Please verify my power series solution to the differential equation $y''-xy'+y=0$. Please verify my answer to the following differential equation:
$$y''-xy'+y=0$$
Let $y = {\sum_{n=0}^\infty}C_nx^n$, then $y' = {\sum_{n=1}^\infty}nC_nx^{n-1}$ and $y''={\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}$
Substituting this to the equation we get
$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-x{\sum_{n=1}^\infty}nC_nx^{n-1}+{\sum_{n=0}^\infty}C_nx^n = 0$$
 $${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $x^n$ term on all the terms
$${\sum_{n=0}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $0$th term from the first and the third summations we get
$$2C_2+C_0 + {\sum_{n=1}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=1}^\infty}C_nx^n = 0$$
Factoring $x^n$ we get
$$2C_2+C_0 + {\sum_{n=0}^\infty}[(n+2)(n+1)C_{n+2}-nC_n+C_n]x^n= 0$$
i.$$2C_2+C_0 = 0 => C_2 = \frac{-C_0}{2}$$
ii.$$(n+2)(n+1)C_{n+2}-nC_n+C_n = 0$$
Therefore solving ii. for $C_{n+2}$
$$C_{n+2}=\frac{(n-1)C_n}{(n+2)(n+1)}, n=0,1,2,3,...$$
If $n = 0$,
$$C_2 = \frac{-C_0}{2!}$$
If $n=1$,
$$C_3 = 0$$
If $n=2$,
$$C_4 = \frac{C_2}{3*4} = \frac{-C_0}{4!}$$
If $n=3$,
$$C_5 = \frac{2C_3}{4*5}=0$$
If $n=4$,
$$C_6 = \frac{3C_4}{5*6} = \frac{-C_0}{6!}$$
Upon seeing the pattern we realize that if $n=2m$ then
$$C_{2m} = \frac{-C_0}{2m!}$$
And if $n=2m+1$ then
$$C_{2m+1} = 0$$
So the final answer would be 
$$y = {\sum_{n=0}^\infty}C_nx^n => {\sum_{m=0}^\infty}\frac{-C_0*x^{2m}}{2m!}$$

You should be able to do a change of variable to transform this equation into Hermite's equation then compare the resulting coefficients to Hermite's. 
Alternatively, you can determine the Rodrigue's Operator formulation of solutions of the equation and verify the operators have the desired effect on your series. 
