0
$\begingroup$

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

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 $2.1 a_2 + a_0$

  • 0
    You can replace each `\Sigma` by `\sum`.2012-11-09
  • 0
    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.2012-11-09
  • 0
    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... :-)2012-11-09
  • 0
    @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.2012-11-10

1 Answers 1