A First Course in Ordinary Differential Equations

6

Introduction

yy and y'. If it is the same as the original differential equation or the equivalent of it, the integral is thereby checked. If equation 1.6 does not contain f it is itself the éliminant. Example 2.

G i v e n the differcntiad equation * + ^ | / + > - 2 x = 0.

T o check the relation xy

\o% y — x"^ ~ c

as its general integral. T h e derivative of the given relation is x y ' + > + - - 2 x = 0. y

T h i s is the original differential equation. checked. Example 3.

T h e integral has thus been

G i v e n the differential equation

^

1

T o check the relation y = log c — tan ' x\

as its general integral. T h e derivative of the given relation is y

=

-1 {1 + x 2 ) { c -

T h e éliminant of c is thus v' = ferential equation.

tan-^jï} —

This is the original dif-

{1 + x-\é' T h e integral is thus checked.

T h e method described may be used to find the differential equation of any 1-parameter family of curves. Thus, let the family have the coordinate equation 1.5. B y differentiation we may obtain equation 1.6. T h e éliminant of c is the differential equation of the family. Example 4.

T o find the differential equation of the family of curves log> + <:x2 -

1 =0.

Checking a General Solution

7

T h e derivative of this equation is -

+ 2cx = 0

y

T h e eUminant of c is the relation logjy - J

-

1 = 0

T h i s is the required differential equation. ASSIGNMENTS In each of the following cases, check the relation given as the general integral of the respective differential equation. 1.

=

for

yy' + X = 0.

2. y = cx^,

for

xy' -2y

3. y = c cos X + sin x,

for

y' cos x + ^ sin x — 1 » « 0 .

+ c tan X,

4. 7 =

5. ^ + log ^

for

f = 0,

—

6. sin^ ^ + 2 tan x = 7.

y - 2

1 + cx \x+cU--\

8.

9. y{x + 4| + 10.

= 0.

=

+

= 0,

1 2 cos X + c sin X

11. log (xy) = cx. -1

Î 2 . sin * X + sin * ^ =*

= 0.

\y' — ^] sin X — {7 —

for

y' -{-ytr" = 0.

for

y' +2

scc^ y esc i2y) » 0. y - 2

for

y'

for

y' +2xy =

for

} sec x •» 0.

= 0, €''\

{x + 4 } / +

{x + 5]y - 2e^ = 0.

for

y' -\- -y cot X — ^ ' CSC X = 0.

for

xy' +y

= y log {xyj. 1

+

for

V i

- = sm * X + sm ^ y.

' V i -

PROBLEMS In each of the following cases, find the differential equation for the given family of curves. i.

2,

y — cx ^ x'.

i „ +I l112 4. U + ^ r2 +I [;r |

3. y — 2 ^ c tan x, 5.

+ Ic^ X

=

33 J

0,

6. 1

+

8, y.cx

1, 1.

2

8

Intrcxiuction 10. -y/y

9. |c + > P + 2«:Jt = 0. 11- >

"i

TT— Jf + 1

12. *

-X -

+ log \cx\ - 1

2 cos

+ c sin X

ANSWERS TO ODD-NUMBERED PROBLEMS

3. / -

\. xy' ~y ~2x*

5.

log X - ^ -

0.

7.

2xy'(/ +

11. 2(x^ -

X+

sm X cos X ' 1 + X -i- V 1

' 9. ^ -

^~^

11-0.

+ {2x -

11/ -

2x + 1.

+

CHAPTER

2 Some Types of Solvable Differential Equations of the First Order 2.1. Equations with separable variables T h e central problem with which we shall be concerned is that of studying the integrals of given differential equations. W e may attempt to find these integrals exactly or approximately, and we may do either of these things analytically by the use of methods of the calculus, or by the use of graphs, direction fields, etc. As we shall see, only certain types of differential equations can be solved explicitly. I n this chapter some such types are considered. T h e i r integrals are applicable to many practical problems. T h e y are also typical of the integrals of many differential equations that cannot be explicitly solved. A differential equation (2.1)

P{x, y) dx +

Q{x, y) dy ^

0,

i n which P(x, y) and Q(x, y) are each a product of a function of x by a function of 7, is said to have separable variables. W e may write such an equation more explicitly as Pi{x)p2{y) dx +

qi[x)q2{y) dy =

0.

If there is a value yo at which ptiyo) ~ 0, the equation is fulfilled by the relation y — >o» since that gives dy — 0. T h u s ^ = is a solution. If there is a value ;ifo at which q\{xQ) = 0, the equation is also fulfilled by the relation x = xo. This, however, does not define ^ as a function of x. For values of x a n d ^ at which ^i(x) diiidpziy) are not zero, we may write the equation thus: q^iy)

— ~ p2{y)

,

P\{x)

^

?iW

dy ^

dx.

10

Some Types of Solvable Equations

T h e variables are now separated, since o n l y ^ appears on the left and only X on the right. E a c h member of the equation is now a differential, namely, the differentizd of its own quadrature. Since functions with equal differentials can differ by any constant, but cannot differ by more than a constant, we conclude that

J p2(y)

J

qi(x)

T h i s is the general integral. Example 1.

T o integrate the differential equation 2{y

-

\ ] dx-\- x^ sinydy

This equation has^ = 1 as an integral.

=

0.

W i t h separated variables it is

- rfy = - - rfx. y — i

x^

/* sin y 2 / y - \ ^ dy = -+€

Thus is the general integral. Example 2.

T o integrate the differential equation xydx -

(* -

2) rfy =

0.

W i t h separated variables this equation is dy

x dx x~2

Thus

log> = 2 log jjv -

2) + X + A

is the general solution, with k as the arbitrary constant. B y replacing each member of the last equation by its exponential, the relation is given the form

T h e differential equation also has the integral y = 0. integrals are included i n the relation , =

w i t h the constant c arbitrary.

21V,

Clearly all these

Exact Differential Equations

11

PROBLEMS F i n d the general solution of each of the following differential equations. i.

{y -2\^dx

3. {x^+

- x^dy ^0.

\]ydy-\-x{y^

5. sec* xsecydx

-

2.

4. {2xV - r f x + 2xd> -

A\dy ^0.

sin ydy = 0.

6. y/1

+ i 1 + x^} (/y = 0.

7.

y dx-\-dy ^ 0.

~ y^ dx -\-3 -s/\

8. |x -

0.

- x^ dy -

0

11 cos )- rfy -= 2x sin > dx.

9. X log Xflfy+ -N/I + ^'^rfx= 0.

10. {1 - 2x y/l

- sin^ x \ dx + V 2 - sin^ x dy » 0.

11. xe"*dx -\-e~'dy = 0. 13.

y/\

12. ^cos^^rfx -\-&ccydy = 0.

•]- y^ dx + xy dy = Q.

= 0.

15. xf.?" + 4} dx 16. x\y -2\dx

14. xy log y dx = sec x dy.

-

\x -

l]\y -\-A\ dy = {2 ~ y\dx.

17. x} dy = xy{a> - (fx!. 18. tan ydx-\- cot x rfy = 0 .

2.2. Exact differential equations T h e diflferential of a function F{x, y) is known from the calculus to be Fx{x,y) dx + Fy{x,y) dy,

where and are the partial derivatives of F with respect to x and ^. T h i s is of the form of the left-hand member of equation 2.1. I n at least some cases, therefore, a differential equation 2.1 is actually of the form dF{x, y) = 0. W h e n this is so, the differential equation is said to be ex£ict. Because a differential is zero only when the function is constant, the exact equation implies that (2.2)

Fix,y)

= c.

T h i s is its general solution. A differential equation 2.1 need not be exact. Before formula 2.2 can be used, therefore, we must determine whether the equation is exact or not, a n d , if it is, how the function F(x, y) is to be found. W e shall show how that may be done provided the functions Py{x, y) and Qx{x, y) are continuous. If the differential equation is of the form dF = 0, it is so because (2.3)

P{x,y)

= F,{x,y),

Q{x,y)

-

Fy{x,y).

12

Some Types of Solvable Equations

B y partial differentiations, then, Pyix.y)

= Fx,y{x,y),

Qxixyy) =

Fj,,j:ix,y),

and, since the cross partial derivatives Fx,y and Fy.^ are equal, we see that (2.4)

Py{x,y)

=

Qx(x,>).

This is therefore a necessary relation for exactness. I f it is not fulfilled the differential equation is inexact. Suppose now, conversely, that relation 2.4 is fulfilled. T h e n let F{Xfy) be defined by the formula (2.5)

F{x,y)

=

r

P{x,y)dx-{-R{y),

J

i n which R{y) is an undetermined function, XQ is any suitable constant, and the quadrature is made as though^ were a constant. T h e first equation (2.3) is then fulfilled, and also =

Fyix^y)

\'Py{x,y)dx-\'R'{y).

But because of (2.4), we may write this last equation FA>^.y) =

r

Q.{x,y)dx^-R'{y),

J

namely,

F^ix.y)

= Q{x,y)

- Q(xo,>) + R'{y).

T h e second equation 2.3 is thus also fulfilled \i R{y) is chosen so as to make R'ij) = Q(xo,>),that is, if m

=

!Qixo,y)dy.

W i t h this evaluation, therefore, the function 2.5 fulfills both equations 2.3, and the general integral 2.2 is given by the relation (2.6)

/J P{x,y)

dx-\- f

Qixo^y) dy = c.

T h e method could have been applied with the roles of P and Q, and of X andy interchanged. W i t h a suitable constant the general integral is so found to be (2.7)

Q{x,y) dy +

j

P{x,yo) dx = c.

T h e two results 2.6 and 2.7 are equivalent. They may, however, be quite different i n form, and the quadratures called for by one may be simpler than those for the other. It is therefore often wise to consider both formulas, and to carry through with the one that is found to be simpler.

13

Exact Differential Equations Example 3.

T o find the general integral of the diflferential equation W

-^y\dx-\-{x-\-^\dy

= 0.

For this equation Py{x,y) = 1, and Qxix^y) — 1. Condition 2.4 being fulfilled, the equation is exact. W i t h XQ — 0, formula 2.6 gives the general solution f e^dy = c,

j^W^y]dx+

that is,

^xy-\-i^

= c.

W i t h ^ o = 0) formula 2.7 gives the solution xy-\-e" - \

-hW

=^c.

This differs from the one first obtained only by having {f: + 11 i n place of c. T h a t , however, is immaterial since c can have any value. Example 4.

T o find the general integral of the differential equation

6xy +

1 V v * - x^

This equation is exact. are, respectively.

3x'

6xy + 2x

Vv'

1

3x^ -

W i t h XQ = 0, and^o =

0 and

dx +

2x

-

x-

1

-

Vy'

dy = 0.

formulas 2.6 and 2.7

dx = c. x'

6x +

dy +

T h e former is clearly the simpler one. solution

1 V l

-x^

dx — c.

F r o m it we obtain the general

+ s i n " ^ \x/y'\ = c.

'hx-y

PROBLEMS Test the following differential equations to determine which ones are exact, and find the general solution for each one that is exact. 19. 20.

21.

{Ixy + 6x! dx y

1

x"-

y

xy

[x"^ rfx +

\ \ dy = 0.

\x

1

1 2 * * 3xY - -1,

X

s dy = 0,

~ X

dy = 0,

14

Some Types of Solvable Equations 1

22,

23,

log > +

dx

1

+

xy

dx-V

-

X

i - + 2>

- !

dy

dy

= 0,

= 0,

24. \xy + log x} rfjf = (xy + log>l dy. cos y dy = xe"^ dy —

25. ^ sin y i/x + x dx -\- y dy 26. ~ + x'

dx.

{Un y -\-x] dx -\-X scc^ y dy = 0. 1

27.

('-:v)'

2

V

dy =

^

28.

{cos

29.

{x + > cos ( x y ) ^ " ' " " U x + rfx

30.

X

tan ^ — sin x sec ^|

dy

xdy

^ + X

y

+ {sin x sec'' y + cos x tan^ y esc y]

dy

=0.

{> + X cos (ry)*-"'f"'>}
— y dx

, X

dx

0,

= 0.

-t y

2.3. Integrating factors A n exact differential equation 2.1 is usually made inexact if a factor is removed from it. T h e restoration of the factor, of course, restores the exactness. W e shall show that i n principle any differential equation 2.1 which has a general solution can be made exact by multiplying it by a suitable function. Such a function (other than zero) is called an integrating factor. Example 5.

T h e differential equation log ydx-\- dy/y = 0,

is inexact. After multiplication either by e^, or by 1/log^, it is exact. These multipliers are therefore integrating factors. Let the general solution of any given differential equation 2.1 be written i n form 2.2. T h e differential equation of which this is the general solution is Fi{xyy) dx + Fy{xjy) dy = 0, a n d , since this must be equivalent to equation 2.1, it follows that Fxjxy y) ^ P(x.y)

Fyjx.y) ~Q{x,y)'

Let ${Xj y) be the common value of these ratios. e{x,y)P{x,y)

= F,(x,y),

e{x,y)Q(x,y)

Then =

Fy{x,y)

Finding Integrating Factors

15

T h i s , however, means that if equation 2.1 is multiplied by 0{xyy) it assumes the form dF = 0, that is, it becomes exact. Thus 0(x, y) is an integrating factor. Suppose equation 2.1 is multiplied by ^{F)d, where 4> is any integrable function. T h e equation becomes dF = 0, and, since the left-hand member of this is also a differential, the equation is again exact. Thus ^{F)d is also a n integrating factor. Since the number of functions i> is unlimited, an equation 2.1 is thus seen to have an unlimited number of integrating factors. Example 6.

T h e differential equation y dx —

X

dy = 0

is inexact. Its general integral isy/x = c. Thus jy) is i n this case —y/x^y a n d , since P{xyy) = we see that 0{xyy) = —\/x^. Thus — 1/x^, a n d i n fact ^{y/x){\/x~)y with any integrable function 4> is a n integrating factor. W i t h appropriate choices of <ï> we see, for instance, that all the functions

xy

y*-

x'

y^

X

x^

y

are integrating factors.

2.4. Finding integrating factors T h e discussion of Section 2.3 has theoretical interest but little practical value. T h e integrating factors it displays are expressed i n terms of the general integral of the differential equation. T o find the integrating factors the general integral would have to be known. T h e n , however, there would be no need to concern oneself any longer with integrating factors. There is no systematic, generally applicable, method for finding a n integrating factor. If there were, a l l equations 2.1 could be solved, and our discussion of this equation could be terminated. T h e finding of an integrating factor therefore remains a problem. Sometimes one is detectable by inspection of the equation directed toward recognizing differentials i n it. However, success in this detection is the exception, not the rule. I n the search for an integrating factor it is useful to observe that (i) A n y function of just one variable, multiplied by the differential of that variable, is a differential; and (ii) A sum of differentials is a differential. W i t h these facts i n m i n d we may try to collect the terms of a differential

16

Some Types of Solvable Equations

equation into groups, each of which is recognizably transformable into a differential by some suitable multiplier. Example 7.

T o integrate the differential equation {xy-\-2 log *} dx +

{x^ + xe^\ dy = 0.

O n grouping the terms i n the manner x{x dy -{-ydx]

+ 2 log x dxxe"

I dy -

0,

it is recognizable that the multiplier X/x transforms the first group and also the last term into a differential. A t the same time it leaves the remaining term a differential. Thus X/x is a n integrating factor. It reduces the differential equation to d\xy\ + d {log;cj^ +rf^*= 0. T h e general integral is thus xy + ( l o g x l ^ + «v = c. M a n y schemes, such as the following, can be invented to help i n the search for an integrating factor. Suppose there is a n integrating factor of the form (2.8)

e{x,y) =

dx+iHv) dv

T h e differential equation, after multiplication by (2.8), is then exact, and, by equation 2.4, -

1P(X, V)J''<'> dx+fb(v) dyj ^ ± {Q(x, V)J«<''rfx+/6(v)dy^

dy

dx

This equation can be reduced to the form (2.9)

P,{x,y)

-

QAx^y)

= a(x)Q{x,y) -

b{y)P{x,y).

T h e reasoning is reversible; hence, if a{x) and b{y) can be found to fulfill equation 2.9, formula 2.8 gives an integrating factor. Example 8.

T o find a n integrating factor for the differential equation — siny\ dx + cosydy = 0.

For this, equation 2.9 is — cosy = a{x) cosy — b{y){e' — s i n ^ j .

A comparison of the terms i n cos^ suggests the trial of a(x) = — 1. T h e equation is then fulfilled if b{y) = 0; hence is a n integrating factor. Example 9.

T o find an integrating factor for the differential equation [y + ^ + / i dx + |2x + 4 / 1 dy = 0.

17

The Linear Equation For this case, equation 2.9 is -1 + X+

= a(x){2x + V )

-b{y)y{\

+x-\-y'^\.

T h e term — 1 on the left must be matched on the right, and this suggests the trial of b{y)y — 1, that is, b{y) = \/y. T o match the term B^"^ on the left, we may then try a{x) = 1. These trials actually do fulfill the equation. Hence, by equation 2.8, the function e^^^°* which can be written, ye^, is an integrating factor. PROBLEMS For each of the following differential equations, find an integrating factor, and, by using it, integrate the equation. 31. ydx

-

dy = 0.

{2x +y\

32.

{2xy2 -y^\dx-\r

33.

\x^

34.

(4*2 ~

35.

[2x^y + > M

-

W

xy^'dy =

dy = 0. 0.

xy -h 7x -\- 2y -

X

dx +

sin y

36.

1

37.

- sec y — tan y

^ ) dx

X

38.

-

+

{2x^ +

I] dx +

xy] dy =

X log X

{x -

6y +

+

~e

0.

0.

x^ COS y

dy = 0.

dx — {x — sec y log x] dy ~

dx

\ ] dy

" + JT sin

^ X

0.

dy = 0

y

2.5. The linear differential equation of the first order A diff"erential equation which is important both theoretically and practically is that of the form (2.10)

y'

+p(x)y

=

q{x).

It is called the linear differential equation of the first crder because it is linear, that is, of the first degree, i n y and y'. For an equation of this type an integrating factor is always easily found. Let the equation be multiplied by ${x) dx. Its resulting form is e{x)

dy +

{p{x)e{x)y

-

q{x)e{x)]

dx =

0.

By equation 2.4, the last equation is exact if 6' = pix)6y that is, if 0'/6 = p{x). This integrates to give log $ = jp{x) dx, which is ^ = ^Jpix) di T h u s tf/p^** dx is an integrating factor, by virtue of which the

18

Some Types of Solvable Equations

differential equation becomes

T h e general integral is thus ^ J p ( x ) dx ^

J^(x)tf/P<^> ^dx

+ C.

This method is easily remembered. It requires no memorization of details. Example 10. T o integrate the differential equation y' -\- y tan x = cos x.

After multiplication by d{x) dx^ the given equation is exact if 6' = 6 tan Xy that is, it 6 = sec x. Thus sec x dx is an integrating factor. T h e general integral found by its use is y sec X =^ X -\- c. Example 11.

T o integrate the differential equation {1 +x^]y'-{-

{1 -x]^y

=

xe-'.

W e begin by writing the given equation i n form 2.10, that is, as I

-

2x

1

+x'

y

=

xe

1 -\-x

T h e method then shows that 0{x) dx is an integrating factor if e' = e

1

-

2x

1 -\-x

T h e variables i n the last equation can be separated to give d0/e =

1

-

2x

1 -\-x^

dx

Thus, log 0 = ;t - log ¡1 + x'^l, that is, = + x^). T h e integrating factor of the differential equation originally given is thus e'dx/il

4-x2).

B y the use of this factor the differential equation is given the form

Changes of Variables

19

Its general integral is, accordingly. -1 1

2{1

+xM

PROBLEMS F i n d the general integral of each of the following differential equations, 39. >' + 2xy = 41. /

-

43. y'

-

xy

-x^

1

X

e-^\

40. y' + y cot

= 3x.

42. xy' log

3x' + 3x'

+ 1

45. y - y - í ^ í l

47. y'

2x

49. y'

=

51. / [ l

46. y'

1.

48. y'

y =

+xM

Ix +4}y'

53. 54.

+

{x + 5)y =

+

11

+

2x|y

=

{2x

1

+

+

X

1 — | - cot X

-

2x

X

sm

X

—cos x csc^ x.

le'.

+ 1 ¡y _ 2xy = |x* + 2x2 + 1} eos y'

+

1 -

X

X

52.

2x2.

50. y' sin X + y sec x =

2 - + 4x

+y

+ y -

1.

44, y' -\- xy = x.

+logx!.

+

X

X =

X.

\ \e-^'.

2.6. Changes of variables O n e of the best modes of dealing with a difíerential equation to which no other method seems applicable is to change one or both of its variables. B y this means we may try to modify its form so as to make it subject to some known method of integration. Let X and y be replaced by new variables s and «, by formulas i

These formulas are to be reversible, so that s and « can also be expressed i n terms of x and y. B y virtue of these formulas dx

= fs {Sy u) ds

dy = gÁ^,

+ / u (j, w)

« ) ds +

du.

gu(sy u) duy

20

Some Types of Solvable Equations

and the functions P and Q are replaced by functions of s and «, say, P(x,>) = i ? ( j . a ) ,

Q(x,y)

= r(j,«).

Differential equation 2.1 is thus transformed into

R[f, ds + / „ ^«1 + T\g, ds + iu du\ = 0. If this differential equation can be integrated, and has the general integral u, c) = 0, the general integral of the original differential equation is obtainable from u,c) = 0 by replacing s and w by their values i n terms of x and)'. Example

12.

T o integrate the differential equation + 11 + > V ' l rfy = 0.

B y the change of variables X

— s,

—

y

e~'Uy

the given equation is transformed into e-'u

ds-\-\\-\-

u'^We-' du -

e-'u

ds\ =

0,

that is, into —u^ ds -\- (1 + u-j rfu = 0. T h i s last equation is solvable since its variables are separable. Its solutions are u = 0 and j - f (1/2«^) -

l o g u = c.

Since s = XfU ~ ^y, the original equation has the solution)» = 0, and the general solution is V2>-) Example

13.

-

log > = c.

T o integrate the differential equation

B y the change of variables x = su, y — u, the given equation is transformed into -u^'du

H- u^s^

+

u]ds

=

0.

W e may drop the factor since u = 0 does not lead to an integral of the original differential equation. W e thus find that (du/ds) — u = s^. For this last equation, which is linear, e~' is an integrating factor. T h e general integral is ue^" = — { + 2j + 2) + Since s = x/y, u = y, the

Changes of Variables

21

general integral of the given equation is X

X

\y

y

ASSIGNMENT 1. Show that the differential equation (2.11) i n which d is a constant other than 1, is transformed into a linear difTercntial equation by the change of variable x " " Differential equation 2.11 is known as a Bemoulli equation, in honor of the Swiss mathematician James Bernoulli (1654-1705).

PROBLEMS Solve each of the following differential equations by use of the indicated change of variables, or as a Bernoulli equation. 55.

56.

57.

(1 + 2JC + 2y|

2 + 4* + 2> +

{2y - xylogx]

1 - e«] dy - 0,

+ 12x + 2y 1

dx-\-{l

1 +x

dx -2x

+2x+>|<6'-0.

{ 3 / - x^y] dx + {x* sin y + x* -

59.

\4xy -

60.

iy +

V +

dx-\-dy0, x^y 1

dx+{x-\-

61.

{sin~* y — cos xjrfxH

62.

{y + V l -

63.

i2> + 2 x V

64. x V dx -

65.

-

y =» u log s. X

3xy^] dy = 0,

= S/Uy

y •= s.

sm u, y = s esc u. X =

x^y\dy = 0,

x '~ y •j==.dy

~ 0,

AT =• J ,

1^ r V „ .

dx + |x + x ' y } dy - 0,

{x» - 2 / + 2y^] dy = 0,

+

y ^ s — 2u.

|* ^ ^'2

Irf^+ xrfy - 0,

.-xiv

u.

X - J ,

log xdy » 0,

58.

2x*

X "

X = J ,

x " s


X =

j ^ - " .

22

Some Types of Solvable Equations e^Hx-\-y-\-

66. (^+'' + 1 -

1)J dx +

{f*+>' -

1)) dy = 0,

1 + ^""(x + ^ -

x = {s + « ) / 2 , y ^ i s - u)/2. 2 67. / - — > 3JC 69. >' + ^

1 = —

cot

75. /

+

77. /

,

X

cos X

- .

= 8v'^ sin^

COS X

X

2x 2\x^ - x

—

1

,

,

COS

=

—

v>.

—

2> 3x

76. y' -\-3y\ogx = 3>^x

X.

I

Ay log

= 2{x + l}

"2- > + - > sin X 2

jy + xy

2)-

+ >

70. y + - ^ > Jf + 1

>^

Sin

X

68.

CSC X .

y

'

X

3xy

jt =

3 71

2 log

log

X

Ma X

2x + 1

+ \\^ ^ 2{x^ -

X

+ \\y

2.7. Equations witti homogeneous coefficients A function fix^y) is said to be homogeneous of the degree k if the effect of replacing x and y by ax and ay, respectively, is merely to multiply the function by a*. Thus /(ax, ay) = a^f{x, y). Example 14.

T h e functions 1 .y/xy tan-" X

X

-yh

/Mog

1 +y

are homogeneous of the respective degrees 0, — 2 , and f .

For

tan~" — == a° t a n ^ " - » ax X

{axp

X

jayP^ log

1 +

ax ay

= fl^-;;'^ log

1 +

A n y differential equation 2.1 whose coefficients P{x,y) and Qix^y) are homogeneous functions of the same degree is transformed by either one of

Equations with Homogeneous Coefficients

23

the changes of variable (2.12)

X = Sj

y = sUf

X = sUf

y =u

or (2.13) into a n obtained obtained changes,

equation with separable variables. T h e differential equation b y the one change of variable is ordinarily different from that by the other. It is therefore often advisable to consider both and to integrate the simpler of the transformed equations.

Example 15.

T o integrate the differential equation \y -

Vx^ + / 1

dx-

xdy = 0.

T h e coefficients of this equation are both homogeneous of the degree 1 T h e change of variable 2.12 transforms the equation into V l

-\-u^ds + sdu = 0.

T h i s has separable variables, and has the solution s{u + V l

= c.

s = Xy and u — y/xy the general solution i n terms o( x andy is y + Vx'+y' Example 16.

= c.

T o integrate the differential equation y'^dx-\- Ix^ -

xy-\-y^] dy = 0.

T h e coefficients of this equation are homogeneous of the degree 2. T h e change of variables 2.12 transforms the equation into [u + u^] ds+

{\ ~ u-^-u^sdu

= 0,

whereas the change of variable 2.13 transforms it into uds -\- {s^ -{- \ } du = 0.

O f these the second is obviously simpler.

Its solution is

tan~^ J + log « = c, and, since s = x/y, and u = >, the general solution of the given equation is tan"^ - + log;" = c.

24

Some Types of Solvable Equations PROBLEMS

Integrate the following differential equations. 79.

{3x + 2y] dx + \2x +>) dy - 0.

80. {x < _ 81.

_ xy ™« _

-_ 2x'y\dx + {2x* + xy* -\- xY

xY

+ x*y] dy - 0

{x^y -{-xy^ - >'} dx + \xy^ ~ x^] dy - 0.

82. y sec^* - dx +

— X sec2 -

y

dy " 0.

y

83.

\2x^y +/}

84.

\x -ye^f^

85.

86.

dx + I V rfx

- COS - tfAT —

X

.

-sin y

M**^* — ^ sin -

2je»} rfy = 0. +

y

-

,

y — h COS X X

rfy

= 0.

dy - 0,


87. x ) r I o g - r f x +

-«Mogj^

2xlog^-x+>|rfx

88.

+

X'

-

+x

dy = 0,

89. fl>^^»'rfx + {7 - flx^''*! dfy = 0. 90.

>-s/x^ + / r f x -

{x V * ^ + /

+>M

= 0,

2.8. Equations with linear coefficients Every differential equation (2.14)

a-iX •\- bxy

ci\ dx +

¡02* +

hiy +

^ 2 ! dy =

0,

whose coefficients are of the first degree i n x and y, can be integrated. Such a n equation m a y be of a type we have already considered. If h\ = 02» it is exact; if ci = C2 = Oj its coefficients are homogeneous of the degree 1; if ¿2 = 0, it is of the linear type 2.10. W e may therefore suppose now that 6 2 ^ 0. T h e change of variables

with a n undetermined constant A, transforms the equation into \A{s -

k) -V Bu-VC\ds-\-udu

= ^.

Equations with Linear Coefficients

25

If, here, = 0, the variables are separable. I n that case we may assign k the value 0. \i A 7^ 0, we may choose h so that —Ah-{-C — 0. T h e equation is then one i n which the coefficients are both homogeneous of the degree 1. Example 17.

T o integrate the differential equation {2y-\]dx^-{ix-y

+ 2]dy = {i.

T h e change of variables, s = x -\equation into [ds-\-u

u = 3 ; f — ^ + 2, transforms the

~ ()h-\-^\ ds ~ udu = 0.

W i t h A = ^, this is a n equation with homogeneous coefficients, whose general integral is [u -\- 2s\'^[u — 7>s\^ = c. I n terms of the original variables, the general integral is thus \Sx — y + 3j^j2jv — I p = c. Example 18.

T o integrate the differential equation {2x -

6> + 3) rfx -

{x -

ly ~ \ \ dy

0.

T h e change of variable, j = x + A, u = x — 3^ — 1, transforms the equation into {5w + 15 j ds -\- udu = 0. Here the variables are separable. W e may therefore take k ~ 0. T h e general solution, i n terms of the original variables, is 2A: — ^ — log fx — Sjy + 2 j = c. PROBLEMS F i n d the general Intcgrsd of each of the following differential equations. 91.

{2x+y

- 2\ dx + [x - y-\-2] dy = 0.

92.

{4x + 2> -

93.

{7y - 3}rfx+ {2x + 11 £/y = 0.

94.

{2x -y

95.

[2x + 3>! dx

96.

I5x '\- 2y -\- \ ] dx

{x -

97.

{3x - 2> + 4} rfx -

{2x + 7> -

98.

{6x + 4> + 3irfx+ {3x + 2> + 2} rfy = 0.

99.

{;-+71rfx+

l]dx -\- {2x - y\ dy = 0.

+ 2,]dx -

{4x - 2y -\- \] dy

{ix - y -

0.

U] dy = 0. \ \ dy = 0. 1}rfy= 0.

{ 2 x + > + 3lrfy = 0 .

100.

\4x-{-y - 2\ dx + {?>x+y - 2\dy = 0.

101.

{5x -\-Ay ~ A] dx -\- {4x -\-5y -

102.

{2x-\-y -

I] dx[2x

- 9y -

S\ dy ^ 0. \] dy = 0.

26

Some Types of Solvable Equations

2.9. Simultaneous equations Just as in algebra a number of unknowns may be given by an equal number of simultaneous equations, so in the calculus a number of functions may be given by an equal number of differential equations. A pair of equations, z'=/2(x,:v,z),

(2.15)

constitutes such a system for two functions > and z. T h e general solution of such a system consists of two equations in x, and and involves two arbitrary constants. T o obtain these equations two quadratures have to be made. (i) A F I R S T I N T E G R A L . If the first equation 2.15 does not involve z, or if the second equation 2.15 does not involve y^ we may integrate the equation in question, and so obtain a first relation in x, y, and z. W h e n neither equation 2.15 permits such a direct integration, we may use undetermined multipliers a , ^3, and 7 , to obtain from 2.15 the equation (2.16)

(a + i3/i +

adx-^^dy^-'idz^

7/2I

dx.

If a , /3, and 7 can be chosen so as to make the left-hand member of equation 2.16 a differential, say dQ (x, >>, z), and at the same time to mzJce the right-hand member a function (possibly zero) of 9 and x times dx^ this choice makes the equation appear as (2.17)

dB =

F{e,x)dx.

T h e general integral of equation 2.17, (p{$, x, c\) = 0, yields one of the required equations (2.18) (ii) T H E S E C O N D second equation.

* ( x , > , z , ^ i ) = 0. INTEGRAL.

T W O

methods are available to obtain a

(iifl) It may be possible to reduce equation 2.16 to a form 2.17 by the use of new multipliers a, /3, 7 that are not proportional to those that were used in obtaining the first integral. A second integral of form 2.18 which involves an arbitrary constant c-z is then obtained. (iiè) It may be possible to eliminate either a n d ^ ' , or z and 2', between equations 2.15 and 2.18. T h e éliminant is then a differential equation for the remaining function. T h e integration of this éliminant then yields the second integral.

Simultaneous Equations Example 19.

27

T o integrate the differential system + 2xz' +

{2x + 2!e -

3 = 0,

x^z' +

Ix^ -{-x}z -

x = 0.

xy' -y-{-

B y algebraic means we may reduce this system to form 2.15 /

= - > - l , X

X

T h e first of the last two equations does not contain z; hence it can be integrated directly to givey = x Cix^. B y the elimination ofy between this equation and the second one of the reduced system, we find that 1 +

1

z.

—X the general integral of which is xz = 1 -\- ci{\ — x] -\~ C2e general integral of the given differential system is, therefore.

y — X Example 20.

cix\

The

xz = 1 -\- ci{\ — x] + C2e

T o integrate the differential system y

=

xy + 2xz

xy + 2xz

T h e choice of multipliers a = 0, |8 = 1 , 7 = 2 reduces equation 2.16 i n this case to dB = (d/x) dx^ with 6 = y -\- 2z. Thus a first integral 6 = c\Xy that i s , ^ -\- 2z = cix, is obtained. T h e elimination oiy between this result and the second equation of the given system yields the equation z' =

2z{cix - 2z\ ClX

T h e general integral of the last equation is ^ = cix^/{4x general integral of the given system is thus y -\- 2z = ciXy Example 21.

z =

+

C2),

and the

CiX

4x +

C2

T o integrate the differential system

y - X

z' =

— \

\y -

x\z

T h e choice of multipliers a = 2>', ^3 = — 2z, 7 = —2A: reduces equation 2.16 for this case to dB = 0, with B = y"^ — z^ — x^. Hence y'^ — z"^ — x^ = is a first integral. W i t h the different choice of multipliers a = x.

28

Some Typ)es of Solvable Equations

/3 = — z , 7 = we obtain for (2.16) the form dB - 0 with B ^ xy - i z T h e generzd integral of the system is thus ~

-

2xy _

x^ = cu

= C2.

r2

PROBLEMS Integrate the following differential systems. 103. xy' -\- z' -2z

= Ixe',

\x^ + 4x\z' -

104.

105. y' - y 106. y'

>' -

r' +

2z

U + 4}^ - y - y = 0,

\x\\z'

{x + 3J,

2z

y - xz' -

3xV -

3x* - xy' + ;r - 0.

xy' - xy -

^ "

X +

2.

z -\- x^ = 0,

X

X

107.

1

X

X

z -

X

X**

- x' -

1 = 0.

{ 1 + 2 log xjv' - xz' -

+ -> - x r ' - ^ + 0,

z=

0.

X

108. 2^5»' + {3 tan X 109.

-2z

/

110./

4yz + l =

2.10.

I

z

.

, z' =

~4xz 3 +4yr

=

2z = -j sec* x.

Ayz + 1 3x 3

+4^ z + 2 ~ Ax 2y

2y - 3z *'{2x -y]

yy' - -fz' -

1

6x - ;y - 3

111. y

112. y

4|z = 3 sec X + sec* x,

-yly

x\y + 2z\

~3z 1

r =

-2z) x\y + 2zi

The linear differential systems with constant coefficient

T h e method of undetermined multipliers can be used to integrate any system (2.19)

>' = fly H- 6z + ^i(x),

z'^hy-\-kz

+ g2(xh

i n which a, bj k, and k arc constants. W e may i n this case choose a and /3 and 7 as constants to fulfill the equations fljS + A7 = mfi,

b^ -\- ky = my.

= 0,

Replacement of a Differential Equation by a System 29 is possible if m ¡5 taken to be a root of the quadratic equation (2.20)

a — m

h

b

k —m

= 0.

T h e multipliers then give equation 2.17 the form de=

with 6 = the integral (2.21)

\me-\- ^gi{x) + yg2(x)] dx,

72. Since this is a linear differential equation for 6, it has

= J { ^ 5 i W + yg2{x)\e-^

{&y + yz]e-^

dx + c^^.

process can be carried out for each value of m i f equation 2.20 has distinct roots. Otherwise the elimination of ^ or z between (2.21) a n d an equation 2.19, yields a differential equation i n the other variable. T h e integration of this éliminant gives the second integral. PROBLEMS Integrate the following differential systenss. 113. y' =ly

-

z' = 3y - 2z.

114. y = 2z - 4y,

z' = -iz - 5y.

115. / = | y + 5r, z' = 116. /

-

~y + h,

117. y = z,

z' = ~h

z' = 4y ~ Sz.

i\9. y' = 2y - z + e',

121. y'

+ h-

z' = -y -\- 2z.

118. y' = 3y - 4z,

120. y -'ly

~ h-

z' = 4y ~

+ Zz'h e^', -2y-\-

3z-{-5.

z' = 2y - Zz - 7**.

2^"'.

122. y' = -2y -\-z-{' 3 « " * ' ,

= 3z + 6> + e~^. z' = 9y - 2z- 2e'.

2.11. The replacement of a single differential equation by a differential system W h e n a differential equation 2.1 is given, it is sometimes advantageous to regard x and y as functions of a third variable tj and to replace the differential equation by the system (2.22)

dx/dt = Q{x,y),

dy/dt =

-P(x,y).

