y'2)y'" = 3y'y"2. (x - y)y" + 2y' + 2y'2 = O. 11. xy" = y' + y'3. 12. y2(1 + y'2) = r2. (x - y)y" = (1 + y')(1 + y'2). 14. (x2 - 2xy - y~)y' = x 2 + 2xy _ y2. 2xyy' = y2. - x 2. 16. yy" + v" = O. 17. 2xy' = y(1 _ y'2). 18. 2xy" + y' = O. 19. (x - a)2 y" = 2(x - a)y' - 2(y - b). 20. y'ez1I'11J + 2-rrxy3 = O. 21. x 3y' = y3. 22. (y - l)y" + y.....= (y - 2)y'2. 23. y'" - 2y" - 5y' + 6y = O. 24. x + y + 1 + 2(3x + 5y + 7)y' = O. 25. y' = cosh [(x/2) + y]. 26. 43;3 y" + 6x2y' - Y = O. 27. (1 + x 2)y" + xy' - y = O. 28. xy" - (x + 2)y' + 2y = O. 29. xy" - 2y' + xy = O. 30. xy" + 2y' - xy = O. 31. yiv + 64y = O. 32. (a) x 2y" - n(n - l)y = 0; (b) xyy'" + 2yy" - xy'y" = O. 33. (a) yy" = 1 + y'2; (b) (1 + y'2)y'" = y'y"2. 34. 2x 2y" - xy' y = O. 2p/d82 2 ) + p = O. 35. (a) 2(xy' - y)(1 + y'2) = (x + y2)y"; (b) (d 36. (a) 2y'y'" = 3y"2; (b) (x - y)y" + 2y'2 + 2y' = O. 37. 9y"2yV - 45y"y"'yiv + 40y"'3 = O. 38. 3y"yiv - 5y'''2 = O. 5. 8. 13. 15.

+

Art. 11

1. (a) y = Cx3 ; (b) sin2 y = C sin x; (c) xVI - y2 2.

6. 9. 11.

+ yVl

- x2

= C;

(d) eZ + e-1J = C; (e) x - 2xy + 2y2 = Cxy2. (a) x 2 = 2[2 - y -log (1 - y)]; (b) x = (l/c)(ae ct / a - b); (c) x = a sin y; (d) x = 3(2 t / 10 - 1)/(2t/l0 - i). 3. i. 4. l;yf!. 5. 0.917. xy + 2(x + y) = O. 7. p2 - 1 = tp sec 9. 8. (4 + x 2)(1 - y2)2 = C. y = e-Z tan x + C. 10. (1 - x - y + xy)e(Z+1J>l2 = C. x(1 + y) = Cel/ 1J . Art. 16 '

1. 2.88ft/sec; 1.05ft. 2. -0.557. 3. (a) 146ft/sec; (b) 945 ft; (c) 3.73 sec; (d) 3.65sec. 4. (a) 6.83V sec; (b) 5.70V sec. 5. (a) 2.16 sec; (b) 67.0 ft. 6. t = [wV/g(w - E)] log [100/(100 - p)] sec. 7. (a) 8.83 It/sec: (b) 1.01 sec. 8. 126 It/see; 3.10. 9. 147 It/sec; 2.67. 10. 16A/7. 231

232 11. 13. 17. 22.

{

Answers

(a) 29.3 per cent; (b) O.415n yr. 12. (a and b) 2.41 min; (c) 27.3°F. (a) 2.36; (b) 2.57. 14. 10.3. 16. 2a/(n 1). 16. 17,900 ft.

+

8.85Ib/in2• 18. 15,800 ft. 19. 344 X 10-7Ib/in3• 20. 1.12 in. 3.45 mi. 23. 64.9 lb/ft"; 680,000 Ib/ft2• 24. 3.11 X 108 cal/daYi 63.4°C. 26. 117 cal/sec; 69.1°C. 26. (a) U = UI - (UI - U2) log (r/n)/log (r2/rl); (b) 86.4°C. 27. (a) U = UI [(UI - U2)(r - rl)/(r2 - rl)](r2/r); (b) 76°C. 28. 5.47 em. 29. klk~(Ul - u2)/(k2h kl~) cal/sec, 30. 2rL(Ui) - Un)/'I,~= 1 (Iller) log (Xr/Xr-l) cal/sec. 32. 10.2 min. 33. 3.93 min. 34. 2.28t. 36. (a) 45.3 min; (b) 54.3 min. 36. (a) 19.2 min; (b) 1.31 ft. 38. t. 39. 56.9 hr. Art. 18

+

2. :c = -y3 + Cy. 3. y = x(log x + C). 4. y = x tan (x + C). 6. y2 = xe'U + C. 6. x 2 - y2 = Cx. 7. xe~/'U = C. 8. 3x2 + 2y2 = Cx4• 9. log (y/x) = 1 + Cz: 10. (x + y)e~+'U = C(x - y). 11. x 2 - 2xy - y2 • C. 12. 6X 2y2 + y4 = C. 13. x 2 + 2y = Cy2. 14. y3 = Cx2 + x 3• 16. x 2 = Cy3esJn 3:. ~~"y = C[l + log (x/y)]. 17. x 2 = C sin (y/x). 18. (x2 + y2)2 = Cxy. 19. 7x 2 + 6xy + 11y2 + lOx + 34y = C. 20. x3 + y3 = xy(x + y + 0). 21. :c + y = Ce~/1/. 22. y2 + yVx 2 + y2 = Cx. 23. xy +Vx2 - 1 V y2 - 1 = cosh (xy + 0). 24. V:c - y(V; = Cevu/("'i-V"iJ). 26. y + V x 2 + y2 = Cx2e"1/:£2+112/11.

