# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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+ Ó[(n+2)(n+1)

an+2 + (a n ) an]x 0.

n=2

Hence the recurrence relation is 22

an+2 = (n -a )an/(n+2)(n+1), n = 0, 1,2,... . For the

first solution we choose a1 = 0. We find that

2 2 2 2 2 2

a2 = -a a0/2-1, a3 = 0, a4 = (2 -a )a2/4^3 = -(2 -a )a a0/4!

2 2 2 2 2

..., a2m = -[(2m-2) - a ]... (2 -a )a a0/(2m)!,

2 ,„2 2. 2 a 2 (2 -a ) a 4

and a2m+1 = 0, so y^x) = 1 - -^x - -—--------------------x - ...

2 2 2 2 2 [(2m-2) -a ]...(2 -a )a 2m

- (2m)! x - ...

where we have set a0 = 1. For the second solution we take

a0 = 0 and a1 = 1 in the recurrence relation to obtain

the desired solution.

2 2 2 10b. If a is an even integer 2k then (2m-2) - a = (2m-2) -

2

4k = 0. Thus when m = k+1 all terms in the series for

2k

y1(x) are zero after the x term. A similar argument shows that if a = 2k+1 then all terms in y2(x) are zero 2k+1

after the x .

11. The Taylor series about x = 0 for sinx is

sinx = x - x3/3! + x5/5! - ... . Assuming that

ro

n2

y = Ó anx we find y + (sinx)y = 2a2 + 6a3x + 12a4x

n=2

345

+ 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

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

n2

nx

n=2 n=2

n2

yields Ó [(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

’Ó {(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 =

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