1. ## Eigenvalues of y''+λy=0

The problem is stated as follows:

"Show that $\displaystyle y''+\lambda y=0$ with the initial conditions $\displaystyle y(0)=y(\pi)+y'(\pi)=0$ has an infinite sequence of eigenfunctions with distinct eigenvalues. Identify the eigenvalues explicitly."

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$\displaystyle \lambda \le 0$ seems to yield the trivial solution, so $\displaystyle \lambda > 0$. The general solution is then $\displaystyle y=A\sin{(\sqrt{\lambda} x)} + B\cos{(\sqrt{\lambda} x)}$. The first initial condition gives $\displaystyle y=A\sin{(\sqrt{\lambda} x)}$, and then the second gives $\displaystyle A(\sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}) = 0.$

I've tried to solve $\displaystyle \sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}= 0$ for $\displaystyle \lambda$ by writing the LHS as a single sine function, and separately by dividing with $\displaystyle \cos{(\sqrt{\lambda} \pi)}$, but neither approach seems to give a good way of giving an explicit formula for $\displaystyle \lambda$.

(The best I've been able to do with the single sine function is to write $\displaystyle \sin{(\sqrt{\lambda} \pi)} + \sqrt{\lambda} \cos{(\sqrt{\lambda} \pi)}$ as $\displaystyle \sqrt{1+\lambda} \sin{(\sqrt{\lambda} x + \delta)}$. Consequently, $\displaystyle \lambda = \left(\dfrac{\pi k - \delta}{\pi}\right)^2, k \in \mathbb{N}$. But then $\displaystyle \delta = \arccos \frac{1}{\sqrt{1+\lambda}} = \arcsin \frac{\sqrt{\lambda}}{\sqrt{1+\lambda}}$, so I wouldn't call this explicit.)

Thoughts?

2. ## Re: Eigenvalues of y''+λy=0

Try this:
$\displaystyle \sin (\sqrt{\lambda}\pi)+\sqrt{\lambda}\cos (\sqrt{\lambda}\pi)=0$
$\displaystyle \sqrt{\lambda}=-\frac{\sin (\sqrt{\lambda}\pi)}{\cos (\sqrt{\lambda}\pi)}$

Your eigenvalues are given by the intersection of this curves.