vY)

Art. 21 1. y = (x C)e~. 2. 15x3y = 3x 5 5x3 + C. 3. y = (2/x) cV;. 4. x 2y = 3 C~. 6. 3y = 1 x 2 C(l x2)-l/2. 6. uv' 1 - x 2 = sin-l X C. 7. 7x 2y4 = .CvY. 8. xy = x-I + Ce-~. 9. 9y = 6x2 log :D - 4x2 cV;. 10. y = Cx - x 2• 11. 2y2 = 2x - 1 Ce-2~. 12. xy3(x + 0) = 1. 13. y = e" C(x - 1). 14. (y - 2x 4)4 = C(y - 3x 6)3. 16. x y 2 = Ce". 16. 2(x - y) log (2x 2y 1) = C. 17. (x - 1)2 + y2 = Ce2 tan - 1 11l/ (Z - l)). 18. i 3 = sin y cos Y..+ ce:». 19. 3y4 = X2(y6 C). 20. (x 3y - 5)2 = C(x 2y - 2). 2 21. y = x (1 Cfil/~). 22. l/y = sin x C cosx. 23. 2x y2 + 1 = Ce,}. 24. (x + y + 1)e(~+2) J(~+1/+1) = C. 26. y2 = Ce~/'U. 26. xel/~'U = C. 27. x 4 = Cy y2. 28. 3y2 = 1 CX2y2. 29. ye:rt-/2,} = C. 30. xy = C(xy 1)e,}/2. 31. y2 = (x - 2)2 fil-(~/2). 32. 4. 33. y = - sin x. 34. y = 2 sin 2 x - 2 sin x; 4, -i.

+ +

+

+ + +

+

+ +

+

+

+ + +

+ + +

+

+

+

+

+

+ + + +

+

+

+

+

233

Answers

36. xyl-n = 3(1 - n)/(2n - 3) + CX S- 2n (n ¢ 1, !); y = Cx2e- S / z (n = 1); 2x = (C - 3 log x)vY (n = i). 36. 3y = 5x - 2 tan-1 [! tan (2x + C)]. 37. 2(x + 4y + 1) = 3 tan (6x + C). 38. x log (yIx) = sin x - x cos x + C. 39. (a) x 2(x2y2 + 2) = Cy; (b) y = Cx~y2/4. 40. (a) x = 1 + Ce(4z-4) f(Z-4 y - 2) ; (b) 6(y - x) + C = log (2x + 2y + 1)(x2 + 2xy + y2 + X + y + 1)4. 41. (a) y = x 2 tan [(x2/2) + C]; (b) y (sin x + C) = sec x + C tan x. 42. Cl = C2 + r/3. 43. Cl = C2 - r/3.

Art. 24 1. 1.32 lb/gal. 2. (a) 45 lb; (b) 39/64 lb/gal. 3. (a) 124 lb; (b) 54 min. 4. (a) 4 hr 25 min; (b) 2 hr 27 min. 6. 2871b; 14.9 min. 6. crV(l - r)(l-r)fr lb; V(l - r)(l-r)fr gal. 7. (a) 0.416 amp; (b) -0.413 amp; (c) 0.432. 8. (a) 0.476 amp; (b) 0.299 amp. 9. (a) q = EC KE- t / R C ; (b) q = [EC/(! w2R2C2)](sin wt - wRC cos wt) KE- t/ R C • 10. i = [wECI(l w2R2C2)J(COS wt wRC sin wt) KE- t/ RC • 11. (a) 1.35 amp, 0.0432 coulomb; (b) -8.50 amp, 0.0131 coulomb. 12. -1.35 amp; 0.00677 coulomb. 13. 0.181 amp. 16. L = [2aILI(1 1L2)J(1 f!''Ir). 16. (a) 21L = (1 - 1L 2) f!''Ir/ 2; (b) 0.732. 17. 2.01 lb. 18. 0.630 ft. 19. 8.38 ft. 20. 19.6 lb; 2.22. 21. 730 4 4 ' from the top of the cylinder. 22. 1 - 101-' 1L2 = 5CIL - (1 - 1L2) ( 0/2)Je 2I£'Ir/ 3. 23. 41 0 4 4 ' . 24. 5.85 ft; 0.13 ft. 26. 13,900.

.

+

+ +

+

+

+ +

+

+

Art. 26

+ y2 = 13. 2. y2 = 4e~. 3. x = 1 + '\1'3 - e2 4. 2x + y2 = 6. 6. x 2 + y2 = Cx, 6. y2 = e2 7. x 2 + y2 = Cx, 8. (a) x 2 + y2 = 5x; (b) x 2 + y2 = lOy. 9. (a) p = 4 cos B; (b) p = (40/3) sin B. 10. y2 = 4(x + 1) + 5eZ. 11. P, 6.93 in.; Q, 13.2 in. 12. 0.697 ft. 13. (b) y = k cosh [(xlk) + C]; area under arc = k times arc length. 14. y = 2 cosh [(xI2) + 0.881]. 1. y2 - x 2 = 5 or x 2

1I•

- (z / y) .

