(Q4) Derive the Green's function for
$\displaystyle \displaystyle \frac{d^2y}{dx^2} - a^2y = \delta(x-x_0) $
with boundary conditions $\displaystyle y(x) \rightarrow 0 $ as $\displaystyle x \pm \infty $
First note that the solution is almost already is its self adjoint form.
$\displaystyle \frac{d}{dx}\left(-1\cdot y' \right)+ay=-\delta(x-x_0)$
So we need to find two different solutions such that each one satisfies one of the Boundary Conditions.
$\displaystyle v_0(t)=c_1e^{-ax}+c_2e^{ax}$
As $\displaystyle x \to -\infty$ this gives $\displaystyle c_1=0$ and
$\displaystyle v_0(t)=c_2e^{ax}$
$\displaystyle v_1(t)=c_3e^{-ax}+c_4e^{ax}$
As $\displaystyle x \to \infty$ this gives $\displaystyle c_4=0$
$\displaystyle v_1(t)=c_3e^{-ax}$
So the Green's function has the form
$\displaystyle g(x,s)=\begin{cases} Ae^{as}\cdot e^{-ax} \text{ for } s \le x \\ Be^{ax}\cdot e^{-as} \text{ for } x \le s\end{cases}$
Now we use the Wronskian and the and the condiditon that
$\displaystyle p(x)W(x)=-1$ so from above we have that $\displaystyle p(x)=-1$ and
$\displaystyle \begin{vmatrix} c_1e^{a(s-x)} && c_2e^{a(x-s)} \\ c_1(-a)e^{a(s-x)} && c_2(a)e^{a(x-s)} \end{vmatrix}=c_1c_2a+c_1_c2a$
Now using this we get
$\displaystyle 2c_1c_2a(-1)=-1 \iff c_1c_2=\frac{1}{2a}$
Now we can pick $\displaystyle c_1$ and $\displaystyle c_2$ to be any real numbers that satisfy the above equation for example
$\displaystyle c_1=c_2=\frac{1}{\sqrt{2a}}$
This gives
$\displaystyle g(x,s)=\begin{cases} \frac{1}{\sqrt{2a}}e^{as}\cdot e^{-ax} \text{ for } s \le x \\ \frac{1}{\sqrt{2a}}e^{ax}\cdot e^{-as} \text{ for } x \le s\end{cases}$