How to solve Sturm-Liouville problem $y'' + \lambda y = 0$ with unknown initial conditions? I am trying to solve the following Sturm-Liouville problem:
$$
\begin{cases}
y'' + \lambda y = 0 \\
y(x_0) = 0 = y(x_1)
\end{cases}
$$
In the interesting case where $\lambda > 0$ I get the following equations:
$$
\begin{cases}
A \sin(\sqrt{\lambda}x_0) + B \cos(\sqrt{\lambda}x_0) = 0 \\
A \sin(\sqrt{\lambda}x_1) + B \cos(\sqrt{\lambda}x_1) = 0
\end{cases}
$$
The main difficulty here is discussing this system. Apparently the solution should be $$y_n(x) = B_n \sin\Big(\frac{n \pi x}{x_1-x_0}\Big).$$
 A: Define $u(x) = y(x + x_0)$, then$$
\begin{cases}
u'' + λu = 0\\
u(0) = u(x_1 - x_0) = 0
\end{cases}
$$
which reduces the original problem to an easier form.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Write your solution as $\ds{a\sin\pars{\root{\lambda}\bracks{x - x_{0}}}}$ which already satisfies the boundary condition at $\ds{x = x_{0}}$.

Then,
\begin{align}
0 & = \mrm{y}\pars{x_{1}} = a\sin\pars{\root{\lambda}\bracks{x_{1} - x_{0}}}
\stackrel{\large\mrm{non\ trivial\ solution} \atop \large a\ \not=\ 0}{\implies}
\root{\lambda}\bracks{x_{1} - x_{0}} = n\pi\,,\qquad
n \in \mathbb{N}_{\ \geq\ 1}
\\[5mm] &
\implies \root{\lambda} = n\,{\pi \over x_{1} - x_{0}} \implies
\bbx{B_{n}\sin\pars{n\,{\pi \over x_{1} - x_{0}}\bracks{x - x_{0}}}}
\end{align}
