Solving differential equation using power series $y'' - xy' -y = 0$ given $y(0) = 1$ and $y'(0) = 0$ I think I made a mistake somewhere:
$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$
so we have:
$$y = \sum_{n=0}^\infty C_nx^n$$
$$y' = \sum_{n=1}^\infty nC_nx^{n-1}$$
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
so subbing:
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$
$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$
$$C_{n+2} = \frac{C_n}{n+2}$$
so 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 $c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$
so the even terms are: $\frac{C_0}{2^n \cdot n!}$
and the odd terms are: $\frac{C_1}{1 \cdot 3 \cdot 5 \cdot (2n-1)}$
and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n-1)} x^{2n-1}$$
but I'm stuck here. If $y(0) = 1$ ... doesn't the equation become 0? I feel like I've hit an impossible condition so I feel like I've made a mistake somewhere.

You did well. Note that the initial conditions $y(0)=1$ and $y'(0)=0$ simply means $C_0=1$ and $C_1=0$.
Also, noting that $C_1=0$, your final solution is
$$y=\sum_{n=0}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}=c_0+\sum_{n=1}^\infty\frac{c_0}{2^n\cdot n!}x^{2n}$$
which dosn't make the equation $0$.
