Here is the problem:
$-y''=\lambda y$ with boundary conditions of $y(0)=0$ and $y(\pi)=-y'(\pi)$
My attempt is below:
General solution is $$y(x) = a \sin(\sqrt\lambda x)+b\cos(\sqrt\lambda x).$$ Considering the boundary conditions $y(0)=0$, we get a general solution of $a\sin(\sqrt\lambda x) $ . With the second boundary condition, I got$$y'(x)=\sqrt\lambda a\cos(\sqrt\lambda x) \Rightarrow -y'(\pi)=-\sqrt\lambda a\cos(\sqrt\lambda \pi)=a\sin(\sqrt\lambda \pi)=y(\pi)$$
I am not sure if the work I have so far is correct and how to proceed from here.
You can normalize the first condition to be $$ y(0)=0, y'(0)=1. $$ There is no loss of generality in doing this because it amounts to a scale factor, which does not affect the second condition $$ y(\pi)=-y'(\pi). $$ The added scale factor results in a unique solution $$ y_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ This solution remains valid for $\lambda=0$ in the limit as $\lambda\rightarrow 0$, and that solution is $y_0(x)=x$, which you can directly verify to satisfy the endpoint conditions at $0$. Now the eigenvalue equation becomes $$ y_{\lambda}(\pi)=-y_{\lambda}'(\pi) \implies \\ \frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}=-\cos(\sqrt{\lambda}\pi). $$ This is a transcendental equation: $$ \tan(\sqrt{\lambda}\pi)=-\sqrt{\lambda}. $$ You can plot these two functions and see that there are infinitely many solutions where the graphs intersect. But there is no explicit expression for the solutions--the equation is transcendental.