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.
 A: Basically what you need is a notation for things like $1\cdot5\cdot9\cdot13\cdots$. These are generalizations of the factorial, where numbers with increment more than $1$ are multiplied. When the increment is $2$ this is called double factorial $n!!:=n(n-2)(n-4)\cdots$, and when the increment is $4$ it is quadruple factorial $n!!\,!!:=n(n-4)(n-8)\cdots$, so $1\cdot5\cdot9\cdot13=13!!\,!!$, $3\cdot7\cdot11=11!!\,!!$. These are all particular cases of the multifactorial. With this notation 
$$C_{n}=-\frac{(2n-5)!!\,!!}{n!}C_0\text{ and } C_{n}=\frac{(2n-5)!!\,!!}{n!}C_1$$
depending on whether $n$ is even or odd. This gives two linearly independent solutions, one even one odd.
As for computing the radius of convergence you don't actually need an explicit formula. Since both solutions have only even or only odd powers the ratio formula for the radius can not be used, but applying the ratio test directly gives $\lim_{n\to\infty}\frac{|C_{n+2}x^{n+2}|}{|C_{n}x^n|}<1$ since the consecutive powers in each series differ by $2$. This only requires knowing the limit of $\frac{|C_{n+2}|}{|C_{n}|}|x|^{2}=\frac{(2n-1)}{(n+2)(n+1)}|x|^{2}$, which is $0$ for any $x$. Therefore both series converge everywhere and both radii are $\infty$.
A: Well, may be I am wrong but it seems for me a general formula can be easily deduced from 
your recurrent one:
$$
C_{2n}=\frac{(2n-2))!}{(2n)!} \prod_{i=1}^n (4i-5) C_0\\ 
C_{2n+1}=\frac{(2n-1))!}{(2n+1)!} \prod_{i=1}^n (4i-3) C_1\\
$$
The product also can expressed in factorials but I let it do for somebody else :). $C_0$ and $C_1$ are free constant. 
A: 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 $$C_{2n}=-\frac{2^{2 n-\frac{5}{2}} \Gamma \left(\frac{1}{4}\right) \Gamma
   \left(n-\frac{1}{4}\right)}{\pi  \Gamma (2 n+1)}C_0$$ $$C_{2n+1}=-\frac{ 2^{2 n-\frac{5}{2}} \Gamma \left(-\frac{1}{4}\right) \Gamma
   \left(n+\frac{1}{4}\right)}{\pi  \Gamma (2 n+2)}C_1$$ 
