Solving differential equation with power series $y'' - xy' - y = 0$ I need a work check

$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$

so we have:
$$
\begin{split}
y  &= \sum_{n=0}^\infty C_nx^n\\
y' &= \sum_{n=1}^\infty nC_nx^{n-1}\\
y' &= \sum_{n=2}^\infty n(n-1)C_nx^{n-2}
\end{split}
$$
so making a substitution into the original differential equation:
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$
$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$
$$C_{n+2} = \frac{C_n}{n+2}$$
so first a few terms:
$C_0 = C_0$ and $C_1 = C_1$ and $C_2 = \frac{C_0}{2}$ and $C_3 = \frac{C_1}{3}$ and $C_4 = \frac{C_2}{4} = \frac{C_0}{4 \cdot 2}$ and $c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$
even terms are: $\frac{C_0}{2^n \cdot n!}$
odd terms are: $\frac{C_1}{1 \cdot 3 \cdot 5 \cdot ...(2n+1)}$
and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$
Where do I go from here? if I do $y(0)$, I don't get 1. How do I take the derivative of this to check the initial condition $y'(0) = 0$

You're given initial conditions so you must define $c_0$ and $c_1$ to satisfy these initial conditions. 
Notice that you can write your series as 
$$y(x) =c_0+ \sum_{n=1}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$
where I have simply taken the first term of the first series out front. Now it is clear, when $x=0$, both series are zero, so you are left with 
$$y(0)=c_0=1$$
so $c_0=1$. Now, take the derivative with respect to $x$ of your entire function, 
$$y'(x)=\sum_{n=1}^\infty \frac{2n}{2^n \cdot n!}x^{2n-1} + \sum_{n=0}^\infty (2n+1)\frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n}$$
Now notice again when $x$ is zero, the only surviving term is the zeroeth term of the second series, giving you
$$y'(0)=\frac{c_1}{15}=0$$
and thus your entire series is 
$$y = \sum_{n=0}^\infty \frac{1}{2^n \cdot n!}x^{2n}$$