16. y = be(z-a)/k. 16. nyn-l = (n - l)(x - a). 17. x - y = Cxy. 19. y = 2n cosh [(xI2n) C] or y = 2n. 20. yS - 3x 2y = k 3 or y = k. 21. Confocal parabolas: (a) x 2 = 2Cy + C2; (b) p(l - sin B) = C. 22. p = C sin" (Bin), or p sin" (Bin) = C. 23. p2 = C sin 2B, or p = C(l - sin B). 24. (a) y2 = 2Cx + 02; (b) p(l - cos B) = C. 26. (a) (lc 2 - 1)x2 + k2y2 = 2Cx C2; (b) p = CI(k - cos B); (c) p = 3/(3 - 2 cos B).

+

+

234

Answers

Art. 26 1. y = Cl:r:&/k. 2. p = CeT fJ/ k. 3. x 2 + y2 - 1 = ey. 4. 3x2 + y2 = Cx. 6. eZ sin y = C. 6. sin x = C sech y. 7. y2 = 4C(x + C). 8. y2 - x 2 = Cx3• 9. x2 + y2 = Ce", 10. x + cos x cosh y = O. 11. 2 sinh y tan (x/2) = C. 12. y2 = i log cos 2x + C. 13. p4 = C cos 2(J. 14. p = 1/(C - sin2 8). 16. p = 2 sin (J + C cos 8. 16. p2 - 1 = Cp sec (J. 17. 2x2 y2 = 17. 18. xS 3xy2 = 13. 19. p = 64 COS1 6«(J/ 4). 20. p = 4(1 - sin 8). 21. v!2 (sin x + sinh y) = cosh y; sinh y = sin x - cos x. 22. ya = C[(a - 2)x 2 - y2], (a ~ 2); ye:rt-/1l' = C, (a = 2). 23. cosh x - a cos y = C sin y. 24. (a) x2 + y2 = 2b log x OJ (b) y = Cz", 26. (a) log p = 1 cos 2(8 + a) + C; 2 (b) (k/p) - (p/k) = (J + C. 26. p(sin n(J)m/n = C. 28. x 2 + y2 1 = Cx. 29. p2 = C(p cos 8 - 1). 2 30. (x + y2 + 1)2 - 4%2 = C.

+

+

+

+

+

Art. 29(b)

+

2. (a) 6ez; (b) -4e-2:l:; (c) 0; (d) (log a - a)naz• 4. y = C1e kz C26-kz; B cosh kx. 5. y = CleGz C2e-z • y = A sinh kx 6. y = C1e(v'i4-S)z + C26-(v'i4+S)z. 7. y = (C1 + C2x)e 2z + (Cs + C4x)e-2z. 8. y = C1 + C"e-(a/2)z Cse-2az. 9. 'II = C1e z + C26-z + Cse(1/2)z. 10. y = Ae(3!2)Z sinh [(V'5/2)x a] Be-(3!2)Z sinh [(V'5/2)x Pl. 11. y = C1 C2X Cp;e2 + C4ev'2z + Cr,e-v'2z. 12. y = (C1 + C2X Cp;e2)e z + C4e-sz. 13. y = C1e(a/b)z + C26(b/a)z. 14. y = C1e-z + C2e2z Cse(2/6)z + C4e-(1/2)z. 15. y = C1e-z + C2e(l/2)z Cse( 6/6)z + C4e-(1/S)z. 16. Y]~=-l = (1 - e)(2 - e)/(3 - e); 4.37.

+

+

+

+

+ +

+

+ +

+

+

Art. 29(c)

1. y = A cos kx + B sin kx. 2. y = eZ(A cos V3x + B sin V3x). 3. y = e-z/2[A cos (x/3) + B sin (x/3)]. 4. 8 = e2t(A cos 3t + B sin 3t). 5. x = e-at (A cos Vk 2 - a2t + B sin Vk 2 - a2t). 6. y = 01 + C26~ + e-z/2(Cs cos v!2x + C4 sin v!2x). 7. y = A sinh (x + a) + Be z / 2 sin [(V3/2)x+PJ+Ce-z/ 2sin[(V3/2)x+"Y] 8. y = C1e6z + e-Sz[C2 cos (x/2) + Cs sin (x/2)]. 9. y = (C1 + C2X)e z + (Cs + C4X)e- Z + Cr, cos V 2x + C6 sin V2x. 10. 8 = A sinh (2'V2t + a) + B sin (2.y2t + P). 11. 8 = Ae 2t sin (2t + a) + Be:" sin (2t + P). 12. y = C1e 2z + C2e-2z + (Cs + C4X) cos .y2x + (C6 + CaX) sin y'2x.

235

Answers

13. y = eCl + C2x)e-Z + eZ(Ca cos x + C4 sin z), 14. y = (Cl + CttX + Caa: 2) cos 2x + (C4 + Cr;t + C6X 2 ) sin 2x. 16. y = Cle- 2z + C2 cos X + Ca sin x + eZ(C4 cos 2x + Co sin 2x). 16. y = Cle b + (C2 + CsX) cos x + (C4 + CoX) sin x. 17. (a) No; (b) yes. Art.30(b)

2. (a) 9x2 + 4v2 = 225; (b) 3 ft, 6 ft/sec; (e) 3.76 sec. 3. 2.63 units. 4. (a) 0040 in.; (b) 13.0 in.zsec downward. 6. (a) 1.77 sec; (b) 2.36 ft above water. 6. 6.67 in. 7. 0.636. 8. 0.783 sec; x = i cos 8.02t. 9. 3.44 ft. 10. (a) 42.2 min; (b) 14.1 min. 11. 2rVLJg; (J = €X cos V g/ Lt. 12. (J = VLJgwo sin v'"iT£t. 13. (a) 3.26 ft; (b) 0.219 rad/sec. 14. i. 1. (a) 0.300a; (b) 0.375a.

