Eigenvalue problem with Robin Boundary Conditions Solve the eigenvalue problem with Robin boundary conditions: $$\phi''(x)+\lambda\phi(x)=0,$$ $$\phi(0)+\phi'(0)=0,$$ $$\phi(\pi)+\phi'(\pi)=0$$
I worked out that both $\lambda=0$ and $\lambda<0$ aren't eigenvalues, and for $\lambda>0$ with $\lambda=+p^2$ I found that $c_1+c_2=0$ and $c_1\cos(\pi p)+c_2p\cos(\pi p)=0$. Something cool was supposed to happen and I'm not sure if I'm doing it right. $$ $$
So continuing with this I got that $p=1$ and therefore $\lambda=1$.
 A: Every non-zero solution of
$$
              \phi''+\lambda\phi =0, \\ \phi(0)+\phi'(0)=0
$$
can be normalized so that $\phi(0)-\phi'(0)=1$ by scaling the solution. If you could not do that, then the solution $\phi$ would satisfy $\phi(0)=\phi'(0)=0$, which is satisfied only by the $0$ function. So, start by solving $\phi''+\lambda \phi=0$ subject to
$$
                \phi(0)+\phi'(0)=0,\;\;\; \phi(0)-\phi'(0)=1, \\
               \implies \phi(0)=1/2,\;\; \phi'(0)=-1/2.
$$
This has the unique solution
$$
                \phi_{\lambda}(x) = \frac{1}{2}\cos(\sqrt{\lambda}x)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}x).
$$
In order for $\lambda$ to be an eigenvalue it will necessary for $\phi_{\lambda}(\pi)+\phi_{\lambda}'(\pi)=0$, which gives the eigenvalue equation
$$
          \frac{1}{2}\cos(\sqrt{\lambda}\pi)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}\pi)-\frac{\sqrt{\lambda}}{2}\sin(\sqrt{\lambda}\pi)-\frac{1}{2}\cos(\sqrt{\lambda}\pi) = 0 \\
        \left(1+\lambda\right) \frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}=0.
$$
The eigenvalues are
$$
             \lambda=-1 \mbox{ and } \lambda = 1^2,2^2,3^2,\cdots .
$$
The corresponding eigenfunctions are
$$
           \phi_{-1}=\frac{1}{2}\cosh(x)-\frac{1}{2}\sinh(x)=e^{-x},\\
           \phi_{n^2}=\frac{1}{2}\cos(nx)-\frac{1}{2n}\sin(nx),\;\; n=1,2,3,\cdots.
$$