T h e two equations that make up the general integral of this system express X and y i n terms o( t. T h e y therefore give a parametric representation of

30 .

Some Types of Solvable Equations

the integral of the original differential equation. T h a t integral i n terms of X and y is obtainable by eliminating / between the pair of equations which constitutes the integral of the system. Alternative to system 2.22 we could equally well replace differential equation 2.1 by the system dx/di = k{x,y)Q{x,y),

dy/di =

-k{x, y)P{x, y),

with any chosen function k{x,y). T h e integration of this system is generally different for different choices oik{x, y). However, such different choices only determine effectively different parameters t. W h e n t is eliminated the same integral for (2.1) is always obtained. Example 22.

T o integrate the differential equation V l

dx-\- {x + 2y] dy = 0.

- /

Equivalent system 2.22 is i n this case dx/dt = x-\-2y,

dyjdi =

-

V l -

y-.

F r o m the second of the last two equations we find a first integral relation y — cos \t + c\\.

W i t h y thus made known, the first equation of the system, which is linear i n Jf, yields the second integral: * = sin j ^ +

(Ti

i

—

cos {( +

(Ti

I + c-ié.

These two integrals together represent the solution of the given differential ' equation parametriccdly i n terms of t. T h e éliminant of t from them is ^^\-y'^-y-\-cé

COB y

1

with c = c^e This is the general integral of the differential equation originally given. PROBLEMS Integrate each of the following differential equations in terms of a parameter. 123.

{2^ -y

124.

13^^ + 2^ + 4} rf* + {1 + 2e-'\ dy = 0. 1 2;- + y

125.

126.

-2}dx+

1

!

7

-

l*i dy = 0.

dx + (6* + 1 0 / -

Ij rf> = 0.

dx + {* + 2y^ + 9} rfy = 0

Answers to Problems 1

127.

dx -

X Xhx + 6x2}

128.

_

{2x -

31

2x^} dy = 0.

j _ _ 3y

dx = 0.

129. 2 V ^ r f x + V 1 - * ((y -= 0. 130.

rfjr

-

V l -

((y =• 0.

ANSWERS TO ODD-NUMBERED PROBLEMS 1. > = 2 + 5. sin^ y

1 -hex

and y = 2.

3. >r» = 4 +

2 tan x = c.

1. y = log 11. fe-v'dy

9. {y + \ / l ~ T ? " | l o g x 13. ^2 „

_ logxp -

1.

—

tan * x}.

=c H-i'il

15. log ( . » ' + 4 1 -

-xj.

{ x - l - l ^ - ' + f.

17. > « X + log {2 - x)2 + c, and> = - 1 . 19. x^y 23. 27.

X

log ^ + log X

X + X

ix^ - y = c.

>

21. Inexact. = c.

+

25. e' sin > + xe~* « c. 29.

+ sin ^ X = c.

- >

1

X

1

^

r

y

33.

39. ye^* =

X

4x^V^* -

37. cos y;

r 3x^^

43. > = |x + I I U ^ +

X

57.

-> « 0,

V^-

X

= c -

49.

53. y = {x^ + l}{sinx

+ Ix +^1 and x + log

59. > « 0, and 2 61.

3*»

= c.

sin ^ ^ — sin

-

= x^ X

{1 -x2}W.

45. :y - < r * ( x l o g x + W •

51. ^-1x2 + x * | = x ' + x + f. 55.

-

x sin y — y log x —

41.

+ c.

47. y = 2x + f

log X X

2/4

35. x-^y-^^;

/'°»'<''+iU2+/|

+

65. 2 t a n - i Ix."} + y \/1

2 log jlog x) « c. 1 + 63. log X

- >^ + s i n " * > = c.

-

xV'*' - f

32

Some Types of Solvable Equations

67. >* = log X

1

75.

+

cx^.

69. y

sin 2x + ^ tan x.

79. 3x* + 4xy +

-

f.

1

%

2 cos JK + c sin X

73.

- ^V T T x

77. y -

{/^"

-

V l

-

+rIogx}*.

81. I o g x + - + - - < : . y X

83. -5 + l o g \xy] - c.

85. ^ sin - — c. X

xM

87. -2

I

J-

x

1 - ^

logy

91. (2x +y

-

89. ae'^v + log > - c.

+ log >

2}* -

3b -

2 ^ = c.

93. 7 log {2x + 11 + 2 log |7> -

n{y+2\^

95. [2x + 3>|2 99.

{> 4 - 7 | 2 l 3 x + ^

103. y = fix* -

xz

109. /

¿1,

111.

X +

X "

2;» + 3z

=

115. > -

119. y =W

|if-*

123. ^ = 1 + 125. X =- 1 -

Ci*"*,

129.

X =

ii4 -

- ^ i d

8x - \0y

C2.

y + z* - ^2. >z =

X -

X* +

^2.

-r = cie' + 3f2*~'. -iVi^'*

- ¿'^2'""'^^

z = Cixe* + [ci + C2l«*. C2e-'^^ « - W 4x

- f^-" + ~

+

cie-*^ + ^2^',

1 127. - « 1 + e i . - " ,

ly"^ + 2y •

{x + 2 }

log X + ci (log x ) ' +

- -k^

z

4xv + 8x -

•« = fi^:^-

z =ri{x*+x} -

c i ^ ^ * + C2e-'^^, z -

117. y - cixe' + f2^,

121. > +

=- C i

ci,

113. y = cie* + ^2«?"',

97. 3x^ -

101. 5x' +Sxy + 5y^ -

2C1X + 2 ^ + f2,

107. jf - ci log X , r +

= c.

+ 11 - ff.

105. r - < r 2 ^ - o U H - 2 | ,

-

3} = c.

> «

6;- -

- i .

+ f2l«~".

[ci - / | V 4 -

[ci -

+ 5 + f i ^ + 4^2«

.-8* z = 6 ; - ' - x^"" + cze

f2«*.

=

-

(1^ + 2 s i n - 1

CHAPTER

3 Applications of Differential Equations of the First Order 3.1. Problems in velocities T o solve a physical problem mathematically, it is commonly necessary to resort to a differential equation. T h e general integral of this differential equation must be found, and from that the particular integral that fits some given condition must be determined. T h a t is to say, the family of integral curves must first be found, and then the curve of this family that goes through a given point must be isolated. T h e differential equation expresses the relevant physical law, and the particular integral applies it to the specific situation. For problems i n velocities and rates, we shall use t to denote the time, and i n the case of a moving particle we shall use s to denote the distance along its path. This distance is to be measured from some fixed point, with one of the directions taken as positive. T h e choice of the origin and of the positive direction may be made as seems most convenient. Once made, however, it must be adhered to throughout the course of the solution. T h e velocity v and the acceleration a of a particle in a straight path are known from the calculus to be given by the formulas (3.1)

V = ds/dty

a = dv/dt,

a = v dv/ds.

T h e first two of these formulas are i n effect the definitions of u and a. T h e third is obtainable as the quotient of the second by the first. T h e fundamental law that describes the way particles move under the action of forces is Newton's second law of motion: (3.2)

/ = ma,

f = m dv/dt,

f — mv dv/dsy

these forms all being equivalent by (3.1). m stands for a constant called the mass of the particle, and / represents the component of the force acting i n the direction of the path (or the sura of such components if there is more 33

34

Applications

than one force). Force / is to be taken as positive or negative depending on whether it acts to increase or decrease s. W h e n a body is allowed to fall freely without resistance under the pull of gravity, its acceleration is the acceleration of gravity, which is denoted by g. This is the same for all bodies. Its numerical value depends upon the units i n which distance and time are measured. I n the foot-pound-second system it is nearly 32.2. Since law 3.2 applies to falling motion as well as to any other motion, we see that w = mg, where w is the weight of the particle. T h u s (3.3)

m —

w/g.

W h i c h of the three forms of Newton's law 3.2 is to be used in any given case depends upon the variables i n terms of which the problem is to be solved. If the problem calls for velocity in terms of time, the second form of (3.2) is appropriate, for the variables are v and t. If velocity is required i n terms of distance, or vice versa, the third form is clearly the one to be used. W h e n a body falls i n a resisting medium like air or water, the law by which the force of resistance is related to the velocity varies with a number of circumstances, such as the density of the medium, the shape of the body, the magnitude of the velocity. I n each of the following examples we shall assume a simple law to take a l l this into account. This would not necessarily be the law that experiment would bear out i n the case of some other body. W e shall be solving differential equation problems, not establishing physical facts. For numerical results, we shall take the evaluation ^ = 32 to be sufficiently accurate for our purpose. Example 1.

A 2-lb. particle is dropped from a balloon at a great height. D u r i n g its fall it is acted upon by air resistance. T o find the velocity of the particle when it has fallen 1000 ft., if it is assumed that the resistance amounts to kv^, with k = 1/20,000, i n the foot-pound-second system. T h e path of the particle is a vertical line. A l o n g this we may take s as mccisured from the point at which the particle is dropped, and as increasing downward. T h e acting forces are then the weight w, which is positive, since it is directed downward like the increasing s, and the air resistance kv^, which is negative, since it opposes the motion and so is directed upward. T h e total force is thus {w — kv^]. T h e problem calls for v i n terms of s; hence we choose the third form of law 3.2, and obtain from it the differential equation w — kv^ = mv dvjds,

with w = 2,k^

1/20,000, and, by (3.3), m = i V -

Problems in Velocities

35

T h e variables i n this differential equation are separable, integral is

T h e general

,2

- 6 2 5 log

2

-

=

20,000

s -\- c.

T o fit the condition ZJ = 0, when j = 0, the value of c must be c — 625 log 2. T h e particular integral that applies is thus - 625 log

1

-

V

2

=

40,000

s,

or, equivalently, V = 200 V\

- e-'^^^^.

W h e n s = 1000, this equation gives v — 178.6 approximately. Example 2.

A m a n w i t h a parachute jumps from a great height. T h e i r combined weight is 192 l b . D u r i n g the first 10 sec, before the parachute opens, the air resistance amounts to kv^ w i t h A: = i n the foot-poundsecond system. Thereafter, while'the parachute is open, the resistance amounts to AID, with k = M. T o find the man's velocity of fall 15 sec. after the j u m p . W i t h the direction of increasing s taken as downward, the acting forces are 192 and —kv. T h e second form of law 3.2 thus gives the differential equation 192 -

yti/ =

6 dv/dt.

which has the general integral (3-4)

log

fl92

I —

V

k-

o fo

A n equation like 3.4 but with different values for k and ¿1 applies to each part of the fall. For the first part of the fall A: = 4, and y = 0, when t = 0. T h e particular integral that fits these conditions is log that is,

V

1

-

= 256(1

V

256

8

-

A t the end of this part of the fall, that is, at / = 10, this gives the velocity Pio = 256{1 —

= 182 (approximately).

36

Applications

For the second part of the fall k = 12, a n d , as we have just found, V = 182, when / = 10- T h e particular integral 3,4 that fits these conditions is log namely,

V -

16

166

-

20 - 2(,

= 16 + 166^^^"".

At t = 15, this last formula gives Pi5 = 16 + 166tf

-10

T h e result is very nearly 16 f.p.s., since 166^~^^ amounts only to about 0.007, PROBLEMS Solve each of the following problems, assuming that the all data given arc expressed in the foot-pound-second system. 1. A ball weighing H lb- is falling. T h e air resistance amounts to u/lOO. p = 15, at / = 0, what is the formula for v in terms of

If

2. A ball weighing 1 lb. is dropped from a great height. The air resistance amounts to p/32. Find the formula for s in terms of /, if f = 0, when j = 03. A bullet weighing H o lb. is fired vertically downward from a balloon, with a muzzle velocity of 1000 ft./scc. T h e air resistance amounts to f^/1,000,000. Find the formula for v in terms of j . 4. A bullet of weight w is fired straight upward with a muzzle velocity of ro ft/.sec, The air resistance amounts to kv^. Find the formula for s (the height), in terms of c, while the bullet is rising. 5. In the case of the bullet of problem 4, find the formula for / in terms of v, 6. A certain particle weighs 8 lb. As it sinks in water its weight forces it downward, but an upward force, its buoyancy, also acts upon it. This buoyancy amounts to 4 lb. T h e water resistance amounts to v'. Find v in terms of j , if :J = 0, when j = 0, 7. When a certain body weighing 16 lb. is placed under water, it rises because of its buoyancy, which amounts to 20 lb. T h e water resistance amounts to Find the velocity with which the body rises / seconds after it is released. 8- A motor boat weighing 200 lb. is going at a speed of 20 ft./sec. at / = 0. T h e motor is then turned off. T h e water resistance amounts to -jj^^. Find the formula for D in terms of (. 9. A 1-lb. particle moves in a straight line on a smooth horizontal plane. A force acting upon it amounts to ^ , and is directed so as to repel it from the point s = 0. Uv = Oy when j = 4, what is the formula for u in terms of s? 10, Find the formula for v in terms of s for the particle of problem 9, if the force amounts to l A ^ , and is directed so as to attract the particle toward the point j = 0.

Frictional Motion

37

11. A certain vehicle weighs 3200 lb. Its air resistance amounts to 2D. If a force of 40 lb. is applied to push it, and starts it into motion, find the formula for the distance s it has gone, when its velocity is v. 12. If the vehicle of problem 11 is moving with the velocity 20 ft./sec. at Í « 0 , and a braking force of 40 lb. is thereafter applied to it, what is the formula for v in terms of

3.2. Frictional motion W h e n a particle moves on a rough surface, its motion is opposed by friction, which acts as a resisting force amounting to fif^fy where (i) fa is the sum of the components of the forces acting upon the particle i n the direction perpendicular (i.e., normal) to the surface; and (ii) / i is a constant called the coefficient of friction. T h e value of M ts a measure of the roughness of the surface. It is zero when the surface is perfectly smooth. W i t h the friction taken into account, the motion of the particle is subject to Newton's law 3.2. Example 3.

A 10-lb. particle is placed upon a plane whose slope is T h e coefficient of friction is and, i n the foot-pound-second system, the air resistance amounts to v/4. T h e particle is at rest at / = 0, and is then released. T o find its velocity at any subsequent time /. Let

V

32 dt

W h e n the variables are separated this differential equation becomes dv

=

idt.

and the condition y = 0, when / = 0, is fulfilled if c = — log (88/5). T h e integral that applies is thus

38

Applications

that is,

PROBLEMS Solve the following problems, assuming all the data to be given in the foot-poundsecond system, 13. A boy and sled together weigh 80 lb. They coast on a hill whose slope is TTJ, and on which the coefficient of friction is -r^T h e air resistance amounts to v/l'i. Find the formula for the velocity, i f = 0 , when ( — 0, 14. What is the formula for the velocity of the boy and sled of problem 13 if the hill is perfectly smooth? 15- A boy on skis weighs 96 lb. He slides down a hill whose slope is which the coefficient of friction is -jfr- T h e air resistance amounts to 2v/l3. p => 0, when J = 0, find the formula for s in terms of v.

If

16. A 16-Ib, particle slides up a plane whose slope is T, and upon which the coefficient of friction is x- T h e air resistance amounts to v/\0. If at / = 0 the velocity of the particle is 64, what is the formula for v at time t? 17. A 100-Ib- boy slides on the ice. T h e coefficient of friction is 7nr> and the air resistance amounts to v/20. If the boy's velocity at / = 0 is 40, what is it at time (? 18. A skater weighing 160 lb. allows himself to be blown along by the wind. T h e coefficient of friction is -j^j-, and the wind pressure upon him amounts to 2 {30 — v]. If his velocity is 14 when t = 0, what is it at any later time /? 19. In the case of the skater of problem 18, find the formula for s in terms of v. 20. A 4-lb. particle moves in a straight line on a horizontal plane whose coefficient of friction is T . T h e air resistance amounts to v^/l6. T h e force acting upon the particle amounts to 4J, and is directed to repel it from the point J = 0. If v = 0^ when J = 4, find the formula that relates v and j .

3.3. Problems in rates I n many applied mathematical problems the immediately known facts concern the rate at which a certain quantity changes. Finding the quantity itself then requires the integration of a diflferential equation. Example 4.

A certain mixing tank of 200-gal. capacity is filled with brine i n which 60 l b . of salt are dissolved. Beginning at / = 0 the solution i n the tank is drawn off at the rate of 5 gal-/sec., and the tank is meanwhile refilled at the same rate with a solution that contains lb. of salt per gal. T o determine the amount x of salt that is i n solution i n the tank at time t. Since at time / there is a total of x l b . of salt i n solution, the amount of salt per gallon at that time is x/200. T h e rate of withdrawal is thus 5x/200. T h e rate of replacement is T ^ - W e thus have the relation

Problems in Rates dx

5

5A:

~ 20Ö

It ~

39

10*

T h i s is a linear differential equation for x. Its general integral is x = 20 + ce^'^^^y and with c determined so that the condition x = 60 when / = 0 is fulfilled, the result is X = 20-\- 40e-'^^^. Example 5.

A certain compound X is formed by the combination of 2 parts of a chemical U with 3 parts of a chemical W. W h e n certain amounts of U and W are placed together, the rate at which X is produced is constantly proportional to the product of the amounts of U and W that are still present at the instant. T o determine the amount of X that is produced i n time t, if 10 lb. of U and 81b. of W are placed together at / = 0, and the amount of X at / = 1 is 2 lb. Let x be the amount of compound X that is produced i n time /. T h e amounts of U and W that have been used are ^x and %x, respectively. T h e remaining amounts are therefore ¡10 — f x j , and ¡8 — f x j . We have thus the differential equation dx/dt =

k{\0-%x]{B-%x].

log 18 - | x l -

log (10 - |x) = -^kt

Its general integral is +

.:.

W i t h c determined so that x = 0, when t = 0, this equation yields log

200 -

15x

200 -

8x

^

-^kt

T h e fact that x = 2, when / = 1, serves to evaluate k. that k = — l o g f I , and that therefore .200 -

15x

• 200 -

8x

W e find from it

. 85 = / log — ^ 92

This result can be written as my\

2oo{i -

15 -

8(M)

PROBLEMS 21, A mixing tank contains 100 gal. of fresh water at / = 0. A solution containing 7 lb. of salt per gal. is then added to it at the rate of 3 g a L / s e c , and the resulting mixture in the tank is drawn off at the same rate. Find the formula for the amount of salt in solution in the tank at time /.

40

Applications

22. In a certain volume of liquid 100 lb. of salt, but no more, will eventually dissolve. T h e rate at which it dissolves is always proportional to the amount that is still dis-> solvable. Find the amount x of salt that will dissolve in time /, if this amount is 5 lb. when 1 = 1. 23. In a certain volume of liquid, 200 lb. of salt, but no more, will eventually dissolve. When a quantity of salt is placed in this liquid, the rate at which it dissolves is T^JTC times the product of the undissolved amount by the amount that is still dissolv* able. If 100 lb- of salt are placed in this liquid at / = 0, what is the amount x that is dissolved at time /? 24. From problem 23, what is the amount of salt that is dissolved in the liquid in time t if 200 lb. arc placed in it at ( = 0? 25. From example 5, if 10 lb. of chemical U and 15 lb. of chemical M^arc placed together at / = 0, how much of compound X is produced at time I? 26- From example 5, if chemicals U and were to combine in equal amounts to produce compound x, what would be the amount of compound produced in time interval / after u lb, of U and w lb. of IV were placed together? 27, T w o chemicals U and IV combine in equal parts to form a compound -V. The rate at which the compound is produced is proportional to the square root of the product of the amounts of U and IV that are still present at the instant. If at ( = 0 there are 8 units of U and 2 units of W^, find the formula for the amount of compound formed at time /, 28- If 2 parts of a chemical U combine with 3 parts of a chemical W and the rate at which the compound is produced is always proportional to the amount of U that is still present at the instant, what is the formula for the amount of compound produced in time ty if this amount is M at / = 0 ? 29- T h e population ^ of a certain insect colony is at ( = 0, In the absence of outside influences the rate of its growth would be kp. T h e colony is, however, exposed to an influence that causes deaths at the constant rate ö. Find the formula for the population at any time t. 30. In the absence of outside influences a certain population p grows at the rate kp. However, it is exposed to an outside influence which induces growth at the additional rate {a + 4 ^ ] , where a and i are constants. If the population is at time /o, what is the formula for the population at time t? 31. T h e amount x of a certain chemical disintegrates at a rate which is always proportional to the amount that is still left. If it diminishes by 1 per cent in 10 years, and its amount is XQ at t = 0, what is the formula for x at time t? 32. T h e amount x of a certain chemical naturally diminishes, because of disintegration, at a rale kx if left to itself. In use there is a further loss by wear at a constant rate Ö . If the amount is XQ at time ( = 0, what is it at time t?

3A. Problems of flow A problem of flow is always based upon a differential equation. This statement applies to a flow of water, or of air, or of some other fluid, and to an electric current, or the flow of heat from a cooling body.

Problems of Flow

41

W h e n a liquid whose volume is V is allowed to escape from its container through an opening at a distance y beneath its surface, the rate at which it flows through that opening is k V ^ , with some constant k. T h e value of this constant depends upon the physical circumstances of the individual Ccise, in particular upon the shape and size of the opening, the character of the fluid, the system of units used, etc. This rate of flow is also the rate at which the volume V diminishes. W e draw from that fact the relation (3.5)

-dV/dt

=

k\fy.

This is not a differential equation such as we have already studied, because it involves too many variables. A differential equation i n just two variables is, however, obtainable from it, by substituting for dV its value i n terms of y and dy. T h i s value is determinable in any case from the shape of the volume F, and the location of the opening through which the liquid escapes. I n a case i n which the figure is simple, the formula for V in terms oiy may be easily discernible. A differentiation then gives dV. It is, however, often simpler to obtain the formula for dV directly from the figure without first finding V. This is done by applying the reasoning of the integral calculus, which says that dV is equal to the area of the surface of the l i q u i d multiplied by dy. Example 6.

A cylindrical tank of length 9 ft. and radius 5 ft. is mounted horizontally and is filled with oil. A t / = 0, a plug at the lowest point of the tank is removed, and a flow results to which relation 3.5 applies with k = T o find the depth ^ of oil i n the tank at any time t while the tank is draining. T h e circumference of the end of the tank is the circle + [y — 5 j ^ = 25.

For this circle, x = +

T h e surface of the oil is thus

a rectangle whose width is 2 V ' l O ) ' — and whose length is 9. Since dV equals the area of this surface times dy, relation 3.5 yields the differential equation

T h e general integral of this equation is 12(10-;'i^ =

^+r.

T o fit the condition that> = 10, when / = 0, the value o f t must be zero. T h e result can be given the form

42

Applications t

V = 10 ^

— 180

^

I n this case it was simpler to find the value of dV from the figure of the liquid, than it would have been to find the formula for V and to differentiate it, for by elementary geometry V is found to have the relatively complicated formula V = 9

25T -

25 cos"^

-

1^ +

0

-

5)

VlO^ -

If the tank had been standing on end the situation would have been different, for then the formula for V would have been recognizable at a glance. W h e n a material body whose temperature is T is immersed in a quantity of water whose temperature is T', there is a flow of heat from the body to the water, or vice versa. T h e law that applies to this is Newton's law of cooling, which states that the rate at which the temperature of the body drops, namely, —dT/dt, is proportional to the difference between the temperatures of the body and water. Thus (3.6)

-dT/dt

= k{T

~

T'].

T h e constant k depends upon the physical values involved, among them the area over which the body is i n contact with the water, the material of which the body is composed, and the units used. T o obtain a differential equation i n two variables from equation 3.6, the value of T' i n terms of T must be substituted. This value can generally be found from the fact that the amount of heat which the body loses is the amount that the water gains. T h e change i n the body's heat content is always obtainable as the product of its weight by its specific heat by the change i n its temperature. Example 7.

A body weighs 45 lb. and is made of a metal whose specific heat is 5 . W h i l e at a temperature of 300 degrees, it is plunged, at ^ = 0, into 100 lb. of water at a temperature of 50 degrees. T o find the formula for the temperature T of the body during its cooling. T h e amount of heat that the body has lost at time / is 45(g) {300 — T]. T h e amount of heat the water has then gained is 100(1) j — 501 (the specific heat of the water being 1). Since these two amounts are the same, we have V-1300 -

T] = 1001 r

-

50},

whence 7"' = 65 — {T/20). O n substituting this value into equation 3.6 we obtain the differential equation

Problems of Flow -dTldi

43

= k

T h e integral which applies is that for w h i c h T = 300, when / = 0. integral is r = -y*f|13 + 50^--t2i/20)fc£j_

This

F r o m this integral we can observe the temperature to which the body w i l l eventually cool. As t increases indefinitely the exponential term approaches zero. T h e limiting value of T is therefore ^ ^ r ^ , that is, 61.9 degrees (approximately). PROBLEMS 33, A cylindrical tank of length 10 ft. and radius 3 ft, stands on end, and is filled with oil. When a plug halfway up its side is removed the oil Rows in accordance with formula 3,5, with k = ^ir- Find the formula for the depth of the oil in the tank while the tank is more than half full. 34, A tank is 6 ft. long, and its end is a square with a 4-ft. diagonal. At / = 0 it b half full of oil, and at that instant a plug in a lower edge is removed. T h e flow obeys law 3.5 with k = At what time is the tank drained, if it lies horizontally on one side? 35, At what time is the tank of problem 34 drained if it is mounted horizontally with a diagonal of its end in the vertical position? 36, A tank is in the form of a right circular cone with altitude 4 ft. and radius 5 ft. At ^ = 0, it is filled with water, and the water is then allowed to escape through a hole to which law 3.5 applies, with k = - j ^ . Find the formula for the depth of water in the tank at time /, if the tank is mounted with its axis vertical and its apex downward, and if the hole is at the apex. 37- If the tank of problem 36 is mounted with its apex upward, and the whole is in the base, what is the formula for the time when the depthj^f water in the lank is^? 38. A tank is in the shape of a sphere with radius 4 ft. At / 0, it is filled with water, and an opening on the level of the center of the sphere then permits the water to escape. Find the formula for the time in terms of the depth k of the water in the tank, while the tank is more than half full. 39. A cylindrical tank with radius 4 ft, stands on end, and has an opening, for which k ^ "ffVj in its lower base. As water excapes through this opening, fresh water is run into the tank at the rate of 1 cubic ft,/sec. Find the formula for the time at which the depth of the water in the tank is A, if A = 9, when / = 0. 40. A 60-lb- piece of metal whose specific heat is -5- is at a temperature of 450 degrees. At f = 0, it is plunged into 300 lb. of water whose temperature is 60 degrees. Find the formula for the temperature of the body at time t. 41. A n 84b. body is made of metal whose specific heat is ^ T J . While at a temperature of 308 degrees it is plunged into 11 lb, of water at a temperature of 53 degrees. Find the temperature to which the body eventually cools.

44

Applications

42. A 10-lb. body whose specific heat is -g- is at 0 degrees when t ^ 0. It is then plunged into 100 lb- of water at 80 degrees. If law 3.6 applies with k — yf* what is the temperature of the body at time t? 43. A 5-lb. body whose specific heat is TnJ" is plunged at / 0 into 50 lb. of water. If the temperature of the body at this time is 200 degrees^ and if it eventually cools to 50 degrees, what is the formula for its temperature at time /? 44. A 50-lb. body whose specific heat is i^f, and whose temperature is 208 dc^ecs, is plunged into 20 lb. of oil whose specific heat is 3^, and whose temperature is 50 degrees. Find the formula for the temperature of the body at time t? 45. T h e differential equation for the current i which flows under an e.m.f. (electromotive force) £ in a simple electric circuit containing a constant inductance L and a constant resistance R is (3-7)

L—i-Ri at

E.

Find the formula for r\ if £ is a constant, and / = 10 when t ^ /Q. 46. F r o m problem 45, find the formula for current 1 that flows in the circuity if I >- 0, when t ^ Oj and E ^ a sin
(3.8)

4 1 + ^

= ^-

F i n d the formula for q, i£ E is constant and q ~ go when / = fo48. From problem 47, find the formula for q on the condenser, if ^ = 0, when / ™ 0, and E = cos at.

3.5. Locus problems.

Rectangular coordinates

A plane curve may often be characterized most directly by a differential equation, for the properties that define the curve may be simply expressible i n terms of its tangent and normal lines. T h r o u g h the slopes of these lines the derivative is involved, and the defining relationship is accordingly a differential equation. T h e integral of this differential equation then yields the coordinate equation of the curve. I n the following statements of problems we shall use the terms " a r e a " and " l e n g t h " to mean the numerical values of these quantities. They refer, therefore, to the number of square units or linear units in the figure referred to. A r e a and length are thus always positive or zero. Example 8.

T o find the equation of the curve that goes through the point (3, 1) and whose tangent and normal lines always form with the AT-axis a triangle whose area is equal to the slope of the tangent line. Since the slope is to be equal to an area it must be positive, and since

Locus Problems.

Rectangular Coordinates

45

the point through which the curve is to go has a positive ordinate, we shall consider > positive too. If {x, y) is any point of the curve, the slopes of the tangent and normal lines are, respectively, y' and — 1 / y ' , and these lines meet the x-axis at the points (x — y/y\ 0) and {x + yy\ 0). T h e triangle therefore has a base of length [yy' + y/y' \. Since its altitude is we have 1

yy'

+

T h i s differenticil equation can be written i n the form y' V 2 — y^ = y^ Its variables are separable, and the integral for which ^ = 3, when * = 1, is - /

-h 2 -

Vaiog

[V2

+

^]y

=

x.

T h i s is the required equation. Example 9.

T o find the equation of the curve that goes through the point (2,1), such that the area under any of its arcs and above the x-axis is equal to the length of the respective arc. F r o m the calculus we know that, for a curve above the x-axis, which extends to the right from a point (XQ, >O)» the area under an arc is given by the integral /

y dx.

T h e length of the corresponding arc is

T h e curve sought is, therefore, one for which

B y differentiation this becomes y y' = + y/y^ — 1.

+ y'^j or, i n another form,

T h e general integral of this differential equation is log \y +

V /

_

ij

=

±x-^c,

and withe determined so that)" = l , w h e n x = 2, the result can be written y +

=

^-

It is possible to improve upon this form. canceling etc., we find that

O n transposing y^ squaring,

We can now drop the ambiguous signs, since the two alternatives give the

I

I

46

Applications

same formula.

T h e equation sought is thus

This is the equation of a catenary. It is the curve i n which a rope or chain hangs when it is suspended from two of its points. I n terms of the hyperbolic functions, which are defined i n the calculus by the formulas (3-9)

svnhx

-

-

r^!,

coshjc =

i l ^

+

^""^h

the result can be expressed i n the form^' == cosh (jt — 2 ) . PROBLEMS 49. Find the equation of the curve that goes through the point (1, 3), and for which the tangent at any point, and the Une joining that point with the origin, have slopes which arc the negatives of each other. 50, Find the equation of the curve that goes through the point (4, 3), and whose tangent and normal lines always form with the x-axis a triangle whose area is equal to the product of { — - T T ) by the slope of the tangent line, 51, Find the equation of the curve that goes through the point (0, 1), and whose tangent and normal lines alw^ays form with the :r-axis a triangle whose area is equal to the negative of the slope of the normal line. 52. Find the equation of the curve that goes through the point (4, 2), such that the segment of its normal line between the curve itself and the >-axi3, is always bisected by the ^r-axis.

53- Find the equation of the curve through the point (1, 5), whose tangent and normal lines always make with the ^-axis a triangle whose area is equal to the slope of the tangent line54. Find the equation of the curve through the point (2, 5), whose tangent and normal lines always make with the >-axis a triangle whose area is equal to 4 times the slope of the normal line. 55. Find the equation of the curve through the point (-J, 0), whose tangent and normal lines cut from the >-axis a segment whose length equals the negative of the slope of the tangent line. 56. Find the equations of the curves through the point (3, 2), for which the tangent and normal lines always form with the and ^-axes a quadrilateral whose area is equal to 2xy. 57. Find the equation of the curve through the point (4, 2), for which the tangent line and the horizontal through the point of contact form with the x- and ^-axcs a trapezoid having the area 6. 58- Find the equations of the curves through the point (4, 3), for which the tangent line and the vertical through the point of contact always make with the x-axis a triangle having the area 9,

Polar Coordinates

47

59. Find the equation of the curve through the point (1, —1), for which each tangent line is at a distance from the origin that is equal to the abscissa of its point of contact. 60. Find the equations of the curves through the point (1, 1), for which any tangent line and the line from the origin to the point of contact make with the>-axis a triangle having the area ^. 61. Find the equations of the curves through the point (0, 5), for which the area under any arc and above the jr-axis is equal to 3 times the length of the respective arc. 62. Find the equation of the curve through the point (-ff, 0), along which the arc length and the function 2x^ increase at the same rate.

3.6. Polar coordinates Polar coordinates are sometimes better adapted to the statement and solution of a given geometrical problem than are rectangular coordinates.

FIG.

3.

T h e two systems are related as shown i n F i g . 3. T h e connecting relations are X = r cos df y — r sin Q. F r o m the calculus, we know that the angle ^ and the differential of arc ds are given by the formulas (3.10)

dB

tan^ = r—I dr

and (3.11)

ds"" =

r''dQ'^-\-dr''.

These are the essential formulas for dealing with the geometrical problems to be considered. Example 10.

T o find the equations of the curves through the point with the polar coordinates (1, ir/2) for which every normal line is at the distance 1 from the pole.

48

Applications

T h e perpendicular from the pole upon the normal line has the direction of the tangent line. It therefore makes with the radius vector the angle ^ or — i^j according as ^ is acute or obtuse. F o r the curves sought we have thus 1 / r = cos ^, or 1 / r = cos ( r — W e can write these relations tan 1^ = ± V ' r ^ — 1. F r o m equation 3.10 we thus obtain the differential equations r d$/dr =

±

V ^ ^ .

T h e i r general integrals are 0 = ±{Vr^

-

1 -

c o s - i (1/r)) + c,

and with c determined so that $ = T / 2 , when r = 1, the required equations are found to be 6 Example 11.

=

T/2

± {Vr^ -

1 -

sec"^ r ) .

T o find the equations of the curves through the point

with the polar coordinates {'s/2y T / 4 ) along which the arc length increases at twice the rate of the vectorial angle. T h e specific property of the curves sought is ds — 2 dd. B y (3.11), this

is

dd^ •\- dr^ — 2 dd.

We

may

write

this i n the

form

r d6 — ±dr. This equation has r = 2 as an integral- However, that integral does not go through the given point. T h e generzil integral is B = i: sin~^ + Cj and when c is appropriately determined this gives 6 = sin~^ and 6 = — s i n " ^ (r/2) + ir/2. W e can write these equations i n the forms r = 2 sin 0 and r — 2 cos 0. T h e curves are circles of radius 2 through the pole. PROBLEMS 63, Find the equation of the curve through the point with the polar codrdinates (2, x/6), for which the angles 0 and ^ are always equal. 64. Find the equation of the curve through the point with the polar coordinates (3, 3ir/2), for which 4" and 0/2 arc always equal. 65, Find the equation of the curve through the point with the polar coordinates (1, r ) , for which 1^ is always the supplement of 0/2. 66. Find the equation of the curve through the point with the polar coordinates (1, ir/4), for which ^ is always the supplement of 2(9. 67. Find the equations of the curves through the point with the polar coordinates (3, T / 2 ) , for which all the tangent lines are at the distance 3 from the pole, 68, Find the equations of the curves through the point with the polar coordinates (5, 0), for which the tangent and normal lines are always equidistant from the pole.

Some Curves of Pursuit

49

69. Find the equations of the curves through the point with the polar coordinates (l/'\/2> ~""/4), for which the distance from the pole to the tangent line is always equal to the square of the radius vector to the point of contact. 70. Find the equations of the curves through the point with the polar coordinates (1, 0), for which the distance from the pole to the normal line is always equal to the square of the radius vector to the point of contact. 71. Find the equation of the curve through the point with the polar coordinates (1, 0), for which ^ is always equal to log r. 72. Find the equations of the curves through the point with the polar coordinates (3, 0), along which the arc length increases at 6 times the rate of 6. 73. Find the equations of the curves through the point with the polar coordinates (1, — T / 3 ) , along which the arc length and the square of the radius vector increase at the same rate. 74. Find the equations of the curves through the point with the polar coordinates (4, JT), along which the arc length increases at a rate that is always 2 tan times the rate at which the radius vector increases.

3.7. Some curves of pursuit W h e n a point moves so as always to be directed toward a certain goal, its locus is a curve of pursuit. T h e goal may be fixed or moving, and outside influences may bear upon the motion of the point itself. A simple example of a curve of pursuit is the path of an airplane that pursues another while always keeping its quarry i n the line of sight. W e shall consider this case here under the following conditions: (i) T h e pursued airplane 0 flies i n a fixed direction with a constant ground speed whose eastward and northward components are a and b, respectively. (ii) T h e pursuing plane A has a constant air speed V. (iii) There is a constant wind whose components toward the east and north are, respectively, Vx and Vy. It need not concern us much that condition (i) is actually quite artificial unless 0 is fixed. W e are presently engaged in developing techniques i n differential equations, not i n obtaining results usable i n aviation. W e shall consider other curves of pursuit in Section 7.9. T o the pilot of the pursued plane 0 of F i g . 4, the path of the pursuing plane A is a curve expressible by an equation either i n the polar coordinates (r, 6) or i n the rectangular coordinates (x — x*, y — y*). With reference to the ground, A has a velocity whose eastward component dx/dt is made up of the air-speed component V cos (x — 0) and the wind component Vx. T h u s dx/dt = — Kcos 0 + Vx.

