I have come across (and indeed, solved) variations of this problem on a finite strip, but not for an infinite strip as in my version of the problem. It is the standard two-dimensional Laplace's equation, with
$$u(a, y) = V,\,\,\,\, -\infty < y < +\infty$$ $$U(-a, y) = -V,\,\,\,\, -\infty < y < +\infty$$ $$\lim_{y \to +\infty}u_y(x, y) = 0,\,\,\,\, -a < x < a$$ $$\lim_{y \to -\infty}u_y(x, y) = 0,\,\,\,\, -a < x < a$$ Now, we are to solve this using separation of variables. My method therefore entails solving the second order ODE's given by $$X''(x) - \sqrt{\lambda}X(x) = Y''(y) + \sqrt{\lambda}Y(y) = 0$$ At this point, I assume that the ODE in $X(x)$ is solved by an equation of the form $$X(x) = C_1\sin(\sqrt{\lambda}x) + C_2\cos(\sqrt{\lambda}x)$$ Then applying the boundary conditions I get that $$X(a) = C_1\sin(\sqrt{\lambda}a) + C_2\cos(\sqrt{\lambda}a) = V$$ and $$X(-a)= C_1\sin(-\sqrt{\lambda}a) + C_2\cos(-\sqrt{\lambda}a)= -V$$ But I can exploit the fact that cosine and sine functions are symmetric, and re-phrase $X(-a)$ in the following way: $$X(-a) = -C_1\sin(\sqrt{\lambda}a) + C_2\cos(\sqrt{\lambda}a) = -V$$ Then taking the difference of these equations I can conveniently note that $$2C_2\cos(\sqrt{\lambda}a) = 0$$ from which I can solve $$\sqrt{\lambda}a = \frac{(2n+1)\pi}{2}$$ and hence $$\lambda = \frac{(2n+1)^2\pi^2}{4a^2}$$ I wonder how to proceed, though? Those conditions that $Y'(y)$ must vanish as $y \to \infty$ seem to suggest a negative exponential for $Y(y)$, but I cannot seem to find one that works.