# Boundary value problems: eigenvalue and eigenfunction

I'm having trouble in understanding eigenvalues and eigenfunctions in BvP

the problem is:

$y''$ + $\lambda$$y = 0 y(0)=0 y(2\pi) = 0. Make characteristic polynomial r^2 + \lambda = 0 r_1,_2 = \pm \sqrt{- \lambda} the general solution is :$$y(x) = c_1 \cos\left(\sqrt{\lambda}x\right) + c_2 \sin\left(\sqrt{\lambda}x\right).$$Applying first boundary condition 0=y(0)=c_1 and applying the second boundary condition 0=y(\pi)=c_2 \sin\left(2\pi\sqrt{\lambda}\right). I know the part how to solve BVP I just wanted to know how get eigenvalue solution:$$\lambda_n = \left(\frac n2\right)^2 = \frac {n^2}{4},\quad n=1,2,3...$$and eigenfunction:$$y_n= \sin \left(\frac {nx}{2}\right),\quad n=1,2,3....$$IF$\lambda > 0$## 1 Answer As your said,$y(x) = c_1 \cos\left(\sqrt{\lambda}x\right) + c_2 \sin\left(\sqrt{\lambda}x\right).$and$0=y(0)=c_1$. The key step is how to solve$0=y(2\pi)=c_2 \sin\left(2\pi\sqrt{\lambda}\right)\$

Indeed, we require

\begin{align}2\pi\sqrt{\lambda}=n\pi, \mbox{ where } n=1,2,\cdots\end{align}

consequently, we deduce
\begin{align} &\sqrt\lambda=\frac{n}{2}\\ &\lambda=\left(\frac{n}{2}\right) ^2=\frac{n^2}{4},\mbox{ where } n=1,2,\cdots \end{align} Thus\begin{align} y_n= \sin \left(\frac {nx}{2}\right),\quad n=1,2,....\end{align}