non-trivial solution boundary problems Given the boundary problem
$$X''=\mu X,\;X(0)=0,\;X'(L)=0$$
We need to find the non trivial solution $X(x)$ that satisfies the above equations. Let $\mu<0$, $\mu=-k^2$ for some $k>0$. The general solution for $\mu<0$ can be described as 
$$X(x)=A\sin(k x)+B\cos(k x)$$
For some constants $A$, $B$. How do I formulate the solution $X(x)$ that satisfies $X(0)=0$ and $X(L)=0$
 A: Hint.
From the boundary conditions
$$
A\sin(0)+B\cos(0) = 0\\
A k\cos(k L)-B k\sin(k L) = 0
$$
or
$$
\left(
\begin{array}{cc}
0& 1\\
k\cos(kL) & -k\sin(kL)
\end{array}
\right)
\left(
\begin{array}{c}
A\\
B
\end{array}
\right)=\left(
\begin{array}{c}
0\\
0
\end{array}
\right)
$$
This linear system has a non trivial solution for
$$
\det\left(
\begin{array}{cc}
0& 1\\
k\cos(kL) & -k\sin(kL)
\end{array}
\right) = -k\cos(k L) = 0
$$
or for $k L = \frac{\pi}{2}+\nu \pi$
hence
$$
k = \frac 1L\left(\frac{\pi}{2}+\nu \pi\right)\ \ \ \nu = 1,2,3,\cdots
$$
now once we have the set of eigenfunctions we proceed with the determination of $A, B$
A: Any non-trivial solution where $X(0)=0$ can be normalized so that $X'(0)=1$, which gives a unique normalized solution
$$
         X_{\mu}(x)=\frac{\sin(\sqrt{-\mu}x)}{\sqrt{-\mu}}.
$$
This works in the limiting case as $\mu\rightarrow 0$ as well, where $X_0(x)=x$. Normalization is a handy way to eliminate special cases for general problems of this nature.
The set of $\mu$ for which there is a non-trivial solution of your problem is the set of $\mu$ for which $X'(L)=0$, or
$$
       \cos(\sqrt{-\mu}L)=0 \\ \implies \sqrt{-\mu}L=\frac{\pi}{2}+n\pi,\;\; n=0,\pm 1,\pm 2,\cdots, \\
        \mu = -\frac{1}{L^2}\left(\frac{\pi}{2}+n\pi\right)^2.
$$
So $X$ can be indexed by $n$:
$$
       X_n(x)=\sin((n+1/2)\pi x/L),\;\;\; n=0,1,2,3,\cdots .
$$
The cases for $n=-1,-2,-3,\cdots$ do not lead to additional solutions.