Art. 30(') 3. 4. 7. 10. 12. 14. 16. 17.

20.

t = [l/(rl - T2)] log [(r2vO - b2xo)/(rlvO - b2xo)]. t = [log (r2!rl)]/(rl - r2). 6. (D2 + 0.0353D + 21.7)x = O.

e-4.68t ; (D2 + 9.37D + 39.5)x = O. 8. -1.56 ft/sec 2. 9. 0.198 sec. K = 1; 0.172 sec. 11. (a) 0.549 sec; (b) 2.16 ft; 0.0895 ft/sec. (a) 0.206 sec; (b) 3.12 ft; -1.74 ft/sec. 13. 1.63 ft; -0.369 ft/sec. (a) 0.223 sec; (b) 4.09 ft; 8.86 ft/sec. 16. (a) 0.154 ft; (b) 0.315 sec. (a) 0.259 in. above equilibrium; (b) 4.94 in. below equilibrium. (a) 1.16 sec; (b) 60.5 per cent; (e) 8.35 in. 18. 0.565 ft above. 19. 40.2'. 93.5 per cent. 21. T = v' ap(47r 2r2 + log2n)/2rV3g(p - 1); 1.15 sec. Art. 32

1. (a) Ely = 20x(x2 - 300); (b) Ely = 20(20 - x)(100 - 40x + x 2); (e) (4 X 104)/EI ft. 2. 102,500/EI ft. 3. (a) Ely = ire -qx 4 - 4Wx3 + (4qL3 + 12WL2)x]; (b) ~W lb. 4. (5PL3 + 6qL4)/48EI ft. 6. (a) wL4/192EI ft; (b) 54.16wL4/104EI ft. 6. 4.22 ft. 7. 1 ft from each end. 8. kwL6/6Eh3 ft.

9. (a) Ely = -(q/48)x2(L - x)(3L - 2x). 10. (a) (6EIL/W)y = (L - e)x(x 2 - 2Lc e2) and e(L - x)(e2 - 2Lx

+

respectively; (b) (eW/3EIL)[(L2 - e2)/3]s/2 ft.

+ x 2)

Art. 34

y = (CI + C2x)e-Z + e-2z. 2.' y = C1e z + C2e-z - x 2 - 2x - 2. y = Cle6z + C2e-z - ie3z - ie-2Z• y = Cle 2z + (02 + 2x)e3z - !ue- 2X. y = C: + C2x + Cse- z + I\X 4 - lxs + i-x2• 6. y = Cse" + e-(S/8)x[C2 sin (y23/8)x + C« cos (v"23/8)x] + Xl - x.

1. 3. 4. 6.

236

Answers

7. y = (01 + 2x)ez + 0~2z + ix2 + ix + i.

8. y = 01e4z + 02e- 3z + lItre9z + !e-z - -h. 9. y = [01 + C2X - (x2/6)]e-z + 03e2z - 1ez + 2. 10. y = Ole(l+-vI2)Z + 02e(l-v2)z + 1(cos x - sin x). 11. y = [01 + (x/4)] sin 2x + O2 cos 2x + f sin x. 12. y = 01 + 02ez + Oae-z - 2 sin x + cos x - 3x2• 13. y = [01 + (x/7)]e z/2 + 02e- 3z - !e- z/2 - i. 14. y = 01 + 0~-z/2 - 4x2 + 20x + ez12 - 2 cos 2x + 1 sin 2x. 16. x = (01 + t) sin t + (02 - t2) cos t. 16. y = Ole2z + (0 2 - Ix - Ix 2 - x3)e-2z• 17. y = [01 - (x/4) + (x2/8)]ez + (0 2 + OaX + ! sin x + ! cos x)e-z , 18. y = 01 + 02X + OaX2 - 120x3 - 30x4 - 6x5 - x 6 + 04ez + (20/9)e3z• 19. y = eZ(Ol sin x + 02 cos X + tx sin x) + ie-Z(cos x - sin x). 20. y = 01 + 02X + e-(4/3)z[03 sin (0/3)x + 0 4 cos (y2/3)x] + (x3 - 6x2 + 12x)e-z• 21. y = 01e4t + 02e-2t + (Iv cos 2t - -h sin 2t]e- t • 22. y = [01 + l(sin x + cos x)]e-z + eZ/ 2[02 sin (0/2)x + 0 3 cos (0/2)x]. 23. y = 01ez + [02 + OaX - (x2/2) - (x3/2)]e-2z - x. 24. y = Aez + B sin (2x + a) - sin (1r/6)] - 1. 2z Z 26. y = e- (Cl sin 20x + O2 cos 20x) + e- sin 20x. 26. y = (01 + 02X + x3)ez + (0 3 + x)e-2z• 27. y = 01ez/ 2 + [02 - (x/4) - (x2/ 8) + ! sin (x/2) + i cos (x/2)]e-z / 2• 28. x = et - fe2 t + le3t• 29. 8y = x 2 + 2x + 4 - 2ez• 30. y = (x + 2)2 + (x 2 - 4)ez • 31. 0.0749.

[vax -

Art. 36

= = 3. y = 4. y = 6. y = 6. y = 1. y 2. y

7. 8. 9. 10. 11.