Applications

50

Similarly, A^s northward velocity component is dy/dt = - F sin

+ Vy.

For the plane 0 we have dx*/dt = a, and dy*/dt = b. these equations into

W e can combine

d{x -

= { - F c o s f l H- Vx - a\ dt,

d{y-y*}

= \ -Vsme

+

Vy-b\dt,

and by eliminating dt find that (3.12)

[VsmB-Vy-\-b]d{x-x*\

-

\Vcos6

~

^ a}d\y -y*\

= 0.

T h r o u g h the use of the relation tan B = {y — y*)/{x — x*), equation 3.12 is expressible either as a differential equation i n r and B, or as one i n [x — x*\ and \y — y*\.

Fio. 4.

Example 12.

A n airplane 0 has a ground speed with constant eastward and northward components of 144 m.p.h. and 108 m.p.h., respectively. T h e pilot observes a pursuing plane .4 at a distance of 10 miles due north. T h e air speed of A is 270 m.p.h., and there is no w i n d . T o find an equation for the path of A as it appears from 0. W e have i n this case a = \ AA, b = \08, = 0, Vy = 0, and V = 270. Equation 3.12 is therefore ¡270 sin ^ + m\d{x - X*] -

{270costf + 144lrf(;' ->'*1 = 0.

Since j j ; A : * ) — r cos B, and {y — y*] = r sin By so that d[x — AT*j = — r sin 6 d9 + cos B dr^ and d{y — y*\ = r cos B dB -\- sin B dr, the equation becomes —

+ 4 cos

H- 3 sin B\TdB +

(4 sin 5 -

3 cos B] dr = 0.

T h e variables i n this are separable, and the general integral is 5 log

5 — 4 cos B — 3 sin 0 4 sin 5 — 3 cos d

-\- log ¡4 sin

— 3 cos B] + log r =

c.

Some Curves of Pursuit

51

T o fit the condition r = 10, when 6 = i r / 2 , the value ofc must be 10 T h e resulting equation can be put into the form r =

10 V 2 ¡4 s i n g (5 -

y/l.

3cosg|

4 cos g - Ssintfj^^

Example 13.

A n airplane A whose air speed is 200 m.p.h. takes off for a fixed destination 0 which is due west and at a distance of 300 miles. A constant w i n d blows with components of 15 m.p.h., and 20 m.p.h., toward the west and north, respectively. T o find the path of A if it is always directed toward 0. Since the point 0 is fixed, we may take it to be the origin. We have, then, X * = 0, = 0, a = 0, i = 0, = - 1 5 , Vy = 20, and V = 200. Equation 3.12 is thus {200 sin e -

20) dx -

{200 cos 6 +

15) dy = 0.

Since tan 0 = y/xj we have cos 6 = V ^ ^ ^ * M - > ^ , and sin d = T h e equation is thus 200;Vx^

+

-

20

dx

200x

-

Vx'-\-y

y^

+ 15

dy =

y/'Vx^

0

T h i s is a differential equation i n which both coefficients are homogeneous functions of the degree 0. T h e method of Section 2.7 therefore serves to integrate it. T h e integral for which x = 300, w h e n ^ = 0, is found to be 10 V x ^

log {4x-\-3y\ + 8

log

+jy2 _

4x

- h

6^ _|_ 8^

= log (1200)

3y

We can write this i n the form 10 Vx^ -\-y^ -

6x + Sy]^ = 1200{4x +

3y]\

A curve of pursuit of a different kind is the locus of one end of a piece of string when the other end is drawn along a given curve. I n F i g . 5 the piece of string is shown as QP. Its length a is constant, and the curve /(x*, y*) = 0, along which point P is being drawn, is given. Since a string can pull only along its own direction, the path of Q has QP as its tangent line. Hence y' = tan 0, and therewith the obvious relations X* — X = a cos 6, y* — y = a sin tf, can be given the forms (3.13)

=

x-\-

ay

V l

+

y*

+

Applications

52

W h e n these values are substituted i n the given equation/(x*,^*) = 0, the result is the differential equation for the path of Q. T w o loci are always to be expected, since point P may be drawn i n either direction along the given curve. W h e n the given curve is a straight line, the locus of Q is called a tractrix. 14. One end Q of a piece of string of the length 3 is placed at the point (2, — 2), and the other end is then drawn along the line y = x. T o find the equation of the locus of Q. Example

FIG.

5.

T h e given curve has the equation y* = x*. values 3.13 transforms^* = x* into

T h e substitution of the

3/ V l

V l

+

+

/"2

which is the differential equation for the path of Q. by making the change of variables x-{-y = SyX—y is thereby transformed into u V l

+

(du/ds)^ =

-3

It can be integrated = u. T h e equation

Vidu/ds,

that is, into Vl8

One integral of this is u = 0. -

V l 8 -

-

u^du

=

±uds.

T h e general integral is

+ V l 8 log

Vis

+

Vis

u

-

=

±s-\-c.

I n terms of the original variables, the integral for which y = — 2 , when X = 2, is

Some Curves of Pursuit ±{x'\'y]

= A/lSlog

V2(x -

53

y) V2

-

V i s -

(x

-y)\

T h r o u g h the two choices of sign, this gives the equations of twotractrices. PROBLEMS 75. A n airplane O whose ground speed is constant at 200 m.p.h. is flying northeast. T h e pilot observes a pursuing plane A southeast of him at a distance of 6 miles. T h e air speed of ^ is 300 m.p.h. There is no wind. F i n d the equation of A^s path as the path appears from O. 76. A n airplane O is flying due north with a ground speed of 200 m,p.h., when the pilot observes a pursuing plane ^ at a distance of 14 miles directly east. T h e pursuer's air speed is 240 m,p.h. There is no wind. Find the equation of A's path as it appears from O. 77- A n airplane 0 is flying with a ground speed whose components are 100 m.p-h. westward and 110 m.p.h. northward. T h e pilot observes a pursuing plane A directly east at a distance of 10 miles. T h e pursuer's air speed is 250 m.p.h., and there is a steady wind of 40 m.p.h, from the north. Find the equation of A*s path as it appears from O, 78, A n airplane O has a ground speed whose components are 100 m,p,h. to the east and 240 m.p.h. to the north. T h e pilot observes a pursuing plane ^4 at a distance of 15 miles due west. The pursuer's air speed is 300 m.p,h. There is no wind. Find the equation of A^s path as it appears from O, 79- A n airplane A whose air speed is 250 m.p-h., takes off^ for a point 0 which is due west and 150 miles distant. A wind of 25 m.p.h. blows from the south. T h e pilot keeps the plane directed toward O. Find an equation for the path of the plane, 80. A n airplane A whose air speed is 250 m,p,h. takes off" for a point O which is 200 miles west and 150 miles south. A wind of 25 m.p.h. blows from the south. F i n d the equation of the plane's path, 81, A n airplane A whose air speed is 200 m,p,h. takes off for a point 0 which is 100 miles due south, A wind of 30 m.p.h. blows from the west. Find the equation of the plane's path, 82. A n airplane whose air speed is 200 m.p.h. takes off* for a point 100 miles due south, A steady wind blows, with components of 20 m,p.h, from the west and 10 m.p.h. from the north- Find an equation for the plane's path, 83, One end Q of a piece of string of length 5 is placed at the point (4, 3), and the other end is then drawn along the x-axis. Find the equations of the loci of Q, 84, One end Q of a piece of string of length 2 is placed at the point (0, 2), and the other end is then drawn along the line ^ = 4. Find the equations of the loci of Q. 85. One end Q of a piece of string of length 13 is placed at the point (5, 2), and the other is then drawn along the ^-axis. Find the equations of the loci of Q.

54

Applications

86. One end Q of a piece of string of length 2 is placed at the point (1, 1), and the other end is then drawn along the line x y = 0, Find the equations of the loci of Q , 87. One end Q of a piece of string of length 6 is placed at the point (3, —1), and the other end is then drawn along the line ^ + 1 ••0. Find the equation of the locus of Q , 88. One end P of a piece of string PQ whose length is «2 is drawn along the circle x^ " b^t for which b > a. Find the differential equation for the locus of and show that + = 6^ — is an integral of this differential equation. 89. A ferry boat whose speed is a 8 m.p.h. plies between opposite points on the banks of a straight river 1 mile wide. T h e river flows at 4 m.p.h. T h e boat is always kept pointed directly at its objective. Find an equation for its path. 90- If the ferry boat of problem 89 takes off for a point that is z niilc upstream on the opposite bank, what is an equation for its path? 91. Show that if the ferry boat of problem 89 had a speed of only 4 m-p-h, it could not reach a point directly opposite the starting point, and find how far downstream its landing would have to be. 92, A ferryboat whose speed is V m.p-h. plies on a river K miles wide. T h e river flows at V m.p.h. If the boat takes off for a point on the opposite bank a miles upstream^ and is kept pointed at its objective, what is the equation of its path?

3.8. Suspension cables A piece of rope or cable, like any other body, is subject to Newton's law of motion 3.2. W h e n it is at rest, therefore, the resultant of the forces upon it, or on any piece of it, is zero. T h e key to the solution of many problems concerning ropes, cables, chains, etc., lies i n this fact. y

FIG.

6.

W h e n a cable (or rope or chain) is supported at two of its points, and hangs between these points under the burden of some distributed load, we call it a suspension cable. T h e cable of a suspension bridge is an example, as is also any heavy chain burdened by its own weight. T h e curve i n which the cable hangs is determined by the distribution of its load. W e shall see how this curve can be found. I n F i g . 6, OQ is any segment of a suspension cable, of which one end 0 is

Suspension Cables

55

the cable's lowest point. T h a t point is called the vertex. T h e tangent line there is horizontal. As i n the figure, we shall take the origin at this point. T h e forces acting upon the segment OQ are its burden or load L (which is not indicated i n the figure, but which acts vertically downward) and the pulls from the adjoining parts of the cable. These pulls are along the tangents of the curve. Thus the pull P at 0 is horizontal, whereas at Q we have tan tp — y'. Since the whole segment of cable is at rest, the sum of the components of all forces acting upon it i n the horizontal, or the vertical, direction is zero. Thus, p cos ^ — P = 0,

^ sin ^ — Z, = 0.

T h e elimination of/* from these equations yields (3.14)

y' = L/P.

If the density of the load as referred to the horizontal is designated by p(x), the value of L is given by the formula (3.15)

fjp{x)dx.

Equation 3.14 is thus a differential equation for the curve i n which the cable hangs. Example 15.

A suspension cable hangs from the tops of two towers that are 200 ft. apart, and its vertex is 50 ft. below the points of support. T h e weight of the cable itself is negligible, but the cable bears a load whose density is ¡100 + x^j. T o find the equation of the curve i n which the cable hangs. Formula 3.15 gives ¡100 + x^] dx = lOOx +

L =

W,

and therewith differential equation 3.14 is /

= ^ | l 0 0 x + ^x3

T h e integral of the last equation for which y = 0, when x = 0, is y =

1

50x2 + - 1 X"'

T h e curve passes through the tower tops, whose coordinates are ( ± 1 0 0 , 50). T o fit this condition, the value of P must be 530,000/3. The curve

56

Applications

thus has the equation y =

10,600

600

If the burden on the cable is not distributed horizontally, but is distributed along the cable itself, say with the density p(j), where s is the arc length of the curve measured from the vertex, the formula for L is =

/; Pis) ds.

W i t h this formula, equation 3.14 is (3.16)

/

= ^

j\{s)ds.

T o obtain a differential equation from (3.16) we proceed as follows: Since ds = V l

dx, the derivative of (3.16) is

(3.17)

dy' =

WT7^

dx.

If it is possible to eliminate s between (3.17) and (3.16), the result is a differential equation for^', the integral of which, with^"' = 0, at x = 0, is the required differential equation for the curve of the cable. Example 16.

A chain is supported at two points on the same horizontal level and 12 ft. apart. Its slope at a point of support is and its density is given by the formula ¡1 + as\~^'^. T o find the differential equation for the curve of the chain, and then, i n the case a = 0, to find the curve itself. Equations 3.16 and 3.17 are in this case y

= -

Wx+as

~ 1),

d/

=

..ZL^

dx.

P V 1 + as T h e elimination of equation

+ asy and thus of J , yields for y' the differential

T h e required integral of the last equation is aP[y/\

+2

log ( Z + ' s / l

=2x/P.

This is the differential equation for the curve of the chain.

Suspension Cables

57

W e are prepared to integrate the last equation only if a = 0. T h e chain is then of uniform density. I n that case the differential equation is log ( y + V i + / 2 }

=v^,

and since it was given that y' = T , when x = 6, we see that P has the value 6/log 2. T h e differential equation can therefore be put into the form

Its integral for which ^ = 0, when x = 0, is thus ^ -

^

f .(i/6)Iog2 I -Cj:^6)log2 _ 9)

Iog2 ^

A n alternative form for the last equation is '

log2'''

^

PROBLEMS 93. A suspension cable of negligible weight is burdened with a load distributed horizontally with a constant density. The supports of the cable are 400 ft, apart, and 80 ft, higher than the vertex. Find the equation of the curve in which the cable hangs. 94, A suspension cable of negligible weight is burdened with a load dbtributed horizontally with the density jlOO + x^/100)}. T h e supports of the cable are 160 ft- apart, and 64 ft, higher than the vertex. Find the equation of the curve in which the cable hangs. 95, A suspension cable of negligible weight is burdened with a load distributed horizontally with the density (24 + Z t ^ ) / \ / 2 4 + x^. A t ;r =^ 1, the slope of the cable is F i n d the equation of the curve in which the cable hangs. 96. A suspension cable of negligible weight is burdened with a load distributed horizontally with the density j2 -j^e'^^^}. T h e point (10, 4) is on the curve in which the cable hangs. Find the equation of this curve. 97, A suspension cable of negligible weight is burdened with a load distributed horizontally with the density 35 sec^ (TAT/SOO). T h e slope of the cable at a point of support is -y, and these points are 150 ft, apart. Find the equation of the curve in which the cable hangs. 98. A cable of constant density hangs under its own weight without any other load. At a point of support its slope is and these points arc 10 ft. apart. Find the equation of the curve in which the cable hangs, 99. A chain of constant density hangs freely from two supports 2 ft. apart, and the slope of the chain at one of them is -j-. Find the equation of the curve in which the chain hangs. 100, T h e density of a certain string varies with the distance from one of its points O, and is p{s) ™ T I ^ ' + T h e string is hung from two pegs so that the point O

58

Applications

is the vertex, and its length is adjusted until the pull upon one of the p ^ has the hori* zontal component 1 lb. Find the equation of the curve in which the string hangs. 101. T h e density of a certain rope at distance s from one of its points 0 is p{s) = sec^ J , T h e rope is hung from two pegs so that the point 0 is the vertex, and its length is adjusted until the pull upon one of the pegs has the horizontal component ^ lb. Find the equation of the curve in which the rope hangs102. In the case of the rope of problem 101, fmd the equation of the curve in which it h a n ^ if its length is adjusted until the pull upon one of the pegs has the horizontal component 2 / V 3

lb.

3.9. The rope around a shaft W h e n a rope is passed over a shaft or around a post, a large pull P on one end of it can be held i n check by a much smaller pull p on the other

T Fio. 7.

end. W i t h a given value of P , the value of p depends upon the extent to which the rope is wound around the post or shaft, namely, upon the angle $ which the arc of contact between the rope and the shaft subtends. It depends also upon /x, the coefficient of friction between the rope and the shaft. Suppose the rope is of constant density p, and that the shaft is horizontal and of radius a. T h e n , when pull p is just enough to keep the rope from yielding to P , the configuration is that of Fig. 7. A t any point of contact Q the gravitational pull due to the weight of the rope has the intensity p, and this has the component p cos ^ in the direction of the rope. Its other component is merged in the pressure that the shaft exerts upon the rope.

The Rope Around a Shaft

59

I f we designate the normal component of the intensity of that pressure by / ( v ) , the corresponding frictional component amounts to nf{(p). These force intensities are indicated i n the figure by dashed vectors. T o obtain the components of the forces which they generate, we must integrate their respective components with respect to s. W e note that ds = a dtp. B y taking components horizontally to the right, and then upward, we thus obtain the equations 9 '0

\p cos ipsvfMp -\- f{
and — P+

^

{

—p

cos
C O S
+ f(ip) sin

/ W COS (p]a

d
+ p{e) cos ^ = 0,

T h e derivatives of these equations are ap sin dcose

-\-

(cos ^ -

ju sin

— ^ sin ^ dd

p{e) cos ^ = 0,

cos0 -

p{e) sin 9 = 0.

—ap cos^ 0 + af{Q) {sin 8 -\- ^cosO} dd

T h e elimination of f(6) from them gives the equation (3.18)

{dp/dS) + up = ap cos 6.

This is the differential equation for/?. T h e requisite integral is the one for which p = Py when ^ = 0. W h e n the rope is i n a horizontal plane, and is passed around a vertical post instead of a horizontal shaft, the effect of the rope's weight is eliminated because it acts along the post and not upon it. T h e differential equation i n that case is accordingly (3.19)

(dp/dd) + fxp = 0.

Example 17.

A rope weighing 2 lb./ft. hangs over a horizontal shaft 1 ft. in diameter upon which the coefficient of friction is ^. O n one side a 3-ft. piece of rope overhangs, and this has a weight of 50 l b . attached to it. O n the other side a 5-ft. piece overhangs, and sustains the pull pi. T o find what pi must be to keep the rope from moving. In this case a = - j , p = 2, and / i = ^. Also P = 50 + 6, and when 6 = IT then p = pi -\- 10. Differential equation 3.18 is (dp/dS) -i-ip

= cos 0

60

Applications

Its integral, for which p = 56, when $ — 0^ is A i l ^—5/3

= -^^ cos 0 + A sin ^ + W ^ O n setting 6 = ir^ and p = pi

10, we find that

pi = rffi557tf"^^^ - 33) = 16-27 lb- approximately PROBLEMS 103, A rope b passed three*fourths of a revolution around a vertical post upon which the coefficient of friction is -ff, A pull of 100 lb, is exerted upon one end of it, What pull must be exerted upon the other end to keep the rope from moving? 104, A man capable of exerting a pull of 100 lb. wishes to hold a pull of 300 lb, in check by passing the rope around a vertical post upon which the coefficient of friction Over how large an angle must he wind the rope around the post? 3 105, By passing a rope just once completely around a certain post^ it is possible to hold a pull on one end of it in check by a pull one*fourth as great on the other end. What is the coefficient of friction on this post? 106, A rope whose weight is negligible hangs over a horizontal shaft upon which the coefficient of friction is -j. One overhanging end has a weight of 100 lb, attached to it, and the other end sustains a weight of W lb. What arc the limits within which fV must lie if the rope is not to move? 107, A rope weighing 1 lb./ft, is wound l-j times around a horizontal shaft whose radius is 6 in, and upon which the coefficient of friction is -j. N o rope overhangs at one end. How much rope may overhang at the other end without causing the rope to go into motion? 108, A rope passes over a large horizontal cylindrical surface whose radius is 6 ft. and upon which the coefficient of friction is y. At one end the rope overhangs by 5 ft,, and to this end a weight of W lb, is attached. T h e arc of contact between the rope and the surface subtends a right angle, and the pull exerted upon the horizontal end is just enough to keep the rope from moving. How much larger must this pull be if the rope weighs 5 lb,/ft, than if its weight is negligible?

ANSWERS TO ODD-NUMBERED PROBLEMS 1, ^ = 50 - 35e tan - 1

5. / = "

7. V = 4

3, V -

9. V = \6 V l o g (j/4),

-At

11, s = 5 0 1 " 20 log [1 - v/20]\. 15, J = 468 log

90e-^'^^^^'r

— tan

I - e—At 1

100 \^\0

24 24 - v\

39 2

v

13. V = 54(1 ^ ^-^'''«^1, 17_ V = 40|2^-^'/^" -

1|

Answers to Problems 19. I - 70 log

23. * - 200

27. * -

28 -

1

-

2

-

10 -

2511

21. X

2

-«-3
150t/

25.

_ 9,-A(/2_

99 1

31. X ^ xo

B

61

6kt + 1

29.P--

+

a

P0--

I

t/io

33. 5 +

^

1001

360x)

35. / = 320 \ / 2 . 37. ( = 5 T 1 H ^ -

- 2y^\.

160;.^ +

39. t = 960ir 13 -

s/h

-

30 -

30 log

VA

27

41. 5 8 t degrees.

43.

4 5 . . - -

47. 9 -

+

49. xy = 3. S3, y =' 6 -

^2 - ^ s i n - » 12.« -

and

65. r =

1 — cos

69. r = cos tf, and r ~ — sin 6.

and r = 3.

(2r) - \ / 3 , and

1 - sec

e = sec"^ (2r) + \/3 6 \ / 2 jcos

-

\/4r^ -

- sin

1 - 2jr/3. 77. r =

| \ / 2 - cos e - sin ^ l ^ ^

90 5 + 4 cos

-

3 sin

81. u + V ^ t ^ + y i 20

79,

83. ±

85.

= i

cos"i

= \^AT^ -

75. r =

{x-\y'-\-y^

- iU'^^ + O^-^^^j^

63. r = 4 sin 6. = 3,

1}. 59.

8.

61. > = i l 9 ^ ^ ' +

73.

-

V 2 - x\

57. 2xy -y

71. log r -

+ 1^0 - £Cl*-'^«*^.

51. ;r » 1 -

55. ^ = - J +

67. r sin

r=50-|-150«-'i*'»^^*'«'*'.

- 4} - •v/25

ii)--

- 5 log

5 2| = V 1 6 9 - .c^ _ 13

5 + V 2 5 -

,13 + \ / l 6 9 ' Sx

-

4

- 12,

1,000,000;- 17

62 87. J -

Applications 89. y -

-1.

93. y = xVSOO.

91. \ mile downstream. 95. ^ « V f f ! [ 2 4 + x 2 ] H _

99

101. ^ -

-

(S cos (3^5) + V Z S cos2 (3JC

log J

103. 100*""'=' lb. 105. M -

[24]'^).

log 9

-4.

^

(20.8 lb. approximately).

( l A ) log 2

107. (3/20) {«"" -1- 1 |ft.

(\/i/2)il - * | .

(0.22 approximately). (3.62 ft. approximately).

CHAPTER

4 Integral Curves Trajectories 4.1. Integrals of a differential equation T h e methods of Chapter 2 for solving differential equations (4.1)

/

=M>)

were found i n Chapter 3 to be applicable to many kinds of physical problems. It remains a fact, however, that all these methods are essentiedly special. It would be easy to write down differential equations 4.1 i n great variety, to which no one of the methods given would apply. A n d that would remain true even if more methods were to be given, as they could be. F o r the fact is, that, among a l l differential equations 4.1, those for which a solving formula, other than one i n infinite series, can (or could) be given are relatively rare. This is by no means due to a mere lack of mathematical proficiency or skill. T h e reason lies deeper than that. A n elementary calculus formula is a combination of certain functions that are more or less familiar. T h e category of these functions is relatively quite restricted. It includes the exponential, the trigonometric, and the power functions, together with their inverses, and the successive quadratures of combinations of these. N o w a differential equation may define its function through a relationship which, i n its very nature, is inexpressible by such means. I n short, it may define a function proper to itself, which is not representable by any combination of elementary functions, just as log X, for instance, is not representable by any combination of polynomials. N o w it is, of course, quite apropos to inquire what one shall take to be an integral of a differential equation, when no formula for one can be given. T h e answer is to be found i n the definition of the term "function." I n the calculus y is defined to be a function of x upon a specified x-interval, if to each X of that interval there corresponds a value of y. This does not require that y be expressible by a formula. It requires only that y be i n some way determined. Such determination may, for instance, be by a 63

64

Integral Cur\'es; Trajectories

graph, or by some process of calculation through which y is computable, either exactly or to as high a degree of accuracy as may be desired. This definition is also adopted i n the theory of differential equations. I n Chapter 1 we have already considered the direction field of a differential equation 4.1. That led to the conception of the streamlines of the field as integral curves, and thus suggested that the graph of the general integral is a family of curves one of which passes through any chosen point of a region. Analytically this statement implies that a differential equation is fulfilled by a family of functions, and that, subject to proper restrictions, there is a member of that family which for an arbitrarily assigned value xo takes on a prescribed value ^o-

4.2. The direction field of a differential system T h e considerations which led to the notion of a direction field for a single differential equation can easily be extended to the case of a pair of simultaneous equations. (4.2)

y

=fi{x,y,z),

z'

=fo{x,y,2).

Consider any region of three-dimensional (jt, y, z) space i n which the functions fi{x, yj z) and f'i{x, >, z) are real and single veilued. A t each point of such a region these functions have numerical values, and the triple of numbers 1, / i ( x , ^, z), f^ix^ y^ z), is thus determined there. These numbers, however, determine a direction in space, namely, the direction whose cosines are proportional to them. By assigning this direction to the point, and doing so for each point of the region, we impose upon the region a three-dimensional direction field. A particle, starting at any point of such a field, and moving thence always i n the direction of the field, describes a streamline. For these streamlines, as for any curve, t/x, dy^ dzy are direction numbers. Because these numbers also give the direction of the field, we see that along a streamline dx

dy

X

f\{x,yy z)

dz fiixyy,

z)

This means that the functions y and z defined by the streamlines fulfilled equations 4.2. T h e streamlines are thus integral curves. A curve i n space is given analytically by two equations i n (v, y, z). E a c h of these equations separately defines a surface. Together they define the curve which is the intersection of the surfaces. E a c h i n d i v i d u a l integral curve is thus the intersection of a surface of one family by one of another. Since each of these families is given by an equation involving a n

The Isoclines of a Direction Field

65

arbitrary constant, the pair of equations representing the family of integral curves involves two such constants. These observations have already been borne out by the actual solutions of differential systems of the form 4.2 that are found i n Sections 2.9 and 2.10. We shall return to this subject again. I n the sections immediately following, however, we shall consider further the integrals of a single differential equation 4.1 only.

4.3. The isoclines of a direction field T h e direction field of a differential equation 4.1 over a chosen region is graphically determinable by calculating the values o f / ( x , y) at a chosen and suitably dense set of points, and assigning to each of these points the direction with the slope respectively calculated. W h e n considerable accuracy is required, the number of points for which the calculation must be made is apt to be relatively large. T h e labor involved may then be substantial. Various methods, such as the following, are known for reducing this labor. For each value k which the function f{x, y) takes on in the region of the field, the equation (4.3)

fix, y) = k

defines a locus that traverses the region. This is the locus at each point of which the slope of the field is k. It is thus a locus along which the field has a single inclination. It is therefore called an isocline of the field. B y drawing i n a number of isoclines, the plotting of a field and the sketching of the streamlines may often be considerably facilitated. W e note especially that the isoclines for which k = 0 are the loci along which the integral curves have their maxima and m i n i m a . Example 1.

T o plot the integral curves of the differential equation ,

2y-2

T h e isoclines are i n this case the loci {2y — 2)/{x — 2) = k. the straight lines

These are

y - \ = ^ U - 2 ] ,

which pass through the point (2, 1). T h e isocline is crossed by all the integral curves at the same slope, and that is here twice the slope of the isocline itself. Figure 8 applies to this example. It shows in full lines some integral curves, and in dashed lines a pair of isoclines.

Integral Curves; Trajectories

Fio. 8,

Fio. 9.

The Isoclines of a Direction Field Example 2.

67

T o plot the integral curves of the differential equation ,

y =

1 -3{x+>P 1 +3lx+^l

T h e isoclines are i n this case the loci 1 -3ix+^p

^

that is, the straight lines

A l l these have the slope — 1 . It is clear that k cannot have any value greater than 1, and hence that no integral curve ever has a slope greater than 1. Some integral curves and a few isoclines are shown i n F i g . 9. ASSIGNMENTS With the use of isoclines, plot the direction field and sketch in some integral curves for each of the following differential equations. 1. y =x/y.

3. /

2. /

= ~2xy.

= ^ ^ -

A. y'

^• X

5. / = - l ^ .

6. / = \ / ? T ? .

4

7, Show that if :to is a value for which /»C^o) = 0, the line ^ = xo is an isocline of the linear equation (4.4)

y'+p(x)y

=q{x).

8. Show that if i3 any value for which p(xi) ^ 0, all the directions of the field of differential equation 4.4 along the line x ^ x\ are concurrent in the point \x\ + l/p(xi)j g{xi)/p(x\)]. This ¡5 shown for the difTercntial equation

in Fig, 10. In the manner of Fig. 10, plot the direction field and some integral curves for each of the following differential cquations4 9.y'^x-2xy.

U.y'=h-y-

, 0 . / =

12./ =

^ ^ + ! .

? ^ ^ .

68

Integral Curves; Trajectories y

X

Fro. 10.

4.4. Trajectories If two families of curves JFi and SF2 occupy a region of the {x^y) plane, one curve of each family passing through any point of the region, the curves of either family are said to be trajectories of those of the other. T h e relation between curves and their trajectories is determined by the law of their intersection. W h e n a family ffi is given, the trajectories which intersect it i n the point {x, y) at the angle i/'(x, y) has a differential equation which can be determined. Let the differential equation of the family SJi be (4,1). T h i s can be found by the method of Section 1.5. T h e inclination of its direction field at (x, y) is tan~^ /(x, y). T h e inclination of the field of the trajectories is therefore {4/{x,y) + xarT^ f{x,y)\. Hence the trajectories have the differential equation /

This equation is

= tan {yp{xyy) + larT^ f{x,

y)].

Trajectories

69

if 1^ is a right angle everywhere, and otherwise, by the addition formula for the tangent, (4.6)

y

f{x, y) + tan yp(x, y)

I

It - f{x,y)

tan^(jf,>)

T h e case i n which is a right angle everywhere is especially important. T h e two families of curves are then said to be orthogonal trajectories of each other. Orthogonal families often have interesting a n d important geometrical properties. Such families also occur i n many connections i n physics. T h e lines of electric flow in a conducting plate, for instance, are the orthogonal trajectories of the lines along which the potential is constant. A n d i n the plane flow of a n incompressible liquid, the lines of constant pressure are the orthogonal trajectories of the streamlines. Example 3.

T o find the equation of the trajectories that cut the hyperbolas of the family — 2x^ — c everywhere at the angle tan~^ (^). T h e diff"erential equation of the family of hyperbolas is ^' = 2x/y. Differential equation 4.6 for the trajectories is thus y

,

=

{2x/y) -f- 1/3 1 -

2x/'hy

which has the general integral [x-^y}'[2x-y\'

This equation gives the family of trajectories. Example 4.

T o find the equation of the orthogonal trajectories of the

curves 4x-\-ly

= cx^'\

T h e differential equation of this given family of curves is , ^

\2x + 25v Ax

T h a t of the orthogonal trajectories is, by (4.5),

^

\2x - h 25y

T h e general integral of this equation, namely log {Ax' + nxy - h 25y^} - - tan 2 gives the family of trajectories.

\

Ay

70

Integral Curves; Trajectories PROBLEMS

1, F i n d the orthogonal trajectories of the family of parabolas

= ex.

2. F i n d the equation of the orthogonal trajectories of the family of hyperbolas

3, Find the orthogonal trajectories of the family of circles x^ + y^ —

= 2cy.

4, Find the trajectories that intersect the lines of the family x + y = c at the angle 45 degrees. 5. F i n d the equation of the family of trajectories that intersect the lines y ^ ax -\- c at the angle tan~^ x. 6. Find the equation of the family of orthogonal trajectories of the ellipses x^/a^ +

7* Find the orthogonal trajectories of the family of parabolas 8. F i n d the equation of the family of trajectories that intersect the lines * + a> = c at the angle t a n " ^ 9. Find the equation of the family of trajectories that intersect the circles + = c at the angle tan~^ (x* + y^\. 10. Find the equation of the family of trajectories that intersect the ellipses y^/2S at the angle whose tangent is 11. Find the trajectories that make with the curves of the family tan~* (y/x) = c the angle { - t a n ~ ^ \ / x * +

+

~

.

12. Find the equation of the family of orthogonal trajectories of the curves 2{1 +3sin"^;'

4.5. Singular points of a direction field F r o m the fact that each point of a direction field lies upon an integral curve, it might seem natural to conclude that each point lies upon only one such curve. That, however, need not be so. Example 5.

T h e differential equation /

= 2{x^ -\-y -

1|'^ - 2x

has the general integral

TTirough the point (2, —3) therefore passes the particular integral curve

But the curve y = \ ~ x^ is also an integral curve, and it also passes through the point i n question.

Singular Points of a Direction Field

71

W h e n two integral curves have a point i n common they are, of course, tangent there, for they must both conform to the single direction of the field. A point of a field through which just one integral curve passes is called an ordinary point. A point which is not ordinary is called a singular point.

If we broaden our definition of a field to permit the inclusion of points at which the function f{xjy) is not continuous, such points may be singular points. However they are not necessarily so. F o r instance, i n the case of an equation 1,4, namely, (4.7)

/

=

Pi",

y)

<2(*,

y)

a point at which Q{xy y) = 0 is one of discontinuity. However, i f at that point P(jf, y) 7^ 0, the field merely has the vertical direction there, as may be seen by writing the differential equation i n the reciprocal form dx/dy = —Q{xjy)/P{xyy). T h e point may then be simply ordinary. I n the neighborhood of a singular point the integral curves may conform to any one of a number of difTercnt patterns. For instance, i n F i g . 8, all the integral curves pass through the singular point (2, 1), and none at all pass through any other point of the line x = 2, although all such points are singular. T h e whole course of the integral curves is often in large measure determined by their character near a singular point. T o show some possible families of integral curves we shall consider the differential equation.

=i n which a, 6, A, and k are constants, and ak — bh ^ 0. F o r this equation the origin is a singular point. T o a considerable extent the integral curves of equation 4.8 are typical of those of a much broader class of differential equations 4.7, namely, of those for which Q{x, y) and y) are both zero at the origin. If P(x, y) and Q(x, y) are replaced by their Taylor's series, such a differential equation appears as ax-\-by-\y

=

kx-^ky-\-

P2ix,

y)

Q2(x, y),

with P2{xy y) and Q2(^, y) having no terms of a degree less than 2. N e a r the origin, therefore, this equation closely resembles equation 4.8 if ak — bh 7^ 0. W h i l e that fact alone does not permit the inference that the integrals of the one differential equation resemble those of the other, that inference can nevertheless often be justified by other means.

72

Integral Curves; Trajectories PROBLEMS

In each of the following cases find for the difTcrential equation the particular integral that fulfills the respective condition. Then show that the function yi{x) given is also such an integral. 13. y

{x^-\-2y -

-

y{-3)

l\^-x,

yi(x)~i[\

=4.

14. y' =• j2** + 4x ;r(l) 15. y yi\)

16.y

>(3)

6.

-

x^].

1 1 « + 4* +

;-i(:c) -

2** +

4,

4:..

-s/yTs -

yiix) » 0.

0.

• \^-2

I

-

yi(x)

-3.

-

-3.

1+ ^(4)

-

y('2)

20.

-

1.

=

>(i)

= i:r.

-iix-V^'-4>i.

18. y =

19. y

yi(x)

1.

>i(x)

=ix2 3x2,

+jrj -

-1.

>iW

=

y = , / ? l ± ^ - 3 . :)'(4) -

-12.

yi(x)

-

-3x.

4.6. Some elementary cases of the differential equation , ^ _ ox + by ^ hx + fcy Section 2.8 shows how any differential equation 4.8 may be integrated. T h e solving formula is, however, often such that the character of the family of integral curves is not easily discerned from it. T w o cases are exceptional i n this respect. If a = A = 0, and ^ = — A, the differential equation is simply y' — y/x. For this equation the family of integral curves is the family of straight lines through the origin. I n this instance, therefore, a l l the integral curves pass through the singular point. If b = h, the differential equation admits {hx -\- ky\ as an integrating

The Integral Curves as Trajectories

73

factor, and the general integral found by means of this factor is ax^ +

+

h\xy +

ky^ = c.

T h e integral curves are thus a family of similar conies centered at the origin and having their axes i n common. T h e y are ellipses (or circles) if [ak — bh\ > 0, and hyperbolas together with their common asymptotes if {ak — bh\ < 0. Figure 11 illustrates these cases. I n (a), no integral passes through the singular point; i n (b), two integral curves do so. A n y differential equation 4.8 remains unchanged if x is replaced by mx, andy by my, with any constant m. Such a replacement can be interpreted

Via. 11.

either as a change of scale, or alternatively, as a magnification (or diminution) of the geometrical figure. T h e family of integral curves therefore always appears unchanged when the figure is magnified. Figures 11 clearly have that property. T h e origin is the only singular point of the differential equation; hence every other point has just one integral curve through it.

47,

The integral curves as trajectories

I n cases of differential equation 4.8 other than those dealt with i n Section 4.6, the form of the integral curves can quite readily be inferred without any recourse to the actual solving formulas. This inference could be made, for instance, by using the fact that the straight lines through the singular point are isoclines. W e shall make the inference by considering the integral curves as trajectories.

