I have the following ODE to solve via Green's function:
$$ \frac{d^2y}{dx^2} - y = F(x) $$
for $0<x<1$ with boundary conditions $y(0)=y(1)=0$.
So far I have done the following:
I try a Green's function $G(x,\xi)$ that satisfies
$$ \frac{d^2G}{dx^2} - G = \delta(x-\xi). $$
For $x \neq \xi$, we have that $\delta(x-\xi)=0$ and so the ODE becomes
$$ \frac{d^2G}{dx^2} - G = 0. $$
This has the solution:
$$ G(x,\xi) = A_1e^x - B_1e^{-x} \text{ for } x < \xi \text{ and } G(x,\xi) = A_2e^x - B_2e^{-x} \text{ for } x > \xi. $$
Applying the boundary condition $G(0) = G(1) = 1$, we get that $B_1 = -A_1$ and $B_2 = -A_2e^2$. Thus the solution is now of the form
$$ G(x,\xi) = A_1(e^x - e^{-x}) \text{ for } x < \xi \text{ and } G(x,\xi) = A_2(e^x - e^{2-x}) \text{ for } x > \xi. $$
We require $[G_x(x,\xi)]^{\xi_+}_{\xi} = 1$. This results in
$$ A_2(e^{\xi} + e^{2-\xi}) - A_1(e^{\xi}+e^{-\xi}) = 1. $$
How can I finish this problem? Thanks