-2
$\begingroup$

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

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 $2.1 a_2 + a_0$

  • 0
    Right thats what I wanted to ask @GerryMyerson thanks for all of the help formatting it2012-11-09

1 Answers 1

0

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.

  • 0
    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 infinity2012-11-09