(01 + log sin x) sin x + (02 - x) cos x. (01 + 1 log tan x) sin 2x + O2 cos 2x. [01 + (x/3)] sin (x/3) + [02 + log cos (x/3)] cos (x/3). [01 + log (sec x + tan x)] sin x + O2 cos X - 2. OleZ + C~-z -! -! sin 2 x. [(01 + x) sin x + (C2 + log cos x) cos x]e-z/2 + iez/2•

+ ce:: + + + + +

y = Olez tf2/2. y = {Ol sin x [02 -log (sec x tan x)] cos x}ez y = [01 (x2/4)] sin x [C2 (x/4)] cos x. z y = Ole(02 2 sin x 3 cos x)e- 2Z Cae3z• y = 01e- z (02 - sin eZ)e- 2z.

+

+

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Art. 37 1. 1.70 ft. 2. (a) x = 1 sin V3Yt - (V3Y/2)t cos V3Yt; (b) 1.57 ft above equilibrium position. 3. (a) 0.138 ft above equilibrium position; (b) ±0.866 ft. 4. (a) 0.171 ft above equilibrium position; (b) 0.792 sec.

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Answers

6. 1.06 ft below equilibrium position. 6. 2.19 ft below equilibrium position. 7. 0.095 ft below equilibrium position. 9. -0.272 amp. 10. (a) 0.638 amp; (b) i amp. 11. (a) 0.909 amp; (b) 1 amp. 12. (a) -0.0868 amp; (b) -0.16 and 0.09 amp. 13. 1.22 amp; 0.00245 coulomb. 14. 0.0323 amp. 16. 0.546 amp. 16. 0.0354 coulomb; -1.03 amp. 1'1. x = Xo cosh wt + [(vo/w) - (g/2( 2)] sinh wt + (g/2( 2) sin wt; Xo = 0, Vo = g/2w. 18. (b) 1.10 in. 19. (b) 1.41 in.

Art. 38 3. y = C1x + C2Xl +va + Car-va + 2x3. 4. y = X(C1 sin log x + C2 cos log x + log x).

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6. y = C1X C2X 2 4x2log x itr sin log x -h cos log x. S 6. y = (C1 log" X - ! log x)X (C2/X) - 2. 7. y = (C1 C2 sin log x C3 cos log x)(l/x) (x2/30) (3x/l0) - 2. 8. y = C1 C2X CaX(5+3V5) 12 C4X(5-3V6) /2 log x To log2 X. 9. y = (C1 2 log x) (l/x) (C2 sin log x Cs cos log x)x. 10. y = C1 C2 log x [C3 sin log x) C4 cos log x)]X S i log" X -h log2 x. 11. y = [C1 C2log x C3log2 X (log" x)/48]x. 12. y = (C1 C2log x i log2 x)0 - /r;x3. 13. y = C1X C2X2+V4.25 + CaX2-V4.25 - -fix log x + log" X 29 log x 475. 14. y = (Cl/X) (C2/X2) (x/6) - (sin x)/x2• 16. y = [C1 C2log (2x - 3)]V2x - 3 2x. . 16. y = C1 C2 log (1 3x) C3(l 3x)3 - i log2 (1 3x) - 3x/2. 1'1. y = C1(2x - 1) C2(2x - 1)1+(V3/2) C3(2x - 1)1-(...;3/2) !(2x - 1) log (2x - 1) - 1. 18. y = ¥ - 3x ¥X2 -lx2log x. 19. U = U1 - [CUI - U2) log (r/r1)]/log (r2/r1). 20. U = U1 - [(U1 - U2)(r - r1)/(r2 - r1)](r2/r).

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Art. 40

1. 3. 6. 7. 10. 14. 17. 20. 26.

y = log (x2 + C1) + C2 • 2. Y = x + C1e-:r?-/2 + C2. x 2 + (y + C2) 2 = C1. 4. Y = x 2 + C1log (x2 - C1) + C2• y = x 3 + C1x2log x + C2X2 + CaX + C4 • 6. Y = (C1 + x)/(C2 - x). y = (C1X + C2)2. 8. y3 = (C1x + C2)2. 9. y = C1 sinh" (x + C2). xy + C1X + C2y = Cs. 12. y = log cos (2x + C1) + C2. 2y = sec [x + (r/3)]. 16. x = 3 - (2/VY). 16. y = 2 sin-1 X + 1. x = 2(e-Y/2 - 1). 18. y = (x - 2)e~ + 2. 19. x = jlog 2 - log sin y. sin y = V2 sin x. 21. -1.73. 22. 27. 23. -1.42. 24. t. 26. 3.21. -1.83; 0.330.

238

Answers

Art. 43

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1. y = ClX - 2x2 - x 3 C2Xe:e. 2. y = C1(2x l)e-:e C2e:e. 3. y = C1(X2 2x 2) [(2x3/3) C2]e:e. 4. y = [Ci/(x 1)] C2X. 6. y = (C1log x C2)e-:e e:e/2. 6. xy = C1 C2e:e - sin x - cos x. 7. 0.843. 8. 0.51 ft; 0.49 ft. 9. 2.39 sec. 10. 116 hr. 11. 9.90 hr. 12. 6.95 mi/seo. 13. 11.6 min. 14. 45.1 min. 16. (a) 15.9 ft; (b) 61.1 ft; (c) 136 lb, 144 lb. 16. 58.4 ft. 18. (a) catenaries; (b) circles; (c) parabolas; (tl) cycloids. 20. y = -log (cos x - sin x); 0,1,0. 21. y = log (cos x sin x). 22. The family of catenaries: y - b = k cosh [(x - a)/k]. 23. 1.38; 3.63. 24. y = ±[(x2 - 1)/4 -llogx]. 26. y = Cl(Sin x)/VX C2(cos x)/0. 26. (7) 0.775; (10) 810 hr; (11) 13.5 hr; (12) 4.91 mi/seo. 28. 0.910. 29. 2.36 sec. 31. (a) 1.02 sec; (b) 0.723 sec. 32. t = VL/g(1 - sin 8) cosh"! (L/a). 34. (a) x = [a- kyl+k/2(1 k)] - [a kyl-k/2(1 - k)] ak/(1 - k2) ; (b) x = 1[(y2/2a) - (a/2) a log (a/y)].

