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
$y''+xy'+y=0$, $y(0)=0$ and $y'(0)=1$
Assume a solution of this form
$y=\Sigma_{n=0}^{\infty}c_nx^n$
$y'=\Sigma_{n=0}^{\infty}(n+1)c_{n+1}x^n$
$y''=\Sigma_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n$
$xy'=x[\Sigma_{n=0}^{\infty}(n+1)c_{n+1}x^n]=\Sigma_{n=0}^{\infty}nc_{n}x^n$
Thus, $y''+xy'+y=0$
or
$\Sigma_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n+\Sigma_{n=0}^{\infty}nc_{n}x^n+\Sigma_{n=0}^{\infty}c_nx^n=0$
$c_{n+2}=-\dfrac{c_n}{n+2}$
Hence, $y=\Sigma_{n=0}^{\infty}\dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}=x-\frac{1}{3}x^{3}+\frac{1}{3.5}x^{5}-...$