The problem states:

Consider wave equation with the Neumann boundary condition on the left and weird b.c. on the right:

$u_{tt}−c^2 u_{xx}=0 , \ \ u_x(0,t)=0, \ (u_x+i\alpha u_t)(l,t)=0, \ \ \ 0<x<l$

with $\alpha$∈ℝ.

Separate variables;

Find weird eigenvalue problem for ODE;

Solve this problem;

Find simple solution u(x,t)=X(x)T(t).

Hint. You may assume that all eigenvalues are real (which is the case).

I used the usual separation of variables: $u = X(x)T(t)$ and made

$\frac{T''}{c^2 T} = \frac{X''}{X} = - \lambda $,

* to then get

$$X(x) = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x).$$

Resolving the "easy" boundary condition gives us

$X(x) = A\cos(\sqrt{\lambda}x) $

OKWhile for T:

T = $C\cos(\frac{\sqrt{\lambda}}{c}t) + D\sin(\frac{\sqrt{\lambda}}{c}t)$

And then taking the "weird" boundary condition gives me something like: $\sqrt{\lambda}\tan(\sqrt{\lambda}l) = i\alpha\frac{T'}{T}$

But I wasn't quite sure where to go from there in terms of solving the problem. One approach that I think brings me closer (hopefully) to the solution is:

Note that $(\frac{1}{c^2}\frac{T'}{T})'$ = $\frac{1}{c^2} (\frac{T''}{T} - (\frac{T'}{T})^2)$

Then knowing that the first term on the RHS is constant ($-\lambda$), we can rewrite the expression wrt to $\frac{T'}{T}$ as follows:

$\frac{T'}{T} = c\sqrt{\frac{T''}{c^2 T} - \frac{1}{c^2}(\frac{T'}{T})'} = c \sqrt{-\lambda - (\frac{T'}{c^2T})'}$

which can be incorporated into the weird BC above :

$\frac{X'}{X}(l) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'}$

and then:

$\sqrt{\lambda} \tan(\sqrt{\lambda l}) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'} $

That being said, I'm worried that I've gone off an a tangent and have failed to simplify the problem. Any hints would be appreciated.