From an old qual exam:
Find the eigenfunctions and eigenvalues associated with the BVP \begin{align} &u_{xx} + \frac{2}{x}u_x + u_{yy} + \theta u = 0,\quad 0 < x < a, \quad 0 < y < b,\\ &u(x,0) = 0,\quad u(x,b) = 0,\\ &u \text{ bounded as } x \to 0, \quad u_x(a,y) = 0. \end{align} ($\theta$ a given real number).
Hint: it may be useful to make the substitution $v = xu$.
My attempt:
Letting $v = xu$ as per the hint, the BVP is transformed into \begin{align} &v_{xx} + v_{yy} + \theta v = 0,\quad 0 < x < a, \quad 0 < y < b,\\ &v(x,0) = 0,\quad v(x,b) = 0,\\ &\frac{v}{x} \text{ bounded as } x \to 0, \quad av_x(a,y) = v(a,y). \end{align} Solving via separation of variables $v(x,y) = \varphi(x) h(y)$, we get (denoting $\lambda$ the separation constant) \begin{align} &h''(y) + (\lambda + \theta)h(y) = 0,\quad h(0) = h(b) = 0\\ &\varphi''(x) - \lambda \varphi(x) = 0, \quad \frac{\varphi(x)}{x} \text{ bounded as } x \to 0, \quad a\varphi'(a) = \varphi(a) \end{align} I find the eigenvalues $\lambda_n = \frac{n^2\pi^2}{b^2} - \theta$, and $h_n(y) = \sin(\frac{n\pi y}{b})$. I also find that for $\varphi$, \begin{align} \varphi_n(x) = \sinh(\sqrt{\lambda_n}x). \end{align} But I can't seem to apply the last homogeneous boundary condition $a\varphi'(a) = \varphi(a)$ at all. It seems like this BC is incompatible with the others - I've made an error somewhere but can't quite pinpoint where.
You seem to have forgotten that $\lambda_n$ isn't an eigenvalue. It's a separation constant, while eigenvalue is $\theta$. Thus your result for $\varphi_n(x)$ is, in expanded view,
$$\varphi_n(x)=\sinh\left(x\sqrt{\frac{n^2\pi^2}{b^2}-\theta}\right).$$
The boundary condition gives you
$$a\sqrt{\lambda_n}\cosh\left(a\sqrt{\lambda_n}\right)=\sinh\left(a\sqrt{\lambda_n}\right).$$
Solving this transcendental equation for $\lambda_n$ (getting $\lambda_{n,m}$ because for each $n$ you'll have a multitude of solutions marked by $m$), you can then find $\theta_{n,m}$ as
$$\theta_{n,m}=\frac{n^2\pi^2}{b^2}-\lambda_{n,m}.$$
Restriction of possible values of $\theta$ is the last action which will finally allow you to satisfy all your boundary conditions.