Find power series solution of $y^\prime=x^2y$ 
Find power series solution of $y^\prime=x^2y$, determine the radius of convergence and identify the series solutions in terms of elementary functions.

So I started with $y=\sum_{n=0}^\infty c_nx^n$
then $y^\prime=\sum_{n=1}^\infty n c_nx^{n-1}$
So rearranging and substituting the series I get $$\sum_{n=1}^\infty n c_nx^{n-1}-x^2 \sum_{n=0}^\infty c_nx^n=0$$
then I changed the index on the first sum $$\sum_{n=0}^\infty (n+1) c_{n+1}x^n-x^2 \sum_{n=0}^\infty c_nx^n=0$$
$$\sum_{n=0}^\infty (n+1) c_{n+1}x^n- c_nx^{n+2}=0$$
$$\sum_{n=0}^\infty x^n\big((n+1)c_{n+1}-c_nx^2\big)=0$$
I'm not sure how to simplify this further.
 A: $$\sum_{n=0}^\infty (n+1) c_{n+1}x^n-x^2 \sum_{n=0}^\infty c_nx^n=0$$
$$\sum_{n=0}^\infty (n+1) c_{n+1}x^n- \sum_{n=0}^\infty c_nx^{n+2}=0$$
$$\sum_{n=0}^\infty (n+1) c_{n+1}x^n- \sum_{n=2}^\infty c_{n-2}x^{n}=0$$
$$c_1+2c_2x+\sum_{n=2}^\infty (n+1) c_{n+1}x^n- \sum_{n=2}^\infty c_{n-2}x^{n}=0$$
$$c_1+2c_2x+\sum_{n=2}^\infty ((n+1) c_{n+1}-  c_{n-2})x^{n}=0$$
You can take it from here ?
As Tristan wrote in the comments you can deduce that:
$$
\begin{align}
c_1=&0 \\
c_2=&0 \\
(n+1) c_{n+1}=&  c_{n-2}
\end{align}
$$
You need to calculate some coefficients and try to find a pattern from the recurrence relation...
I ended with $$c_{3n}=\frac {c_0}{3^nn!}$$
$$\implies y(x)=c_0\sum_{n=0}^\infty \frac {x^{3n}}{3^nn!}=c_0e^{x^3/3}$$
A: You need to write the result in the form$$\sum_{n\ge0}a_nx^n=0.$$If we define $c_{-1}:=c_{-2}:=0$,$$0=a_n:=(n+1)c_{n+1}-c_{n-2}.$$Since $c_0=1$, $c_n\ne0$ only when $3|n$. So define $b_n:=c_{3n}$, viz.$$b_0=0,\,b_{n+1}=\frac{b_n}{3(n+1)}.$$By induction, $b_n=\frac{1}{3^nn!}$.
