I have the following problem:
Find all Eigenvalues and associated Eigenfunctions for
$Y'' + LY = 0$,
$Y(0) + Y'(0) = 0$,
$Y(\pi) + Y'(\pi) = 0$,
$x \in (0,\pi)$.
So far, I have tested $L$ less than zero and found a contradiction because I believe it forces a $(L = a^2)$ to be zero. $L = 0$ yields only the trivial solution. When I test $L>0$ I find that a must be an integer, but I am lost as to how I find the associated Eigenfunction for this value.
The solutions of $$ Y''+LY=0 \\ Y(0)+Y'(0)=0 $$ are simplified by adding normalization such as $Y(0)=1$. The solutions are $$ Y_L(x)=\cos(\sqrt{L}x)-\frac{\sin(\sqrt{L}x)}{\sqrt{L}}. $$ These solutions satisfy $Y(0)+Y'(0)=0$, including the limiting case where $L=0$, which is $Y_0(x)=1-x$. $L$ is a valid eigenvalue iff $Y_L$ satisfies the required endpoint condition at $x=\pi$, which holds iff $L$ satisfies $$ Y(\pi)+Y'(\pi)=0, $$ or equivalently, $$ \cos(\sqrt{L}\pi)-\frac{\sin(\sqrt{L}\pi)}{\sqrt{L}}-\sqrt{L}\sin(\sqrt{L}\pi)-\cos(\sqrt{L}\pi)=0 \\ \left(\frac{1}{\sqrt{L}}+\sqrt{L}\right)\sin(\sqrt{L}\pi)=0. $$ There are solutions $\sqrt{L}=n\pi$ or $L=n^2\pi^2$ for $n=1,2,3,\cdots$, as well as $L=-1$. The solution for $L=-1$ is $$ Y_{-1}(x)=\cosh(x)-\sinh(x) = e^{-x}. $$