Use Taylor Series method to solve $y''-2xy+y=0$ I am doing some practice problems for solving second order ODEs, and I am a bit stuck on this one.
Here is what I have:

$y''-2xy'+y=0$

Let $y = \sum_{n=0}^{\infty} C_nx^n  \implies y' = \sum_{n=0}^{\infty} nC_nx^{n-1} \implies y'' = \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2} $ 
Substituting this into the ODE, and I get:
$$ \sum_{n=0}^{\infty} n(n-1)C_nx^{n-2}  -2\sum_{n=0}^{\infty} nC_nx^{n}+ \sum_{n=0}^{\infty} C_nx^n = 0$$
Then getting each term to $x^n$ and starting each sum at $n=0$, I have:
$$ \sum_{n=0}^{\infty} [(n+2)(n+1)C_{n+2}-2 nC_n+ C_n]x^n = 0 $$
$$ \implies C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$
I notice that this decouples into two series' for odd and even terms, but I am having trouble with determining the general formula for $C_n$ for each series:
For $n$ even:
When $n=0: C_2 = \frac{-C_0}{2} $
When $n=2: C_4 =  \frac{3C_2}{4 \cdot 3} = \frac{-3C_0}{4!} $
When $n=4: C_6 =  \frac{7C_4}{6 \cdot 5} = \frac{-7 \cdot 3C_0}{6!} $
When $n=6: C_8 =  \frac{11C_6}{8 \cdot 7} = \frac{-11 \cdot 7 \cdot 3C_0}{8!} $
For $n$ odd:
When $n=1: C_3 = \frac{C_1}{3 \cdot 2} $
When $n=3: C_5 =  \frac{5C_3}{5 \cdot 4} = \frac{5C_3}{5!} $
When $n=5: C_7 =  \frac{9C_5}{7 \cdot 6} = \frac{9 \cdot 5C_1}{7!} $
When $n=7: C_9 =  \frac{13C_7}{9 \cdot 8} = \frac{13 \cdot 9 \cdot 5C_1}{9!} $
I am mainly finding it difficult to determine the closed formula for the numerator in each series, so that I can calculate the radius of convergence of each one.
Thanks so much, any help is greatly appreciated.

Starting from your last relation $$ C_{n+2} = \frac{(2n-1)C_n}{(n+2)(n+1)}$$ it seems that $$C_n=\frac{2^{n-\frac{9}{2}} \Gamma \left(\frac{n}{2}-\frac{1}{4}\right) \left({C_1}
   \left((-1)^n-1\right) \Gamma \left(-\frac{1}{4}\right)-2 {C_0}
   \left((-1)^n+1\right) \Gamma \left(\frac{1}{4}\right)\right)}{\pi  \Gamma (n+1)}$$ which effectively separates the $C$'s for odd and even values because of the $\Big((-1)^n\pm 1\Big)$ terms which multiply the $C_0$ and $C_1$ terms.
This so gives $$C_{2n}=-\frac{2^{2 n-\frac{5}{2}} \Gamma \left(\frac{1}{4}\right) \Gamma
   \left(n-\frac{1}{4}\right)}{\pi  \Gamma (2 n+1)}C_0$$ $$C_{2n+1}=-\frac{ 2^{2 n-\frac{5}{2}} \Gamma \left(-\frac{1}{4}\right) \Gamma
   \left(n+\frac{1}{4}\right)}{\pi  \Gamma (2 n+2)}C_1$$ 