74

Integral Curves; Trajectories

T h e trajectories that cut the straight Hues through the origin at the angle ^(x, y) have the differential equation y

,

O'A) + tan ^

=

1

—

•

tan 4f

{y/x)

T h i s is equation 4.8 if ax -\- by hx

ky

(y/x) + tan ^ 1 —

0»/*) tan ^

that is, i f k(y/x)^{b-\-k\(y/x)a b(y/^)^-\-{a-k\{y/x)-h

T h e integral curves thus cut any line y = \x at the angle iff given by the formula •^"''^

bX'^ +

[a -

k]\ - h

B y virtue of equation 3.10, we may write (4.9) alternatively i n the form U 10Ì ^ ^

b\^+la~k\\-h

^I'og'-I _

^

de

jtX2+

{b-\-k]\-\-a

where (r, 0) are the polar coordinates of the intersection of the integral curve with the line y = }ix. Thus X = tan 0. T h e discriminants of the quadratic forms i n the numerator and denominator of formula 4.9 are Di=

{b + h]^-4ak,

D2 =

{a -

k]^-{-4bh.

I n this case the numerator of formula 4.9 is not zero for any value of X; therefore the angle ^ is never zero. A l l the integral curves thus cut across every straight line through the origin and so encircle this point completely. If any one of them closes, all do so, for all are similar. T h e y are then a family of ovals such as were shown in Fig. 11 (a). If they do not close they are spirals. A l o n g these the distance from the origin always increases or always decreases if D2 ^ 0, for the numerator of formula 4.10 then maintains its sign. W h e n D2 > 0, the distance from the origin alternately increases and decreases along the spirals. Figure 12 shows families of integral curves of these respective types. C A S E 1.

Di < 0.

I n this case the numerator of formula 4.9 is a perfect square. It therefore vanishes at just one value of X, say X i . T h e straight line y = Xix is an integral curve. This line is therefore not crossed by any other integral curve except at the singular point. A s X —> X i , t h e numerical value given by (4.10) increases indefinitely. Thus r C A S E 2.

Di — 0.

as Trajectories

,

4x-f 9y • x-Ay

^ "

75

8jc+y 7x+2y

(6)

(o)

Fio. 13.

either approaches zero or becomes infinite. If D2 ^ 0, formula 4.10 maintains its sign, and log r always increases or always decreases. If D% > 0, there are 2 lines through the origin upon which ^) is a right angle. O n these lines r has a relative maximum and m i n i m u m . Figure 13 shows families of integral curves of these respective types.

Integral Curves; Trajectories

76

3x+> w Bx-2y y -

—X

6x+5y

(a)

f6) Fio. 14.

2x-Zy X

F I O . 15.

Answers to Problems

77

3. Di > 0, A N D {ak — bk] > 0. I n this case the numerator of formula 4.9 vanishes for 2 values of X, say X i and X2. T h e lines y = \\x and y = X2* are integral curves. They are therefore not crossed by any other integral cur\'es except at the singular point. A l l the integral curves are tangent to one of these lines at the origin, and approach parallelism with the other as r increases indefinitely. Figure 14 shows families of integral curves of this type, (a) applying in an instance in which D-z < 0, and (b) i n an instance i n which D2 > 0. CASE

4. {ak — bk\ < 0. I n this case Di is necessarily positive, a n d there are therefore again two straight integralsjy = Xix andy = Xox. T h e right-hand member of formula 4.9 has a derivative whose denominator is a squared quantity, and whose numerator is a negative definite quadratic form i n X. Thus tan ip always decreases as X increases, that is, a l l the integral curves, except for the straight ones, are convex toward the origin. T h e y have the straight integrals as asymptotes. Figure 15 shows a family of integral curves of this type. CASE

ANSWERS TO ODD-NUMBERED PROBLEMS 1. T h e family of ellipses 2x^ + 3. T h e family of circles 1

+ a

X

-\-+

7. The family of parabolas V y 2 tan ' X- = r. 1

9. X- +—y" 2 -

11. T h e straight lines y = ex. 13. y = 15. 17.

{>

-

+ X

5 , 2 + 4;,

+|.

I p = 4x.

-

\y^\\^^x.

19. y =^\\

-

2x^^ -

= 2cx.

log 11 + ax\ = c.

a2

1

= ^.

8x'}.

-\-

= c ~ x.

CHAPTER

5 Approximate Solutions Infinite Series Existence and Uniqueness of a Solution 5.1. Polygonal graph approximations A graph to approximate the integral curve of a differential equation (5.1)

y

=My),

for which (5.2)

= ^-o,

y{xo)

can be constructed i n the following way: Let XQ, xi, X 2 , " ' * , be any set of abscissas such that XQ < xi < X2 < • ' ' . F r o m (XQ, ^O) draw the straight-line segment with the slope /{XQ, yo) until it meets the line x = xi. Let the coordinates of this point be (JTI, ¥{), and from the point draw the straight-line segment with the slope f(xi, Yi) until it meets the line x = X2. Give this point the coordinates (x2, K 2 ) , and from this second point draw the straight-line segment with the slope / ( j f 2 , J^z). Continue i n this way. T h e graph so constructed is a chain of segments, that is, an open polygon, w i t h vertices at the points (AT,-, Yi), a n d is called a polygonal graph. W i t h YQ = yof we have for this graph (5.3)

Yi+i

=

Yi^fixi,

Yi){xi+,

-

Xi],

I = 0, 1, 2,

. . •.

T h e direction of advance along this graph coincides at each vertex w i t h that of the field of the differential equation. T h e graph may therefore be expected to diverge but little from a streamline if the intervals {xi^i — Xi) are short. It then gives an approximation to an integral curve; moreover, the shorter the intervals taken, the better the approximation. T h e approximation is always best near the point {XQ, yo). A s the construction 78

Polygonal Graph Approximations

79

advances fron^ this point the approximation becomes poorer because the divergences accumulate. A polygonal graph extending to the left from the point (;ifo, ^o) can be constructed in a similar way. Example 1.

T o construct for the differential equation 1

7J

the polygonal graph to the right from the point (0, 0) with [xij^i ~ Xi\ = ^. Also to compare this graph with the integral curve over the interval In this instance = = 0, and formula 5.3 gives Yi = — T ^ . This, together with x\ = -y, yields f{xi, Yi) = —jsF o r m u l a 5.3 thus gives ^2 = — i ^ ' Proceeding in this way, the values of T a b l e 1 may be calculated: Table 1 1

0 Yi fi^i. Yd

1

0 1

3

1 1 7

1 17

3

e 07

—

1

— T7

1 7

— 7^

1 17

1S43

T5T

•ffO 7 3

2 15 4 3

5

5 14 1

7144

20S 5

"807T 6 14 1

^144

The graph obtained is that designated by Y i n F i g . 16. Approximate solutions are primarily useful when the differential equation cannot be integrated. I n this example the differential equation is linear and therefore can be integrated. We may, therefore compare the approximation with the actual integral curve. T h e relevant integral of the differential equation is y =

that is.

-1

.-«V2

ds.

ea:V4

y =

where I{x) stands for the probability integral I{x)

=

ds.

T h e values of !{x) and of e" are obtainaljle from standard mathematical tables. Thus the numerical values of y can be computed. T o three significant figures, they compare with those of K,-, as Table 2 shows.

80

Approximations, Series, and Existence Table 2

y

1

3

T

2

5 T

-0.083

-0.177

-0.305

-0.502

-0.837

-0.087

-0.197

-0.368

-0.676

-1.30

0

1

0 0

T h e graph of > is included i n F i g . 16. T h e approximation is, i n this instance, at most moderately good, because the intervals taken were relatively large. W i t h smaller intervals the number of computations needed

F I G . 16.

to cover the given range would have been larger, but the approximation would also been better. ASSIGNMENTS 1. Construct a polygonal graph for the differential equation

y'-i-y-

h

through the point (0, —1), with [x,-+i — xi] = T , and plot also the integral curve. 2. Construct the polygonal graph for the diflTcrential equation

through the point (0, 1), with {xi+i -

x, and plot

Xi\ =

also the integral curve.

3. Construct the polygonal graph for the differential equation

/

= y +

-

- Xy

through the point (0, 1), with \xi^\ — Xi] ^ \^ and plot also the integral curve. {Note: Fig. 10 applies to this differential equation.)

Approximations for Simultaneous Equations

81

4. Gomtruct the polygonal graph for the differential equation >' =- sin {x +y] through the point (0, T / 2 ) , with

-

— x,} = -j-.

1, (The integral in this case is

y = 2 t a n - l [e^] - x.)

5.2. Polygonal approximations for simultaneous equations T h e method of polygonal graphs can be applied also to approximate the solution of a differential system (5.4)

y=fi{x,y,z),

z'

^f2{xyyyz\

for which (5.5)

z(x(,) = ZQ.

y{xQ) = yoy

T h e integral curves of such a system are found i n Section 4.2 to be three dimensional. T h e y may, however, be studied through their projections upon the (x, y) and the (x, z) coordinate planes. These projections are plane curves, one defining as a function of x and originating at the point (jffl, ^o)) the other defining z{x) and originating at the point (xo, ^o). W e may construct polygonal graphs to approximate these curves. W i t h abscissas xo, x i , X2, • * • , chosen as i n Section 5.1, draw the straight-line segment with the slope fi{xo,yoj ZQ) from the point (xo, >o) to the point where it intersects the line x = x i . Give this point the coordinates (xi, K i ) . Similarly, draw the straight-line segment with the slope fzixot yoi ZQ) from the point (XQ, ZQ) to the point with the abscissa x i , say (xi, Z i ) . F r o m the two points thus determined, draw the straight-line Table 3 1

1

Xi

0

Yi

0.5

0.37

0.19

Zi

1.5

1.58

1.75

/i(x.-, Vi, Z . )

-0.25

-0.36

/2(xi, Yi, Zi) .

0.17

0.34

Yi+i

0.37

0.19

Zi+i

1.58

1.75

-0.52 0.46 -0.07 1.98

3

-0.07 1.98 -0.63 0.70 -0.39 2.33

-0.39 2.33

82

Approximations, Series, and Existence

segments with the respective slopes Yu ^i) corresponding points {x2y Y^) and {x2, -^2)» etc. Example 2.

and / 2 ( ^ i , ^ i , ^ i ) to

T o construct for the differential system y' ==2y + z' =

z-{-

- 3 > -2z

ie^'^ -

h'^^

3,

+ 5,

the polygonal graphs for which ^(0) = ^j, and z(0) = -f, with (xi^i — *,•) T h e method described calls for the computation of the values which T a b l e 3 sets forth.

X

Fro. 17.

T h e graphs Y and Z as they may be plotted from these, and as they appear upon a single plane, are indicated in F i g . 17. T h e actual integration of the differential system is in this instance possible. B y the method of Section 2.9, the relevant solution is found to be y ~

I — se

Z =

I -t e^

-\- se —

—

, •

T h e graphs of these relations based upon the computed values set forth i n T a b l e 4, are also shown i n F i g . 17.

Power Series, Polynomial Approximations

83

Table 4 1

0

3

1

2

0.5

0.34

0.14

-0.16

-0.54

1.5

1.62

1.81

2.12

2.52

5.3. Solutions in power series.

Polynomial approximations

W h e n the function f(x,y) in differential equation 5.1 is representable by its Taylor's series about the point {XQ, yo), the solution fulfilling condition 5.2 can be obtained as a power series. T h e sum of the first {n + 1} terms of that series is an approximating polynomial of the degree n. By the change of variable S

=

X

—

XQ.

u = y -

yo;

condition 5.2 is replaced by w(0) = 0, and the differential equation is transformed into (5.6)

du/ds = / i (s, u).

T h e function f\{s, u) is then expressible i n a series in powers of s and u, namely. (5.7) k.3 - 0

and it is the supposition that this series converges for some range of values of s and u. F r o m the calculus, the coefficients Aj^k are known to be given by the formula L

ds' du

Ju = 0

I n practice it is, however, often easier to find series 5.7 by combining known power series for terms or factors of / i {s, u). A m o n g the most useful of such series are the following: 1 4-r

-

+

(5.8)

¿' = 1 + ^ + 1 2!

^2

+ i r3 _j_ 3!

84

Approximations, Series, and Existence sin z —

r

6_

3!

. . .

^5!

(5.8)

log \a ^ z\ = log a + Let u be taken i n the form (5.9) with undetermined coefficients ÛI, (22, as, * • • . W h e n this is substituted into series 5.7 and the requisite multiplications are made, all terms i n like powers of s may be collected, and their sum equated to the term of that power in the series for du/ds. W e obtain thus a system of equations which successively determine the values of a i , a^, as, * * " . The solution is then obtained by expressing equation 5.9 in terms of the original variables. A n assurance that this procedure actually does give a solution can be obtained only by proving that the power series which it yields for the solution is convergent. T h a t can be done. W e shall not, however, do it here. T h e whole matter of convergence is irrelevant if only the first n terms of the series, instead of the whole series, are considered, that is, if the method is applied to obtain a polynomial approximation. T h e method is applicable also to systems of simultaneous differential equations, to obtain either power series solutions or polynomial approximations. Example 3.

T o find for the differential equation

1 \ — X — y

the polynomial of the fourth degree that approximates the solution for which _y(l) = - 2 . T h e change of variables s = x — \, u = y -\- 2 transforms the given differential equation into form 5.6, with his,

u) =

{2-

is+u)\ —I

By the first equation of series 5.8, therefore,

• | . »

•

•

4

•

Power Series, Polynomial Approximations

85

F r o m (5.9), -1- «

=

(ai +

1)J +

{s +

u)^ =

(aj. +

2.2 + 1) V

(^ +

«)' =

(ai +

l ) 3_3 V +

J

025^ +

OiS^

2^2(^1 +

+

1).^

+

and thus. a2

{ay

4 az

+ 1)" 8

, a^iax +

1)

4

, (¿11 +

+

1)

16

O n equating this to the series for du/ds, and comparing the coefficients of like powers of j , we see that 1

+ 1

2ao = a2

3(Ja = -

4^4

=

,

1)'

(ai +

^3

^2(^1 + 1)

4

4

(^1 +

1)'

16

T h e first of these equations gives a\\ the second then gives «2, the third thereupon 03, etc. Thus, « =

+ A ^ ' + irs^

+

T ^ ^ '

+

• • • -

I n terms of the original variables, the approximating polynomial of the fourth degree is thus

Example 4.

T o find for the differential equation y' = e''y j^x

-

2,

the polynomial of the sixth degree that approximates the integral for which T h e change of variable u = y — \ gives form 5.6 with s = x, and / i = ^*'C«+i) -\-x-2. By the second equation of 5.8, with z = x\u + 1),

86

Approximations, Series, and Existence

therefore,

T o terms of less than the sixth degree this last equation would be /i

^

=

j( ^

^

-\- ^x* -\- x*u -\- • • •

O n substituting form 5.9, and equating the result to the series that

we find

fli = — 1 , 2^2 = 1, 3^3

= 1,

6^6 =

+ ai,

fl3

«

Thus

and the polynomial sought is

Example 5.

T o find for the difi'erential system /

= ^'^^^'\

z' -

(x+jy)» +

z,

the polynomials of the fourth degree that approximate the solution for which ^(0) = 0, and ¿(0) = 0. T h e second and third equations of 5.8 give ^io(v+*) = 1 + s i n ( y + ^) + : i s i n 2 ( y + z) 4 - : ^ s i n ' 0 + z) + sin (y + 2) = (y + ^) - i ( y + z)3 + Sin^

+

^ (y +

- ^(y +

sin^(y + z) = (y + ^)' + whence

••• ,

•••, +

. . . ,

•• •,

Power Series, Polynomial Approximations

87

T h e substitutions y = aix -\- a2X^ + a^x^ + a^x^ + b,x

+

¿2*2 ^ ¿^3 ^ ¿^^4 +

• ••, . . .

therefore yield 2 02 + ¿2 +

and {x+yy

+ z = bix

-f

(¿2 +

(01 +

1)^1*'' + ¡¿3 + 2^2(01 +

1)|X«+

• ••.

O n equating these series respectively to the series for u' and z', we obtain the equations ai =

1,

bi =

0,

2a2 = a i +

¿1,

2b2 = bi, 3^3 = ^2 + ¿2

+

i(oi

+

+ 1)',

3i-3 = ¿2 +

(«1

4a4 = «3 +

63 +

4^4 = ^'3 +

2d2(ai +

[ai +

6 i ] [ o 2 + ¿2],

l),

T h e required approximations are thus found to be yA = x-{'^x^-^

W ix'

+

+ %x\

PROBLEMS Find for each of the following differential equations or systems the polynomial approximation of the fourth degree for the solution specified. 1. v' = > - 6e-',

3. /

=

1 - 2x

y(0) = 3.

y(0) = 0.

2./

= ^ f ' ' y(0)

4. / - 2 ^ + > ,

1

y(\) = 0 ,

88

Approximations, Series, and Existence

5. y'

——? 3x — > — 3

7. y'

>2+8U+2}

8. y'

«2""<'+i'>,

9. y'

10. y' z' 11. y'

-

—

-

12. y'

\ -

>(1) -

3.

6. /

+log il

-

3 + 2 sin X,

:K ~

y(0)

= 0

:v(-2)=0.

j(0) =- 0.

z + Axy,

2 + 2> + 2>2 + xz^ + 2x^, 3 + 4x -x+y-2z,

2> + 4z + Axz, y{0) - 0 .

y{0) = 0,

z(0) - 0 .

:y(0)=0, 2 + log 11 + ¿1.

z(0) - 0.

r ' = I + icy,

^(0) - 0. ;f(0) =- 0,

z(0) -

0.

5.4. The method of successive approximation T h e solution of differential equation 5.1 which fulfills condition 5.2 is a function for which / W

~f{x,y{x))

^0.

B y a quadrature, therefore, y{x) -yQ-

f'f(x,y(x))dx

^ 0.

T h e last equation says that>'(j:) is a solution of the equation (5.10)

y=yo

+

f^'jix,y)dx.

Conversely, any continuous solution of equation 5.10 is a solution of 5.1 and 5.2, as is easily checked by differentiating (5.10) and setting x = XQ. T h e single equation 5.10 therefore effectively stands for the two relations 5.1 and 5.2. W h e n ^ i n the function /(x, y) is replaced by ^IQ, the resulting function /(*> yo) depends only on x and can be integrated. T h e formula yiix)

= >o + j^,/(^.>o) dx

thus defines the function yi{x). W i t h this formula for _>'I(A:), now, fix, yi{x)) is a function of x that can be integrated. Hence the formula yiix)

= >o +

fixyyiix))

dx

defines the function ^2(x). This process can (theoretically) be repeated again and again, and thus the functions of the sequence yMyy2ix)yyzix),

•••,

Successive Approximation

89

can be successively defined by the formula (5.11)

=>o + fjj{x,yn-i(x))

yM

n = 1, 2, 3, • • - .

dx,

W e shall prove i n Section 5.6 that, when suitable hypotheses on /{x, y) are fulfilled, the functions yn{x) converge to a limit when «>, and that this limit function is the integral of the differential equation. T h a t proof will show that the functions^i(x),^2(j^)j>3(^)) * * " , are approximations to the integral. T h e y are called successive approximations. Example 6.

T o find for the differential equation y' =

2
the successive approximations to the integral for which ^(0) = 1. I n this instance XQ = 0, and^o ~ 1- T h u s /{x, yo) = le" - 1, and, by (5.11), w i t h n = 1, fj{2e'-\}dx^ 2e'

>iW = 1 +

-

X -

1.

W i t h this, n o w , / ( x , >'i(x)) = \ -\- x, and by (5.11), with n = 2, y2{x)

= 1+

¡1 +

rfx

= 1 + X + ix\

Continuing i n this way, f{x, yzix)) = 2^* (5.11), w i t h n = 3, yM

= 1 +

2^ -

1

-

A:

-

1 ,

=

dx

V

(1

—

2e'

+

A: + - I X 2 ) ,

-

1

I n this instance, which is simple, the process can be continued. that 2e' yn{x) =

\\-\-x +

1

\-\-x-\--x^-\-

2!

^x'^'j-^,x^-\'

3!

1

-\--xni

^

*

f

*

a n d , by

—

when

! ? when

1

W e find n is odd,

n is even.

It is easy to see from this general formula how the approximation takes place. T h e polynomial part oiynix) is the sum of the first (n -{- 11 terms of the power series for Hence, as n is taken larger and larger, ^ „ ( . 1 : ) converges to e^, which is the solution of the given differential equation and condition. It is only rarely possible to make all the successive quadratures as shown in example 6. A few of them at most are ordinarily feasible, and formulas

Approximations, Series, and Existence

90

can then be found for only the first few successive approximations. Nevertheless, the method is of considerable practical importance, for even the first few approximations are often quite good, and further quadratures, even if they are impossible, can still be calculated numerically by a number of different methods. Example 7.

T o find for the differentied equation y

=

2 -

e-v

\ -\-2x'

the successive approximations to the integral for which ^(0) = 0. Since (xo, yo) is (0, 0), we have /(x, yo) = 1/(1 + 2x), and thus dx

\

; ^

,

= 2log|l+2.1.

Using the last equation,

1 y2(x)=log{l+2x| + | ^ - p ^ -

hence.

1.

W i t h this i n turn. Kx, yzix))

=

1 l+2x

exp

11+2x}

1

-

1 VI

+2x.

and yz(x) = l o g ( l + 2 x 1 -

1 +

1 V I + 2x

exp

1

-

1 V l

+ 2x.

+ 2

T h e quadratures that are required to carry the process farther are difficult or impossible. L e t us, however, compare the approximations we have found with the actual integral, which is^ = log ¡1 + x|. O n the interval — M ^ X ^ 2, the respective values, to three significant figures, are those of T a b l e 5. Tabic 5 — T

0

1 T

\

yi

-0.347

0

0.202

0.347

yi

-0.279

0

0.223

-0.289

0

0.288

0

.If

y

1

5 T

3

7 T

2

0.458

0.549

0.627

0.693

0.752

0.805

0.400

0.550

0.676

0.788

0.886

0.975

1.05

0.223

0.405

0.559

0.692

0.809

0.913

1.01

1.09

0.223

0.405

0.560

0.693

0.811

0.916

1.01

1.10

Successive Approximations for Systems

91

It appears from these values that even ^ 2 W is a tolerable approximation near (ATQ, JVO), and that yz{x) is quite a good approximation over the v/hole interval. PROBLEMS 13. Find the general formula for yn(x) for the differential equation y' ^ y with the condition ^(0) = 3,

6^"^,

14. Find the general formula for ^nW for the differential equation y = ^ + ^, with the condition >(1) = 0. 15. Find the formula for y%{x) for the differential equation y' — > + 2 sin with the condition >(0) = 0. Find adso the integral of the difTcrcntial equation and condition, and compare its Taylor's series with y^{x), 16. Find the formula for > A W for the differential equation y' = cos x — y^ with the condition ^(0) = 1. Find also the integral, and compare its Taylor's series with

17. Find the formula for y\{x) for the differential equation > + — ^ , with the condition ^(0) = 3. F i n d also the integral, and compare its Taylor's series with

18. Find the formula iov yz{x) for the differential equation >' = 7^ + x^y + 1, with the condition ^(0) = 0. 19- Find the formula for yz{x) for the difTcrcntial equation y' = \y — the condition ^(1) — J-^.

A i with

20, Find the formula for^'sW for the diflrcrentialcquationy =^ \y^ — 2y}/{2 with the condition y(0) = 1.

—

2^1,

5*5* Successive approximations for difFerentiai systems T h e method of successive approximations is also applicable to a system of simultaneous differential equations. I n the case of a system 5.4 with the conditions 5.5 it shapes up as follows: T h e solution of the system consists of a pair of functions y{x) and z{x)y which fulfill the equations of the system. By quadratures they are therefore seen to fulfill the equations

z = ¿0+

I'f2{x,y,

z) dx.

JXa

T h e functions yo, ZQ) and fzix, yo, integrated. T h e equations :yiW = >o + zi{x)

=

ZQ

-\- f

ZQ)

involve only x a n d can be

vo, ^0) dx, f2{x, yo, zo) dx

92

Approximations, Series, and Existence

therefore define the functions )'i(x) a n d zi{x). yzix) =>o

+

zi{x)) dx,

j'ji{x,yi{x),

2 2 W = ^ 0 + l'ji{x,yi{x),

define the functions Example 8.

W i t h these the equations

zi{x)) dx

and Z2W, etc.

T o find for the differential system y' = x-\- z,

z' = \ -

y,

the successive approximations to the integral for which ^(0) = 2, a n d z(0) - 1. T h e formulcis that are successively found arc 2H-X,

zi = 1 +

\)dx

jy2 = 2 +

2 + x,

^2 = 1 + j '

\ -\

= 2 +

1

>3

^3=1

jy4 = 2

Z4

= 1

- x\

-

x\ dx = \

-

dx^2

+

-x-ix'^\

x-^^x\

x] dx = \ — X — ix^\

+ 1 +

X

-

3! 1

— X — —

2!

' 1 X* +

— X*.

4!

T h e actual integral is i n this instance ^ = 2 + sin X ,

z — cos X

— X.

Existence of a Solution

93

I n power series form these formulas are y

=2-[-X-~X^-\-~-^X^

-

2!

^4!""

'

• '

6!

,

^

It is easy to see that approximations to these series were obtained. PROBLEMS 21. Find the formulas for yi{x) and Zi(x) for the differential system

y - ^ - 1 ,

z'-2>-z-2,

with the conditions ;'(0) — 2, z(0) =• 0. 22. Find the formulas for;'8(x) and zi{x) for the differential system

with the conditions >'(0) = 1, z(0) = - 1 . 23. Find the formulas for ys{x) and z^^x) for the differential system

z'^y-2,

-z-1, with the conditions ;r(0) =» 3, z(0) = — 1 .

24. Find the formulas for yh(x) and zeW for the differential system y' = —z/x,

z' ^ y -~ z,

with the conditions >(0) = 1, z(0) — 0.

5.6. A proof of the existence of a solution Thus far we have relied upon the geometrical evidence of the integral curves for an assurance that a differential equation 5.1 has an integral fulfilling a condition 5.2, even when no formula for such an mtegral c a n be given. Subject to suitable hypotheses upon the function f{x, y) we shall now prove that such an integral does i n fact exist. Such a proof coidd be given by several different methods. W e shall give it by the method of successive approximations. T h e hypotheses upon /(x, y) shall be the following ones : that i n some region of the (jt, y) plane of which (XQ, yo) is an interior point, (5.12)

(i) (ii)

\fix,y)\<M, {fix,y) -

fix, y*)\

^N\y-y*l

O f these, relation (i), i n which M may be any positive constant, says .that /(x,y) is bounded. Relation Çii), i n which N may be any positive constant, shall hold whenever both (x, y) and (x, y*) are points i n the region.

94

Approximations, Series, and Existence

Equation 5.12(ii) is known as a Lipschitz condition, i n honor of the G e r m a n mathematician R . Lipschitz (1832-1903). It is worth observing that any function having a partial derivative w i t h respect to y fulfills a Lipschitz condition i n any region i n which fy{x, y) is bounded, but does not fulfill such a condition i n a region i n which/^{Ar, y) becomes infinite. This can be seen at once from the mean value relation of the calculus /(^.>)

-=fy{x,yx)\y-y*\,

i n which^1 is some value between^ a n d ^ * . T h e relations (5.13)

X

—

X(i\ ^

A,

y - ya\ ^

Mh,

confine the points {x, y) to a rectangle centered at (XQ, ^'O) - T h e size of the rectangle is determined by h. It is small enough to lie within the region i n which relations 5.12 apply, if A is small enough. We shall suppose such a value of A to have been chosen, and shall henceforth refer to this rectangle as R. I n it the range of variable x is (5.14) W e begin the proof by showing that all the graphs of the functions _y„(x) given by formula 5.11 remain within rectangle R for all x of interval 5.14. This is easily done by mathematical induction. T h e fact is obviously so for^o- W e need therefore only show further that if it is so f o r ^ „ _ i ( x ) it is also so {oTyn{x). Suppose it is so for^n_i(x). T h e n \f{x,yn—\{x))\ < A f , by (5.12), and, by (5.11), (5.15)

yn{x) - yo

f^^ f{x,yn-i{x))

<

dx

M

X ~

XQ

Thus which means that the graph of^n(x) is within R. W e proceed to show now, again by mathematical induction, that for every n (5.16)

yn{x) -yn-x{x)\

<

MN--^

\X

—

XQ

This relation is valid when n = 1, by (5.15). Suppose it is so when n = p. Since both the points (x, yp{x)) and (x, >'p_i(x)) are in R, it follows from (5.12) and (5.16) that J{x.yp{x)) -fix,yp-i{x))\

^ A ' i » ( x ) ~yp-M\

< MN^

' "

"

Existence of a Solution

95

F r o m formula 5.11 taken with n — /» -f- 1, and with n = yp+i{x) -yp{x)

= j^J{x,yp{x))

dx -

we see that dx.

j^J{x,yp^i{x))

Thus yp-\-\{x) -yp{x)

-J{x,y^-i{x))]dx

(MJVPW) X

MN''

—

XQ

Herewith relation 5.16 has been established. range 5.14, and for every n. (5.17)

ynix) - yn-i(x)\

dx

<

= MNP

X

~

XQ

It follows that for all x o n

<

B y virtue of this result, the infinite series of functions ^"0 +

^

bnix)

-yn-l{x)\,

is seen to converge uniformly on interval 5.14. Let us denote by y{x) the function to which the series converges. Since the sum of p terms of the series reduces by cancellations to^p_i(jif), we have then the relation l i m yp_i{x) = y{x)i

(5.18)

p—•

uniformly on (5.14).

CO

W e shall show that this function ^(x) is an integral of differential equation 5.1 and fulfills 5.2. Since each function ^p_i(x) has a graph within rectangle R and is continuous, it follows from (5.18) that the graph of^(j:) is also within R and is continuous. Therefore, by (5.12), fix,yix))

- / ( x , : v p _ ] ( x ) ) | £ N\y{x) - j p - i ( x )

F r o m the relation jy(^) - yo -

rfix,y{x))

dx = y(x) ~ y^ix) -

r

{f(x,y{x))

-

fix,yp^i{x))]

dx.

which reduces by cancellations to (5.11), it follows that > W -yo

-

f^Myix))dx

>W

- ypix)\ 4X

y(x)

-yp-iix)\dx

96

ApproximationSj Series, and Existence

T h e right-hand member of the last equation is arbitrarily small when p is sufficiently large, by (5.18). T h e left-hand member, which does not depend on/>, is therefore zero. Thus^(x) fulfills equation 5.10, and, as we have already seen, it therefore fulfills differential equation 5.1 and condition 5.2. By producing the integral ^(x) we have shown that the integral exists. T h i s same method of proof can be used to show that a differential system 5.4 has an integral fulfilling conditions 5.5, provided the functions fi{x,yy z) and fiix, y, z) fulfill hypotheses analogous to (5.12).

57,

Uniqueness of the solution

For a differential equation that fulfills hypotheses 5.12, the integral for which>(xo) = yo is unique. T o prove this, let both^(x) and rj(x) be such integrals. Both are then functions that fulfill equation 5.10, and therefore y{x) -

nix)

- /J

{f{x,y{x)) - / ( x , i,(x)) | dx.

{x, yix)) and (x, ri{x)) are points for which relation 5.12 (ii) applies upon

some X interval centered at XQ. fix.y(x))

Thus

-f{x,v{x))\

^ N\y{x) -

v{x)

hence, for x ^ XQ, \y{x) -r,{x)\

^ N

fj^\y(x) -r}{x)\dx

This relation implies that the function

Ll \yi^)

-

vix)\ dx

has a non-positive derivative. T h e function therefore does not increase w i t h X . A t xo it has the value zero, and it clearly cannot be negative. It is therefore identically zero, which means that |^(x) — 7i(x)| = 0. T h u s ri{x) is the same asjy(x). Example 5 of Section 4.5 showed that the differential equation y' = 2{x^

-

I P ' - 2x

has at least two integrals for which ^(2) = — 3. T h i s differential equation must therefore fail to fulfill hypotheses 5.12, in a region containing the point (2, — 3). T h e function/(x, y) has in this instance a partial derivative

97

Answers to Problems

This partial derivative becomes infinite as (x, y) approaches the locus -\- y — \ = 0; hence the function f{xyy) fulfills no Lipschitz condition in a region containing points of this locus. Since (2, —3) is such a point, the failure of the hypotheses is evident. ANSWERS TO ODD-NUMBERED PROBLEMS 1. yi{x) -

3 -

3. y,(x) =2x 5. yi(x)

-

3x +

+ 1_4 ix\

-

+ 4x' + i ^ ; c » + ^ l V .

3 -

3U -

1| -

3U -

-

5{x

7. y^ix) - 4{* + 2}^ - I j x + 2}' + ^2 + x3

9. ^-i

i\. yi = X + x^ +

1 13.

ix^ + ix\

- X

yM 6*-' -

15.

2x

3

1 2! 1 -

+

-

;c +

1}" - - V U -

1|'

2\\

-\-x^

z i = x +

x^

-

X*.

1..3 ix\

1

1

3!

n!

1

1

2!

n!

,

x'

when rt is even.

when

« x2 + |. ^3 _^ 2 ^6 _^ ^ ,7^ ^ • ^ — sin jc — cos jc. 7 m x\

17. yi(x) = l i + V ' + Vx' > »• « ^ + 12 — x ) ^ .

T h e Taylor's scries of this coincides with ^8(;r) to terms

+ i^x' + ix* +

- 4*«.

T h e Taylor's series coincide to terms in x*.

19. yz(x) = - + - log X - — log^ X + — log* x - — ^ — l o g ' x. / 2 4** 48 480 ^ 16,128 ^ 1 21. >4 = 2 + x + :^x2 + - X ' ' + - X\ 3! 2! 41 23.

n is odd.

= 3 -

^ x" + x^ 2 4!

26 =

24 =

2X +

- X'. 3

CHAPTER

6 Further Studies of Differential Equations of the First Order The Riccati Equation 6.1. Singular solutions A diflferential equation 5.1 may have a locus of singular points, and this locus may fulfill the differential equation without being included i n the general integral. T h e locus is then called a singular integral, or a singular solution. O n the other hand a locus of singular points may not be a n integral curve at all. Whether it is or not can be determined by testing i t to see whether it fulfills the differential equation. T h e function / ( x , y) i n the differential equation y' — [y — x]^^ has a partial derivative fy{x, y) which becomes infinite o n the locus ^ — X — 0. It therefore does not fulfill any Lipschitz condition i n any neighborhood of this locus. Since the condition for uniqueness of a solution of the differential equation is thus unfulfilled, the locus may be suspected of containing singular points. T h e locus, however, is not, an integral curve, for it does not fulfill the differential equation. Example 1,

Example 2.

F o r the differential equation y' = 2x ~ Sly — x^]^, the locus y — x^ = 0 may be suspected of containing singular points. It is not included i n the general integral, which is y = x'-

Wl"

-

<:]'.

Nevertheless it is an integral curve. It is therefore a singular integral. T h r o u g h each of its points (xo, XQ^) passes, besides this singular integral itself, also the particular integral y =

x ' - W \ x - x o ] '

98

Singular Solutions

99

Figure 18 shows a number of these particular integral curves P and the singular integral F . Since two integral curves passing through a single point must be tangent there, it is clear that a singular solution of a differential equation is an envelope of the curves of the general solution. PROBLEMS For each of the following differential equations find a singular solution, and show that there b also a particular integral through each of its points.

2y

y

3. y = i { x - V ' * ' - 4 y l

5. y

' \ / 3 7 + 7 - 3.

7. y = \/{x + 4 } b - 3 i .

9.

y =

,0. / =

-1

i

-

2

y=

4-

y =

6. >

8.

2 \ / c o s (x + sin (x + ^)

1 - 2 cosM > +

tani > +

y

+1

- VJ

1.

r i

100

Further Equations of the First Order

6.2. Factorable equations A difTerential equation of the first order need not be of the simple form 5.1, but may be more generally (6.1)

F{x,y,/)=0.

T h i s may not be explicitly reducible to form 5.1 at all, or it may be reducible to a number of such equations. I n the latter case it assigns a number of different slopes to a point (x, >), and the integral curves therefore constitute a net of intersecting loci rather than just a simple family. Example 3.

T h e differential equation y a _ 2yy' -

1

-1-^2

= 0

reduces to the two equations y' = y -\- I and y' = y — \. It therefore defines two slopes at any point, and all curves of the two intersecting families y ~ ce' -\- \ and y = ce' ~~ \ are integral curves. W h e n the function F(x, y, y') is factorable, and has the factors Fi{x, y, y'), ^zixj y, y'), • ' • , ^^(x, y, y'), a l l the solutions of the several differential equations (6.2)

Fiix,y,y')

= 0,

F2{x,y,y')

=0,

•- • ,

Ft(x,y,/)

= 0

are solutions of equation 6.1; and conversely, every solution of (6.1) is a solution of some equation 6.2. If the general integrals of equations 6.2 are, respectively, 'Pi{x,yyc) = 0,

'P2{x,yyc) = 0 ,

••• ,

tpk{x,y,c) « 0,

the general integral of (6.1) can be expressed as (6.3)

Wi{xyy,c)\{
=0,

• • • \*Pkix,yyC)\

for this product is zero if and only if one of its factors is so. Example 4.

T o find the general integral of the differential equation yy'^ -

{xyhy^-\-

x]/

+ x^ + xy -

0.

T h e left-hand member of this equation is factorable into the form {/

-x-y]\yy'

-

x\

=0.

T h e given equation is therefore reducible to the equations y^ — x -\- y, yy' = X . Its general integral is, accordingly, {e-^lx -^y'\-\\~c\\x^-y^-c\=

0.

Singular Solutions Again Example 5.

