Solve the given differential equation by using Green's function method I am really struggling with the concept and handling of the Green's function. I have to solve the given differential equation using Green's function method
$$\frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0$$
 A: For these sorts of problems, knowing the properties of the solution you are looking for is just as important as computing the solution. Generally, we assume that $y$ is continuous, $y$ is differentiable except at $x'$, and that 
\begin{align}
\lim_{\epsilon\to0} \int_{x'-\epsilon}^{x'+\epsilon} y'' +k^{2}y \, \mathrm{d}y = \lim_{\epsilon\to0} \int_{x'-\epsilon}^{x'+\epsilon} \delta(x-x')\, \mathrm{d}y.
\end{align}
This latter condition implies that
\begin{align}
\lim_{\epsilon\to0} y'(x'+\epsilon) - y'(x'-\epsilon) = 1
\end{align}
We'll write this condition as "$y'(x^{'}_{+}) -y'(x^{'}_{-}) = 1$."
For $x<x'$, our DE reads $y''+k^{2}y = 0$ with $y(0) = 0$, giving $y(x) = A\sin(kx)$.
For $x>x'$, our DE reads $y''+k^{2}y = 0$ with $y(L) = 0$, giving $y(x) = B\sin(k(x-L))$.
Since $y$ is continuous, $A\sin(kx') = B\sin(k(x'-L))$, so that 
\begin{align}
y(x) = 
\begin{cases}
A\sin(kx), & x<x', \\
A\frac{\sin(kx')}{\sin(k(x'-L))}\sin(k(x-L)), & x\ge x'
\end{cases}
\end{align} 
The condition $y'(x^{'}_{+}) -y'(x^{'}_{-}) = 1$ specializes to 
\begin{align}
\frac{Ak\sin(kx')\cos(k(x'-L))}{\sin(k(x'-L))} - Ak\cos(kx') = 1,
\end{align}
so that
\begin{align}
A = \frac{1}{k\left(\sin(kx')\cot(k(x'-L)) - \cos(kx')  \right)}
\end{align}