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Art. 44

+ X)2 + (z - 3x)2 = 1; (b) (3y + Z)2 = 3x - z. 2. y = x log x + 2x, z = x log x + 3; [1, 3, 1]. 3. y = 0 cos [x 1 + (71'/4)], z = 0 sin [x 1 + (71'/4)]. 2 2 4. C1e:e + ce:: and C1e:e - C~-:e. 6. x = t + Cl, Y = C2(X + t). 6. (1 + y)2 = (1 + X)2 + Cl, Z = C2(x + y + 2). 7. y = C1e-:e + C2e:e, Z = C1e-:e - C~:e + sin x. 1. (a) (y

2

2

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8. (a) x ;:= ee-at , y = [ac/(a - b)](e-bt - e-at ) , Z = e + [e/(a - b)](be-at - ae-bt ) ; (b) x = ce- at , y = acte-at , Z = e[1 - (1 + at)e- at ] . 9. 161 min, 4.51 lb. 10. 18.3 min. 11. 9.43 lb. Art. 46

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1. y = C1e2:e C2e-2:e l sin 2x - 1 cos 2x, Z = C1e2:e - 3C2e-2:e - i cos 2x. 2. y = C1eb C2e-:e - 2 cos x - sin x, 3:e Z = -2Cle 2C2e-:e - cos x 8 sin x. t(C1 3. x = esin t C2 cos t) (t/2) - 1, t[(C2 y = e- C1) sin t - (C2 C1) cos t] 1. 4. x = C1 sin t C2 cos t sin 2t, y = C1 cos t - C2 sin t cos 2t. 6. y = C1 sin x C2 cos X e:e -lx sin x, Z = [C2 - C1 l (x/2)] sin x - [C2 C1 - (x/2)] cos x - e", 6. y = C1 sin 0 x C2 cos 0 x 5 sin x - cos x, Z = [(C1 20C2)/3] sin 0 x [(C 2 - 20C1)/3] cos 0 x sin x - 3 cos x.

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239

7. y = 01e- 2z + ! sin 2x - ! cos 2x, Z = - (Ol/2)e-2:e + i cos 2x - t sin 2x. 8. x = 01 sin t + O2 cos t + et + (t/2) sin t; y = [01 - O2 - t + (t/2)] sin t + [01 + O2 + (t/2)] cos t + 2et• 9. y = (01 + 02x)eZ + (Os - x)e-Z, Z = 1 - 02ez - (20 3 + 3 - 2x)e-z• 10. y = (01 + 02X)eZ + Ose-z + 3e-2z - 1, Z = - 02ez - 20se- z - 8e-2z. 11. x = C1 + 02t2 t, y = 01 - 02t2 + t. 12. x = [01 - 02 + (01/2) log t]t - (Os/t) + log t + 1, y = [02 - (01/2) log t]t + (Os/t) - 2 log t - 2. 13. ~ = 0: y = 1 - 2ez , Z = eZ + x-I. n = 1: y = x - eZ, Z = Oe- z + !eZ + x - 2. n = i: Z = 01e sz + 02e-(3/2)z + -/rrez + x - t, y = -t01e3z - 102e-(3/2)z - fez + (x/3) + i. 14. x = t sin t - if: sin 3t, y = -1.,p- sin t + -A sin 3t; 0.955, 1.96. 16. x = A sin (t + a) + B sin (9t + (j) + (at/9) + (h/9) + (8b/81), y = -A cos (t + a) + B cos (9t + po) + (bt/9) + (k/9) - (&/81). 16. y = 01 + 02e t, x = -01 + (02t + ,03)et, Z = -01 + (Ott 02 + 03)et•

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Art. 46 1. (a) y = - [gx2/(2v~ cos2 a)] + x tan ai (b) (v~ sin2 a)/2g ft, (v~ sin 2a)/g ft. 2. (a) [(wvo sin a)/kg] - (w2/ k 2g) log [(kvo sin a w)/w] ft. 3. (a) 1.24, 1.33; (b) - i < x < i, -0 < Y < 0. 4. x = -cos (t/y'2) cosh (t/0), y = sin (t/y'2) sinh (t/y'2); (3.83, 3.70). 6. (b) 2.37 amp, 0.061 sec; (c) 2.15 amp, 2.85 amp. 6. (c) R = [(I/ro) - (k/h 2)] cos 8 + (k/h 2).

