# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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n—2

3 4 5

+ 20a5x + 30a6x + 42a7x + ...

3 . 5 . 2 3 4

+ (x-x /3!+x /5!-...)(a0+a!x+a2x +a3x +a4x +...)

23

— 2a2 + (6a3+a0)x + (12a4+a1)x + (20a5+a2-a0/6)x +

45

(30a6+ a3-a1/6)x + (42a7+a4+a0/5!)x + ... — 0. Hence

a2 — 0, a3 — -a0/6, a4 — -a^/12, a5 — a0/120, a6 — (a1+a0)/180, a7 = -a0/7! + a1/504, ... . We set

a0 — 1 and a1 — 0 and obtain

y1(x) — (1 - x3/6 + x5/120 + x6/180 + ...). Next we set

a0 — 0 and a1 — 1 and obtain

4 6 7

y2(x) — (x - x /12 + x /180 + x /504 +...). Since

82

Section 5.3

p(x) = 1 and q(x) = sinx both have p = ro, the solution in this case converges for all x, that is, p = ro

x 2 3

18. We know that e = 1 + x + x/2! + X/3! + ..., and

x2 2 4 . 6 .

therefore e = 1 + x + x/2! + x/3! + ... . Hence, if

y = Xanxn, we have

2 2 4 2

a! + 2a2x + 3a3x + ... = (1+x + x /2+...)(a0+a1x+a2x +...'

2

= a0 + a1x + (a0+a2)x + ...:

Thus, a1 = a0 2a2 = a1 and 3a3 = a0 + a2, which yield the desired solution.

ro

20. Substituting y = anxn into the D.E. we obtain

n=2

n-1 n 2

> nanx - > anx = x . Shifting indices in the summation

n2

nx

n=2 n=2

n2

yields > [(n+1)an+1 - an]x = x . Equating coefficients of

n=2

both sides then gives: a1 - a0 = 0, 2a2 - a1 = 0, 3a3 - a2 = 1

and (n+1)an+1 = an for n = 3,4,... . Thus a1 = a0,

a2 = a1/2 = a0/2, a3 = 1/3 + a2/3 = 1/3 + a0/2-3,

a4 = a3/4 = 1/3-4 + a0/2-3-4, ..., an = an-1/n = 2/n! + a0/n! and

hence

2 n 3 4 n

x x x x x

y(x) = a0(1 + x + — + ...+— ...) + 2( +— + ...+— + ...).

0 2! n! 3! 4! n!

Using the power series for ex, the first and second sums

x x 2

can be rewritten as a0e + 2(e - 1 - x - x /2).

22. Substituting y = anxn into the Legendre equation,

n=2

shifting indices, and collecting coefficients of like powers of x yields

[2-1-a2 + a(a+1)a0]x° + {3-2-a3 - [2-1 - a(a+1)]a1}x1 +

ro

{(n+2)(n+1)an+2 - [n(n+1) - a(a+1)]an}xn = 0. Thus

n=2

a2 = -a (a+1)a0/2!, a3 = [2-1 - a(a+1)]a1/3! = -(a-1)(a+2)a1/3! and the recurrence relation is

Section 5.4

83

(n+2)(n+1)an+2 = -[a(a+1) - n(n+1)]an = -(a-n)(a+n+1)an,

n = 2,3,... . Setting a1 = 0, a0 = 1 yields a solution

with a3 = a5 = a7 = ... = 0 and

a4 = a(a-2)(a+1)(a+3)/4!,..., a2m = (-1) ma(a-2)(a-4) ...

(a-2m+2)(a+1)(a+3) ... (a+2m-1)/(2m)!,... . The second

linearly independent solution is obtained by setting a0 = 0 and a1 = 1. The coefficients are a2 = a4 = a6 =

... = 0 and a3 = -(a-1)(a+2)/3!, a5 = -(a-3)(a+4)a3/5-4 = (a-1)(a-3)(a+2)(a+4)/5!,... .

26. Using the chain rule we have:

dF(^) dF[$(x)] dx ? ,w , 4T/, 2

= ----------------- — = -f (x)sin6(x) = - f (x)V1-x ,

d^ dx d^

2 ,______________________

d F(6) d , ./ 2 dx 2 ,

-------- = — [-f (x)V1-x ] — = (1-x )f (x) - xf (x),

d^2 dx d^

which when substituted into the D.E. yields the desired

result.

22 28. Since [(1-x )y']' = (1-x )y" - 2xy', the Legendre

Equation, from Problem 22, can be written as shown.

Thus, carrying out the steps indicated yields the two

equations:

2

Pm[(1-x )Pn] = -n(n+1)pnpm 2

Pn[(1-x )Pm] = -m(m+1)PnPm.

As long as n ^ m the second equation can be subtracted from the first and the result integrated from -1 to 1 to obtain

1 2 2 1 I {Pm[(1-x )Pn] -Pn[(1-x )Pm] }dx = [m(m+1)-n(n+1)]I ^PnPmdx

The left side may be integrated by parts to yield

2 f 2 ' 1 f 1 ' 2 f ' 2 '

[Pm(1-x )Pn - Pn(1-x )Pm]-1 + J_1[Pm (1-x )Pn - Pn(1-x )Pm]dX, which is zero. Thus J1 Pn(x)Pm(x)dx = 0 for n ^ m.

Section 5.4, Page 259

1. Since the coefficients of y, y' and y" have no common

factors and since P(x) vanishes only at x = 0 we conclude that x = 0 is a singular point. Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we obtain p(x) = (1-x)/x and q(x) = 1. Thus for the singular point we have

84

Section 5.4

12.

17.

2

lim x p(x) = lim 1-x = 1, lim x q(x) = 0 and thus x = 0

x—0 x—0 x—0

is a regular singular point.

Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we

22 find p(x) = x/(1-x)(1+x) and q(x) = 1/(1-x )(1+x).

Therefore x = ±1 are singular points. Since

lim (x-1)p(x) and lim (x-1) q(x) both exist, we conclude

x—1 x—1

x = 1 is a regular singular point. Finally, since lim (x+1)p(x) does not exist, we find that x = -1 is an

x—— -1

irregular singular point.

Writing the D.E. in the form y + p(x)y' + q(x)y = 0, we

see that p(x) = ex/x and q(x) = (3cosx)/x. Thus x = 0 is

a singular point. Since xp(x) = ex is analytic at x = 0 2

and x q(x) = 3xcosx is analytic at x = 0 the point x = 0 is a regular singular point.

Writing the D.E. in the form y" + p(x)y' + q(x)y = 0, we

x4

see that p(x) = ----- and q(x) = -----. Since lim q(x)

sinx sinx x—0

does not exist, the point x0 = 0 is a singular point and since neither lim p(x) nor lim q(x) exist either the

x—±nn x—±nn

points x0 = ±nn are also singular points. To determine whether the singular points are regular or irregular we must use Eq.(8) and the result #7 of multiplication and division of power series from Section 5.1. For x0 = 0, we have

2 2 2 x x x

xp(x) = ----- = = x[1 + — + ...]

sinx x3 6

x- — +...

6

3

x

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