Answer
$y=c_{1}xe^{\frac{x^2}{2}}+c_{0}+c_{0}\Sigma_{n=1}^{\infty}\dfrac{2^{n-1}(n-1)!}{(2n-1)!}x^{2n}$
Work Step by Step
$y''-xy'-2y=0$
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, $y''-xy'-2y=0$
or
$\Sigma_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n-\Sigma_{n=0}^{\infty}nc_{n}x^n-2\Sigma_{n=0}^{\infty}c_nx^n=0$
$c_{n+2}=\dfrac{c_n}{n+1}$
Hence, $y=c_{1}xe^{\frac{x^2}{2}}+c_{0}+c_{0}\Sigma_{n=1}^{\infty}\dfrac{2^{n-1}(n-1)!}{(2n-1)!}x^{2n}$