101

T o find the general integral of the differential equation i / ^

_^2j,v'

_ xy'^ +

xy^ =

0.

T h e factors of the left-hand member of this equation yield the equations /

- : v = 0,

y'-\-y

gy-

= 0,

X ^

0.

T h e last of these may be put into the form y' = log x. integral is thus — c\ lye' — c\\y — x\og X

X

T h e general

— c] = 0 .

PROBLEMS Find the general integrals of the following differential equations. 11. xy'^ 13.

{Ax'' + 1 1 / + 4x = 0.

y « + y -

15. 2 x V ' 16.

X^

'^^x

12. e'yy''^ + 14. y - +

0.

+ 12

21..' + 2x^> + V t y +

[xb

+/I

U= + 2>}y2 + |2x> + [x= + 2>l[y - r']]y'

17. y'^ + 2xy' +x^l 19. y ^ -

-4/}

log

|4*^ - y'^]y' X

xlogx

y

-

-\ = 0,

= 0. + 2;o'l> " ^'""1 = 0.

18. y ^ - 4y2 -

= 0.

1

Ay = 0.

2 i y = o.

5 / = 0.

20. y i o g y - x y ' +

[> - / ! l o g y - x ; - ! ! -y\

=0.

6.3. Singular solutions again A diff"erential equation 6.1 may have a singular solution, for its general integral, or some part of it, may have an envelope which is not itself a curve of that net. A t the points of a n envelope the differential equation is fulfilled by fewer directions than at some neighboring points. This can be seen from F i g . 19, in which the circles are the curves of a general solution. T h e i r two common tangents are envelopes, and hence are singular solutions. A t a point Pi the differential equation is fulfilled by two directions which make the angle 6i with each other. A t a point P2 nearer the envelope the angle 62 is smaller, and as the point approaches the envelope at P the angle between the two directions reduces to zero; that is to say, the two directions reduce to one. T h e envelope is thus included among such loci as there may be upon w h i c h the differential equation defines a reduced number of slopes, namely, upon which it has a reduced number of roots y'. T h e condition for that is (6.4)

dF(x,y,y')

ay

0.

102

Further Equations of the First Order

T h e éliminant oiy' between this equation and (6.1) includes any existing envelope. It cannot be assumed that every locus given by the éliminant is an envelope, for there are other loci also upon which the differential equation defines a reduced number of directions. T h e line of centers of the circles in F i g . 19 illustrates this. O n this line only the perpendicular direction fulfills the differential equation. Each locus obtained from the

X

FiO. 19. éliminant must therefore be tested to determine whether or not it is a solution. Example

6.

T h e differential equation for the family of circles i n F i g . 19

is \x -

Viy\ny'^

+

\\ -

{x+yy'\^

= 0.

For this equation, equation 6.4 is =0,

{x-Vly]^y'-y\x+y/]

and the y' éliminant is y\y -

Vlx][Vly

-

x]'' = 0.

T h e éliminant thus gives the three l o c i ^ = 0, )" = y = (l/\/3)x. B y test in the differential equation the third of these loci is found not to be a solution. It is the line of centers. T h e other two loci are solutions.

Integrations by Differentiation

103

T h e y are singular solutions, because the general solution

does not include them. Example 7.

F o r the differential equation { / - ^ 2 i y + / + i =0,

y 2 _

the y' éliminant is = 0. This does not fulfill the differential equation. Thus there is no singular solution. PROBLEMS Find a singular solution for each one of the following differential equations. 21. / y ' - W - ' + T b ' - ' l

22. {3y -7\y^

=0.

23.

{xy' - ; y p - y 2 -

25.

{x^ - y\y''^ - xy' -\-x^ = ^.

-4{y

-2]

=0.

24. xy'* - 2yy'^ + 2x^ = 0.

1 =0.

26. jc^"' - /

+ ^ -

2 = 0

X

27. y 2 -

4xy' + 4> = 0.

28. >' -

29. xy' - y' cos~' (y') + V l 30. f i + / i

-

log \xy' - y\ - 1.

> =• 0.

log {1 + y i - y i i -t-xj 4 - y -

6.4. Integrations by differentiation.

1 = 0,

The Clairaut equation

B y definition a solution of a differential equation 6.1 is a function ^(x) tor which the expression F{x, y{x), y'ix)) is identically zero. T h e derivative of this expression is therefore also zero, that is, F x ( ^ , > W , > ' W ) + F^{x,y{x-),y'{x-))y'{x) + F^{x,y{x),y'{x))

"

^

= 0.

This signifies \S\aXy{x) is a solution of the equation (6.5)

FÂ^yy.y')

+ Fy{x,y,y')y'

- f F^{x,y,y')

= 0,

and that it is therefore a solution of simultaneous equations 6.1 and 6.5. T h i s fact can sometimes be used to accomplish the integration of equation 6.1. If >' is determinable i n any manner from equation 6.5, the substitution of that value for^i' i n equation 6.1 converts that equation into an integral. W h e n equation 6.5 involves too many variables to permit a determination of^', we may eliminate^ between equations 6.1 and 6.5. A value of y' determined from the;) éliminant also converts (6.1) into an integral.

104 Example 8.

Further Equations of the First Order F o r the differential equation 2/ + > -

2 / logy = 0,

equation 6.5 is y' -

0.

2 log/idy'/dx)

This is a differential equation f o r ^ ' whose integral is * —

= jlog^y')^.

F r o m this r e s u l t / = e^^^^^'y and / = e~^^^''. These values convert the given differential equation into the respective integrals

and Example 9.

2e^'l\

-

2e~^^'{\

+

Vx

-

c]y = 0

-

c\y = 0.

F o r the differential equation 2yy'^ + 2x^ = 0,

xy'^ -

equation (6.5) is =0.

[4xy'' -6yy''\^+{6x^-y"] dx

There are too many variables in the last equation to permit a determina tion oîy'. T h e éliminant of y between the two equations is dy' 'Tx

T h e factors of the last equation give, respectively, y' = ex, and y'^ = 6x^. These values convert the given differential equation into, the general integral c^x^ -

2c^y + 2 - 0 ,

and the singular integral 9y^ = 4 V 6 ; f 3 _

There are two types of differential equation 6.1 to which this method is peculiarly well adapted. (i) T h e equation that can be made explicit for y, namely, (6.6)

> =/(^,/);

(ii) T h e equation i n which x and y occur only i n the combination [xy' - y\, namely, (6.7)

f(x/

-y,y')

=0.

A n equation 6.7 is called a Clairaut equation i n honor of the French mathematician A . C . Clairaut (1713-1765).

Differentiation with Respect to y

105

For a Clairaut equation, equation 6.5 is simply {dy'/dx) • Fj/(xyy, y') = 0. T h e first factor of the last equation gives y' = c. A Clairaut equation is thus converted into its general integral by merely replacing y' by c. Example 10.

T h e differential equation \ =0,

W - y \ ^ - y ^ -

is a Clairaut equation. Its general integral is therefore \cx — y\^ — 1=0. Equation 6.5 is i n this instance

dx

Uy

-y'

-"y]

—

=0.

T h e second factor of the last equation gives ^' = xy/jjc^ ~ this value of y' the differential equation is converted into the singular integral x'^^y''

= 1.

PROBLEMS Solve the following differential equations. 31. xy' -y

+ / 2 = 0.

32. 1 6 / -

—

{xy' -y\^

33. 2y - A ~ log {y^+ 1} = 0.

34.

35. xy'^ - yy'"^ + 2 = 0 .

36. / 2 Jr\x-

37. 2y - 1

1

38. y

V^y"" + 1 39. y< - x.2..»2 y + "^.xyy' - y

0.

y — log y' =» 0. \\y' -y-\.

1 = 0.

-Ixy' .'2 -

1

40. \zx - 1 l y ^ - 2{> -

41. 2x>./2 = 4 x V ' + 1.

43. 16{:c + 1

X

- 0.

1 l y + 4 =• 0,

42. i l -\-r^']y = (y - 1]*.

- 4{4> - ! } / - !

= 0.

44. ^ + 2 + 2 1 o g (y/2} - xy' = 0.

45. Axy - 2xy'^ + ^

= 0,

46. y * + l / + > } { 4 y 2 + l | - 0 .

6.5. The method of differentiation with respect to / If we think of ^ as the independent variable, and recall that dx/dy = 1 / y ' , the rccisoning of Section 6.4 shows that any solution of a differential equation 6.1 is also a solution of the equation obtained by the differentiation of (6.1) with respect t o ^ , namely, of the equation (6.8)

FAx,y,y')

~ + Fy{x,y,/) y

+ FAx^y,/)

7^ = 0. ^y

A n y value o f y that can be determined from (6.8) converts (6.1) into an integral.

106

Further Equations of the First Order

If y' cannot be obtained from (6.8) because too many variables are involved, we may eliminate x between equations 6.1 and 6.8. A value of ^' obtained from the x éliminant also converts (6.1) into a n integral. T h i s method is peculiarly well adapted to a differential equation that can be made explicit for x, namely, (6.9)

^ =/(>,>').

Example 11.

T o integrate the differential equation ilog{l

- l o g / - x

+ 2 = 0.

T h e derivative of the given equation with respect tojy is 1

dy' r 2 - ~

1

dy

+ i

and integrates to give y' = tan {c — y\. differential equation into its integral log sin

^1

(c —

+

=0,

This value converts the given

X —

2 = 0.

Example 12. T o integrate the differential equation

| x 2 - 2 x l { y 2 + l) + 1 = 0 . T h e y derivative of the given equation is . . lv'^ + 1 {x^-2xiy-<- + | x - i !

= 0,

dy

and the éliminant of x between the two equations is '2

y

+ 1

d/

-, + V y 2 + 1 = 0

F r o m this we obtain y' = ± v {1 + (y + f^) ^ Ì / 1 ^ + c}, and these values convert the given differential equation into its general integral x^ - 2x +

+

= 0.

PROBLEMS Solve the following differential equations. 47. 2xy' = y -

yy'\

49. 2xyy' = y'^ + 4y~. 51. xy' + 2 log /

+ 2 = 0.

48. x = 25>V" + ~ 5y 50. x = 5 + log \y' + y/ì

+ y'^\

The Riccati Equation

107

52. 3 / V l - JV'^ + 3 s i n - i 1/1 + 2JC = 0. + Sxy' -

53.

2> = 0.

54.

-

dAxy' + 32> = 0.

Ó.Ó. The Riccati equation A differential equation (6.10)

/

= io(^) + ii(^)> + ? 2 ( x ) / ,

i n which q2{x) ^ 0, is called a Riccati equation, i n honor of the Italian mathematician Count Riccati (1676-1754). T h e reciprocal of any solution of this equation, w = l / y , is a solution of the associated equation (6.11)

w' =

-q^ix)

— qi{x)w — qo{x)w^.

T h u s the integrals of either one of these differential equations c a n be obtained by accomplishing the integration of the other equation. A n equation 6.10, or 6.11, may be of a type that is integrable by a method of Chapter 2. W h e n put into form 2.1, it may be linear, or it may have separable variables, or it may have coefficients that are homogeneous and of the same degree. Except for these, and a few other rare elementar)' cases, the Riccati equation is not integrable by any systematic method other than that of power series. It is sometimes possible to find a solution of a Riccati equation by trial. One may try such functions as ^ = ax^, or y = a^^, with undetermined constants a and h, or some other function that the appearance of the differential equation itself suggests. Success i n finding a solution is rewarding, for, if any integral can be found, the general integral is obtainable from it. Suppose ^I(x) is any particular integral of diñ"erential equation 6.10. T h e change of variable (6.12)

y =yi{x)

-

u

with an arbitrary constant cu transforms the equation into A _L ' ^ i " ' _L r \ yi {x) = qo-\- qiyiix) u

'^^ii ^ 2/ ^ + 9 ^ 1 [x) Ü

2ciq2yi{x) a

i

^• u

B y such cancellations as can be made because yi{x) fulfills (6.10), this transformed equation reduces to +

{qi + 2q2yi\u = ciq2.

This differential equation for u is linear. of a form

Its general solution is therefore

U = Ci
108

Further Equations of the First Order

i n which tpi{x) and
y ^ y i i x ) -

Cl(pl{x) + C2
This is the general solution. A n y particular solution obtainable from it is determined by the ratio of ci to C2* This ratio is therefore the single arbitrary constant which formula 6.13 actually involves. Example 13.

T o integrate the R i c c a t i equation

y = j1 +

X +

2x^ cos A: I

{1 + 4* cos

—

+ 2y^ cos x.

By the trial of> = ax^ it is found that the given equation h a s ^ i = * as a solution. T h e change of variable y = x — Ci/u transforms the equation into u' — u = 2c\ cos X . T h e general integral of the last equation is u = ci{sin X — cos x\ + C2^\ Hence the general solution of the given equation is c\ >

Example 14.

=

X

-

f T l j s i n X — cos

x|

+

T o integrate the R i c c a t i equation /

=e^'+

|1

+^e'\y^y\

By the trial oiy = ae^' i t is found that general solution as found by the use of is

— —^e' is a solution.

1

c,e^

2

1^1 + c^e-^*'^^

The

A solution found by trial may be complex, say, y, = a{x) + ;-^(x), with i = is then

— 1.

(6.14)

T h e general solution, expressed wholly i n real terms,

y = a(x) + ^(x) tan {jq2{x)0{x)

dx-\-c].

This may be proved as follows: Since a + i/3 is a solution, we have a' +

i^' = ?o + qiW

+ m

+ 921«^ + 2ia^ -

^'}.

109

The Riccati Equation

F r o m the last equation we see, by equating real and imaginary parts, that (6.15)

a' =

90 +

qxct + q^la^ -

/S^),

^' = q,^ + 2 ^ 2 « ^ .

T h e result of substituting evaluation 6.14 into equation 6.10 is a' + /3' tan 6 + yaS^ sec^ 0 = ?o +

+ q\& tan B + q^oi^ + 2?2a/3 tan B + jaS'^ tan^ ^,

where B = {fq2{x)^{x) dx -\- c}. 6.15. Example 15.

This is fulfilled by virtue of relations

T o integrate the Riccati equation /

= 1 -y

+

e^y.

By the trial of ^ = ae^' the given equation is found to have the solution yi = i>~*. T h i s is of the form above with a = 0, and /3 = e~'' T h e general solution, by (6.14), is thus y — e~^ tan {e^ + c\.

PROBLEMS Find the general integrals of the following differential equations. 55. /

V 1 5 + 4 ^ + - ^ .

=

3^ cos x\ — Ixy cos x

56.

cos x.

= 3f*' -Vly - 12>^

57.

58. xy' => [ 6 - 4 log x} -

[5 -

4 log x\y + (1 - log x j / .

59. _y' = 3x' + - + 3x).2. X

61. /

2 - 2x'

= 2x* +

y -Vy

^ - 1

60. ^'

=

62.

=> sin^ X + _)> cot X

X

+

+

Find the integrals of the following differential equations that fulfill the given conditions. •

63. y'

1

2 +X

64. y = 9x + {3 65.

y

66. >' -

X

I S x b + (9x -

•X

tan

-7

X +

sin

y{-\)

= ^

..2

y

1 +x^

31/,

x l l +x^j

X

cos X

'

>(1) = 0.

- y"^ cot X ,

;y ( -

0

=

1

110 67.

Further Equations of the First Order - 5** + 6*^7 + 2**/,

68. y' "x'

69. /

-\-

= tan

-f.

X

X + S i n jr

70.

;r(0) -

cos Jc

+ /

^-(0)

cot x,

( - ^) = \ 4/

-

-1.

6.7. Some properties of the integrals of Riccati equations If xo is any point of an interval (6.16)

a-^x-^b,

upon which ?o(*), ?i(*)» and 5 2 ( ^ ) are continuous functions, the point ( ' 0 ) ^ 0 ) with any ordinate ^yo lies within some region i n which the function i{x,y)

= qo(x) +

qi(x)y + q2{x)y^

and its partial derivative y) are bounded. T h e function therefore fulfills a Lipschitz condition 5.12, and accordingly every point (xo, ^o) has just one integral curve passing through it. A Riccati equation therefore has no singular solutions. A R i c c a t i equation may have some integrals that are continuous over the whole of an interval 6.16. However, it always has some that are discontinuous, a fact that is easily seen. For, with any chosen XQ, differential equation 6.11 has an integral «;o(^) for which wo{xo) = 0. T h e reciprocal of wo{x) is an integral of equation 6.10, and it obviously has the vertical line X = xo as an asymptote. Example 16.

T o find the solution of the Riccati equation

which is discontinuous at x = 1. T h e differential equation has constant coefficients. constant solution, and by trial this is found to bejJi solution, as found by the use of^"! = 2, is y = 2

—

It therefore has a 2. T h e general

-

T h i s is discontinuous at x = 1 if ci + 12^ = 0. thus -

1

T h e required solution is

Integrals of Riccati Equations

111

T h e R i c c a t i equation has important relations with the equation (6.17)

pMu"

+ pi{x)u' + p2{x)u - 0,

which is the linear differential equation of the second order. Thus if i;(x) is any solution of the R i c c a t i equation

Po{x}^ PQ{X)

on a n interval upon which po{x) ^ 0, the formula (6.19)

u = el"^'^'^'

gives a solution of equation 6.17. Therefore equation 6.17 can be solved by solving equation 6.18. Conversely, if v{x) is any solution of the differential equation (6.20)

v" -

v' + qo{x)qi{x)v = 0,

qiix) H — 7 qiix)

which is of type 6.17, the formula (6.21)

=

-

q2{x)v{x)

gives a solution of Riccati equation 6.10. Example 17.

T o find a solution of the differential equation u" -

{x +

l j u '+

{x -

lju =

0.

T h e related Riccati equation 6.18 is r,' = ~{x-\]

+

{x-\-\U

- ri\

It has 71 = X as a solution. By formula 6.19 the given differential equation therefore has a solution u = e'^^^. Example 18.

T o find a solution of the R i c c a t i equation

,

2H-X x[l -\- x\

2 x\\

X

—

x^

12

X

T h e related equation 6.20 is

^

x{\ -\-x]

x\\ + x l

T h i s has f = as a solution. By formula 6.21 the given equation therefore has a solution)' = — 1 / | 1 +

112

Further Equations of the First Order PROBLEMS

71. Find the solution of the equation 1

2 x

which 15 discontinuous at * = 4. 72. F i n d the solution of the equation 3 4 - -

1 2 - -

x

which is discontinuous at x

x

1

73. Find the solution of the equation

for which y{0) — -a".

Determine its discontinuities, if any.

74. Find the solution of the equation x'y + y\ for which ^(1) = 0.

Determine its discontinuities, if any,

75. Find a solution of the differential equation

76. Find a solution of the differential equation u " + u' log X + -

=0.

x

11. Find a solution of the differential equation u " — u' cos X + u sin

0. x

=

78. Find a solution of the differential equation u"

—

2u' tan X

— u —

0.

ANSWERS TO ODD-NUMBERED PROBLEMS 1, y' = X-

T h e general solution is 5

3- x^ — 4^ = 0. 5. ^ = — 3x, 7. ^ + 3.

—

x}^^ = x^ + f.

T h e general solution is \ / x ^ — 4)^ = x + c-

The general solution is 2 ' s / 3 x + ^ = log x +

T h e general solution is 3 y/y

— 3 =

9. ^ = (n + ^jw — X, with any whole number n. The general solution is

cos (A: +

= x + c.

+ {x + 4 p

Answers to Problems {y - 2x^ - c][y - log X - c\

11. 13.

xy --x^

log

V-^-c

15.

— h 4 log X y

- c

2x X

+

y -

113

0. = 0,

— c

= 0

— c

X

11 J *

19. y\21y ~ 5(x 23.

= 0.

2S.y

= 1.

x2

- > = 0.

31.

— ^ + tf* " 0,

and ^ =

^ = 2 — log cos \x + ¿1,

4;- + Ix -

= 0,

39. c* - c^x^ 4- 2£xy 41. ^

-2^^

+ ^

= 2cx + c\

51.

59. jvi

IX,

61. yi

{1 + i l x ,

and 2y^ = x. 49. 16^ = c(2x - c\\ 2\/y

y =

3x -

X

57. :y fi +

tan {x^ + c]. jv = x 2

X

+x2tan

{-i*' 65. > =

11

- 3

69. y

X

~ X

* + l tan

X

cot

X +

-8x X -

4' 1

•> = ' ' + ? T - i =* cxp

and 27^ = 2x*.

+ c + 2] = 0 .

2

67. > - 7 tan f*' -

u

and 14;^ + 1 p + 4|x + 1} = 0.

and 27>'* + 125x' = 0.

3 x - x 2

75.

0,

Scix^

-2x -

71. >

1 -

+ 2 H x * - v ^ -

=0,

55. y -

63.

1} -

= 0,

f ^ f V ^ + 2 \/>

3 +

and > = 1

and 1 6 / = xV

and > - 0.

53. 5 « - 2 / ^ +

73

- 0,

4c{4y -

43. 16c^{x + 1} -

47.

and y = 0,

K> and 2;»' = 27x.

2c - s / x

45. 4c^ Vx

and y = 2.

and 2y^ = 27x^.

35. c^x - c > + 2 = 0, 37. 4 / -

= x ' ~ l

29. y = sin x.

27.

33.

=x-V\.

21.

It has no discontinuitiefl. 77. u

lin X

2

1:2^'''*'

CHAPTER

7 Some Solvable Differential Equations of the Second Order Approximations.

Applications

7.1. DifFerential equations of the second order A relation (7.1)

which actually involves^", is an ordinary differential equation of the second order. T h e simplest type of such equation is that of the form y"

=

fW-

B y successive quadratures the last equation yields

and i n which ci and C2 are arbitrary constants. A solution of a differential equation of the second order is called the general solution, or the general integral, if it involves two arbitrary constants i n such a way that these constants are not replaceable by just a single constant. A n y differential equation of the first order that is obtainable from the given equation of the second order by an integration, and that involves a n arbitrary constant, is called a first integral. W h e n a first integral c a n be found, the problem of integrating the original equation is reduced to that of integrating the first integral. F o r that, the methods and observations of the preceding chapters can be drawn upon. W e shall consider a few types of difi"erential equations 7.1 for which first integrals can be found. 114

Equations in Which Either ^ or ;c are not Present

115

7.2. Equations in which either / or x are not present A differential equation 7.1 in which y is not present, namely, F{x, y\y") = 0, is i n effect F(x, y\ dy'/dx) = 0, that is, it is a differential equation of the first order for^'. A n integration of the equation is therefore possible and yields a first integral f{x, y\ c) = 0. Example 1.

T o integrate the differential equation \x + 2]y''-\-y'^

-

1 - 0 .

As a differential equation for^' the given equation has separable variables. A n integration gives

y + 1

that is,

>» = 1 +

T h i s is a first integral. integral

{x-\-2]'-ci

B y a quadrature of the last equation the general

[x -\- 2 + ci

+ C2

is obtained. A differential equation 7.1 in which x is not present can be treated quite similarly. Since dy dy' dx

dy

,dy' dy

the equation Fiy, y\y") = 0 is, i n effect, y\y' dy'/dy) = 0. This is a differential equation of the first order for^' as a function of;'. A n integration of it as such yields a first integral f{y, y', c) = 0. Example 2.

T o integrate the differential equation y"2

y

yy" y

By setting^" = y' dy'/dy, the form of this equation is changed into dl dy

dy

This is a Clairaut equation 6.7 f o r ^ ' as a function oiy. It therefore has the first integral c\ + c i j — >' = 0, and the singular integral^' —i:)'^.

116

Solvable Equations of the Second Order

Integrations of these relations yield the general integral y and the further integral)» = 4/¡jt + f j .

= Í2Í^^' —

íi,

PROBLEMS Find the general integrals of the following differential equations 1. xy" + y - * -= 0.

2. xy" -

3. y" - /

4. yy" + {1 + 2 / | / 2 -

V l +

5. xy'' + xy''' + / 7.

= 0.

'

+

2yy' = 0.

y \y' = 0.

8. 1 + xy" - y' ~ y"^ = 0 . /

A\ - 0.

6. {;y + 1 b " + {> + 2 b ' 2 - 0.

cos ^ + sin ^ \y" + {2 cos y — y

10. y 3 _ ^ y y ' +

{1 + x'^Wy'

/

9. ¡1 "

- / ^ J - 4 / / = = 0.

= 0.

7.3. The differential equation of a 2-parameter family of curves. Checking a general solution A family of curves whose equation (7.2)

^{x,y,cuC2)

-

0

involves two arbitrary constants is called a two-parameter family. A n y curve of this family characterizes)' as a function)'(x), for which the expression ^{.x, y{x), c\y C2) is identically zero. T h e first and second derivatives of this expression are therefore also zero. Hence y{x) also fulfills the equations ^x{xyyy ci, C2) + ^y{x,y, c i , C2)y' = 0,

(7.3) *z.x(^,:y, ci, C2) + 24>x.v(^,>, cu C2)y' H- %,v{x,y,

€2)/' = 0,

that is, y \s a solution of simultaneous equations 7.2 and 7.3. T h e éliminant of ci and C2 from these three equations is the differential equation of family 7.2. T h e final step of elimination i n this finding of the differential equation can be avoided by arranging the several equations so that the differentiations themselves bring about the eliminations. Thus if equation 7.2 c a n be made explicit for one of the constants, that constant is eliminated by the differentiation, a n d does not appear i n the first equation 7.3. If that equation can then be made explicit for the remaining constant, the second differentiation completes the elimination. T h e final equation 7.3 is then the differential equation of the family. Example 3.

T o find the differential equation for the family of a l l circles having the radius 3.

The Differential Equation of a Family of Curves

117

T h e coordinate equation of this family is (7.4)

= 9,

\y-C2]'

and the derived equations 7.3 are thus

1

= 0.

{y-c2]y"

T h e éliminant of ci and cz from these three relations is the differential equation (7.5)

9/'2_ {l+y2j3^o

If equation 7.4 is first made explicit for C2, thus,

the first differentiation yields

V 9 -

{x-

ci\

If the last equation is now made explicit for ci, thus, x ± 3y'/y/l y'^ ci, the second differentiation yields, without any further step, the result

(1

+y'T

=1.

T h i s is, of course, the same as result 7.5. T h e graph of the general integral of a differential equation 7.1 is a twoparameter family of curves. T h e correctness of an integration of such a n equation m a y therefore by checked by using the integral obtained to re-derive the differential equation. Example 4.

T o check that y^ of the differential equation

—

cix^y

—

cz = 0 is the general integral

T h e derivative of the relation given is 2y/

-

ci{xy2xy}

= 0.

If this is first made explicit for c i , the second differentiation yields

118

Solvable Equations of the Second Order

T h e denominator may be dropped. Since the result is then the given differential equation, the questioned relation hcis been checked to be a n integral. PROBLEMS F i n d the differential equations of the following families of curves. 11. ylogy

=cix +C2.

13. y -

12. y =- {cix

+ C2\^.

15. ciy + Ci+

log \xy\ + 0.

14. y -cie'

+ C2X^ -

16.

2

17. log 19. 7 -

+/} X -

21. [xy]*

= ci t a n - i (y/x) + C2.

18.

+ ci\x - y\ + <:2 = 0.

2cix -

0.

2ciy ~ 0.

2

- ^

20. y =

cie'**.

+^2}^

+ 1 -

0.

,«1-io•««

22. Aaxy-\-c^y/x

"

0,

ASSIGNMENT 1. Check the general integrals of the differential equations of odd-numbered problems 1 to 9.

7.4. Exact equations A differential equation 7.1 is said to be exact if its left-hand member is exactly the derivative of some function y, y'). Because such a derivative is linear i n y", only a differential equation of the form Pix,yyy)y"

-hQixyy,/)

=0,

that is, (7.6)

P(xyyy/)

dy' + Qix^y^y') dx = 0,

can be exact. W e shall derive the condition under which such an equation is exact. Let
iP(x,y,y')dy'

^v{x,y,y'),

i n which the quadrature is made as though y were the only variable, that is, as though x and y were constants. T h e n Pix^y^y') =
dy' = dip {x,y,y')

-

iPx(x,y,y') dx -

tpyix, y,y') dy.

Equation 7.6 may then be written as (7.8)

d,p(x,y,y')

+

{Q(x,y,y')dx

-

tp,ix,y,y')

dx -


0.

Exact Equations

119

This is exact if, and only if, the expression within the brace is a differential, that is, if, when y' is replaced by dy/dx, the differential equation of the first order obtained by omitting the first term of (7.8) is exact. T h e test for such exactness, and the integration of the equation into the form V'i(^) y) = c if it is exact, are given i n Section 2.2. If differential equation 7.6 is exact, therefore, it has the first integral w {x,y,y') +
5. T o integrate the differential equation \2y -

\ }y" +

+ /

- 2 > / - 2 « * - 2e'] = 0.

F o r m u l a 7.7 gives (p(xy y, y') = 12>' — 1 \y\ and with this evaluation the given differential equation can be written i n the form d\{2y -

\)y'] + ! / ~ V

- 2xe' ~ 2e'] dx = 0.

B y omitting the first term of the last equation and replacing)"' by dy/dx^ we obtain the equation -

{2xe' + 2e'\ dx -

\2y -

1\ dy ^ 0.

This is exact and has the integral — 2xe'

+

y

—

y^

= ci.

T h e given differential equation therefore has the first integral {2y ~

\]y' -

2 x e ' y

-

y^ = ci.

This c a n be integrated again. T h e change of variable u = y^ — y transforms the last equation into u' — u = ci 2xe'j which is linear. T h e general integral of the given differential equation is thus found to be y

~

y

—

X

e* -\- c%e

— C\.

PROBLEMS Determine which of the following differential equations are exacts and find the general integrals of those that arc. 23. {1 -x^\y" 24. —

— +

^\\

- 3 x } y - : y = 0. ^

{1 + x\e = 0.

25. xy" + {3x/ + 2 i / + { 2 / + 2 }

=0.

26. y" sin x + Isec x + 2> cos x]y' + tan x = 0. 27. {x co&y + sin x\y" — xy''^ s i n ^ + 2{cos^ + cos x\y' — ^ sin x + 0.

120

Solvable Equations of the Second Order

28. xyy" + xy'"^ + yy' - - - 0. X

1 + xy 30.

2y" yy

31.

y' log y

y

1+ _\

y"

—

1

xy

xy

= 0,

+ - - 2 = 0 . y

32. y" cos ^ + {cos^ •- y' tin. y\y' — 2x =• 0.

7.5. Simultaneous equations A differential equation (7.9)

y"

= Six, y, y'),

is replaceable, i n an unlimited number of ways, by a pair of simultaneous differential equations of the first order. L e t / i ( x , y, z) be chosen as any function that actually involves z, and set^' = y, z). T h i s last equation, its derivative equation, and equation 7.9 together make up a system of three equations from w h i c h and y" can (theoretically) be eliminated. T h e éliminant is an equation z' = /2(x, y, z). Equation 7.9 has thus been replaced by the differential system (7.10)

/

^fiix,y,z),

z'

^h{x,y,zY

O n e particular way of making such a replacement is to take / i = z. T h e system obtained is then y' = z, Example 6.

z' ^ f{x,y, z).

T o replace the differential equation y" cosy + {cos;; — y' sin;»)^' — 2xy = 0

by a differential system. T h e choice y' = z gives the derivative equation y'' = z'. and the elimination of^' andy" between these and the given differential equation yields the éliminant z' cosy + z cosy — z^ sin y — 2xy = 0.

T h e system obtained is thus y' = z,

z' — 2xy sec y +

tan y — z.

T h e alternative choice, y' = z sec y^ replaces the differential equation by the system y' = z sec y,

z' = 2xy — z.

Simultaneous Equations Example 7.

121

T o replace the differential equation

1+

1 + xy

xy

by a differential system, by setting ¡1 + x^y'] = z. T h e derivative of this assigned relation is {x'^y" + 2xy'\ = z', and the éliminant oî y' and y" from the three equations is {yz'/xz) — log z = 0. T h e system obtained is thus y

/

-

^

= — ^ » X

/

xz

z = — log z. y

Conversely, any differential system 7.10 is replaceable by a single differential equation of the second order. T h e two equations 7.10 and the derivative of the first of them constitute a system of three equations from which z and z' can (theoretically) be eliminated. T h e éliminant is a differential equation 7.9 for y. If this equation is integrated, y and y' are thereby determined. T h e first equation 7.10 is then an equation for 2. A system 7.10 can thus be integrated by replacing it by a differential equation 7.9 and integrating that equation 7.9. Example 8.

T o integrate the differential system y = 2 - 3,

z' =

-z^

3x+> + l

T h e derivative of the first of these equations is j " = z', and the éliminant of z and z' from the three equations is l 3 x + : v + l i / ' + ( 3 - / | 2 = 0. This differential equation of the second order is integrable because it is exact. A first integral of it is !3x+>4-ii!3-y!

=cu

and its general integral is {3x +JV +

1)2 =

2cix^-C2-

B y these relations )? andjy' are determined. Therewith the first one of the given equations determines z. T h e solution of the given system so found is

122

Solvable Equations of the Second Order PROBLEMS

I n t r a t e the following d i y - z.

33.

34. y' 35. / 36.

z' - 2^-" - z\ y sin X — z cot

sin X - z - 2*, -

systems.

z' -

- ( l / x ) { 2 ; f + 2«* + 4 « )

>+2

-

37. y = z

í ' « ^ - ^ + s e c {z - > Í .

38. ; r ' = e\

z' = í " " .

Replace each of the following differential equations by a differential system, by using the respectively assigned relation. x'-y'

39. y

+

40.

-

41.

— / 2 + (2> - 1 ly** - y {1 + í ^ ' ) -

y' -\-xy'' -2x

y j 1 + tan

42. y" -y

' - ^ -^

43. y" - yr^' 44. y" ~y

X

y' = x^ - yz.

+ cot x\y' — y tan x » 0,

=0,

- 2 = 0,

V i - ,

= 0,

/

^' = z tan x.

= 0, y - y/* + z.

=;r + z + l.

=. log z.

y

j'''^ eos > «=" 0,

—

sin z.

7.Ó. Solutions in power seríes.

Polynomial approximations

W h e n the function /(x, y') is representable by its Taylor*s series about the point (xo, >o, >o'), the integral of differential equation 7.9, for w h i c h (7.11)

y(xij) =:>o,

y'(xo)

yoy

can be obtained as a power series, Polynomial approximations of the integral are thus also obtainable. T h e method is that of Section 5.3. W e seek an integral i n the form 7

= flo +

ailx

—

XQ] 4-

azlx

—

XQ]^ +

asix

—

XQ]^

+

•

#

*

Conditions 7.11 require that Jo = >o, and ai — yo'. T h e remaining coefficients are determinable by substituting the series (ory into the differential equation, a n d equating the coefficients of like powers of {x — xo]. Example

9.

T o find as a power series the integral of the differential

equation y"

=

, y

_

y

/2

Approximate Integral Curves

123

which fulfills the conditions y{0) = O.>'(0) = 1. Since XQ =

ya = 0, and yo' — 1, the series sought is i n this instance

T h e substitution of this series into the differential equation converts the equation into 2(i2 H" (ia-^ +

\7.a^x^ +

* ' '

=

— 1 — 4a%x

+

|1 -

6^3 + 4 ^ 2 ^ ! : ^ ^ +

•••.

T h e equating of coefficients of like powers of x yields the relations 202 = - 1 . 6fl3 =

—4(22)

12^4 = 1 - 6fl3 + 4a2^

T h u s ai = —^, ^3 =

• • • , and accordingly,

= —

PROBLEMS Find (up to terms in ;c^) the integrals of the following differential equations that fulfill the given conditions.

45. y" = xy' sin y,

46.

y{^) = 0, >'(0) -

2.

= jy + cos {xy), ).(0) - 0,y(0) = - 1 .

47. y =

+ /

+ 4, >(o) = o,y{o) = 0.

48. / ' = { / ^ x y p . 49. y = >' log:y + X .

:v(o) - o , y ( o ) = 2 . :yCO) = 1,>'(0) = 3.

50. y =. 1 + {jy - 2i^!'^2|^

^(oj ^ 2.y(0) = - 2 .

7.7. Approximate integral curves W e shall give two methods by which a graphical approximation to the integral of a differential equation 7.9 fulfilling conditions 7.11 may be obtained. 1. T h e differential equation may be replaced by a differential system 7.10. T h e relation _)'o' = fiixo, yoi ZQ) then determines ZQ. T h e polygonal graphs that approximate the^- and z of the resulting system METHOD

124

Solvable Equations of the Second Order

are then constructable upon the basis of any chosen set of abscissas x „ b y the method of Section 5.2. O f the two graphs so obtained, the one through the point (xo, yo) approximates the designated integral curve. Example 10.

T o construct a polygonal graph to approximate the integral of the equation / ' + { 2 x - h 2 ! y + l 2 x - l ) > = 0, for which

>(0) - 0,

jv'(O) = 2.

