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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 $\lambda > 0$

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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}

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