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 $\lambda > 0$
Best 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}