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Art. 49 3. x = Ae tlt + Be-tltl2 sin [( V3/2) at + a], y = Ae tlt + Be-tltl2 sin [( y'3/2) at + a + (21r/3)], Z = Ae tlt + Be-tltl2 sin [(V3/2)at + a + (41r/3)]. 4. x = Ae3 t + B, y = Ae St + 0, Z = Ae St - (B + 0). 6. x = Ae t/2 + Be- t, y = Aet/ 2 + Ce:', Z = Ae tl2 - (B 6. x = Ae t + Be-tl2 - t/2 -!, y = Aet + Oe-tl2 - t/2 Z = Ae t - (B + C)e- t / 2 - t/2 - !. 7. x = At2 Bt-I , y = At2 + Ot-l, Z = At2 - (B + C)t-1• 8. Xl = Acz + Bt-tl + K sin (a log t + a), X2 = Acz - Bt-tl + K cos (a log t + a), Xs = Acz + Bt:» - K sin (a log t + a), X4 = Attl - Bt-tl - K cos (a log t + a). 12. Ji' = (1 + wG)/(1 + w2G).

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240

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Art. 60

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1. y = ao(l x) CZ2X2 • 2. y = ao(l - x) lZ2(X2 - x3 x4 - ••• ) = A(l - x) B/(l x). 2 3 li 3. y = ao(l - 2x ) aleX - ix - (1/2·4)x - (1·3/2·4·6)x7 - ••• ] . = ao(l - 2x2) alxVl - x2• 2 2x4/2·4) 3x 6/2·4·6) 4. y = ao[l - (4x / 2) - (1·4 - (l·3·4 (l·3·S·44x8/2·4·6·S) _ ... ] alX = aoVl - 4x2 alX. 6. y = ao(l x 2) a4[x4 (x6/3) (x 8/ 3 ·4) (xl°/3·4·S) 2 = A(l x ) B~. 6. y = ao[l - (x4/21) (xSj4f) - ••. ] lZ2[x2 - (x 6j3f) (xlOjSI) - ... ] 2 2 = ao cos x lZ2 sin x • 2/2) 7. y = ao[l (x + (x4/2·4) (x6/2·4·6) +

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+ + al[z + (z3/3) + (zij3·5) +...] =

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+...] 6""10(0
8. y = ao[1 = ao{ 1 9. y = ao(l - 2) (3· S· 7xU/II I) _ ... ] = (1 - z2) {O
+ al

f

3/3f)

+...]

I::

dz). _

(3·5x'/91) -

[6""1°/(1 - z2)2] dz}.

+ (x2j21) - (3x4/ 4 f) + (5x6/61) - (7x8jSf) +...] + ~[(2x3/3f) - (4xIi/SI) + (6x 7/ 71) - (8x 9/ 91) +...] = A(cos x + x sin x) + B(am x - x cos z). 11. y = ao[l - (x2/l6) - (x + (5x4/l536) +... J + alex - (x / 24) - (x4/l92) + (x +.'..]. 4/SI) 2/3f) 12. (a) y = ao[l + (x + (x +...] = ao(sinh x)/x; (b) y = Cl(cosh x)/x + C2(sinh x)/x. 10. y

=

ao[l

3/96)

li/384)

3

Art. 62

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1. y = (A/x) (B/x 2 ) . 2. Y = (A/x) (Bez/x). 2 3/6f) 3. y = ao[l (x/21) (x j 41) (x 2/51) a~1/2[1 (x/31) (x (x3/71) J = ao cosh Vi a~ sinh Vi. 4. y = ao[l - (x/2) - (x2/2·4) - (1·3x 3/2·4·6) - (1·3·5x 6j 2 .., 4·6·S) a'r;x1/2 = aoVl - x a~Vi. 2 6. y = ao(l 4x ) a~/2[1 Ix (3·1/2·4)x2 (3·l·1/2·4·6)x3 (3·l·1·3/2·4·6·S)x4 - ... ] " = ao(l 4x 1x2) a~1/2(1 X)3/2. 2/2·5) 4/2·4·5·9) 6. y = ao[l - (x (x - (x6/2·4·6·S·9·l3) ~-1/2[1 - (x 2/2·3) (x4/2.4.3.7) - (x6/2.4.6.3.7.ll)

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7. y = eZ{Cl C2[(I/x) log x - (x/21) (x2/2·31) - (x3/3·41) S. y = ao{x 1) ~[X-l (x- 2/ 3) (x- 3/ 3 ·4) (x- 4/ 3 ·4 ·5) = A(I x) Bxet1z • 9. -0.2601, 0.5767. 11. (a) y = aox2 [1 - (x 2/ 2·4) (x 4/ 2 ·42·6) - (x 6/ 2·42·62 ·8)

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= AxJl(X).

12. y = 2.23yxJo( Vi). 13. (a) y = 7.69xJo(x) ; (b) 8.87. 14. (a) y = aox[1 + (x/1121) + (x2/2131) + (x3/3141) + ...] = ~ViJl(2iVi)·

+ ]} + ]

+ ...]

INDEX (The numbers refer to pages.) Amplitude, of emf, 80 of simple harmonic motion, 117 of steady-state current, 81 Arbitrary constants, essential, 5 Archimedes, Principle of, 121 Areal velocity, 201 Atmospheric pressure, 41 Beam, deflection of, 133, 163, 164 elastic curve of, 133 Bedell and Crehore, 79 Bending moment, 135 Bernoulli's differential equation, 62 Bessel functions, 219, 226 jJ. Bessel's equation, 183, 226 Cable, suspended, 178 Cassinian ovals, 102 Catena~, 93, 95, 179 Cgs system of units, 36 Chemical reactions, 40, 191 first order process, 40 second order process, 40 Chemical solutions, 75 jJ., 192 Coefficients, undetermined, method of, 142jJ. Cohen, 212 Constants, isolation of, 19 Curves, determined from geometric properties, 89, 183 family of, differential equation of, 23 in space, 187 n-parameter, 23 Cyclic systems of differential equations, 202 Damped oscillatory motion, 125 Damping factor, 125 Deflection of beams, 133 jJ. Degree of d'fferential equation, 3 243

