Heat Equation with Weird Boundary Conditions

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 $\displaystyle \lambda$?

Re: Heat Equation with Weird Boundary Conditions

If you sub. in your BC's you get

$\displaystyle -A \sin 2 \lambda + B \cos 2\lambda = 0,$

$\displaystyle A \sin 2 \lambda + B \cos 2\lambda = 0,$

or

$\displaystyle A \sin 2\lambda = 0$ and $\displaystyle B \cos 2 \lambda = 0$.

This gives rise to two cases

$\displaystyle (1)\;\;A = 0,\;\; \cos 2 \lambda = 0,$

$\displaystyle (2)\;\; B = 0, \;\; \sin 2 \lambda = 0.$