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