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So far I have worked out the the power series is

$\Sigma_{n=0}^{\infty} (n+2)(n+1)a_{n+2} x^n+ 2 \Sigma_{n=0}^\infty a_n x^n -4 \Sigma_{n=2}^\infty a_{n-2} x^n$

but I don't know how to take out the first two terms to get the whole thing into the form of $\Sigma_{n=2}^\infty$. I know its something like $2.1 a_2 + a_0$

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You can replace each \Sigma by \sum. – Did Nov 9 '12 at 12:24
One solution to the diffEQ is $y=e^{-x^2}$, but there should be two independent solutions. Nevertheless, this shows at least that the series for $e^{-x^2}$ should give coefficients solving your recurrence. – coffeemath Nov 9 '12 at 17:26
It looks like the other independent solution is an antiderivative of $e^{2x}$, which is not known in closed form. I guess that's why one uses the power series approach to get all solutions... :-) – coffeemath Nov 9 '12 at 17:40
@coffeemath I'm attempting to get the recurrence relation to show that the solution is $e^{-x^2}$ but I'm not sure how to do that if you check Dennis Gulko's answer you can see how far I've got. – Adam Nov 10 '12 at 12:53

If $y=\sum_{n=0}^\infty a_nx^n$, the equation turns into: $$\begin{align*} 0=& y''+(2-4x^2)y=\sum_{n=0}^\infty (n+1)(n+2)a_{n+2}x^{n}+2 \sum_{n=0}^\infty a_n x^n -4 \sum_{n=2}^\infty a_{n-2} x^n \\ =& 2a_2+2\cdot3a_3x+2a_0+2a_1x+\sum_{n=2}^\infty[(n+1)(n+2)a_{n+2}+2a_n-4a_{n-2}]x^n\\ =& 2(a_0+a_2)+2(a_1+3a_3)x+\sum_{n=2}^\infty[(n+1)(n+2)a_{n+2}+2a_n-4a_{n-2}]x^n\end{align*}$$ So you get a system of equations (assuming you know $y(0),y'(0)$ you can find all the coefficients): $$a_0+a_2=0\\ a_1+3a_3=0 \\ (n+1)(n+2)a_{n+2}+2a_n-4a_{n-2} =0 \hspace{5pt}\forall n\geq2$$

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I have been given y(0)=1 and y'(0)=0 and figured out that $$a_0=1$$ $$a_1=0$$ and $$a_2=-1$$ and have the recurrence relation for $$a_n+2= 4a_n-2 - 2a_n/(n+2)(n+1)$$ but can't figure out how to write all of the even terms in terms of $$a_0$$ – Adam Nov 10 '12 at 12:43
That should be $$4a_{n-2}$$ – Adam Nov 10 '12 at 12:49

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