Chapter 17.4, Problem 6E

### 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.6. y" = y

To determine

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

Explanation

Given data:

The differential equation is,

yâ€³=y (1)

Consider the expression for y(x) ,

y(x)=âˆ‘n=0âˆžcnxn (2)

Differentiate equation (2) with respect to t,

yâ€²(x)=âˆ‘n=1âˆžncnxnâˆ’1 (3)

Differentiate equation (3) with respect to t,

yâ€³(x)=âˆ‘n=2âˆžn(nâˆ’1)cnxnâˆ’2

yâ€³(x)=âˆ‘n=0âˆž(n+1)(n+2)cn+2xn (4)

Substitute equation (2) and (4) in (1),

âˆ‘n=0âˆž(n+1)(n+2)cn+2xn=âˆ‘n=0âˆžcnxnâˆ‘n=0âˆž(n+1)(n+2)cn+2xnâˆ’âˆ‘n=0âˆžcnxn=0

(âˆ‘n=0âˆž(n+1)(n+2)cn+2âˆ’âˆ‘n=0âˆžcn)xn=0 (5)

Equation (5) is true when the coefficients of xn are 0. Therefore, the required expression is,

(n+1)(n+2)cn+2âˆ’cn=0

Re-arrange the equation,

cn+2=cn(n+2)(n+1),â€‰â€‰n=0,1,2â‹…â‹…â‹… (6)

Equation (6) is the recursion relation.

Solve the recursion relation by substituting n=0,1,2,3â‹…â‹…â‹… in equation (6).

Substitute 0 for n in equation (6),

c0+2=c0(0+2)(0+1)c2=c02Ã—1

Substitute 2 for n in equation (6),

c2+2=c2(2+2)(2+1)c4=c24Ã—3

Substitute c02Ã—1 for c2 ,

c4=c02Ã—14Ã—3=c04Ã—3Ã—2Ã—1

c4=c04! (7)

Substitute 4 for n in equation (6),

c4+2=c4(4+2)(4+1)c6=c46Ã—5

Substitute c04! for c4 ,

c6=c04!6Ã—5

c6=c06! (8)

Similarly, like equations (7) and (8), write the expression for c2n ,

c2n=c0(2n)! (9)

Substitute 1 for n in equation (6),

c1+2=c1(1+2)(1+1)c3=c13Ã—2

Substitute 3 for n in equation (6),

c3+2=c3(3+2)(3+1)c5=c35Ã—4

Substitute c13Ã—2 for c3

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