I'm trying to solve a non-homogeneous second order ODE. I've read similar other questions, but all use the method of Laplace transformations, which I've not seen/used before.
The ODE is: $$y''(x) - a^{2}y(x)=\delta(x-p)$$ subject to the boundary conditions $y(0)=0$ and $y(b)=0$ for $0<p<b$.
I've established the homogeneous solution to be
$$ y(x) = C_{1}e^{ax}+C_{2}e^{-ax}, ~\text{if} ~~x<p \\ y(x) = D_{1}e^{ax}+D_{2}e^{-ax}, ~\text{if} ~~x>p $$
My professor suggested the best way to proceed was to apply the boundary conditions, to get rid of some of the constants. I've done this and obtained
$$ y(x) = 2C_{1}\sinh(ax), ~\text{if} ~~x<p\\ y(x) = -2D_{1}e^{ab}\sinh{(a(b-x))}, ~\text{if} ~~ x>p\\ $$
Getting rid of the last two constants is where I'm struggling. My professor said that for a $\delta(x-p)$ ODE, you need to ensure continuity for $y(p)$ and then have a unit displacement for $y'(p)$. I'm unsure about how to apply this. I have not done Laplace transformations before. Any help would be greatly appreciated!
Continuity of $y$ at $x=p$ implies $$(D_1 e^{ap} + D_2 e^{-ap}) - (C_1 e^{ap} + C_2 e^{-ap}) = 0.$$
A unit step of $y'$ at $x=p$ implies $$(D_1 a e^{ap} + D_2 (-a) e^{-ap}) - (C_1 a e^{ap} + C_2 (-a) e^{-ap}) = 1.$$
Using these two conditions you can eliminate two of the constants $C_1, C_2, D_1, D_2.$
But you probaly want another pair of boundary conditions: $$\lim_{x \to \pm\infty} y(x) = 0.$$ What do these imply?