So If I'm looking at a boundary value problem with boundary values of $U_x(0,t) = 0 = U(2\pi,t)$.
I get to $\frac{T'}{kT} = \frac{X"}{x} = \lambda$ and try $\lambda = -\alpha^2$.
This leaves me with $X(x) = A\cos (\alpha x) + B\sin (\alpha x)$ & $X'(x) = -A\alpha\sin (\alpha x) + B\alpha\cos (\alpha x)$
$X(2\pi) = 0 = A\cos(\alpha 2\pi) + B\sin(\alpha 2\pi) = A(1) B B(0) = A$ so $A = 0$.
$X'(0) = 0 = -A\alpha\sin (\alpha 0) + B\alpha\cos (\alpha 0) = A(0) + B(1) = B$ so $B = 0$.
Thus $A = B = 0$, which leads me to believe I have done something wrong here. Any hints or tips you can give me as to what I have done wrong here? I feel confidant I know how to proceed with the problem once I get past this point, however I've been looking at this step for a while now and am not seeing an issue.