Finding a reccurrence relation We have the following equality: $y''-2zy'-2y=0$, (the derivatives are according to the variable z). 
With the initial conditions: $y(0)=1, y'(0)=0$
Let us suppose that the solution can be written as: $\Sigma_{n≥0} a_n z^n$ with a radius of convergence $r>0$
I am asked to find a reccurence relation on the coefficients $a_n$.
Attempt at finding the solution:
We can notice the three following equalities:


*

*y = $\Sigma_{n≥0} a_n z^n$

*y' =$\Sigma_{n≥0} a_n n  z^{n-1}$

*y''=$\Sigma_{n≥0} a_n n(n-1)z^{n-2}$
We can thus rewrite:
$y''-2zy'-2y=0$
$\iff \Sigma_{n≥0} a_n n(n-1)z^{n-2} - 2z\Sigma_{n≥0} a_n n  z^{n-1}-2\Sigma_{n≥0} a_n z^n = 0$
$\iff \Sigma_{n≥2} (a_n (n-1)nz^{n-2}-2za_n z^n - 2a_nz^n) - 2a_0 -4a_1z=0$
$\iff \Sigma_{n≥2} (a_n (n-1)nz^{n-2}-2a_nz^n(n+1))-2a_0-4a_1 z =0$
Now I'm not sure in what direction I'm going and I don't know how I can find a recurrence relation on $a_n$
 A: Using 
$$ y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n $$
then
\begin{align}
0 &= y'' - 2 \, x \,y' - 2y \\
&= \sum_{n=0} a_{n} \, n(n-1) \, x^{n-2} - 2 \, \sum_{n=0} n \, a_{n} \, x^{n} - 2 \, \sum_{n=0} a_{n} \, x^n \\
&= \sum_{n=2} n(n-1) \, a_{n} \, x^{n-2} - 2\, \sum_{n=0} (n+1) \, a_{n} \, x^{n} \\
&= \sum_{n=0} (n+1)(n+2) \, a_{n+2} \, x^{n} - 2\, \sum_{n=0} (n+1) \, a_{n} \, x^n \\
&= \sum_{n=0} \left[ (n+1)(n+2) \, a_{n+2} - 2 \, (n+1) \, a_{n} \right] \, x^{n}
\end{align}
Equating the coefficients yields
$$a_{n+2} = \frac{2 \,(n+1) \, a_{n} }{(n+1)(n+2)}$$
The first few coefficients are:
\begin{align}
a_{2} &= a_{0} \\
a_{3} &= \frac{2a_{1}}{3} \\
a_{4} &= \frac{a_{0}}{2} \\
a_{5} &= \frac{4a_{1}}{15}.
\end{align}
By using $y(0) = 1$ and $y'(0)=0$ ( which leads to $a_{0} = 1$ and $a_{1} = 0$) then the general form of the coefficients can be determined, which is:
\begin{align}
a_{2n} &= \frac{a_{0}}{n!} = \frac{1}{n!} \\
a_{2n+1} &= 0
\end{align}
and
$$y(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} = e^{x^{2}}.$$
