After a few trivial steps I reached:

$\displaystyle X'' + \lambda^2 X = 0$

$\displaystyle X(x) = Asin(\lambda x) + Bcos(\lambda x)$

But my boundary conditions are:

$\displaystyle X(-2) = 0, X(2) = 0$

How am I supposed to solve for A and B and $\displaystyle \lambda$?