Chapter 17.4, Problem 11E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Use power series to solve the differential equation.11. y" + x2y' + xy = 0, y(0) = 0, y'(0) = 1

To determine

To solve: The differential equation by the use of power series.

Explanation

Given data:

The differential equation is,

yâ€³+x2yâ€²+xy=0 (1)

With y(0)=0 and yâ€²(0)=1 .

Consider the expression for y(x) .

y(x)=âˆ‘n=0âˆžcnxn (2)

Multiply x on both sides equation (2),

xy=xâˆ‘n=0âˆžcnxn

xy=âˆ‘n=0âˆžcnxn+1 (3)

Differentiate equation (2) with respect to t.

yâ€²(x)=âˆ‘n=1âˆžncnxnâˆ’1 (4)

Multiply x2 on both sides equation (4).

x2yâ€²=x2âˆ‘n=1âˆžncnxnâˆ’1

x2yâ€²=âˆ‘n=0âˆžncnxn+1 (5)

Differentiate equation (4) with respect to t.

yâ€³(x)=âˆ‘n=2âˆžn(nâˆ’1)cnxnâˆ’2

yâ€³(x)=âˆ‘n=âˆ’1âˆž(n+3)(n+2)cn+3xn+1

yâ€³(x)=2c2+âˆ‘n=0âˆž(n+3)(n+2)cn+3xn+1 (6)

Substitute equations (3), (5), and (6) in (1),

2c2+âˆ‘n=0âˆž(n+3)(n+2)cn+3xn+1+âˆ‘n=0âˆžncnxn+1+âˆ‘n=0âˆžcnxn+1=0

2c2+âˆ‘n=0âˆž[(n+3)(n+2)cn+3+ncn+cn]xn+1=0 (7)

Equation (7) is true when the coefficients are 0. Therefore, the expressions are,

c2=0

And

(n+3)(n+2)cn+3+ncn+cn=0

Re-arrange the equation.

cn+3=ncnâˆ’cn(n+3)(n+2)

cn+3=âˆ’(n+1)cn(n+3)(n+2),â€‰â€‰n=0,1,2â‹…â‹…â‹… (8)

Equation (8) is the recursion relation

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