T h e given differential equation and conditions are replaceable by the system y = z, z' = - ) 2 x - l b - { 2 x - | - 2 l z , w i t h the conditions ^(0) = 0, ^(0) = 2. O n the basis of abscissas w i t h the differences {x.+i — x,j = i , the method of Section 5.2 yields for Yi and Z,- the calculated values of T a b l e 6. Table 6 1 T

0

1

8 T

5 T

1

3

7 T

9

2

1

6

Yi

0

0.5

0.75

0.88

0.91

0.89

0.83

0.76

0.68

0.59

0.51

Zi

2

1

0.54

0.14

-0.09

-0.23

-0.30

-0.34

-0.35

-0.34

-0.30

Yi'

2

1

0.54

0.14 - 0 . 0 9

-0.23

-0.30

-0.34

-0.35

-0.34

-0.30

Zi' - 4 y

0

-2.25

-1.62

-0.93

-0.55

-0.30

-0.16

-0.03

0.06

0.15

0.39

0.61

0.70

0.74

0.71

0.67

0.60

0.54

0.47

0.41

T h e polygonal graph Y thus obtained is shown i n F i g . 20. T h e actual solution is i n this instance^ = 2xè~'. Its values at the points x,- have been included i n the table for purposes of comparison, and the integral cur\^e^ is also shown i n F i g . 20. 2. A differential equation 7.9 associates a numerical value y" w i t h each triple of values x, y, and y'. It thus associates with these values a curvature K, given by the calculus formula METHOD

(7.12)

K

=

¡1 + y 2 | « '

and a radius of curvature,

(7.13)

p

=

fl

+y''\^ ft

Approximate Integral Curves

125

A graph approximating the integral curve through the point (xo, yo) with the slope ^^o' may therefore be constructed as follows: A t (xo, ^o) the normal to the integral curve has the slope — 1/yo'A l o n g this normal, with the direction of increasing y taken as positive, lay off po, the value obtained from (7.13) by assigning to x, y, and y' the values xo, yo, and yo'. This locates the center of curvature Co of the integral curve at (xo, ^o)- W i t h Co as center, draw a circular arc through (XQ, yo). Let Vi and Yi be the ordinate and slope of this arc at any suitable chosen

X

FIG.

20.

point xij where x\ > XQ. T h e construction may now be repeated with the point (xi, Yi) and the slope Yi i n place of the original ones, etc. T h e graph obtained is a chain of circular arcs which at each abscissa Xi advances i n a manner that fulfills the differential equation. I n a certain way it therefore approximates the integral curve. T h e approximation is best near the point (xo, yo), and the smaller the differences {xi+i — are taken, the better is the over-all approximation. Table 7

y

1

1

3

5 T

S

0.54

0.32

0.07 - 0 . 1 8

0.71

0.50

0.23 - 0 . 0 4

-0.55

-0.78

-0.96

-1.72

-3.36

-5.33

0

T

1

0.97

0.87

0.73

0.96

0.86

0 -0.26 -1.13

Yi Yi' Pi - 1

1

T

1

-1.07 -13.6

3

7 T

9

6

2

1

-0.42

-0.63

-0.80

-0.27

-0.62

-0.86

-1.09

-1.07

-1.07

-1.07

-0.99

-0.81

78.5

11.6

5.07

3.24

1.95

126

Solvable Equations of the Second Order

Example 11.

F o r the differential equation and conditions y+JV-O,

;-(0) = 1. / ( O ) = 0 .

the solution is ^ = cos x. This solution and its approximating g r a p h based upon the calculated values of T a b l e 7 are shown i n F i g . 21. F o r the

X

Fro. 21.

construction of this table the values of ¥{ and Y'i were determined by measurements of the figure, and the values of Pi were computed b y formula 7.13. ASSIGNMENTS 2. ConstrucE by Method 1 a graph to approximate the integral curve of the difTerential equation

/ ' + {2* + 2 } /

+ (2x -

l b - 0.

for which >(0) = 1,>'(0) = 1. 3. Construct by Method I a graph to approximate the integral curve of the differential equation I2x - x ^ i y for which

= hy'

¡2 ^ x^\y + J2 - 2xly = 0,

(0 = - A -

4. Construct by Method 2 a graph to approximate the integral curve of the differential equation for which ^(0) = l , y C O ) = 1.

Some Geometrical Applications

127

5, Construct by Method 2 a graph to approximate the integral curve of the differ ential equation /' for which >(0)

=

-T,/CO)

-

3>' + 2;- = 0,

"0-

7.8. Some geometrical applications W h e n a curve is specified i n terms of its curvature, it is in fact characterized by a differential equation of the second order. Its coordinate equation can then be found by integrating that differentiéd equation. Example 12.

T o find the equation of the curve whose curvature is always equal to the sine of its inclination, and that goes through the point ( — 1 , 0) with the slope 1. T h e inclination has the tangent^'. Its sine is therefore ) ' ' / \ / l +>'". Since the curvature is given by formula 7.12, the curve is one for which /'

y'

|i + y ' i ^

(1 4-/^1^^

This is a differential equation i n which ^ is not present, and which is therefore of the first order for^'. Its variables are separable. A first integral is thus log

1

= 2x + ci.

/2 +y

T h e condition y* = 1, when x = — 1 , is fulfilled by W i t h ci thus evaluated, the first integral yields

= 2 — log 2.

e'+' dy = —.

—

dx.

and by a quadrature of the last equation y = sin~^ {tf*'*"VV^} + a. W i t h C2 determined so that y — 0, when x = — 1, the equation of the curve sought is found to be ^ = sin ^

TT

V5

PROBLEMS 51. Find the differential equation of the curves whose centers of curvature are always on the x-axis. Integrate this differential equation, 52. Find the differential equation of the curves whose centers of curvature are always on the^-axb. Integrate this differential equation.

128

Solvable Equations of the Second Order

53. Find the equation of the curve that goes through the point (0, 3) with the slope zero, and whose curvature is always equal to twice the cosine of its inclination, 54- Find the equation of the curve that is tangent to the ;r-axis at the point (1, 0), and for which the product of the slope by the curvature has the constant value — 1 , 55, Find the equation of the curve along whose normal the center of curvature and the jr-axis are equidistant from the curve, and are on opposite sides of the curve. 56, Find the equation of the curve that goes through the point , 0) with the slope •ff, and along whose normal the center of curvature and the ^--axis are equidistant from the curve, and are on opposite sides of the curve, 57, Find the equation of the curve that goes through the point (1, 0) with the slope zero, and for which the product of the curvature by the sine of the inclination has the constant value -j, 58- Find the equation of the curve that goes through the point (0, 1) with the slope 1, and for which the product of the curvature by the cosine of the inclination has the constant value —

7.9. Curves of pursuit W h e n an airplane P flying along a certain path is pursued by another airplane Q that is always directed toward it and that flies at k times its speed, the path of Q is a curve of pursuit. T h e equation of this curve is determinable as follows: Let the coordinates of Q be (x, y) and those of P be (x*, y*), as i n F i g . 5. T h e n because Q is always directed toward P we have y' = {y — y*)/{x — X * ) . T h i s relation and its differential give us the two equations {x~x*\y'

(7.14)

jx - x*| d/ -y'dx*

= =

[y-y*], -dy*.

T h e equation of the path of P and its differential, and the fact that the speed ds/dt is k times ds*/dt give us, further, the three equations f{x*,y*)

(7.15)

fAx^y*)

= 0,

dx* + / „ * ( x * , > * ) dy* = 0, Vdx^

+ dy^ = k Vdx*^

+ dy*\

F r o m equations 7.14 and 7.15 the four quantities x*, >•*, dx^, and dy* can (theoretically) be eliminated. T h e éliminant is the differential equation of the curve of pursuit. Example 13. A n airplane P flies northeast.

twice its speed.

A pursuing plane Q has T o find the equation of the pursuer's path.

Curves of Pursuit

129

If the origin is chosen on the path of P , equations 7.15 are y*

=

X*,

dy* =

dx*,

dx =

2V2

Vl

dx*.

T h e éliminant of x*, >*, dx*, and dy*, between these and equations 7.14 is -

1P Vl

+

-

= 2V2{y

x]y".

This is the differential equation of the path of Q. T o integrate this equation we set ^ — x = u, which transforms it into a'2 1)2 = 2 "N/Z UU", and this equation does not contain x. Hence, when u" is replaced by u' du'/du, the differential equation appears as an equation of the first order i n u' and u. T h e variables i n this are separable. A first integral is thus V2

+

2u'

+

1

+ V2

u

V~2

Vu

which can be put into the form ci Vu

rr - V i -

Vu

2(r 1

«' = 2 V 2

T h e general integral is accordingly 2<:i \ 4 -

\/2 u -

—

u'^ = 2 V 2 X + «:2,

3c 1

and, i n terms of the original variables, is thus 2ci

Vy

-

X

^

V2\y

-

x]

-

^

\y -

x]^^ = 2

x + cj.

3«; I

T h e constants C\ and are determinable when the positions of the two planes at some instant are known. PROBLEMS 59. A n airplane flies due east. It is pursued by a plane that has twice its speed. Find the differential equation for the curve of pursuit. Integrate this differential equation. 60. Find the differential equation for the curve of pursuit if the pursued plane flies cast and the speed of the pursuing plane is only one-half as great. Integrate the difi"erential equation.

130

Solvable Equations of the Second Order

61, A pursued airplane flies north. T h e pursuing plane has the same speed. the differential equation for the curve of pursuit, and integrate it,

Find

62, A pursued airplane flica north. T h e pursuing plane has three times the speed. Find the differential equation for the curve of pursuit^ and integrate it. 63. A n airplane flies north on the line JC — 4, A pursuing plane has 50 per cent more speed, and passes through the point (12, 0) in the northwest direction. Find the equation of the curve of pursuit, 64. A n airplane flics east on the line ^ = 3. A pursuing plane has the same speed and passes through the point (5, 13) in the direction with the slope f-. Find the equation of the curve of pursuit.

ASSIGNMENTS 6, Describe the method for obtaining the differential equation of the curve of pursuit, if the pursuing plane has k times the speed of the plane P, and is always directed toward the point that is at a certain fixed distance directly ahead of/*.

7.10.

The motion of a particle on a curved path

T h e motion of a heavy particle that slides without friction along a fixed curved path is in accordance with Newton's law. This law again has form y

X

FIG.

22.

3.2, if s is taken to be the arc length of the curve, and / is the tangential component of the active force. W h e n the only force acting is the weight of the particle, it is seen from F i g . 22 that / — — zi; sin that is, / = —z/j dy/ds. T h e n , since m = w/g^ by (3.3), the law of motion is dv/dt =

~gdy/dSy

Motion on a Curved Path

131

which, after muUiplication by 2 ds, appears as 2v dv =

—2g

dy.

A n integration thus gives v"^ = a -

2gy,

and when ci is determined by a condition v = VQ, w h e n ^ = ^o, the result is (7.16)

[ds/dt\^ = vo"

•\-2g\yo-y].

By expressing ds i n terms of^ and dy, (7.16) can be converted into a differential equation for as a function of t. Alternatively, by expressing both ds and y i n terms of x and dx, (7.16) can be converted into a differential equation for x as a function of /. Example 14.

A heavy bead slides along a smooth wire i n the shape of the curve y = —x^^. A t time / = 0, it is at the point (1, —1) and its velocity is f V T 3 . T o find its position at any later time. W i t h the given values oivo,yo, a n d ^ , and with g = 32, equation 7.16 is ds = ^ y/^ + 9J:^^ dl. ds =

Because we also have V l

dx = V 4 + 9x^^

it follows that dx/x^ = 8 dt. Hence = 8/ - j - C2- T h e condition X = 1, when t = 0, requires that C2 = |. W i t h x thus determined, the equation of the curve gives y. T h e result is

PROBLEMS 65, A smooth wire has the shape of the curve y = —x"^, A bead sliding on it is at the point (0, 0) at ( = 0, and has at that time the velocity 4. Find the position of the bead at any later time t. 66, A smooth wire has the shape of the curve ^ = 25 — ;c'/16.

A bead sliding on

it is at the point (4, 24) at / = 1, and has at that time the velocity 8 \ / 5 , position of the bead at any later time 67, A smooth wire has the shape of the curve ^ = ^ {1 ~ x^^}.

A bead sliding on it

is at the point ( j , f ) at / = 0, and has at that time the velocity 2 \ / 3 0 , position of the bead at any later time 68, A smooth wire has the shape of the curve y =

— f x^^-

= h \/8

- r

+ 4 sin^^ {y/2

V2|-

Find the

A bead sliding on it is

at the point (1, — f ) at / = 0, and has at that time the velocity \/^3'^position of the bead at any later time 69, A smooth wire has the shape of the curve X

F i n d the

Find the

132

Solvable Equations of the Seconcf Order

A bead sliding on it a at the point (0, 0) at / ~ 0, and has at that time the velocity 8\/3.

F i n d the ordinate of the bead at any later time t

70. A smooth wire has the shape of the curve i y V 9 - / + | 8 i n - '

{>/3|.

A bead sliding on it is at the point (0, 0) at t = 0, and has at that time the velocity 8\/5.

F i n d the ordinate of the bead at any later time L

71. A smooth wire has the shape of the curve. jr -

8{e -

sin

fl},

y = 8[cos e -

\],

A bead sliding on it is in the position for which d = 0, at / = 0. of the bead at any later time

Find the coordinates

72. A smooth wire has the shape of the curve 2\e + sin e],

X ^

jK = 2(1 "

A t>ead sliding on it is in the position for which 9 — of the bead at any later time t

cos 9 .

at f » 0.

Find the coordinates

ANSWERS TO ODD^NUMBERED PROBLEMS \. y = \x^ + a log X + C2.

3. >

2 tan ^ \ci€^] + C2

5- > — log {log X -\- a] -\- C2y and y " c. 9. y=

7, y %in y = c\x + rj-

-

C\X + f2 - 1 ClX + C2

./2

11.

{1 + l o g ; - î / ' + — y

15. : r y V + / y

+

27. 31.

{1

+ 0.

= 0.

19. | ^ _ ; , i y ' _ ( y _ i j 2 23.

n. yy" - yy' -

= 0.

= o.

= O - ^ i l o g 11 -

X sin y + y sin x =

cix

21. 2 U + : y l / ' + {1 - / M i l + / i 25.

x].

Inexact.

1 29. ^ = - + C2 -

+ cj.

X

1^ -

[ x + c i l y e ' + c i .

33. > = log [x^ + cix + C2], 35. y = 37. y -

z=

{ci+

2xU-''.

CiX + C2 —

X'

.3

Ix + c i l s i n - ^ {x + C 1 Î + - \ / l -

z = ^ 4- sin

+^2,

+ i'll-

39. y' = x^ ~ yz, 43. :y' = log ^,

U +f,j2

z' = xj-.

2' = > + 2z.

41. /

z' - x ^

45. ;v = 2x + i x * +

4 «

+ /

-0,

Answers to Problems 47. y ~ 2x' -{-^x* -\- • • • . SI. yy"y'^

I -0;

49. ^ =- 1 + 3;t +

{x +

a]^-\-y

-

Ci

53. ^ = 3 — -J log cos 2x. 57. y

-

59. V l

{V('

-

+

65. x «> 4r, 67.

X

=

69. ^ = 71.

X

1 w

;

"

-

4)^^ -

-

2x)\\

cos-» (3 -

2M

^

63. y = 6 [ V 2

'

- i

= 2»"

+

61. v * i

-x)

1)(2

— log X -

2} -

_u

^—1-

4*1

^^'^ + 40

-l6/^ ?5

16

{12;

= 16( -

+3

V3|** -

8 sin (2/),

3.

y =

-8

«2.

+ 8 cos (2/)-

-

4)« -

32}

fx*

+

CHAPTER

8 The Linear DifFerential Equation with Constant Coefficients 8.1. The complete equation T h e differential equation (8.1) i n which a,

y"

+

ay' + i> =

or its derivatives, is called the linear differential equation of the second order because each of its terms is of, at most, the first degree in y and its derivatives. If wi{x) and W2{x) are, respectively, solutions of equations of this type w i t h the same coefficients a and by thus (8.2)

h, f{x)

do not involve

/(x),

y

wi'

+

awi

+

bwi

= fi{x)y

W2"

+

aw2

+

bw2

=

/2(jf),

we see, upon multiplying these equations respectively by any constants ai and 02 and adding them, that the combination (8.3)

y =

aiwiix)

-\- a2W2(x)

is a solution of equation 8.1 with (8.4)

/(;.)

=

aifi(x)

+

a2f2{x).

Clearly a corresponding statement can be made if there are more than two equations 8.2. Hence a differential equation 8.1 in which the function fix) consists of a number of terms, as in (8.4), can be integrated by integrating each of the simpler equations 8.2, and combining their solutions. Several important conclusions may be drawn from this observation. T o begin with, let us consider the differential equation (8.5)

u" + au' -\-bu

= Oy

which is obtainable from (8.1) by replacing/(x) by 0. If ui(Af) and are any two particular integreils of this equation, the combination (8.6)

U

ClUlix)

+

134

C2U2{x)y

U2{x)

The Complementary Function

135

w i t h arbitrar)' constants ci and C2 is also an integral. It is the general integral if the constants ci and are not replaceable by just a single arbitrary constant, that is, if the ratio of ui{x) to « 2 ( ^ ) is not a constant. W h e n differential equations 8.1 and 8.5 are considered together, (8.1) is generally referred to as the complete equation, and 8.5 as the reduced equation. T h e genera] integral 8.6 of the reduced equation is called the complementary function.

If i n equations 8.2 we take fi(x) = f{x), and fiix) = 0, we may choose for wi{x) any particular integral yi{x) of equation 8.1, and for W2{x) the general integral of equation 8.5. T h e combination (8.7)

y = yi{x)

+

ciUi{x) + czuzix)

is thus also an integral of equation 8.1. T h e general integral of a complete equation can therefore be found by finding a particular integral, and adding the complementary function to it. Thus the task of integrating a complete equation is resolvable into two parts, namely, the determination of the complementary function, and the determination of a particular integral. T h e simplest type of equation 8.1 is that in which the coefficients a and b are constants. Such an equation is always integrable by quadratures. T h e case i n which ¿ = 0 is especially simple. A quadrature can then be applied directly to obtain a first integral, and this first integral is a linear differential equation of the first order.

8.2. The complementary function One method of integrating the reduced equation 8.5 is already known to us. W e found, in Section 6.7, that by setting

this equation is transformed into the Riccati equation v' =

—b

— av — v^.

This Riccati equation, since it has constant coefficients, has a constant solution. B y the trial of y = m, it is found that this is a solution if m is a root of the equation (8.8)

m^ +

am -\- b =

0.

This is called the auxiliary equation. If its roots mi and mz arc distinct, we obtain i n this way two integrals «i = e^^' and «2 = e^^'. W i t h these integrals, formula 8.6 gives the general integral. If mo = m i , we obtain only one integral, «i = Í " * ' ^ in this way. However, the Riccati equation then also has the integral = m\ -\- \/x, and &2 leads to u^ = xe^^'. T h e

136

Linear Equations with Constant

Coefficients

complementary function is thus (8.9)

u = CKT"*!' H- C2f"'^'y

if

mu

u = citf"'* + C 2 « " ' * ,

if

m2 = m i .

W i t h this result once established, a procedure for recovering it quickly i n any instance is easily formulated. By substituting e"*' for u i n the given differential equation, we are led at once to auxiliary equation 8.8, the roots of which must be found. W i t h these roots, formulas 8.9 yield the general integral. T h e particular integral for which (8.10)

u{xo) = « 0 ,

U'(XQ)

uo'

=

is obtainable, as usual, by determining the constants ci and C2 i n (8.9) to fulfill these conditions. Example 1.

T o find the integral of the differential equation u" -

2u' -

2« = 0,

for which w(0) = 1, and u'(0) := - 1 . T h e substitution u = e"' converts the differential equation to f"'|m2 -

2m -

2)

=0.

B y striking out the exponential factor we obtain the auxiliary equation, the roots of which are m i and m2 = 1 — T h e general integral is thus B y differentiating the general integral we find that «' = ¡1 + \/3)tfitf!i+V3I*+ |1 - \ / 3 | c 2 ^ U - v ^ l « .

T h e conditions to be fulfilled are therefore ci-\-C2

=

These yield ci = ^ integral is therefore

1, —

{1

+

I/'N/S,

V3]ci

and

¿"2

+

{1

-

V3]C2

= i + i/'s/Sy

-

-1.

a n d the required

u = Example 2.

T o find the integral of the differential equation « " - 1 « ' + iftr« = 0,

for which u ( l ) = 0, and u'{\) = 2.

Complex Roots.

Euler's Formulas

T h e auxihary equation is i n this instance ~ - f ^ = 0. mi = f as a double root. T h e general integral is therefore u —

cye

137 It has

-\- Czxe

By differentiating the general integral we find that

T h e conditions are thus = 0,

{ci-\-C2]e^

F r o m these conditions, ci = —2e~^^ and integral is therefore u =

= 2.

Ihi-^^c^U^ C2

= 2e~^^. T h e required

2{x-\]e^^^'-^K

PROBLEMS Find for each of the following differential equations the integral that fulfills the respective conditions. 1. u" -

3u' + 2u = 0,

2. u" + u' - TU = 0,

u(0) -

9u = 0,

4. u " -

7u' +

5. u" -

4«' -

u = 0,

6.

2

a' + 2u = 0,

7. a " -

2u' -

80u = 0,

8. u" -

2 V ^ u ' + 3« = 0,

9. u " -

7u' + lOu = 0.

10. 9 « " -

6u' + u = 0 .

11. u" + 3u' -

u(l) = 1, u'(0 = 1= 0,

u =0,

12. 2u" + u' = 0,

1.

uCO) = 1, u'(0) = 0.

3. u " -

u" -

0, o'(0) -

u ( - l ) = 0, u ' ( - l ) = 2.

u(l) = 2, u'(l) =

-1.

u(0) = 1, u'(0) = 2.

uC-1) = 0, u ' ( - l ) = 0.

uCO) => 0, u'CO) = 0.

u(0) = 7, u'(0) = 14. u(3) = - 1 , u'C3)

u(0) = 2, u'(Q) =

4. -3.

u ( - 4 ) = 1, u ' ( - 4 ) = 1.

8.3. The case of complex roots.

Euler's formulas

Formulas 8.9 are valid for complex as well as for real values of mi a n d mi. However, they do not give the integral i n a real form when these roots are complex. Alternative real formulas are therefore desirable i n that case. If mi = a + 1^3 with /3 0, the R i c c a t i equation has a + as a complex solution. Its general integral, as given by formula 6.14, is then V

= a — 0

tan

{0x +

c].

138

Linear Equations with Constant Coefficients

F r o m this value of v we obtain the general integral of equation 8.5: w — - „ - « + I o 8 c o « ItfaH-cJ

T h i s result can be put into any one of the following forms: u — C2^"* cos {j8x + c i l ,

(8.11)

u = CiC^'sm

{/3x + c i | ,

sin

u —

+

C2 cos ^x\.

T h e first of these forms is obvious. T h e third is obtainable from the first by using the cosine addition formula, a n d renaming the arbitrary constants. T h e second is obtainable from the first by replacing c\ by ci — i r / 2 . W h i c h one of forms 8.11 is the most convenient to use w i l l ordinarily depend upon the circumstances of particular problems. Example 3.

T o find the integral of the differential equation u" -

6 « ' + 13« = 0,

for which u(0) = 2 and u'(0) = - 1 . T h e roots of the auxiliary equation are i n this instance 3 ± 2 i . T h u s a = 3 a n d j3 = 2. B y the third formula 8.11 the general integral is therefore u = e^'{ci sin 2x + ^2 cos 2x\.

T h i s general integral has the derivative u' = e^'{{3ci — 2c2) sin 2x + (3^2 + 2ci) cos 2x\.

T h e given conditions are accordingly fulfilled if ^2 = 2 and 3c2 -\- 2ci — 1. T h e required integral is thus u = e^'{ - i sin 2x + 2 cos 2x Example 4.

T o find the integral of the differential equation M" +

2u' +

4« = 0,

for which U(T/2) = 0 a n d U'(T/2) =- 1 . I n this instance the roots of the auxiliary equation are — 1 ± i/2. T h e first formula 8.11 therefore gives u = C 2 ? ~ ' c o s l(x/2) -\- ci]. This has the derivative w' = —cze *cos

+ ci

—

1

- f 2*

•

X

Complex Roots.

Euler's Formulas

139

and the conditions therefore take the form cos 5 + ,

C2e

cos

c^e

-

1

^ C2e

= 0,

72 sm

=

-1

T h e first of the last pair of equations is fulfilled if ci = 7r/4, and the second is then fulfilled if t:2 = 2e'^^. T h e required integral is therefore

2^4 Some important formulas of mathematical analysis follow from the results we have obtained. T h e integral of the equation u — 0, for w h i c h u(0) = 0 and u'(0) = 1, is found from (8.11) to be sin and from (8.9) to be

-

— e"*'].

Similarly, the integral for which u(0) = 1

and a'(0) — 0 is found to be both cos x and i\e*' + that (8.12)

1 sinx = - {^"-^-"1,

W e see thus

1 2

2i

I n their converse form these evaluations are (8.13)

e^' =

cos

X -\- i

sin

tx

x,

=_ cos

A: —

4 z•sm

x.

Relations 8.12 and 8.13 are called Euler^s formulas, i n honor of the Swiss mathematician Leonard Euler (1707-1783). PROBLEMS F i n d for each of the following differential equations the integral that fulfills the respccdve conditions.

13. u" - 4u' + 5u = 0,

u{0) = 4, u'(0) = 2.

14. u" + 2u' + lOu = 0,

u(l) = 0, o'(l) = 4.

15. u" - 3u' + V-u = 0,

u(0) = 1, u'(0) = 0.

16. u" - lu' + 2u - 0, 17. u" + 25u = 0, 18. u" + 2 y/l

u(2) = 1, u'(2) = 0.

u(3) = - 1 , u'(3) = 0.

u' + 4u = 0,

aI

VV2/ 19. « " + 2 V 3 «' + 4u = 0,

J = 0, u' (

)

V V i /

u(0) - 0, a'(0) = 0.

1

140

Linear Equations with Constant

20. u" -

12 log 2]u' +

21. u" -

4 « ' + 53« -

22. u " - 2 { 1 23. « " -

u

o(l) - 1, u'(l)

2.

0,

-

=

2ru' + 2^^*« = 0,

u(l) = 0. «'(1) =

«(1) = 1, «'(1) =

- 1 , u'

uC-3)

+ \ / 2 l u ' + | Y H - 2 \ / 2 | U =0,

24. u" + 17« = 0,

8.4.

|4 + log^ 2\u = 0,

Coefficients ^ 0.

= 0, u ' ( - 3 )

=0.

2T.

1.

Differential operators

W h e n a differentiation is denoted by D instead of by the less simple d/dxj and is used to denote the second derivative, such expressions as fp'{x) 4- ctp(x)j and
ip"(x) 4- V W

4-^«^« =

{D-\-c]
4- b
+

a£) 4-

fik.

Herewith the symbols {D -\- c], and {D^ ••{- aD -\- b\ have been introduced. They do not stand for quantities, as is the case with literal expressions i n algebra. Rather, they indicate operations that are to be performed, namely, differentiations, additions, and multiplications that are to be carried out upon the function to whose symbol they are prefixed. For that reason they are called differential operators. D itself is the simplest differential operator. T h e effect of a differential operator upon a function is to convert it into another function, precisely as the operator D converts a function into its derivative. Thus, for example, ( D 4- 3) cos

A:

[D^ 4- 4 D 4- S]xe'

sin

=

—

=

de'.

X

4- 3 cos

x,

T h e function which is obtained as the net result of first applying the operator {D -\- c\\ to ^ ( x ) , and following that w i t h an application of the operator \D -\- C2\,\s denoted by (8.14)

\D +

C2\[D

ci]ip.

This is by definition the same as {D-^C2\W{x)^ciip{x)\.

It is thus further the same as W{X)

+

Ci,p{x)Y +

C2W{X) +

.i^(x)).

141

Finding a Particular Integral which is the function

But we have agreed to write this last as (8.15)

\D''•\-{ci +

C2)D-ir

C1C2W.

Because expressions 8.14 and 8.15 are thus the same for every function ^(jf) we shall write (8.16)

\D + C2][D

+ ci}

=

\D''^{cx

+

C2)D

+

ciC2\.

(8.16) is not an algebraic equation between quantities because D is not a quantity. F r o m the way it was derived it is shown to be the statement, expressed i n symbols, that the net effect of the two operators appearing on the left is the same as that of the single operator appearing on the right. F r o m relation 8.16 we now obser\'e a most important fact, namely, that the right-hand symbol is correctly obtained if the ordinary rules of multiplication are applied on the left, even though the letters do not all stand for algebraic quantities. T h e usefulness of differential operators is largely due to the fact that they do combine with each other and with numbers by the laws that are already known to us from algebra. I n particular, if mi and m2 are again the roots of auxiliary equation 8.8, they have the sum —a and the product b. Hence relation 8.16 shows that (8.17)

-^aD + b] = {D -

mi\{D -

m^].

This is precisely as if the left-hand expression were "factored." A n application of the left-hand operator of 8.17, which is of the second order, is thus replaceable by the successive applications of the right-hand operators, each of which is only of the first order.

8.5. An operational method for finding a particular integral B y the use of relation 8.17, differential equation 8.1 may be given the form I D - mi]{D

-

m2]y

=

f(x).

This is equivalent to the successive equations (8.18)

{D -

mi\Y = f(x\

and (8.19)

{D-m2\y

=

Y{x).

N o w these are two linear differential equations of the first order. tion 8.18 has a solution (8.20)

Y(x)

= e'^''Sf{x)e-^^' dx,

Equa-

142

Linear Equations with Constant

Coefficients

and therewith equation 8.19 has a solution (8.21)

y{x) = í * " í * / r ( x ) í - ™ « ' í / x .

T h e substitution of (8.20) into (8.21) gives the integral (8.22)

y(x) = e'"^'je^'"'-^'^'\Sf(x)e-^^'dx]

Example 5.

dx.

T o find the general integral of the differential equation y"

-

3/

']-2y

= x.

T h e roots of the auxiliary equation are i n this instance m i = 1, a n d m2 = 2. T h e given differential equation can therefore be written as {D^i}{D^2\y^x.

T h e successive equations by which it can be integrated are \D -

l | r = X,

\D-2]y

=

Y(x).

O f these the first has a solution Y{x) = — (x + 11. Therewith the second has a solution v = (x/2) + and this is a particular integral. T h e complementary function is cie' + Í 2 Í ^ ' . T h e general integral is therefore > = Example 6.

+ 1 + cie' + C2e^.

T o find the integral of the differential equation + 2y' + 10;r = i « + 2,

/'

or which y(0) = 1 and y'{Q) = 0. T h e roots of the auxiliary equation are — 1 ± 3i, therefore the given equation is \D-\-i

3i] ¡Z) +

-

1 + ?>i]y =

+

2.

T h e successive equations by which this equation can be integrated arc (Z) + 1 -

3i]Y

=

+ 2,

+

3i]y -

Y(x).

T h e first of these has an integral 2 2 — 3i

1

—

3i

and therewith the second has an integral y = - J ^ Í * -|- i. T h e complementary function is e~'lci cos 3x + f2 sin 3x|, T h e general integral is therefore y = + i + ^"'Ui cos 3x + cz sin 3x1.

Finding a Particular Integral

143

F r o m the general integral and its derivative we see that the conditions to be fulfilled are T^

+ i + '^i = 1,

3^2 = 0.

T V - ^ J +

These determine ci and C2, and the integral required is thus found to be > = TV«* + i + e-'{U Example 7.

cos 3x-\-U

sin 3x\.

T o find a particular integral of the differential equation iy = xe' + sin 2x + 5x.

y" -

I n place of this equation, we may consider the simpler equations 8.2, which i n this instance are wi"

~ iwi =

W2"

—

-1^2

~

sin 2x,

^z" — i«^3 = Sx.

These, respectively, have the integrals Wi — ^xe' — -V"^*>

1^2 = ~ A" sin 2x,

wz = — 20x.

Their sum, 4^ — 3

321 4 . —\ — — sm X 9I 17

— 2\Jx.

is a particular integral of the given diflferential equation. PROBLEMS Find the general integral of each of the following differential equations. 25. y" - y' -2y

= 3«^

26. y" ~ Ay = x^.

27. >" + > = Ae'.

28. y" + 2y' + 5y = 3. e~'

29. v " + 2y' + y =

31. y" -

2y' -

x+2

•

30. y " -

8> = 9**^ + l O * " ' .

2v' + 2y = sin x.

32. y" - y =^ - ; f + 2 —

+

x""

Find for each of the following differential equations the integral which fullills the respective conditions. 33. y" +2y 34. y" -

= 4,

y{Q) = 3 , / ( O ) = 1.

2y' •\-2y =

:y(0) = 0, ^-'(0) - 0.

144

Linear Equations with Constant Coefficients

35. y

+9> = 8sin:c,

36. y" + 2y' -f-

= -1,

> » 2 cos^- -

1 +

37. >" - h ' + h = 38.

+ 2;.' -

1.

2T

sin^'

= 1, / ( I )

+ 2 U ^ yiO) = 3. / ( O ) - 4,

+

>(0) - 0, /(O)

3;^ -= - 3 ; c - + x + 7.

= 2.

8.6. A second operational method A n alternative method by differential operators can be used whenever the roots of the auxiliary equation f re unequal. T h i s method is suggested by the following analogue from algebra. If k is any number not equal to either mi or m2, the value of ^ that fulfills the equation [k y =

is

mi\{k

-

m2\y

=/

{k — mi) (A — m 2 )

A resolution into partial fractions gives the last equation the form {//('"I - ' " 2 ) 1

y = -

Thus

3» = ^1 + > 2 »

{k -

k — m

+ ,

|//(m2-mi)j

k — m2

where

mi]yi

=

m2l>2 =

{k -

mi — m2

/ m2 — mi

Consider now the analogue of this procedure for differential operators T h e relations (8.23)

{/) - m i i j v i =

fix)

{D -

mi — m2

m2]y2

=

fix) m2

— mi

are linear differential equations of the first order. T h e results of operating upon equations 8.23, respectively, by \D — m2l a n d [D — m\] are the equations yi"

+ ayi' + hyi =

{fix)

-

m2f{x)\,

7fl\ — y2

+

ayi + ¿^'2 =

-1

l/'W

- m i / W ) .

T h e sum of the last two equations shows that \y\ + ^ 2 } is an integral of differential equation 8.1. W e can therefore find a n integral of equation

A Second Operational Method

145

8.1 by finding and adding integrals of the simpler equations 8.23. T h e formula obtained i n this way is (8.24)

y(x) =

/

J

Example 8.

-^^^^

¿-'"1'

dx +

/

mi — m2

J

e-"^^ dx. m^ — mi

T o find the general integral of the differential equation jy" + >' - 2V

T h e roots of the auxiliary Equations 8.23 are therefore

=

e'

1

equation are mi = 1, and m2 — — 2

3(1+^)

'

3(1-1-^}

T h e y have, respectively, the integrals y\{x) =

1

(

log ^ll

3

e'

y2{x) = - 7 + J ^ - ' - ^ ^ ^ l o g ¡1 + . ' 1 . 6

3

3

T h e sum of these is a particular integral of the given differential equation. A d d i t i o n of the complementary function yields the general integral > = i (T* log

\ +en

-\e'' 3

log{l + °'

+ ' 3

6

+ cie'

Example 9.

+

C2e~^''

T o find a particular integral of the differential equation y"

-ly'

^-ly

^ e sin^ x.

T h e roots of the auxiliary equation are 1 + i . Equations 8.23 are therefore \D-

—t

\ ~ i\yi = ^

^ sin^ X , 2 '

|/) -

'

I

1 + i\y2 ^ ^ • •"• 2

2 sin^ x.

T h e y have as solutions yi = >2

—

{— sin^ * + 2i sin x cos x + 2}, — sin^ X

— 2i

sin * cos j ; + 2 ) ,

from the sum of w h i c h , we obtain the desired particular integral y = h'{2

- sin^xj.

146

Linear Equations with Constant Coefficients PROBLEMS

Find the general integral of each of the following difTerential equations. 39. y" -

40. y" + 5 / + 6> = 12x».

8>' + 15> =

42. y" — y — sin x.

41. / ' +:y = 43. y" + ly' - 2y = e' + 45.

+4/

47. y" -

+ 13;- =

5>' -

=

44. :y" 46.

X.

-

2> = 3*^'

-

+ 4;- = cos 2x.

48. > " + > ' - 6> = X + «2x

7;> = 1.

49. y" - ly' -\-2y

51.y'-:y

r'.

x\

50. / ' + /

5

1

52. :v

1

-2> =

3^

4^

i l + x ^

8.7. A comparison of formulas Formulas 8.24 and 8.22 each give a particular integral of differential equation 8.1. But 8.24 does so by two independent quadratures, whereas 8.22 does so by two successive quadratures. Although i n any given instance the evaluation by one of these formulas may be simpler than that by the other, the content of the two formulas is the same. T h a t can be shown as follows: A n integration by parts applied to formula 8.22 changes its form into

m\

—

J

mi

—

m^

f{x)e-^^'dx

It is Ccisy to see that this is formula 8.24. It is sometimes desirable to choose for the indefinite quadratures i n formula 8.24 those that reduce to zero at a given point JCQ. T h e y are the definite quadratures from XQ to x. W i t h the variable of integration denoted by s they make the formula appear as lis)

y{x) =

Jo mi

ds -

—

Jxo mi — m2

ds.

Since X is now not the variable of integration, the factors and may be put under the integral signs. T h e formula can then be written compactly with a single integral sign as (8.25)

yM

=

fis) xo

^1

—

ffl2

T h i s gives the particular integral for which >'(xo) = 0 and>'(xo) = 0,

The Method of Undetermined Coefficients

147

ASSIGNMENTS 1. Consider formula 8.22 when nti the form (8.26)

yix) "

f'f{s){x

mu and show that it can then be reduced to

- .r}*'"'^^') ds,

if m2 = mi.

