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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 $\sum_{n=0}^{\infty} c_n x^n$

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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This is not a second order differential equation. It's only a first order differential equation, because it only involves first derivatives. – alex.jordan Apr 26 '14 at 1:28
Got something from my answer? – Did Apr 26 '14 at 6:38

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 $n\geqslant0$.

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How did you get (x^2)y term? And can you add the actual start index to your riemann sums? – Billy Thompson May 5 '13 at 10:17
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

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