De Moivre's theorem, 201 Differential equation (ordinary), 1 degree of, 3 equivalent solutions of, 8, 68 formation of, 15jJ. by elimination, 15 general solution of, 4 verification of, 29 Laplace's, 3 linear with constant coefficients, l03jJ. of a family of curves, 23 of efflux of water, 49 of first order, 28 jJ. Bernoulli's, 62 exact, 52 homogeneous, 53 jJ. integrable by suggested substitution,66 integrable combinations, 50 integrating factor of, 52 linear,58jJ. reducible to homogeneous, 64 separable, 28 jJ. of motion of pendulum, 123 order of, 3 partial, 1, 2, 206 particular integral of, 6, 141 particular solution of, 6, 31 particular values of variables, 31 , Riccati's, 75 series solution of, 212 jJ. simultaneous, 185 jJ. solution of, 3 special higher order, 165 jJ. dependent variable absent, 170 general linear of second order, 175 independent variable absent, 171 reducible to linear with constant coefficients, 165

244

Index

Differential operator, 103 Dynamics, 35

Hooke's law, 123, 134 Hyperbolic functions, 9

Efflux of water, differential equation of, 49 Einstein's equations, a special case of, 206 Elastic curve of beam, 133 Electric circuits, 78 if., 160 if., 200 Equation of motion, 118 Equipotentials, 102 Equivalent solutions of a differential equation, 8, 68 Essential arbitrary constants, 5 Euler's relation, 113, 206, 210 Exact differential equation, 52

Impedance, 81 Inee, 212 Integrable combinations, 50 Integrating factor, 52, 60 Inverse square law, 176, 201 Isolation of constants, 19 Isothermal surface, 44

Family, of curves, differential equation of, 23 in space, 187 n-parameter,23 of surfaces, 187 Flexible chain, motion of, 85, 184 Forced vibrations, 157 if. Formation of differential equations, 15 if. by elimination, 15 if. determinants, 20 differentiation and combination, 15 finding differential equation of a family of curves, 23 Forsythe, 193 Fps system of units, 36 Frobenius series, 220 if. Functional dependence, 69 Functions, Bessel, 219, 226if. hyperbolic, 9 Goursat-Hedrick,69 Goursat-Hedrick-Dunke1, Z1

Halphen, Z1 Heat flow, steady-state, 43 Homogeneous differential equations, 53JJ.

Jacobian, 69, 70, 211 Kasner, 206, 209 Lagrange, 151 Laplace's equation, 3 Lens, 96 Limiting velocity, 37 Linear differential equations, 58 of first order, 58 if. simultaneous, 192 with constant coefficients, 103 if. complementary function of, 141 general solution of, 141 particular integral of, 141 Modulus of elasticity, 134 Young's, 134 Motion, equation of, 118 simple harmonic, 117 Newton's law of cooling, 46 Newton's second law of motion, 35 cgs system of units, 36 fps system of units, 36 Oceanic pressure, 41 Ohm's law, 82 Order of differential equation, 3 Ordinary differential equation, 1 Orthogonal trajectories, 96 if. Parameter, 23 Parameters, variation of, 142, 151 if. Partial differential equations, 1, 2, 206

Index Particular integral of a differential equation, 6, 141 Particular solution of a differential equation, 6, 31 Pekce, 13,39,40,63,67,80,92,181 Pendulum, 123 Period, of damped oscillatory motion, 125 of simple harmonic motion, 117 Power series, 212 if. Pressure, 41 atmospheric, 41 oceanic, 41 Projectile, motion of, 198 PUr.&rit, curve of, 184 Rectilinear motion, 116if. attractive force proportional to displacement, 116 resistance proportional to velocity, 124 . repulsive force proportional to displacement, 117 Reddick and Miller, 3, 13, 44, 79, 143, 219 Reflector, 96 Refraction, law of, 96 Resonance, 158 Riccati's differential equation, 75 Roots of unity, 201 Rope wound on a cylinder, 83 if. Separable differential equations, 28 if. Series, power, 212 if. of Frobenius, 220 if. Shift, exponential, 106 reverse exponential, 106 Simple harmonic motion, 117 amplitude of, 117 period of, 117 Simultaneous differential equations, 185 if.

245 Simultaneous differential equations, cyclic system of, 202 Einstein's, a special case of, 206 integral of system of, 209, 211 linear, 192 of first order, 185 Steady-state current, 81 amplitude of, 81 maximum value, of, 81 Steady-state ~ea/flow, 43 Streamlines, 10 Substitutions differential equations solvable by, 62 if. Bernoulli's equation, 62 equations reducible to homogeneous, 64 suggested by form of equation, 66 Surface, isothermal, 44 Surfaces, family of, 187 Suspended cable, 178 Temperature gradient, 43 Thermal conductivity, 44 Thomson, 201 Tractrix, 92, 95 Trajectories, orthogonal, 96 if. Transient, 81 Trial and error, 38 if. Undetermined coefficients, method of, 142 if. Unity, roots of, 201 Variation of parameters, 142, 15111. Velocity, areal, 201 limiting, 37 of escape, ·182 Vibrations, forced, 157 if. free, 158 natural, 158 Young's modulus, 134