## Calculus 8th Edition

Published by Cengage

# Chapter 17 - Second-Order Differential Equations - Review - Exercises - Page 1221: 17

#### Answer

$$y=\Sigma_{n=0}^{\infty}\dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}=x-\frac{1}{3}x^{3}+\frac{1}{3.5}x^{5}-...$$

#### Work Step by Step

As we are given that $y''+xy'+y=0$, $y(0)=0$ and $y'(0)=1$ Need to assume a solution of this form: $y=\Sigma_{n=0}^{\infty}c_nx^n$ $y'=\Sigma_{n=0}^{\infty}(n+1)c_{n+1}x^n$ $y''=\Sigma_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n$ $xy'=x[\Sigma_{n=0}^{\infty}(n+1)c_{n+1}x^n]=\Sigma_{n=0}^{\infty}nc_{n}x^n$ $y''+xy'+y=0$ After simplifications, we get $\Sigma_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n+\Sigma_{n=0}^{\infty}nc_{n}x^n+\Sigma_{n=0}^{\infty}c_nx^n=0$ Thus, $c_{n+2}=-\dfrac{c_n}{n+2}$ Hence, the result is $$y=\Sigma_{n=0}^{\infty}\dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}=x-\frac{1}{3}x^{3}+\frac{1}{3.5}x^{5}-...$$

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