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I want to find eigenvalue and eigenfunction of this problem: $$ y''+ \lambda y=0, 0<x<l \\ y(0)=0, ly'(l)+ky(l)=0 $$ And $y'$ stands for $\frac{dy}{dx}$ and similar for $y''$.

I get the following results: After solving this homogeneous ODE, $$ y=C_1 \cos(\sqrt{\lambda} x) + C_2 \sin(\sqrt{\lambda} x)\\ y(0)=0 \implies C_1=0\\ \therefore y=C_2 \sin(\sqrt{\lambda} x)\;\;\; \text{and} \;\;\; y'=C_2 \sqrt{\lambda} \cos(\sqrt{\lambda} x)\\ ly'(l)+ky(l)=0 \implies C_2 \bigg( l \sqrt{\lambda} \cos(\sqrt{\lambda}l) + k \sin(\sqrt{\lambda}l) \bigg)=0 $$ For non-trivial solution $$ \bigg( l \sqrt{\lambda} \cos(\sqrt{\lambda}l) + k \sin(\sqrt{\lambda}l) \bigg)=0 $$

Now, I do not think any $\lambda$ satisfying the above equation. Can someone comment?

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For simplicity let me call $\lambda=\mu^2$. The equation for the eigenvalues can be written as $$ \mu=-\frac{k}{\ell}\tan(\ell\,\mu). $$ $\mu=0$ is a solution that gives$y\equiv0$, so it is not an eigenvalue. If you draw the graph of $\mu$ and $-(k/\ell)\tan(\ell\,\mu)$, it becomes cleat that there is a solution $$ \mu_k\in\Bigl(\frac{(2\,k-1)\,\pi}{2\,\ell},\frac{(2\,k+1)\,\pi}{2\,\ell}\Bigr),\quad k=1,2,3\dots $$ To find the numerical values of $\mu_k$ you will have to resort to numerical methods. From the picture is also clear that $$ \lim_{k\to\infty}\mu_k-\frac{(2\,k-1)\,\pi}{2\,\ell}=0. $$

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