$y''+(2-4x^2)y=0$
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 $2.1 a_2 + a_0$