Basically I got a simple wave equation with an extra twist. The PDE is $$\frac {\partial^2 y}{\partial t^2} = c^2\frac {\partial^2 y}{\partial x^2} + L$$ with homogeneous boundary condition. As usual, I use the ansatz $ Y(x,t) = F(x)G(t) $ and I have $\frac {\partial^2 y}{\partial t^2} =F''G $ and $\frac {\partial^2 y}{\partial x^2} =FG''$. I substitute this in, and now I have $$FG''= c^2F''G +L$$ I don't know how to make this separable--obviously if I divide both sides with $FG$ as usual, then I will have that annoying constant $\frac {L}{FG} $ and I don't know how to make it separable then.
Can anybody help me? I know I have to separate it somehow, but I don't know exactly how to.
{The actual problem is an inhomogeneous boundary condition type of problem, but I have found the steady state solution, so I just have to find the transient solution, thus the homogeneous boundary condition}
The actual question:
- PDE: $$\frac {\partial^2 y}{\partial t^2} = c^2\frac {\partial^2 y}{\partial x^2} + L,\quad 0 \leq x \leq J,\quad t>0$$ where $L$ is constant and $c$ is constant wave speed.
- Boundary Condition: $ u(0,t)=0$, $t>0$ and $u(J,t)=h$, $t>0$
- Initial Condition: $u(x,0) = 0$, $0<x<J$ and $\frac {\partial u}{\partial t} (x,0)=0$, $0<x<J$
I did the steady state solution bit, and got it as $w(x)= \frac{-Lx^2}{c} + \frac {Hx}{L} +\frac{GLx}{c} $. So I modified the boundary condition and initial condition to get the transient part, and I got the boundary condition homogeneous. However when I try to solve it, the question above arises.