I'm trying to solve an S-L problem, but I'm finding it difficult to find any solved examples of equations that aren't in the form of $y''=\pm \lambda y$.
Instead, it's in the following form: $$\frac{d}{dx} \bigg[-e^{ax}\frac{dy}{dx}\bigg]-e^{ax}y\bigg[1+\lambda\bigg] = 0$$
(I'm leaving out the exact value of a, and the given conditions, because I don't want an exact solution, but can someone please explain how I'm supposed to go about this? What I'm trying is setting $\lambda = 0$, $\lambda > 0$, and $\lambda < 0$. (I've already checked for $\lambda = 0$, and got a trivial solution, but I have yet to try both other conditions). Anyway, is this enough? Or is there a different way to actually get a solution?
I would really appreciate any help! Thank you in advance!
$$ -e^{ax}\frac{d^2y}{dx^2}-ae^{ax}\frac{dy}{dx}-e^{ax}y\left[1+\lambda\right]=0 \\ \frac{d^2y}{dx^2}+a\frac{dy}{dx}+(1+\lambda)y=0. $$ This is a linear ODE with constant coefficients.
From your comment, you want conditions $$ y(0)=0,\; y'(\pi)-y(\pi)=0. $$ You have solutions $e^{mx}$ where $m$ satisfies $$ m^2+am+(1+\lambda)=0 \\ (m+a/2)^2+(1+\lambda)-a^2/4=0 \\ m+a/2 =\pm\sqrt{a^2/4-(1+\lambda)} \\ m=-a/2\pm\sqrt{a^2/4-(1+\lambda)} $$ A solution with $y(0)=0$ is $$ y(x)=e^{-ax/2}\sinh(\sqrt{a^2/4-(1+\lambda)}\,x) $$ Then the eigenvalue equation becomes $$ y'(\pi)-y(\pi)=0. $$ The $\sinh$ becomes $\sin$ if the argument of $\sinh$ is purely imaginary, and that's what I would expect here.