Find the normalized eignefunctions for $$y'' + \lambda y = 0$$ $$y(0)=0, y(\pi)-2y'(\pi)=0.$$
My teacher gives me this hints:
Consider$$(py')'+qy+\lambda ry=0$$ where $p, p', q, r$ are real-valued continuous functions. Let $\{ \phi_n(x)\}^{\infty}_{n=1}$ be the set of eigenfunctions such that $$m\neq n \Rightarrow \int_0^{\pi} \phi_n(x) \phi_m(x) r(x) dx = 0$$ and $$m= n \Rightarrow \int_0^{\pi} \phi_n(x) \phi_m(x) r(x) dx = 1$$ then $\phi_n(x)$ are the normalized eigenfunctions.
I am lost in the question. I have the following three cases:
1) If $\lambda=0$, we get the trivial solution.
2) If $\lambda <0$, we let $\lambda= -k^2$, we get $y(x)=0$ (trivial solution) or $k=0$ (Contradiction).
3) If $\lambda >0$, we let $\lambda= k^2$, we have $y(x)=c_2 \sin kx$, where $\tan k \pi = 2k$.
It seems that I can get some eigenfunctions from the last case. But I don't know how to proceed.
Now, suppose I can find the eigenfunctions, I use the hint given by my teacher. But I rewrite the Sturm-Liouville problem to $$py'' + p'y' +(q+\lambda r)y=0.$$ From this, I see $p(x)=1$, but what about $r(x)$? Can I still find the normalized eigenfunctions without knowing what $r(x)$ and $q(x)$ is?
Thanks in advance.