2, Consider formula 8.25 when the auxiliary equation has the complex roots a ± tß^ and show that it can then be reduced, by the use of Euler's formulas 8,13, to the form ,W -

(8.27)

äs.

8.8. The method of undetermined coefficients M a n y problems i n engineering and science depend upon differential equations of type 8.1 in w h i c h / ( x ) is a polynomial, an exponential, a sine or a cosine, or some product of such factors. F o r such an equation a n integral can be found without making any quadratures. Suppose f(x) is a polynomial of the degree n. W e shall then take y as a polynomial of the same degree (8.28)

yix"-^

y^g^^--\-

with undetermined coefficients. T h e substitution of (8.28) into the given differential equation converts the latter into a n equation between polynomials. B y equating the coefficients of like powers of x we obtain a set of relations from w h i c h the values of ^o, ?i> * ' ' » ?n can be successively found. Example 10.

T o find a n integral of the differential equation y" H- jy' -

6> = 2x^ +

-7x-\-2.

Substitution 8.28 with n = 3 converts the differential equation into the equation -6?6*» +

1 - 6 ^ 1 + 3qo\x^ +

1 -6q2 - f 2qi + 6qo\x +

{-6qz

+

?2 +

2qi] = 2x^ + Sx^ -

7x + 2.

By equating coefficients of like powers of x we obtain the relations -6qo

= 2,

—6^1 + 3^0 = 5, —6q2 + 2qi + 6qo = — 7 ,

— 6^3 4- 72

+ 2qi =

2.

148

Linear Equations with Constant Coefficients

F r o m these equations qo = — i ; qi — — 1 ; q2 =

and q^ =

—

Thus

is an integral. A n equation 8.1, i n which f{x) is of the form €*Yi(x), is transformed by the change of variable y = e^v into v" +

(a + 2k\v' + | i +

If is a polynomial, method above. Example 11.

+ k^\v =

can be determined as a polynomial by the

T o find an integral of the differential equation /'

e-'^^\\2x^ -

- W

9x -

V-l-

T h e change of variable y = e~'^'^v transforms the given equation into 2v' = 12JC^ -

v" -

9x

-

I n the transformed equation the coefficient h is zero. W e may therefore perform a quadrature upon the equation directly, to give it the form = 3x* -

v' -2v

\x^ -

^x.

If we now substitute for v a polynomial 8.25 with n = 4, we are led to the equations =

3,

4qo =

0,

-2qo -2qi

+

-2q2

+ 3^1 =

-2?3

+ 2?2 -f

-2?4

F r o m these

v = -h*

Hence

-

?3

=

= 0.

-

- e'^^H ~ix^

-

-f,

+ 2x + 1-

3x^ -

^x^2x-\-I],

is an integral. Example 12.

T o find an integral of the differential equation y"

1 x ^ - 1 ) ^ ^ + {3A:+

-2y'-^y=

4)^'.

A pair of equations 8.2 are i n this instance wi" W2"

-

2wi'

+

«;i -

e^^'lx^ -

2w2' -\- W2 = e'i^x

Ij,

-\- 4].

Undetermined Coefficients Continued

149

T h e respective changes of variables wi = e^vi and W2 — them into + 2oi' +

v{'

=

-

1,

transform

v^' = 3* + 4.

T h e first of these has an integral v\ = — 4x ~\r 5. T h e second, w h i c h can be solved by two quadratures, has an integral v% — ^x"^ + 2x^. By multiplying v\ and v^. by the appropriate eponentials to obtain w\ and KJ2, and then adding w\ and w^, we find the integral V =

-

4x +

51.2^ +

\kx^ +

2^2|^.

PROBLEMS Find, by the method of undetermined coefficients, an integral for each of the following differential equations. 53. / ' + / - ! > ' = 3*2 54. y" + / 55. y" -

-

6> = Jf.

ly' = - 3 5 * ^ + 76x^ -

56. y" -^-y = \2x 57. >" + y

-6y

58. y" + 2y' 59. y" - h ' 60. y" -y' 61. y" -

-

By = 9e^^. \20x^ + 24^2 + 5x -

iU^\

= 12*V^-.

3 / +|>

= 27«'^^^

6y = 2e^ + 3^' + 5JC + 1-

Ay' + 2> =

-2y=

65. 66. y" -

42x + 6.

= \x^ -\-\U'.

-\-h

64. / ' ^y-

lAx'^ -

2}*"^

- y=

62. y" + Sy' 63. / '

8x + 1.

123*2 _ 43^ _,_ \2\r'^^ +

\\^x^ + 4x + S}*** + + 24 + 3e^ +

2/ -

3> = - i4x + 4}^' -

\\\x +

19

{6x + Sjir^''.

4f-^ {8x -

2)«-' -

6^^' -

(5x -

6|f-K

8.9. The method of undetermined coefficients continued T h e method of undetermined coefficients can also be applied to an equation 8.1 i n w h i c h i s of the form /(x) = P i W sin

¡7:^

+ 5) +

^ 2 «

cos

(7X

+ 51,

i n which 7 and 5 are constants, and F\{x) and Pi{x) are polynomials. I n this instance we shall take y i n the form (8.29)

y - Qi(x) sin (7^: + 6) + Q2(x) cos ¡7^ + S j ,

Linear Equations with Constant Coefficients

150

w i t h Q i and Q2 as polynomials with undetermined coefficients

Q2W

= ?2,0X"

?2.1«""

+

+

•••+

?2.n.

T h e degree n shall be the larger of the degrees of P i and Pz, unless the instance is one i n which a = 0 and b = 7^. I n this exceptional instance n must be taken greater by 1. Substitution 8.29 converts the differential equation into an equation between sums of functions of distinctive types. B y equating the coefficients of like functions, we obtain relations from which the values of qtj can be determined. Example 13.

T o find an integral of the differential equation y"

+ 2>' + 5> = 3 cos

T

X

4

I n this instance n = 0, 7 = 1, and 8 = — T / 4 . Substitution 8.29 converts the given differential equation into the equation H^i.o

— 292.0) sin

X —

ir 4

{2qi,Q +

+

492.0I

T

cos

" - 4

3 cos ' * ~ ^ By equating the coefficients of like functions we obtain the relations 4?i.o — 2^2,0 = 0,

2^1,0 + 4^2.0 = 3.

These simultaneous equations give 71,0 — ^ 0 and 72,0 = %• , 3

T

X

+

4

Thus

T

-COS

Jf — 4

is an integral. Example 14. y"

+>'

T o find a n integral of the differential equation +

87

=

IIOJC^ +

21*

+

9!

sin

3x

+

x cos

3x.

I n this instance n = 2, 7 = 3, and 5 = 0. Substitution 8.29 with these values converts the differential equation into the equation l(-?i.o -

3^2.0)-«^ +

(-

qi.i

-

372,1 4- 271.0 -

1272,0)*

+

( - 7 1 . 2 - 372.2 H- 71.1 - 672,1 + 271,0)) sin 3x + {(371.0 - 72.o)x^ + (371.1 - 72.1 + 1271,0 + 272.o)x H- (371,2 — 72.2 + 671.1 + 72,1 + 272.0)) cos 3x =

\\0x^ + 21jr + 9| sin 3x + x cos

3x.

Undetermined Coefficients Continued

151

By equating the coefficients of like functions we obtain the relations — ?i,o —

3^2,0

= 10,

3?i.o -

?2,o

0;

— ? i . i — 3 ? 2 . i + 2yi.o — 12^2.0 = 21,

— 6^2,1 + 2yi,o = 9 ,

~ ? 1 . 2 — 3^2.2 + — ?2,2

391,2

+

6^1.1

+

?2.1

+

2^2,0

= 0-

T h e first pair of these relations is a simultaneous system which yields fji.o = — 1 and 9 2 , 0 = — 3 . Therewith the second pair yields ^ i . i = 7 and q2,i = 2. T h e third pair then yields 9 1 , 2 = — 1 3 and ^2,2 = ~ 1 T h e integral thus obtained is y =

l-x"^

Example 15.

-\.Jx

13j sin 3x +

-

{-2,x^ +

2x -

1 i cos

3x.

T o find an integral of the diff"erential equation _

^ e, 2 x

4y' -\-Sy

^) + 2 cos

sin(. +

I

A:

+

-

T h e change of variable y — e^^v transforms this diff'erential equation into v" -\-

V

= sm \x

+ 2 cos

~

'+5

Although the polynomials i n the right-hand member of the transformed equation are of the degree zero, we take n = 1, because this is an instance i n which a — 0 and b = 7^. B y the method described, we then find that 2?i,o ~ 2, and — 2 ^ 2 . 0 = N o relations are found for 91,1 and 9 2 , i T h e values of these coefficients are therefore arbitrary. W e may assign to them the value zero. Thus V

=

X

1

sin \x -\- -

- X

2

cos x-^^n

and the integral found is 2x

X

sm I X + ~ I

1 ~

-

X

cos

PROBLEMS Find, by the method of undetermined coefficients, an integral for each of the fol lowing difTcrcntial equations.

152

Linear Equations with Constant Coefficients

67. / ' + 3 / +7> 68. y" -

2)-' -

-

15 cos

3;- = sin X

X.

12 cos x.

-

IT

69. y" + 2;-' + 5/ -

3 sin

70. y" -

-*'cos«.

2/ + > -

71.

X

4

+ 2) sin X +

72.

-

4 / + 4^ = sin X -

73.

+ 6y' + 16> = sin 2x + cos

74. y

-

ix^ + 4x| cos

3/^'. 2JC

+ 4 sin 4x.

4y' + 5> =- «2"^ sin x

75. y" -\-Sy = - 1 5 ^ ' sin 3x -

29^"' cos x.

76. / ' -

2 / + 2> = e'{2 sin x + 4 cos x\ -

77. y

2 / + 6> = «*{4 sin x + sin 2x -

-

jf.

78. / ' + 3y' + 4;. = } Jx^ + x| sin 2x + 79. / ' -

3 / + 3> = x ' + ^'1 x^ -

80. / ' -

4 / -\-Sy = e'\{x'^ -

lOx cos 2x.

4 sin 3xi. \x^ + 2x2]

4xj sin X -

2x.

2x cos x|.

\2x + 1) sin 3x + ( - I B x * + 12x + 18) cos 3x1 + «2'{4sin X + 2 cos x|.

8.10. DifFerential equations of higher order than the second A l l the methods of this chapter are applicable to linear differential equations of higher order if these equations have constant coefficients. T h e general form of such a differential equation if it is of the reduced type is u^"l + aiKt"-^l + 0 2 « ' " " ' * ' +

• • • + a„u = 0,

with u'"^ standing for d^u/dx^. T o obtain the complementary function, we substitute e"^' for w, and thereby convert the differential equation into the auxiliary equation

For each root rrij of the auxiliary equation, the function e"^'' is a particular integral. If all the roots are distinct, the complementary function is

and involves the n arbitrary constants c i , (^2) * ' ' , ^n- F o r any root m that is multiple, say of the order r, the functions r " " , x^"**, • • •, x'^^^"" are particular integrals. If the coefficients of the differential equation are real, as we have been supposing, a complex root m = a + of the auxiliary equation occurs only i n conjunction with its conjugate ih =

Equations of Higher Order

153

a — i/3. B y the use of formulas 8.13, the integrals e"^' and e^' can be replaced by the real ones, e"' cos 0x and e"' sin jS.t. T h e complete linear equation of the order n is

and is expressible by the use of differential operators i n the form {D -

mi] ID -

m2\ - • • {D -

m^ly

=

f{x).

A particular integral of this equation is therefore obtainable by integrating successively the equations -milri =

f{x\

— 7722 Y2 =

Viix),

\D — tn^^Yz =

y2{x), »

{D-~m,]y

T h e method of undetermined and 8.9, can be applied without higher order if its function f{x) kind mentioned in Section 8.8. particular integral. Example 16,

=

coefficients, as described in Sections 8.8 modification to a differential equation of is made up of terms and factors of the This method can then be used to find a

T o find the general integral of the differential equation

T h e auxiliary equation is m^ -

2m^ + 6m=^ -

32m + 40 = 0.

Since this equation can be given the form [m -

2\^{m^ +

2m +

iO\ =

0,

it is seen to have m = 2 as a double root, and to have also the roots — 1 ± 3i. Thus the reduced equation has the integrals i " * and xe"^^, along with cos 3x and sin 3A:. T h e complementary function is u = cie'^ + c^xe'^ + Czé~^ cos "hx + c^e"'

sin 3A:.

T o find a particular integral of the complete equation we make the change of variable y = e^w. This transforms the given equation into ^ [ 4 ] _^ 2«^'^' + dw" -

22«;' + 13«J = 2x^ -

3x + 1.

154

Linear Equations with Constant Coefficients

T h e substitution w = qox^ + qix'^ + qzx + 73

is now found to fulfill this equation if qo = 2, qi = — 6, 72 = " " 3 , a n d 73 = 4. Thus y = tf*{2x^ - 6x^ - 3x + 41 is a particular integral. B y adding the complementary function, the general integral is obtained. PROBLEMS Find the general integral of each of the following differential equations. 81. u''> - 2u" - u' + 2u = 0.

82. «1^1 + u " -

83. u^^l + 6 u " + 5u' -

84. uf'I -

85.

-

86.

-

12u = 0.

2ul''J _ 9 „ " + 2u'

4«' -

4u = 0.

6 u " + 3ii' + lOo = 0,

+8u=0.

- 21u" + u' + 20u = 0.

87. a'^J + 2uf^] + l O u " + 18u' + 9u = 0, 88. uf^J -

2u'^' + 5u" -

89.

2u" -

8u' + 4u = 0. 90.

+ u" + 8u' -

91. u'^l - f 6 u " + 8u = 0.

92.

+ 6 u " + 5u = 0.

93.

94, y^^i + 3>" -y'

-

+ 2/' -

3u' + lOu = 0.

3>' =

X.

95. y [31

3;-" + 3^' - ;r = 5x -

97. y^^^ -

7>" + \9y' -

98.

3y" +y'

-

96. ^1^1 -

lOu == 0.

-3y

=* H

Sy" + 16>' = 4x + 2,

13)- = tf2'{2 cos x + 6 sin x}.

-\-5y = ^'(4 cos 3x + 33 sin 3x}.

99. y^*i + 4^f^l + 8;-" + Sy' + 4;» = 1. 100.

-

8>'31 + 26>" -

40;-' 4- 25> =• 5,

ANSWERS TO ODD-NUMBERED PROBLEMS 3. u = if2.='f--'> + * - ^ f ' - i M

I. » = 5. u = [(2 -

/2:

'\/2)x + l U " ^

9. u = i2 - x ^ 3x-9 13. u = f-'|4 cos

X —

7. w = 0. 11.

6 sin x}

15. u = ^''^-{cos 2x -

f s i n 2xj.

17. u = —cos [5x — 15

19. u a 0.

21. u = ^2^^^^' cos {7x - 7 } .

23. u = - Z i ' t ' ^ - ^ ' cos i^rx -

25. y =

27. y =

+ ,i,2r

2g'

ci

-

sin x + ¿2 cos x.

Answers to Problems 29. > -

155

U + 2 } « - ' log {x-\-2\ + {cix -h czU-'.

31. ^ -

+ C2f.-2*

2*-' +

-

2 + cos V 2 * + — ^ sin

33. jy

-N/2 35. ) r - sin

+ ^ cos 3x + 2 sin 3x.

X

37. > ^ x V + S i x + l l * ' ^ ^ 41. ^ = ^e' +

39. > = 43. y = e- —

X +

^2 cos x.

1.

+ f2*

45. y - T^x —

sin

+ e~^^\ei sin 3x + ci cos 3x)

47. ;y = 49-

y = ^x^ +

51, J- =

X

+ -s

e^{ci cos jf +

-

-1 + ^ log {1 +

sin

x\.

log |1 + ^} + cie' + C2fi~'.

- 4*2 _ 12.

55.

>

57. >

59.

> =

61. y

63.

53. ^ =

65. y 69. >

4x' + 24x +

Sin

X

—

T

1

4

2

+ 2x

71. >

cos

7 j sin

73- > = Tff{si" 2x -

{2x*-i-x^

=i

67, > = sin

+ 4*-'.

2x* + 3*^

=- X * -

-

+ \2x + \U bx

_ X +

2 cos

x.

4

X -

•

5x + 8} cos

X.

2 cos 4x1.

75. _y ^ {sin 3J: + 2 cos 3x)f* + [2 sin x — 5 cos x]e 77. >

5^(sin x + sin 2JC + sin 3jej.

79. y

ijjt' +

+ 4JC + 21 + ^{2 sin X +

81. u

cos x).

83- tf - cid^ + c»r-^^ + €^^^^

85. « * + C2^e * + ^3 cos 3x + Ci sin 3x,

87. a ™ 89.

.2x sin x. + C2«"'2x ^^ cos X + cz€^^

u =- ci^

91. u ^ ci cos 2x + ^2 sin 2x + f3 cos ^2x 93. :y -

- - i x ^ - |x + o + -:2^ + -:3^"^^

95. > - x^ + 97- > =

+ ^4 sin \ / 2 x .

X

sin

X +

3 + cie^ + ^2«^^ + c ^ x V . ^itf* +

C2fi^^ cos 2x + c^^^ sin 2x,

99, ^ = T + ci€~' cos X + c^fi"^ sin X + czx$~^ cos x + axe^^ sin x.

CHAPTER

9 Applications of Linear Differential Equations of the Second Order 9.1. Simple harmonic motion A particle moving on a straight line is said to be in simpU harmonic motion if its acceleration is constantly directed toward a fixed point of the line and is proportional to the distance (displacement) from that point. T h e fixed point is called the central point of the motion. T o be directed toward this point the acceleration must be negative when the displacement is positive, and vice versa. It is therefore a restoring acceleration. W h e n the origin is taken at the central point, and the displacement is denoted by the acceleration is d^y/dt^ and is proportional to y with a negative constant of proportionality, say —ß^. T h e differential equation for simple harmonic motion is thus (9.1)

d^dt^-^-ß'^y

-

0.

This is a differential equation of type 8.1, with / i n the place of x. Its auxiliary equation is + = 0, and has the complex roots +ßi. By (8.11), therefore, its general integral is given by any one of the formulas y = C2 sin {ßt H- ci\,

(9.2)

y = C2Cos {ßt + ci\, y — c\ sin ßt + C2 cos ßt.

T h e constants ci and cz can be determined to give the solution w h i c h fulfills the conditions (9.3)

:v(/o) =>o,

v{t) = vo,

w i t h any assigned values of /QJ >0J and VQ. It is clear from formulas 9.2, for instance from the first one, that, as / increases, y goes through a cycle of positive and negative values, and that 156

Simple Harmonic Motion

157

this cycle is repeated again and again. T h e motion is therefore oscillatory, that is, vibratory, and periodic. Since the largest and smallest values of a sine or a cosine are 1 and — 1 , the largest value of ^ given by either of the first two formulas 9.2 is \c2\. This m a x i m u m displacement is called the amplitude of the motion. T h e motion completes a cycle i n the time it takes the value [0t -\- ci\ to increase by lir. This interval, 2ir/0, is called the period of the motion. Thus Period = Irr/^.

(9.4)

T h e number of cycles completed in unit time is the reciprocal of the period. This is the frequency of the motion. Thus Frequency =

(9.5)

27r'

T h e character of the variation of y w i t h time / is shown by F i g . 23. indicated values cj and C2 are those of the first formula 9.2.

The

Fio. 23.

Example 1.

A particle is i n simple harmonic motion with y = "3, and y = 0, at / = 0. These same values apply to y and v again 3'^ second later, but not before then. T o find the formulas for y and v at any time. If the second formula 9.2 is used to describe the motion, the conditions given take the form C2 cos

Ci =

3,

—C2^

sin

ci =

0,

C2

cos

= 3.

T h e second of these three equations is fulfilled if ci = 0. T h e first is then fulfilled if C2 = 3, and the last if/3 = AT. T h e required formulas are thus = 3 cos 4ir/,

v =

— 127r sin Airt.

Applications

158 Example 2.

A particle in simple harmonic motion passes through the central point at ( = 1, and at intervals of second thereafter. W h e n it does so at / = 1, its velocity is 10, T o find the amplitude of the motion. T h e interval between passages through the central point is a half period. T h e period is thus % second; hence, by (9.4), /3 = 5ir, T h e first formula 9.2 therefore gives y = €2 sin I Sirt +

],

v ~ Sirc2 cos {Sirt +

1•

T h e conditions^(1) = 0, v{\) = 10 require that €2 sin {5ir + €i]

= 0,

T h e y are fulfilled by ci = — 5T, and

5ir^2 cos {Sir + ^ri) =

= 2/x.

y = (2/T) sin Sirj/ -

10.

T h e formula is thus

Ij.

and from this the amplitude is seen to be 2/T. PROBLEMS 1. A particle in simple harmonic motion with the period T has the position > « 2, and the velocity P = 4, at ( — 0, Find the amplitude of the motion, 2. A particle in simple harmonic motion with the amplitude 8 passes through the central point with the velocities ± 1 0 , F i n d the frequency of the motion, 3. A particle is in simple harmonic motion with a frequency of 10 cycles/sec. and the amplitude 7, What is its speed as it passes through the central point? 4. A particle is in simple harmonic motion. Its velocity as it passes the central point is twice its velocity as It passes the point y = Find the amplitude of the motion, 5. A particle is in simple harmonic motion with the amplitude 6 and the period x / 4 . With what velocity does it pass the point ^ = — 3? 6. A particle is in simple harmonic modon with the period 4. At f 0 its position is given b y ^ ^ and at / =» 1 it is given by = ^ i Find the formula for the position at any time. 7. A particle is in simple harmonic motion with the period p. When its position is given by J' = its velocity is ±.v\. What is the amplitude of the motion? 8. A simple harmonic motion has the frequency 10. At / = / i , the velocity is and ji second later the velocity is Find the formula for the velocity at any time, 9. The motion of a particle is along a straight line. Its acceleration is always directed away from the origin on this line, and is equal to the distance &om the origin. Find the general integral of the differential equation for this motion. 10. If in the motion described in problem 9, ^ = 2 , formula for the distance at any time?

and r = 2, at ^ =* 0, what is the

Free Vibrations

159

11. Ifj in the motion described in problem 9, the position and velocity of the particle at ( = 0 are y S and :^ = — 5, find the position which the particle approaches as a limit when the time increases Indefinitely12, If, in the motion described in problem 9, the position and velocity of the particle at f = 0 are 7 = 3 and v ^ 0, and at / = 1 the position is ^ = 5, find the formula for the position at any time.

9.2.

Free vibrations

M a n y mechanical systems include a heavy particle which vibrates with respect to some coordinate when it is displaced from its equilibrium position. W i t h the varying coordinate taken to be^, Newton's law of motion 3,2 and the relation 3,3 give - 5 = / ,

m

g dt-

where / is the force acting i n the jf-direction. T h e following discussions a, bj and c apply to such mechanical systems. a. A coil spring (or an elastic string or wire) resists stretching by a force T which is proportional to the amount by which it is already stretched. Thus, if its unstretched length is /o, its backpull when it is stretched to the length / is (9.7)

/o(,

T = k\l-

with some constant k. This value T is also the pull which must be applied to stretch the spring to the length /, for the stretching ceases when this pull just balances the resistance of the spring. If the spring hangs vertically from a point of support and a particle of the weight w is suspended from its lower end, it stretches to a length / i , where (9.8)

w -

k[h

-

/o|.

I n equilibrium, then, l\ is the length of the system. F r o m w and l\ the value of k is determinable through relation 9.8. T h a t is the value of k in (9.7). Consider the particle now when it is displaced by the amount^ from its equilibrium position, as shown in F i g . 24. T h e resultant of the upward force of the spring and the downward pull of gravity is T ~ w. B y (9.7) and (9.8), this resultant is k[l - h], that is, {w/{h - h)]\l - h\. Since li ~ I = y, the resultant is —wy/{lx — /o), and the equation of motion 9.6 is therefore g__

160

Applications

Because this equation is 9.1 with = g/{l\ ~ lo)j the vibration of the suspended particle is seen to be simple harmonic, with the equilibrium position as the central point. Example 3.

A certain spring, shown i n F i g . 24, stretches by 6 i n . under a pull of 4 lb. A 1-lb. particle is attached to it, and is released at / = 2, with the velocity 0, from a point 2 i n . above the equilibrium position. T o determine the formula for the position of the particle at any later time. I n the foot-pound-second system, equation 9.7 holds with 7" = 4, and {/ — lo] = M ' T h e spring is therefore one for which k = 8. F o r the S

1

I 1

•

y

i Fio. 24.

1-lb. particle the equilibrium position is, by (9.8), that i n which {/i — lo] = -ff. T h e equation of motion (taking g = 32) is thus d'^y/dt^ + 256;» = 0,

Its integral, for which y{2) = i , v{2) = 0 , is = i cos ¡16/ - 32}. F r o m this evaluation of y the amplitude of the motion is seen to be -B- ft., and the frequency is 8/ir cycles/sec. b. A body floating i n a liquid is acted upon by the usual downward pull of gravity, but also by an upward force called its buoyancy. This buoyancy is due to the supporting action of the liquid, and is equal to the weight of the liquid which the body momentarily displaces. T h e position i n which the body Hoats in equilibrium is that in which the buoyancy just balances the weight. If the body is displaced from the equilibrium position it bobs up and down with an oscillatory motion. For a body which is so shaped and ballasted that the amount of liquid it displaces is proportional to the depth s to w h i c h it is immersed, the buoyancy is given by a formula B = ksj with some constant k. T h e weight w and the depth so of immersion in equilibrium are thus connected by the equation w = ksQ. W h e n the body is displaced so that its depth of immersion is as in F i g . 25, the resultant upward force is B ~ w.

Free Vibrations

161

This is k\s — SQ], that is, {w/so){s — SQ]. It is therefore —{w/so)y, y = SQ — Sj and the equation of motion is thus d'y/dt^

=

-

if

(g/so)y.

This determines the motion of the body to be simple harmonic.

^

• —j

t

'

^ Fio. 25.

F r o m an analysis of Fig. 26, the differential equation for the motion of a simple pendulum of length / is found to be d'0

(9.9)

—

dt

g

+ ^ sin I

= 0,

where 6 represents the angle between the pendulum and the vertical. Equation 9.9 is not of form 9.1. T h e motion of a pendulum is therefore

FIO.

26.

not simple harmonic. However, it is nearly so if the angle of swing (measured in radians) is small, as it is, for instance, in a pendulum clock. T o show that the motion is nearly simple harmonic we draw from the third of equations 5.8 the relation » - sin ^ = — 3!

-

—

5!

+ i : 0^ - T T 5^ + • 7! 9!

É «

162

Applications

This is an evaluation of the change that is made if sin 9 is replaced by $. T h e relative change, namely, the ratio of {$ — sin 0] to $ itself is thus ^ (9.10) ^

^

e—

smB $

1 ^ = — ^2 3!

1 5!

. e*

42

I n (9.10) each quantity within a brace is positive when 0 ^ir. W e see, therefore, that the left-hand member of (9,10) is less than the first term on the right. T h e relative change made i n replacing sin 6 by $ is thus less than 6^/31 This is less than 0-5 per cent when \d\ < 0.175 (i.e., 10°), and less than 2 per cent when 6 < 0.35 (i,e-, 20^). W i t h only such relative errors, therefore, equation 9.9 is replaceable by the differential equation d^e g dr

I

T h e variation of 0 i n accordance with this is simple harmonicPROBLEMS 13. A coil spring ha5 the unstretched length 2 ft. When a 4-lb. particle hangs from it the equilibrium length is 4 ft. T h e particle is pressed down until the spring is 6 ft. long, and is released from that position, while it is at rest, at time i = 0. Find the formula for the position of the particle at any later time /, 14. A piece of steel wire 100 ft. long stretches by 6 in. when a 50-lb. particle is suspended from it. What is the frequency with which this suspended particle oscillates when displayed vertically from the equilibrium position? 15. A 1-ft. cube of wood is ballasted so that it floats with a horizontal base in water. In equilibrium it is just half immersed. If the block is raised 1 in. from the equilibrium position, and is there released while it is at rest, what are the amplitude and period of its resulting oscillation? 16. If the block described in problem 15 is pressed down until its upper face is at the water surface, and is then released while it is at rest at time f = 15, what is the formula for its velocity at any later time i? What b the highest position to which it will rise? 17. A simple pendulum swinging in a small arc without resistance completes the cycle of its motion in 1 sec. What is the length of this pendulum?

}^

18. A certain coil spring is stretched 1 in. by a pull of 10 lb. A particle of weight w lb. suspended from it oscillates with a frequency of 5 cycles/sec. What is w? 19. A buoy in the shape of a right circular cylinder is ballasted so that it floats with its bases horizontal. Its period of vertical oscillation is 1 sec. What is the depth of immersion at which it floats in equilibrium? 20. A simple pendulum swinging in a small arc without resistance passes through the vertical position at t ^ 2, It has at that instant an angular velocity {dB/di) ofta radians/sec. What is the formula for its position 6 at any later time /?

163

Damped Vibration

21. T h e length of a simple pendulum is ft. At ( = ti, the pendulum has the position 5 = "5 radian, and its angular velocity is zero. What is the value of 6 at the instant % second later? 22. A pull of 5 lb. stretches a certain coil spring by 2 ft. A particle of weight 5 lb. hangs from it in equilibrium, and while in this state it is given a hammer blow which gives it a velocity of 6 ft./scc. Find the formula for the position of the particle / sec. later.

9.3. Damped vibration W h e n a mechanical system whose free vibrations would be simple harmonic moves in the presence of a force which resists the motion, the motion is said to be damped. If the resistance is proportional to the velocity it appears i n the right-hand member of the equation 9.6 as a term —rv, namely, as —r dy/dt, w i t h some positive constant r. W e shall call r the coefficient of resistance. If the restoring force of the system is —ky^ the equation of motion 9.6 is (9.11)

-•—2=

-ky

g dt'

r

'

dt

which is d^y

(9.12)

dy

^ 2 +^^^

+ ^^ = 0^

with (9.13)

w

w

T h e general integral of equation 9.12 is given by formula 8.9 or 8.11, w i t h ^ and / in the place of u and x. I n these formulas mi and m2 are the roots of auxiliary equation 8.8, and a and /3 are the real and imaginary parts of m i . As we shall see, the character of the motion depends upon the relative sizes of a and 6, that is, of r and k. CASE

(9.14)

1.

a^ < 4b.

I n this case the roots mi and

m2

are complex, with

/3 = i -N/46 - a^.

a = -a/2,

The general integral of the equation of motion is therefore given by any one of formulas 8.11, that is, by y = ae"'

(9.15)

cos {^t

y = ca^'^'sin

+

+

y = e"^{ci sin ^t +

C2 cos

164

Applications

T h e first of these formulas shows, through the factor cos {^t + ¿^1!, that the motion is oscillatory with the frequency ^/2ir. It has i n the place of an amplitude constant the factor cj^"'. This decreases as t increases, since a is negative. T h e motion is therefore a vibration w i t h a fixed frequency, but w i t h an amplitude that diminishes toward zero with incresising time. T h e graph of such a motion is shown i n F i g . 27. T h e exponential tf"', that is, is called the damping factor. A n increase i n r produces a n increase i n a and a decrease i n ^. Increasing the resistance therefore hastens the damping, and decreases the frequency of the motion.

Fio. 27. Example 4.

A certain coil spring is stretched 1 ft. by a pull of 3 f l b . A 5-lb. particle is suspended from it, and is set to vibrating. T h e air resists the motion w i t h a coefficient of resistance ^. T o find the frequency of the motion. T h e value of k for this spring is T h e equation of motion is accordingly _5^^__29 ndt^

~

~

8

_ 1 _ ^ ~ Sdt'

T h e general integral of this equation is y = ^26-2^/5 cos {^t + Ci]. F r o m this formula for> the frequency of the motion is seen to be 12/5T. CASE

2.

a^ ^ 46.

I n this case the roots of the auxiliary equation are

Damped Vibration

165

T h e y are, therefore, real and negative. Hence (9.16) V =

micif"*!* + m^2t^^\

if a > 4Ô,

and (9.17) V ^

{C2 + mici + miC2/)^'"*',

if a'^ = 4b.

Since a l l the exponentials i n these formulas decrease toward zero as a limit when t increases, the values of y and v are damped out. N o right-hand member of a formula 9.16 or 9.17 vanishes for more than one value of T h e moving particle therefore passes through its equilibrium position once at most, a n d changes its direction once at most. T h e motion is clearly non-oscillatory. Example 5.

A certain coil spring is stretched 8 i n . by a pull of 1 l b . A 1-lb, particle is suspended from it, and the system is immersed i n oil for which the coefficient of resistance is -y. T o determine the motion for w h i c h ) ' = 1, and v = 0, when ( — 0. T h e equation of motion (with g — 32) is

Its general integral \&y = given conditions is

\ d^y

3

1 dy

32 dt^

2^

2dt

-1- C2e~^^\

y = "^e

and the integral that fulfills the

— se

I n the right-hand member of this formula the first term is obviously always larger than the second when / > 0. T h e y both approach zero as t increases. T h e moving particle thus subsides, without any change i n the sign of y, to the equilibrium position. Example 6.

T o determine the motion of the system of example 5, for which)* = 3, and v = — 4 4 , when / = 0. T h e integral of the equation of motion that fulfills these conditions is y -

-e-*'

+ 4e-'^',

v = Ae-'' -

48^-*^'.

T h e moving particle passes through the equilibrium position )» = 0, i n this instance when ~e~^^ + 4^~'^' — 0, that is, when t = ^]og2. It changes its direction when Ae~^ — 48^^'^' = 0, that is, when / = ^ log 12. Thereafter it subsides to the equilibrium position.

Applications

166

PROBLEMS Solve the following problems, assuming the units to be feet, pounds, and seconds. Take g - 3223. A certain coil spring is stretched 2 ft. by a pull of 1 lb. For a 1-lb. particle suspended from it the coefficient of resistance is x V F i ^ d the formula for ^, if > 3, and p = 0, when / = 0, 24. For the system of problem 23, find the time when when ( = 0,

= 0, if;' = 0, and v = \2,

25. A certain coil spring is stretched 3 in. by a pull of 1 lb. A 2-lb. particle is suspended from it, and for this the coefficient of resistance is 1. Find the formula for i[y(0) = ^'o, and v(0) — VQ. 26. A certain coil spring has a 3-lb. particle suspended from it. This stretches it ^ ft. T h e coefficient of resistance is x- Find thcformula for y^ if >(0) = 0 and v{0) = 7. 27. For the system of problem 26, find the formula for v if >(0)

3 and r(0) = 2,

28. For the system of problem 26, find the frequency of the motion, and find what the frequency would be if the resistance were not present. 29. Derive the differential equation for the approximate motion of a simple pendulum swinging through a small arc, if there is a resistance of which the coefficient is r. 30. A simple pendulum is 32 ft. long, and has a bob weighing 4 lb.

What Is the

smallest coefficient of resistance for which this pendulum will not oscillate? 31. A simple pendulum has a bob weighing 1 lb. It is to swing in a small arc with a coefficient of resistance xtIts frequency is to be -y. What must its length be? 32. A particle moves on a horizontal line against a resisting force which is proportional to its velocity. N o other force acts upon it. Find its position at any time if y ^ yoi and v ^ VQ, ahen ( ^ 0.

9.4. Forced vibration without clamping.

Resonance

A mechanical system i n w h i c h a l l the forces are internal, as i n those considered i n Sections 9.2 a n d 9.3, is called a free system. Its motions are said to be free motions. A system upon w h i c h an outside force acts is called a forced system. W h e n a force /(/) acts upon a system w h i c h i n the free state obeys equation 9,11, the dilTerential equation for the resulting forced motions is g dt

dt

W i t h a and b determined by relations 9.13, this equation is (9.18)

g

at

+ a

dt

J+

w

Forced Vibration without Damping.

Resonance

167

This is a complete differential equation. It can therefore be solved by finding a particular integral and adding the complementar)' function to it. T o find a particular integral, any one of the methods of Sections 8.5, 8.6, 8.8, or 8.9 may be used. T h e case in which j{t) is a pulsating force which varies harmonically in accordance with a formula (9.19)

/(/) = Ksin \yt + 5}

is of special interest. F o r a system without damping, whose free motion would accord w i t h equation 9.1, the equation of motion is then (9.20)

-J+^^y dt

=

w

~sm{yt-{-d

T h e character of this motion depends, as we shall see, upon the relative values of/3 and 7, and especially upon whether or not ^" and 7^ are equal. 1. 7^ jS^. A particular integral of the differential equation (9.20) , may be found i n this case, by the method of undetermined coefficients of Section 8.9, to be CASE

T h e general integral is thus (9.21)

y =

^ s i n ¡7^ + 51 +

sin {^t + d}.

