Solve the following differential equation $ u_{xx}-m^2u=\delta(x-x_0)$ Find the solution of following equation
$$ u_{xx}-m^2u=\delta(x-x_0),$$
$u(0)=0=u(L),\ x\in\mathbb R^2$
Actually, I don't know how to solve. Is there someone to help?
 A: Let us temporarily assume that the solution can be extended over $R^+$. Next, apply the Laplace transform to the DE;  
$$(s^2U(s)-su(0)-u_x(0))-m^2U(s)=e^{-sx_0}$$  
Where, $$U(s):= \int_0^\infty \! e^{-sx}u(x) \, \mathrm{d}x.$$
$$\Rightarrow(s^2U(s)-u_x(0))-m^2U(s)=e^{-sx_0}$$ $$\Rightarrow U(s)=\frac{u_x(0)+e^{-sx_0}}{s^2-m^2}$$  
Applying the inverse Laplace transform yields:
$$u=\frac{u_x(0)}{m}\sinh(mx)H(x)+\frac{1}{m}\sinh(m(x-x_0))H(x-x_0)$$ 
In which $H(.)$ denotes the Heavisde step function. Imposing $u(L)=0$ with the assumption $0<x_0<L$ gives:  
$$u_x(0)=-\frac{\sinh(m(L-x_0))}{\sinh(mL)}$$ 
Therefore, the solution is  
$$u=-\frac{\sinh(m(l-x_0))}{m\sinh(mL)}\sinh(mx)H(x)+\frac{1}{m}\sinh(m(x-x_0))H(x-x_0)$$
A: Write $u=G+y$, where $G=\frac{1}{4\pi |x-x_0|}$ solves $-u_{xx}=\delta(x-x_0)$ in the whole $\mathbb R^1$. Then plugging this in your equation gives   $-y_{xx}+m^2G+m^2y=0\quad (*)$ with new boundary conditions: $y(0)=-G(0)$ and $y(L)=-G(L)$. Now you have to solve the regularized equation $(*)$ for $y$.
