How would i go about solving this equation
$$\frac{-b''(y)}{b(y)}+\frac{-a''(x)}{a(x)}=\lambda$$
I've attempted to set $$\frac{-a''(x)}{a(x)}=\mu$$ where $\mu$ is an arbitrary constant. and then solve using separation of variables, however, i can't seem to solve for $\mu$. My plan is to use this to then solve
$$\frac{-b''(y)}{b(y)}=\lambda +\mu$$
Could somehow explain if im going about solving this the right way and if yes help me, if no explain how i can do this to solve for $\lambda$
If i've made any major errors or missed information in the above or if someone just wants to know where this is coming from i'm trying to solve The Eigenvalue Problem for the Rectangle where $\lambda$ is an eigenvalue for the Dirichlet Laplacian
You want to solve
\begin{align} \left\{\begin{array}{l} \Delta f(x,y) = - \lambda f(x,y) \qquad \text{in} \,\, \Omega \\ f(x,y) \Big|_{\partial \Omega} = 0 \end{array}\right. \end{align}
where $\Omega$ is a rectangle. By assuming that $f(x,y)$ is separable, i.e.
$$f(x,y) = a(x) b(y)$$
you get
$$\frac{\partial^2 f(x,y)}{\partial x^2} + \frac{\partial^2 f(x,y)}{\partial y^2} = b(y) \frac{\partial^2 a(x)}{\partial x^2}+a(x)\frac{\partial^2 b(y)}{\partial y^2} = - \lambda a(x) b(y)$$
Separating the variables yields (now in your notation)
$$\frac{a''(x)}{a(x)} = -\frac{b''(y)}{b(y)} - \lambda$$
The left side solely depends on $x$, the right side solely depends on $y$, since $\lambda$ is an eigenvalue in $\Bbb R$. So this equation can be fulfilled only if both sides are constant, i.e.
$$\frac{a''(x)}{a(x)} = -\frac{b''(y)}{b(y)} - \lambda = \mu \qquad \mu \in \Bbb R$$
So you get two equations
$$a''(x) - \mu a(x) = 0 \qquad \text{and} \qquad b''(y) + (\lambda + \mu) b(y) = 0$$
These are well-known second-order ODE's. So basically you're more or less right in your attempt. You can then determine $\lambda$ and $\mu$ by solving these ODE's and thus obtain $f(x,y) = a(x) b(y)$. Applying the boundary conditions on the obtained $f(x,y)$, you'll get conditions for the constants $A$, $B$, $C$ and $D$ of the solutions of these two ODE's. Keep in mind that you probably seek solutions, where $f(x,y) \not\equiv 0$ holds. Finally, you'll get equations out of the boundary conditions, where, in order to be fulfilled, it must hold that either $f(x,y) \equiv 0$ (which you probably don't want) or $\lambda$ (or $\mu$) is of a particular form.