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Eigenvalue problem:
$y''+ \lambda y = 0$ subject to $y'(0) = 0$ and $y(1) + y'(1) = 0$

The question is:

show that the eigenvalues are given by $\lambda = \mu^2$ with $\mu$ any root of $ \mu \tan \mu$

I actually have some solution which says that for $\lambda =\mu^2 >0$ we find solutions $y= A\cos\mu x + B\sin \mu x$.

I can continue the problem by substituting the boundary conditions into the the solution equation....

Can anyone enlighten me with how they got to the answer?

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are you sure you've copied everything correctly? –  chemeng May 12 '12 at 14:47

2 Answers 2

I believe that there is a misprint in the problem statement. Just as you said, the general solution of $$ y''+\lambda y=0 $$ has the form $$ y=A\cos\sqrt{\lambda}x+B\sin\sqrt{\lambda}x\,, $$ (we assumed here that $\lambda>0$). However, this only satisfies the differential equation, not boundary conditions. Therefore, we need to insert this general solution into the boundary conditions and choose parameters $A$, $B$ and $\lambda$ from there. The described substitutions yield $$ y'(0)=B\sqrt{\lambda}=0\,, $$ $$ y(1)+y'(1)=A(\cos\sqrt{\lambda} -\sqrt{\lambda}\sin\sqrt{\lambda})+B(\sin\sqrt{\lambda}+\sqrt{\lambda}\cos\sqrt{\lambda})=0\,. $$ First of these equations means that $B=0$ (remember, we assumed that $\lambda>0$). The second equation then simplifies into $$ A(\cos\sqrt{\lambda}-\sqrt{\lambda}\sin\sqrt{\lambda})=0\,. $$ $A=0$ is not interesting, because it would give us a trivial solution. Hence, we must ensure that the second factor is equal to zero. It can be rearranged as $$ \sqrt{\lambda}\tan\sqrt{\lambda}=1\,. $$ If you introduce $\mu=\sqrt{\lambda}$, then $\lambda=\mu^2$ and $\mu$ solves $\mu\tan\mu=1$.

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Let $\phi (x)=e^{(\mu x)}$ be a solution to your ODE. Substituting in your equation:

$$\phi''(x)+\lambda \phi(x)=0 \Leftrightarrow\mu^2e^{\mu x}+\lambda e^{\mu x}=0$$and since $e^{\mu x}\ne0$ it should be: $$\lambda+\mu^2=0$$

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