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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 $\lambda=1$.

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    $\begingroup$ I think you made a mistake with applying your boundary conditions $\phi(x) = c_1 \cos(p x) + c_2 \sin(p x)$, $\phi'(x) = p[ -c_1 \sin(p x) + c_2 \cos(p x)]$, would lead to $c_1 + p c_2 = 0$ and similar for the other one. $\endgroup$ – Gregory Jun 8 '20 at 18:16
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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 $$ \phi_{-1}=\frac{1}{2}\cosh(x)-\frac{1}{2}\sinh(x)=e^{-x},\\ \phi_{n^2}=\frac{1}{2}\cos(nx)-\frac{1}{2n}\sin(nx),\;\; n=1,2,3,\cdots. $$

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