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$$y''+(2-4x^2)y=0$$

So far I have worked out the the power series is

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

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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This is close to unreadable. The faq has pointers to help with formatting mathematics on this site. –  Gerry Myerson Nov 9 '12 at 11:41
    
@GerryMyerson I tried looking at them and I don't understand how to format these questions. The equation is straight forward and the summations are from 2 to infinity for the first one and 0 to infinity for the next 2. I shall add some brackets to try and split up what I mean but if someone could try and help me format it that would be great and I can tell you what i'm trying to write if you format anything wrong. Thanks –  Adam Nov 9 '12 at 11:46
    
The an+2, an-2 and an are a's with subscript n+2, n-2 and n's –  Adam Nov 9 '12 at 11:48
    
and obviously x^n is suppse to be x to the power of n –  Adam Nov 9 '12 at 11:49
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You can click on edit to see how I did it, so you can do it next time. Or, this time --- I didn't know what to do with 2x1. –  Gerry Myerson Nov 9 '12 at 11:55
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You simply need to take out the first two terms manually. You'll end up with a recursive equation, and so naturally the first terms will need to be treated separately since they are the initial terms for the recursion. That said, I think you have an error. The final expression I obtained (hopefully I'm right), using the series expansion $y(x) = \sum_n a_n x^n$, was $$ 2a_0 + 2a_2 + (2a_1 + 6a_3)x + \sum_{n=2}^{\infty} \left[-4a_{n-2} + 2a_n + a_{n+2}(n+1)(n+2)\right] x^n = 0 $$ Please check this to be sure. Regardless of what the actual answer is, you'll end up with a recursion that you can't solve unless you have initial conditions such as $y(0) = y_0$, $y'(0) = y_1$, or something, because otherwise you cannot determine what the initial terms $a_0, a_1$ are.

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@ChristoperA.Wong Yeah i have the initial conditions y(0)=1 and y'(0)=0 so I can work it out from here I was just confused, where do the first two terms come from? –  Adam Nov 9 '12 at 12:24
    
yeah I had got two of my limits the wrong way round so your solution is correct. So to take the terms out manually you just set n to 0 and then n to 1 for both of the summations between 0 and infinity –  Adam Nov 9 '12 at 12:31
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