I want to solve the equation $$\nabla^{2}\,{\rm P}(x, y) = \frac{k}{c^{2}}{\rm P}(x, y)$$ or $$\frac{\partial^{2}{\rm P}}{\partial x^{2}}+\frac{\partial^{2}{\rm P}}{\partial y^{2}}=\frac{k}{c^{2}}{\rm P}(x, y)$$ I want to do this by separation of variables, let ${\rm P}(x, y)={\rm R}(x){\rm S}(y)$ so we then have $${\rm R}''(x){\rm S}(y) + {\rm R}(x){\rm S}''(y) = \frac{k}{c^{2}}{\rm R}(x){\rm S}(y)$$ Now dividing by ${\rm R}(x){\rm S}(y)$ yields $$\frac{{\rm R}''(x)}{{\rm R}(x)}+\frac{{\rm S}''(y)}{{\rm S}(y)}=\frac{k}{c^{2}}$$
Now at this point i'm not really sure how to serparate this any further. Can I let $k=\alpha+\beta$ then say $$\frac{{\rm R}''(x)}{{\rm R}(x)}+\frac{{\rm S}''(y)}{{\rm S}(y)}=\frac{\alpha}{c^{2}}+\frac{\beta}{c^{2}}$$ So we have $$\frac{{\rm R}''(x)}{{\rm R}(x)}=\frac{\alpha}{c^{2}}, \frac{{\rm S}''(y)}{{\rm S}(y)}=\frac{\alpha}{c^{2}}$$
OR is this invalid?
Rewrite $$\frac{{\rm R}''(x)}{{\rm R}(x)}+\frac{{\rm S}''(y)}{{\rm S}(y)}=\frac{k}{c^{2}}$$ as $$\frac{{\rm R}''(x)}{{\rm R}(x)}=\frac{k}{c^{2}}-\frac{{\rm S}''(y)}{{\rm S}(y)}$$ The left side is independent of $y$. The right hand side is independent of $x$. Since they are identically equal, they are in fact constant. Let the constant be $\lambda$. Then $$ R''(x) = \lambda R(x) \quad \text{ and }\quad S''(x) = (k/c^2-\lambda) S(x) $$ which are differential equations you can solve. This is where you need to use boundary conditions (not specified in your question).