# Solving the second order differential equation: y' =( x^2)*y as a power series

What is the proper way to do this problem as a power series? The way I'm doing it, I end up with a very complicated term.

how I'm doing it:

take the series for y and assume it's of the form series(from n =0 to inf) cn*(x^n)

derive it to get the y' power series. Then multiply the y series by x^2. From here, I try to do some index shifting but I think I'm doing it wrong because I reach a form that I can't do anything with.

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If $y(x)=\sum\limits_{n\geqslant0}c_nx^n$ then $y'(x)=\sum\limits_{n\geqslant0}(n+1)c_{n+1}x^n$ and $x^2y(x)=\sum\limits_{n\geqslant2}c_{n-2}x^n$ hence $(n+1)c_{n+1}=c_{n-2}$ for every $n\geqslant0$, with the convention that $c_{-1}=c_{-2}=0$, a recursion which does not seem so hard to manage.
Another approach, maybe more direct, is to solve $y'(x)/y(x)=x^2$. This yields $\log|y(x)|=\frac13x^3+c$, hence $y(x)=y(0)\exp\left(\frac13x^3\right)$. Since $\exp(t)=\sum\limits_{n\geqslant0}t^n/n!$, one gets $y(x)=\sum\limits_{n\geqslant0}c_nx^n$ with $c_{3n}=\left(\frac13\right)^n\frac1{n!}$ and $c_{3n+1}=c_{3n+2}=0$ for every $n\geqslant0$.
Well, if $y=\sum\limits_nc_nx^n$ then $x^ay=\sum\limits_nc_nx^{n+a}=\sum\limits_nc_{n-a}x^n$. // I fail to understand "Riemann sums". –  Did May 5 '13 at 14:50