eigenfunction and eigenvalue how can I find the eigenfunction for this BVP?
$$y'' + (\lambda)\, y = 0, \,\, y(0)=0, y(2\pi)=0$$
with this case $\lambda >0$.
I got the general solution: $$y(x)=c_1\cos(\sqrt{\lambda}x) + c_2\sin(\sqrt{\lambda}x)$$
 A: Your general solution is correct. Now you just have to play with the boundary conditions.
You have that
$$
\left\{
\begin{array}{l}
        y(x) = c_1 \cos(\sqrt{\lambda}x) + c_2 \sin(\sqrt{\lambda}x)\\
        y(0) = 0\\
        y(2\pi) = 0 
\end{array}
\right.
.$$
From the boundary condition at $x=0$:
$$y(0) = 0 \quad \Rightarrow \quad c_1 \underbrace{\cos(0)}_{=1} + c_2 \underbrace{\sin(0)}_{=0} = 0 \quad \Rightarrow \quad c_1 = 0,$$
so the general form of the eigenfunctions is
$$\tag{1} y(x) = c_2 \sin(\sqrt{\lambda}x).$$
Now, plugging in the boundary condition at $x=2\pi$:
$$y(2\pi) = 0 \quad \Rightarrow \quad c_2 \sin(\sqrt{\lambda}2\pi) = 0.$$
Since we want $y$ to be an eigenfunction, the last equation must have a nontrivial solution, so we require $c_2 \neq 0$, and this forces us to choose $\lambda$ such that $\sin(\sqrt{\lambda}2\pi) = 0$ holds. Therefore,
$$\tag{2} \sqrt{\lambda}2\pi = n\pi \quad \Rightarrow \quad \lambda = \left( \frac{n}{2} \right)^2, \quad n = 1, 2, \ldots $$
are the eigenvalues and their corresponding eigenfunctions are
$$\tag{3} y_n(x) = \sin{\frac{nx}{2}}, \quad n = 1, 2, \ldots \quad .$$
(I took the liberty of setting ${c_{2}}_n=1$, as it is costumary for eigenfunctions.)