For each possible motion, c\ and have specific values. A motion 9.21, for which C2 = 0, is simple harmonic. It has the same frequency and phase as the impressed force, but ordinarily it has a diff'erent amplitude. A motion for which C2 7^ 0, on the other hand, is compounded of two motions with different frequencies. It is therefore less simple. If the two frequencies arc commensurable, namely, if ^/y is a rational number, the motion is periodic and executes a certain pattern i n the course of a period. T h e pattern is, however, more intricate than that of simple harmonic motion. Figure 28 shows the graph of such a motion for which ^/y = 4. T h e period of this motion is 27r/7. A motion for which 5^ 0, and ^ and 7 are incommensurable, is not periodic but has an aspect of being erratic. However, even such a motion is bounded, for neither of the sine functions in formula 9.21 takes on a value numerically greater than 1. Hence l^^l never exceeds the value

168

Applications

T h e range of values w h i c h ^ may take on is, however, large when {/3^ — 7^) is small. 2. 7^ == RESONANCE. A particular integral of differential equation 9.20 is in this case found to be CASE

2ßw

cos {ßt + a).

and the general integral is accordingly (9.22)

y

=

~gKt

2ßw

cos \ßt + 51 + C2 sin {ßt + Cl]

T h e motions which accord with this formula are quite dissimilar from those of Case 1, for the multiplier of the cosine term in (9.22) contains / as a factor. T h e amplitude of the motion therefore increases indefinitely as

Fro. 28. t increases, which is to say that the motion becomes increasingly violent. A n actual mechanical system is i n this case pushed to the bounds of its physical limitations, the point where it must break down. This phenomenon is called resonance. It results when the impressed force is of the same frequency as the free vibrations of the system. A marching army is said to break step i n crossing a bridge to forestall such strains as may result from actual or approximate resonance between the frequency of the footbeat and that of the free vibrations of the bridge. T h e application of a vibrating force 9.19 to a particle suspended from a spring, as i n F i g . 24, affords a simple instance of a system in forced vibration. A way i n which this can be realized practically is to make the point of support S of the spring itself move vertically in a simple harmonic manner. T h u s if S is raised by the amount F, the length of the spring w i l l be

Forced Vibration with Damping { y + / i — y]then be

169

T h e equation of motion for the suspended particle will

w at

¿1 — ' 0

If y = A sin ¡7^ + 5 j , the previous equation becomes

0

This is clearly of form 9-20, PROBLEMS Solve the following problems, assuming the units to be feet, pounds, and sccondsTake g = 32, 33, A particle of weight 2 is subject to a restoring force which is numerically equal to of its displacement. It has also the outside force / = sin 2l impressed upon it. Find the motion for which ^ = 0, and v = -^^ when / = 0. 34- Find the motion of the particle in problem 33 for which)' = ^-^andv

= 0, when

/ = 3x/4. 35, A particle of weight 32 is subject to a restoring force which is numerically equal to 9 times its displacement. It has also the outside force / = 2 cos 3f impressed upon it. Find the motion for which ^ = 2, and = 0, when t = 036, A particle subject to a restoring force vibrates freely with a frequency v greater than 1, When the outside force / = cos / is impressed upon it, the motion for which y — yu and v = 0, when ( = 0, is simple harmonic. Find the value of ^ i , 37, Find the general integral of the difTcrential equation fy , „ . . . . . 7 - ~ + /37 = 2 sin 7' + 5 sin ^ I,

(9.23)

assuming that the motions arc not those of resonance. 38, Show that there are two values of y for which the motions defmed by differential equation 9.23 are resonant. Find a particular integral of the differential equation for each of these values of 7.

9.5. Forced vibration with damping W h e n a damping resistance is present i n a forced vibrating system the law of motion is given by (9.18) with (9.19), that is, by the differential equation (9.24)

^

at

+ ^ ^ 4. at

= ^ sin ¡7/ + Ô|, w

170

Applications

with a > 0, T h e roots of the auxiliary equation are then either real and negative, or complex with a negative real part. T h e complementary function therefore approaches zero as a limit when t increases indefinitely. Every motion of the system accordingly subsides into the motion that is represented by the particular integral. This is called the steady-state motion of the forced system. As it is found by the method of undetermined coefficients, its formula is y =

rl2

,

-

2 2>

-y') sin (T¿ + 5) -

cos (7¿ +

5)1.

W e can apply to this formula the general trigonometric relation (9.25)

^ sin i> ±

fíeos

=

+

fí^sin

± t a n - ^ (5/^4)1,

to give it the form (9.26)

= ^ s i n | 7 i + ( 5 - 61)!,

with (9.27)

K =

,

^'^

=y

« ; V ( e - T ' ) ' + «'7'

5i = t a n ' ^ ^--^

F o r m u l a 9.26 shows that the steady-state motion is simple harmonic, w i t h the same frequency as the applied force. T h e motion is, however, not i n phase with the force, for, by (9.19) / = 0, when / = (nx — 5)/7, w i t h n any interger; whereas, by (9.26) = 0, when / = \m — (6 — 5i)l/7, that is, at times w h i c h are later by the amount 61/7. T h e amplitude K of motion 9.26, though it is given by 9.27, can also be put into the alternative form (9.28)

K = w

If — ^ 0, the value of K always decreases when 7 is increased, that is, when the frequency of the impressed force is raised. O n the other hand, if — 6 < 0, the value of K first increases to the m a x i m u m (9.29)

-

^'^

w

and then decreases. T h e m a x i m u m is attained when 7 = V 6 — For this value of 7 the system is said to be in resonance. T h e resonance

Forced Vibration with Damping

171

amplitude 9.29 is evidently large if a is small. Unless sufficient damping is present, therefore, the breakdown of a mechanical system at resonance results. T h e frequency at w h i c h an applied force is i n resonance w i t h a damped system is 1

(9.30)

^TZl.

This is somewhat less than the frequency of the free vibration of the systern, w h i c h is (1/2T) '\^b — PROBLEMS 39. F i n d the formula for y in the steady-state motion given by the differential equation dh

dy + 2 ^ + 10, = cos 2/.

40. Find the formula for y in the steady-state motion given by the difTercntial equation + 2-~ + II7 = sin yt. dt^ • -dt if 7 has a value for which the system is in resonance. 41. Find the general integral of the differential equation

^ + 3 -

+

2,=sm

and put it into the form that displays the phase difference 2i between the force and the steady-state component of the motion. 42. Find the formula for y in the steady*state motion given by the differential equation

when y has such a value that the phase difference b\ between the force and the motion has the value ir/2,

43. Find the general integral of the differential equation

Observe that all the motions here represented subside completely-

172

Applications

44. Find the general integral of the differential equation ^ at with k > 0. vibration.

+

sin ßt,

=

Observe that as / increases all the motions subside into a simple harmooic

9.6. The forced vibration of a system with a repelling force If a particle that is subject to a central repelling force that is proportional to the distance is acted upon by an outside force 9.19, the differential equation of motion is again of form 9.24. However, the value of è is now negative. T h e auxiliary equation therefore has real roots, and of these one is positive and one is negative. A particular integral is again given by

Fio.

formulas 9.26 and 9.27. formula (9.31)

y

=

29.

T h e motions of the particle are thus given by the

Ksm{yt-\-

8 -

5i\

-\- cie""^* +

Cie""^'.

N o w one of the exponentials i n this formula increases w i t h t. A n y motion whose formula 9.31 includes that exponential is therefore unbounded. U n d e r it the particle flies off in the positive or the negative direction. T h e motions whose formulas 9.31 do not include the increasing exponential are different. They subside into the simple harmonic steady-state motion given by (9.26). A simple example of a mechanical system of this type follows. A smooth straight rod is mounted to rotate i n a vertical plane about one of its points 0. It carries a heavy particle which is free to slide eilong i t without friction or resistance. T h e system is shown i n F i g . 29, where y denotes the angular velocity of the rod in radians per second. T h e rotation of the r o d generates a centrifugal force upon the particle, w h i c h amounts to {ivy'/g)yj and repels it from 0. Gravity is a n outside force

Motion Under an Intermittent Force whose component along the rod \s —w sin {yt + 6]. motion of the particle along the rod is therefore ID d\

17.3

T h e equation of

wy^ y ~ w sva. [yt -\- h\J g

gdt'

that is, (9.32)

= - 5 s i n ( 7 / + 5!.

dt

Some of the motions defined by this equation are unbounded. I n them the particle flies off from one or the other end of the rod. T h e other motions subside into the simple harmonic steady state (9.33)

y -

i

- % s i n [yt ^-h]

2y

T h e polar coordinate equation r — ¿4 sin ^ represents the circle with its lowest point at the pole and with the radius {}'i)A. Equation 9.33 isof this form. I n the steady-state motion the particle therefore describes a circle in the vertical plane, the lowest point of the circle being at 0, and its radius being g/Ay \ PROBLEMS 45. Find the formula for^ in the motion defined by equation 9.32, for which ^ = 0, and 0 = 0, when ( »» 0. 46. Find the formula for D in the motion defined by equation 9.32, for which y — g/3y^, and v = g/^y^ when / = 0. 47. Find the formula for y in the general motion defined by the differential equation (9.34)

+ 6

-

16> = - 3 2 sin 4t

48. Determine what the relation between ^(0) and v{0) must be in order that the motion defined by differential equation 9.34 may subside into the steady state. 49. Find the formula for y in the general motion defined by a differential equation 9.24, in which b ^ 0 and a > 0. Observe that every motion in this case subsides into a simple harmonic steady state, but that these do not all have the same central point. 50. Find the formula for y in the general motion defined by a differential equation 9.24, in which 6 0 and a » 0. Observe that some motions are unbounded and the others are simple harmonic.

97.

Motion under on intermittent force

A force /(() that must be described by different formulas i n different parts of the total time interval is said to be intermittent. A particular

174

Applications

instance of an intermittent force is one which acts during only a part of the time. T o such a force may be assigned the value zero while it does not act. A n intermittent force may well be discontinuous. E v e n when / is discontinuous, however, formulas 8.25, 8.26, a n d 8.27 give an integral for diflferential equation 9.18, w h i c h is continuous and has a continuous derivative. T h e general integral is obtainable, as usual, by adding the complementary function. Example 7.

F o r a certain particle the differential equation of motion is

at

+

2 - + 5> = -fit), at w

w i t h fit) = e^ from / = 1 to / = 2, and otherwise /(/) = 0. T o find the integral of this equation for w h i c h ^(0) = 0 and p(0) = 6. T h e auxiliary equation has the roots — 1 ± 2i\ hence formula 8.27, w i t h X and fix) replaced by t and ig/w)f{t), gives a particular integral. I f we choose to = 0, since the conditions apply at that time, the formula is

2w Jo

N o w , because/(j) = 0 when j < 1, and when j > 2, this formula is more explicitly 0,

(9.35)

^

yp(t) =

\2w

when

0 g / ^ 1,

ds,

when

1 ^ / g 2,

e^'-^ sin 2it - s) ds,

when

( § 2.

sin 2it~~s)

j'^

^

2w

T h e general integral is y = ypiO

+

sin 2t + C2 cos

2t],

and gives p = ^

dt

+

e-^{i-ci

-

2c2) sin 2t +

(-^2

+

2ri) cos

2t].

Since yp(t) a n d its derivative are both 0 at / = 0, the conditions to be fulfilled are = 0, —C2 + 2ci = 6. T h e solution required is therefore y = >p(0 + 3(r~' sin 2t.

I n the instance of this example quadratures 9.35 for^p(/) can be made. T h e resulting formulas are

Motion Under an Intermittent Force 3tf~' sin 2/,

when 0 ^ Z g 1,

^2-'[sin 2(t -

-f- {g' -

Sw

175

1) + cos 2(/ -

1)11 + 3
when 1 ^ Z ^ 2,

y = g

8a;

{e*''[sm 2{t -

2) 4- cos 2{t -

2)] - «'-'[sin 2{i -

cos 2{t Example 8.

1) +

1)11 + 3«~' sin 2/, when t > 2.

T o find a particular integral of the differential equation ^ - y - m .

i f / ( 0 = tjromt = OtoZ = l ; / ( 0 = l , f r o m Z = 1 toZ = 2; a n d / ( 0 = 0, when t > 2. F o r m u l a 8.25 is i n this instance VW = i

pis)

-

ds,

W-

-<:-'+•] ds,

more explicitly, i pU'~' i /„'

s{e'-' -

when

ds +

i

/ ; (.'-' - .-'+•!

yi') = <

0 g < g 1,

ds,

when

1 g < S 2,

when

/ > 2.

Thus, by (3.9), sinh / — t, v(/) = I sinh Z - sinh (Z - 1) -

1,

sinh Z — sinh (Z - 1) — cosh (z - 2),

when

0 ^ Z ^ 1,

when

1 ^ / ^ 2,

when

Z > 2.

PROBLEMS 51. In the difTcrcntial equation

+ 2

I+

2> = /(,).

the function f(t) has the value e"^ from / = 0 to / ^ 1, and is otherwise zero. the formula for ^, if ^ «= 0 and y' = 0, when / = 0,

Find

176

Applications

52. In the difTcrential equadon dS

dy

the function /(/) has the value 1 from i 1 to / = 2, and is otherwise zero. There is a solution for which ^ =* ^, and ^' =» 1, when ( = 1, Find the formula for this solution when ( > 2. 53, Find a particular integral of the differential equation

dt2 if / ( O has the value sec / from ( = 0 to / = ir/4, and is otherwise zero, 54. In the differential equation d'^y

dy

+ i :-

dt^ ' dt

6> = / ( / ) ,

the function f{t) has the values

fit)

= { 0,

when

0 ^ r ^ 1,

when

1 < / < 2,

when

/ ^ 2,

e~^\ Find a solution of the equation.

55, T h e differential equation of problem 54 has a solution for which > =• 0, when J » 2, and which docs not increase indefinitely in numerical value as t increases. F i n d the formula for this solution when 1 ^ / ^ 2. 56. In the differential equation

2 + 3 T + tv=/W.

dt^ ^ ^ dt

the function /(/) has the value / when ( > 1, and is otherwise zero. T h e equation has a solution for which > = 0, when / = 0, and for which also y ^ 0 when t ^ 2. Find this solution.

9.8. Simple electric circuits W h e n an electric circuit has an electromotive force E impressed upon it, a current i flows and a charge q accumulates i n it. A simple circuit is diagrammed i n F i g . 30. Its elements arc a resistance an inductance L , and a capacity C. T h e values of these are constants. I n terms of an appropriate system of units (t i n seconds, E i n volts, i i n amperes, q i n coulombs, R i n ohms, L i n henrys, and C i n farads) the laws that apply are:

(9.36)

I =

do

Simple Electric Circuits

177

and di

(9.37)

g

^ 7 / + ^ ' + ¿ = ^'

These are derived i n the elementary theory of electric flow. If the value of i , as given by (9.36), is substituted into relation 9.37, the result is the equation 1

dq

(9.38)

This is the differential equation for the charge q. If, alternatively, equation 9.37 is differentiated and dq/dt is then replaced by its value 9.36, the result is the equation (9.39) This is the differential equation for the current i. Differential equations 9.38 and 9.39 are of the same form as equation 9.18 for the motion of a particle. T h i s reflects an analogy between the

WW R

L

30.

FIO.

phenomena of electric flow and those of mechanical motion, particularly between those of alternating currents and those of vibrating particles. Under an electromotive force E which pulsates in simple harmonic fashion, there may be a manifestation of resonance, of a steady-state flow, etc. Example 9.

F o r a certain electric circuit L = ^Vj ~ 4, and C = -r^ffT o find the formula for q,ifq = go, and i = to at Z = 0, and £ = 0, when t > 0. Differential equation 9.38 is i n this instance 1

do

25 dt 2 + 4 dt + 676? = 0.

178

Applications

Its general integral is q =

s i n l 2 0 / + <:2cosl20z),

and from this, by (9.36) = ^-6o<j[_50(:i -

12OC2I

sin 120/ + [120ci -

5OC2I

T h e conditions are fulfilled if ^2 = ?0) 120iri — 50^2 = formula is therefore '°

cos 120/i. T h e required

sin 120/ + qo cos 120/

Example 10.

F o r a certain electric circuit / , = 1, i f = 40, and C 1/20,000. T o find the steady-state current when E = 120 cos 150/. Differential equation 9.39 is i n this instance d^i

di

dt

dt

V 5 + 40 y + 20,000z = - 1 8 , 0 0 0 sin 150/. It has the particular integral . -36 . z = — s m

150/ +

tan-i^

and the complementary function e~^'^^[c\ sin 140/ + cn cos 140/j. Since the complementary function approaches zero as a limit when / increases indefinitely, the steady-state formula for i is that given by the particular integr2il.

PROBLEMS 57. For a certain electric circuit L =

R ™ \i and C = y ^ ^ .

Find a particular

R = 40, and C = TDVO"-

Find the formula

formula for g if £ ™ —5 sin 140(. 58. For a certain electric circuit L = for ^ if 7 = iTTOirt

I = 10, at ( = 0 ,

and E — 0, when t > 0.

59. For a certain electric circuit L = TTJ-* R ^ tor »• if £ - 10^-><" + 50 sin 60t.

and C «• T O -

Find the formula

60. For a certain electric circuit L = ^¿11, R = i i r , and 1/C = 0.

Find the formula

for I, if £ = 5 cos 200/. ^, R = 0, and C = TtnnT-

Find the formula

62. For a certain electric circuit L «• 5, i? = 0, and C = u^nnj"for i' if £ => 10 cos 1100/ + ir/A\.

Find the formula

61. For a certain electric circuit L for I when £ = 10 cos 180/.

63. For a certain electric circuit Z. = ^, and C = 5a ^.

How large must R be in

order that the current may merely subside to zero without oscillating when £ is reduced to zero?

Answers to Problems

179

64. For a certain electric circuit Z. = y , i2 = 20, and C = T S U value of the steady-state current when E = 5 cos 60/.

Find the maximum

ANSWERS TO ODD-NUMBERED PROBLEMS 1. 2 V 2 .

3. ±140ir.

5. ± 2 4 V i .

7. Vyi" + 0'=^^IV4T2)11. > = 0.

9. > = cie' + C2e~'.

13. ^ = - 2 cos V(g/2) t. 17.

15. Amplitude inj, period ir \ / 2 / ^ ,

19. i f t

/ = ^ / T T ' ^ ft.

21. e = 0.1.

23. V = 4^-2* _ ^-8i_

25. H8>o +1-0]/ +>o|^-''.

27. & _= ,-4(/3 |2 cos At -

29. - 2 + - + f e = 0. dt^ w dt I

31. / = 32/(T2 + 1).

33. y = - Y s i n 2/ + 11 sin /.

35. y = it sin 3/ + 2 cos 3/.

sin yt +

37. jy

39. )- =

41. > =

1

14 sin 4/j.

sin - ( + C2 sin {j3; + ci|.

2/ + t a n " ' I

sm

2 -\/l3

1

VlO

stn

/ -I- ' 4

—21

tan-'3

cos | i + Uit + c s l ^ - ^

43.

45. > =

g 4T

¡2 sin yt -

+ ^-T'l'

47. > = I sin |4f + tan-^ J} + CK^' + C2^-^'.

49. y =

y^ + -

51. y =

sm 7/ + 5 + tan when

cos ¿1,

^ '|cos (/ — 1) — cos t\y 53. y

-

—at 0 S / ^ 1,

when

i > 1-

M i n / + cos / log cos /,

when

/ ^T/4,

(ir/4) sin i — -ffOog 2) cos /,

when

i > 7r/4.

55, y = 59. i" = 5 sin 60/ +

57. q = TshslA sin 140/ + 3 cos 140/}

sin 10

-\/llr+

61. j" = 40/ cos 180/ + ci sin 180/ + 63. R ^ AO.

fa cos 10 -^/u

cos 180/.

t

CHAPTER

10 The Linear Equation with Variable Coefficients 10.1. Ordinary and singular points T h e general linear differential equation of the second order, i n its complete form, is (10.1)

po{x)y" -hPiix)/

+P2(x)y

=/«,

and i n its reduced form is (10.2)

Po{x)u" + piixW

+

p2{x)u =

0.

W e shall suppose that the functions po{x)j piix), and p2(x) have no common factor. W h e n such a factor is present, it shall be divided out of the equation. A point X at w h i c h (i) the coefficients poix), pi{x), and p2{x) are continuous and (ii) poix) ^

0

is called an ordinary point for the differential equation. A point which is not ordinary is called a singular point. Throughout this chapter we shall suppose that the differential equation is being considered upon a n x-interval that does not contain any singular point. As we have already observed i n Section 8.1, if ui{x)y and U2{x) are any integrals of equation 10.2, the linear combination {ciUi(x) -\-czU2{x)\ with arbitrary constant coefficients ci and C2 is also an integral. If this combination cannot be expressed with the use of only a single arbitrary constant, it is the general integral. F o r any equation 10.2, u{x) ^ 0 is a n integral. This, however, is rarely of any use, and is therefore called the trivial integral. Other integrals are non-trivial. If yi{x) is any particular integral of the complete equation 10.1, the general integral is obtainable from^i(j>f) by adding the complementary function. Aside from the equations with constant coefficients, there arc only a few minor classes of equations 10.1 or 10.2 that are solvable by methods other than that of power series. 180

The Euler Equation

181

10.2. The Euler equation T h e differential equation (10.3)

xol Y

\x -

+ a\x - xo\y' + by = f{x),

in w h i c h a, b, and xo are constants, is called the Euler equation. It has a singular point at XQ. Hence we suppose that ;i: > J:O, or J: < XQ. The change of variable

and

X

= xo -\- e",

X

=

—

XQ

e*,

if

X

if

x <

>

XO, XQ,

transforms differential equation 10.3 into d^y , , + ¡a ds^ • '

,,dy 11 - f +by=f{xo ' ds

±

e')

This equation has constant coefficients and can therefore be integrated. Its integral, when expressed in terms of the original variable x, is an integral of (10.3). Example 1. \x -

T o find the general integral of the differential equation \ \Y

-

4{x -

l|y -

Uy

= x^ ~

3x2 +

3;^ _

8.

T h e change of variable x = \ + e' transforms the given equation into d^y ds^

-

^ dy

5-f " ds

14> = - 1 ± e^\

By the method of undetermined coefficients the general integral of the last equation is found to be y = s + sjie

+

cie

-\- cze

T h e integral i n terms of x is therefore > = i Example 2.

íVí:^ - 1 r +

- 1 r + C2\x ~~

1

r'.

T o find the integral of the differential equation x^u" -

3xu' + 13u = 0,

for which u(l) = — 1 , and u'(\) = 1. I n this instance xo = 0, and the imposed conditions apply at a point where x > 0. T h e appropriate change of variable is therefore x = e'. It transforms the equation into

182

Linear Equations, Variable Coefficients

F r o m this equation the general integral is found to be u — x^{ci sin (3 log x) + C2 cos (3 log x)\.

T h e assigned conditions are fulfilled by C2 = — 1 , and required integral is thus

= 1.

The

u = x^{sin log x^ — cos log x^]. PROBLEMS Find for each of the following differential equations the integral fulfilling the given conditions, or the general integral for the indicated values of x. 1. fjf + 2 | V ' -

20a -

2. \x + 2} V

-

A\x + 2 } « ' -

3. \x -

3} V

+

U -

4. |x -

5}V'+7{jf -

5. xY

-

2xy' +ly

6. x-y"

-

*y' + 5^ -

8. , Y

+^'+

0,

* > 0. 14« = 0,

^\u' + a = 0,

2i,

X < 3. x
5 ] « ' + 9u = 0,

= flog ;r}^ 3,

X

o(0) = 5, u'(0) = 13.

;c > 0.

<0.

= ' ^ ' ° ^ " + \

, > 0.

X

9. \x -

\\y" + 3 / + —

L

-

, -

1,

,(2) -

\x ~ \\ 10. 6 r V ' -Ky'

-yy=

X 12. :ry" +

ly'

6x-^

y{\) -

O.y'CD =

^

sin log

\' 4

--ft-

13

X

--y

^,y(2) 4

13

jt > 0.

10.3. Exact equations A differential equation 10.1 is easily tested for exactness, and is readily integrated if it is exact. After being multiplied by dx the equation can be written i n the form d\poy' -

poy]

+

d[piy\ +

\po" -

pi

+ p2\y dx = f{x) dx.

T h e term \po" — pi -\- p2]y dx is not a differential unless it is zero, for it involves)! but not dy. T h e condition for exactness is thus (10.4)

po"(x) -pr'(x)

-\-p2{x)

=0.

The Adjoint Equation

183

I n the event that the equation is exact, a first integral is po{x)y + pi{x)y =

po{x)y' Example 3.

J / W dx +

a.

T o find the general integral of the differential equation

X

1

2

X

X-

y

f1

2

x^

x'

Condition 10.4 is fulfilled, and the equation is therefore exact, first integral is 1 X

X

y =

X

The

e'-\-c 1

F r o m the last equation the general integral is found to be y =

\xe'

+

c\x + C2xe~''.

PROBLEMS Find the general integrals of those of the following differential equations that are exact. 13. * V ' 14.

+

{3^2 JfxW

|1 + 2e' + e'^'W

-{-u = 0. +

¡1 + li" + Ze'^'W + e^u = 0.

15. u" cos X + u'(csc X — sin jt) — u esc x cot ;c = 0. 16. >" cos X + 17. xy"

+

cos

{ J : ^ sin J : ) ; ' ' +

2 +

+

I. [x^ cos JT + a: sin J : ) ^ = cos x.

6

6JC

1 20. y"

Jf =

y =

4x

X,

0.

2)2 Ax^

{1 -

1 + log ^ X log

{x log x|-2 y = 2x.

X

21. u" + w' tan x -\- u sec" 2JC

,

^" ^ lT7^^'

AT

2|1 -

=

0.

x^\

_

11 + x 2 p ^ "

4x |1 -Tx^i-

10.4. The adjoint difFerentiol equation.

Integrating factors

After a differential equation 10.1 is multiplied by a function v{x), it is exact, by (10.4), if (10.5)

\pQv}" -

[pivY ^-Piv

=

0.

184

Linear Equations, Variable Coefficients

T h i s is a differential equation for f, which is, more explicitly, (10.6)

po{x)v" +

\2po'{x) -

Pi{x)W

+

Wix)

-pi'ix)

-\-p2{x)\v

= 0.

Equation 10.6 is called the adjoint of the given differential equation 10.1 or 10.2. A function v{x) is thus a n integrating factor for a given differential equation, if and only if it is an integral of the adjoint equation. T h e adjoint of equation 10.6 is i n turn found to be equation 10.2; thus each is the adjoint of the other. Example 4.

T o find the general integral of the differential equation \x^ -

x\/'

+

¡2*2 + 4;f -

3 1 / -h Sxy = 1.

T h e adjoint of this equation is {x"" -

x\v" -

\2x^ -\]v'

-\- |4x -

2]v = 0,

B y the trial of A:"*, this equation is found to have x^ as an integral. Thus x^ is an integrating factor of the given differential equation. T h e general integral, as found by the use of this factor, is 1

y =

1 7 — T +

- 5 + *^2«

PROBLEMS Find an integrating factor for each of the following differential equations, and b y the use of it find the general integral. 4x

23. u" +

Bx -S

7 «' + 2x - \

2x - \\

u - 0.

24. {3x-\-x-\u" + |12 + 4* - X 2^ I..' H' -

25. {x

-

26. u " 27. u" +

|4 + Sx\u - 0

1 X

u' tan x -

1

u{2 +

+x

tan x + tan^ x\ = 0.

— ~ u - 0. x{\ +x\

28. ¡2 sin J: — cos x\u" + j? sin x + 4 cos x\u' + lOu cos x = 0, 29. y

+

1 -

1

y

-

1

1

> -

30. / ' + I2x + 11/ + i2x + 2\y -

|1 - x\e-2JC

2*-^

The Existence of a Solution 10.5.

185

The existence and uniqueness of a solution. equation

The Riccati

It was observed in Section 6.7 that any solution of the R i c c a t i equation , • ^^^H

p2{x)

pl{x)

po{x)

po{x)

^

^^^^^^^^^^^^^^

2 ^?

yields, through the formula

an integral of differential equation 10.2, and vice versa. W e can use this fact to show that, when XQ is any ordinary point, differential equation 10.2 has an integral for which (10.7)

U{XQ)

=

Mo,

u'{xo) = ttu',

and that this integral is unique. If U(j ^ 0, the Riccati equation has an integral UO'/UQ. T h e formula u{x) = Wo exp /

T)(X),

for which

TJ{XO)

=

7i(x) dx

Jxo

gives a n integral of (10.2) which fulfills (10.7). If UQ = 0, let ai(x) and U2(x) be integrals for which «i(xo) = 1, «i'(xo) = 0, and «2(^0) = — 1 , "2'(^o) — Wo'have just shown that such integrals exist. T h e i r sum, u(x) = {ui{x) + "2(^)1, is a n integral that fulfills (10.7). Suppose now that two integrals fulfill any given set of conditions (10.7). Their difference is an integral C/(x), for which (10.8)

U{xo) = 0,

U'ixo) = 0.

Therefore both the integrals U(x) + u\{x) and «i(x) fulfill the conditions u{xo) = 1, u'{xo) = 0. N o w each of the functions + ui'(x) '—' — Lh{x) Uy{x)

V'(x)

+

^ and

ui'(x) U,{X)

is a n integral of the Riccati equation, which at XQ has the value zero. There is only one such integral; therefore the two functions are equal, that is, U(x) + ui(x) = kui{x). Since this equality applies at XQ we see that A — 1; hence that U{x) = 0. A n y two integrals fulfilling a specific set of conditions 10.7 are thus identical; the integral is unique. W e note especially that the integral fulfilling condition 10.8 is merely the trivial one. It was shown in Chapter 6 example 17, that an integral of a differential equation 10.2 is sometimes obtainable through the use of the Riccati

186

Linear Equations, Variable Coefficients

equation. T h e method can even be given a somewhat more general form. If i}{x) is an integral of any Riccati equation (10.9)

7} - <Tix)7)-,

po{x)


in which iT(.if) is a function that is not zero anywhere on the given interval, the formula (10.10)

u = exp lj(T{x)ijix) dx]

yields an integral of equation 10.2. Example

5.

T o find an integral of the differential equation xu"

-

\ \u' -

xh'u

= 0.

Riccati equation 10.9 is i n this instance xe^ -\-\

-

W h e n a = X, this equation has an integral e' given differential equation has the integral

Thus by (10.10), the

ui{x) = e

PROBLEMS By the use of a Riccati equation, find an integral for each of the following differential equations. 31. u" -

-u' x

-

9x*u = 0.

33. 2 u " + \5e' -2]u'

+ le'^^u = 0.

35. u " + I2x cos Jf + tan x\\i' + 36. x u " log X — u' — xu|Iog x| 38. xu" +

A

32. u " -

{cos x + 0.

34. xu"

+

+ +

.8 X -

0, 3. = 0 2 ! « ' + x'u

cos" x\u = 0. 37. u

2u' -

36f*'« = 0

11 — X sin x\u' — jsin x + x cos x|u = 0.

10.6. The Wronskian.

Linear independence of solutions

If, on a given interval, two integrals «i(x) and U2{x) of a differential equation 10.2 are such that (10.11)

aiui{x) + azuzix) = 0,

with some constants a i and 02» which are not both zero, the integrals are said to be linearly dependent. O n the other hand, if relation 10.11 is fulfilled only when ai and a^ are both zero, the integrals are linearly independent.

The Wronskian

187

Suppose ui{x) and uzix) are linearly dependent. A t any x, equation 10.11 a n d its derivative then yield a pair of homogeneous simultaneous equations for the " u n k n o w n s " ai and 02. Since these unknowns are not both zero, the determinant of the system, namely, «lW

U2{x)

«l'W

U2'{X)

is zero. This is so for every x. This determinant is called the Wronskian ofu\ and «2, i n honor of the Polish mathematician H . Wronski (1778-1853). The symbol for it is W^(«i, U2\ x)\ thus (10.12)

W{U,,

U2\ x)

= «i(x)«2'(x) -

u{{x)u2{x).

We have shown that, when the integrals u\{x) and W2(^) are linearly dependent, then W{u\, U2\ x) = 0. Suppose now, conversely, that the Wronskian of u\ and «2 is zero at a point XQ. T h e system of equations a\ti\{x^) + a2«2(*o) = 0,

ai"i'(^o) + a2"2'(*o) = 0 is then fulfilled by some certain pair of values ai and f72, of which at least one is not zero. T h e combination aiui(x) - f

a2U2{x)

with those values of a\ and «2 is an integral, and it fulfills conditions 10.8. Therefore it is the trivial integral; that is, relation 10.11 is fulfilled. Hence, if W{u\, U2', x) vanishes at any ordinary point xo, the integrals «i(x) a n d «2(x) are linearly dependent. Example 6. {2x

T h e differential equation +

\ ]u"

-

( 4 x 2 +\\u'

has a singular point at x = «2 = e^ . F o r these integrals,

-

¡4x2 H- 2x +

2|« =

0

and it has the integrals u i = e~^ a n d

W{uu U2\x) = . * " - ^ { 2 x + 1}. This value is not zero on any interval that does not contain the singular point; therefore the solutions are linearly independent on that interval. Example 7.

T h e differential equation u" -f- {tan X -

2 cot X]M' = 0

188

Linear Equations, Variable Coefficients

has the integrals ui = sin x — -I sin 3x, and «2 = s i n ' x.

H'Xwi,

« 2 ; x) = 2

F o r these,

sin^ * cos x — sin^ xjsin 3x cos x — cos 3x sin x].

By trigonometric reductions this value is found to be sin^ x sin 2x — sin^ X sin 2x, that is, zero. T h e solutions are linearly dependent. By differentiation of formula 10.12 we see that W'iuu « 2 ; x) = uiixWix)

-

u,"{x)u2{x).

W h e n Uz^ix) a n d ui^x) are replaced by the values assigned to them by differential equation 10.2, this is found to be pi

W'iui, « 2 ; x) = ui{x)

po

that is, W = -{pi/po)W. (10.13)

«2

,

p2

po

pl

«2

po

,

p2

po

«2(x),

Thus

^/'i(x)

W{ui, U2\ x) = ¿1.2 exp

xo being any suitable point of the interval i n question, a n d being a constant. This result shows, among other things, that M^(ui, « 2 ; x) is either identically zero (if ¿1,2 = 0), or else is not zero at any ordinary point (if ¿1.2 ^ 0).

10.7. The general integral W h e n formulas for the integrals of a differential equation are obtainable, as they were i n Chapter 8, the question whether the combination (10.14)

Cmix) +

C2«2(jf),

i n which Cl and C2 are arbitrary, is, or is not, expressible with the use of only one arbitrary constant can be answered by inspection. T h a t is not so when no formulas are available. Suppose ui{x) a n d U2{x) are linearly dependent integrals, a n d that ai and fl2 3re constants (not both zero) for w h i c h identity 10.11 maintains. T h e n let u^{x) be the integral «3(x) = azUlix)

—

aiuzix).

W e can easily verify the fact that expression 10.14 is now simply with _ <:ia2 — f2^1 ~~ 2 I 2 ' a i + 02

c^usix),

The Change of Dependent Variable

189

It is thus expressible with the use of only one constant, namely, Cz, and is thus is not the general integral when ui(x) and U2{x) are linearly dependent. O n the other hand, when a n d U2{x) are linearly independent, every integral of the differential equation is obtainable from (10.14) by giving appropriate values to the constants. T h e relation

i n which Uz{x) is any integred whatever, is fulfilled because all its terms cancel. However, it is recognizably expressible as W^(«i,

« 2 ; x)uz{x) +

W{u2,

Us; x)ui{x)

+

W^("3,

« i ; x)u2{x) = 0;

hence, by formula 10.13, it is {*1.2"3(^) +

iC2,Ztil{x)

4- k3,lU2{x)]

CXp

= 0.

W e may divide this by Ai,2 and drop the exponential, since neither of these factors is zero. It follows that Uz{x)

=

-

7 — « l ( ^ ) - 7 — "2(JC), «1,2 Al.2

that is, uz{x) is included i n (10.14).

10.8. The change of dependent variable It is at times desirable for various purposes to transform a differential equation 10.1 by a change of variable of the form (10.15)

y = ip{x)w.

T h e transformed equation is then (10.16)

poW

+ {2po
p2
fix).

T o preclude the introduction of singular points, it is clearly requisite that the function
1

f

Piix)

2 J Po{x)

190

Linear Equations, Variable Coefficients

Example 8.

T o transform the differential equation xu" - f 6u' +

u=

4x-h

0,

into one w h i c h lacks the first derivative term. T h e change of variable u — *p(x)w transforms the equation into (up

T h e choice of
+

6^? =

0,

w" + 4«; =

i£j =

that is, ^ = l / x ' , reduces the

0.

Since the general integral of this equation i n w is ci sin 2x + the general integrad of the given equation is

u=

1

0.

cos 2x,

c\ sin 2x + C2 cos 2x\.

PROBLEMS Transform each of the following differential equations into one which lacks the term in the first derivative. 39. jjc" + 1 }u" + Axu' 4- {9x2 + 11 }u = 0. 40. x V +

{2x2 _ 2x\u' +

41. / ' + 4 /

+

4 +

42. {x" + 31u" -

{x" -

1

-2*

4xu' + 6 u = 0.

43. y" -|- 2y' sec x esc x

2y tan" x =

44. y" — 2y tan x +;?{sin x 45. xY

46. xY

-

+

4xb' + 2x *

^

log*

2x + i ) u = 0.

I4x* -

y +

1)

cot

X.

=1.

2x2 + 51^ = X

1 logx

1

y = x^,

10.9. Solution of an equation when an integral of the reduced equation is known There are at least two methods by which a differential equation 10.1 can be completely integrated when a particular integral of the reduced equation is known. M E T H O D 1. If «i(x) is the known integral, change of variable 10.15 w i t h
Solution by a Known Integral

191

T h i s change transforms the differential equation into form 10.16. But i n this form the term i n w is lacking because of the choice of