   Chapter 17.4, Problem 6E

Chapter
Section
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,

y=y (1)

Consider the expression for y(x) ,

y(x)=n=0cnxn (2)

Differentiate equation (2) with respect to t,

y(x)=n=1ncnxn1 (3)

Differentiate equation (3) with respect to t,

y(x)=n=2n(n1)cnxn2

y(x)=n=0(n+1)(n+2)cn+2xn (4)

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

n=0(n+1)(n+2)cn+2xn=n=0cnxnn=0(n+1)(n+2)cn+2xnn=0cnxn=0

(n=0(n+1)(n+2)cn+2n=0cn)xn=0 (5)

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

(n+1)(n+2)cn+2cn=0

Re-arrange the equation,

cn+2=cn(n+2)(n+1),n=0,1,2 (6)

Equation (6) is the recursion relation.

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

Substitute 0 for n in equation (6),

c0+2=c0(0+2)(0+1)c2=c02×1

Substitute 2 for n in equation (6),

c2+2=c2(2+2)(2+1)c4=c24×3

Substitute c02×1 for c2 ,

c4=c02×14×3=c04×3×2×1

c4=c04! (7)

Substitute 4 for n in equation (6),

c4+2=c4(4+2)(4+1)c6=c46×5

Substitute c04! for c4 ,

c6=c04!6×5

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=c13×2

Substitute 3 for n in equation (6),

c3+2=c3(3+2)(3+1)c5=c35×4

Substitute c13×2 for c3

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