Power series for differential equation

The idea is to solve the differential equation $y'=x^2y$ by power series.

it is easy to arrive at the expression $\displaystyle\sum^{\infty}_{n=0}\left((n+1)c_{n+1}-c_nx^2\right)x^n=0$ and therefore $c_{n+1}=\frac{c_nx^2}{n+1}$, then

$y=\displaystyle\sum^{\infty}_{n=0}c_nx^n=c_0+c_1x+c_2x^2+\cdots=c_0+c_0\frac{x^3}{1!}+c_0\frac{x^6}{2!}+\cdots=c_0\displaystyle\sum^{\infty}_{n=0}\frac{x^{3n}}{n!}$

but the real solution of the differential equation is $c_0e^{x^3/3}=c_0\displaystyle\sum^{\infty}_{n=0}\frac{x^{3n}}{3^n n!}$.

The truth I have not been able to get the factor $\frac{1}{3^n}$. I appreciate the help you can give me

Solution 1:

Hint: The coefficient of $x^{n}$ is $(n+1)c_{n+1}-c_{n-2}$. You have to equate this to $0$.