How to solve nonhomogenous 2 dimensional wave equation using separation of variables? I've been trying to solve following 2-dimensional nonhomogenous wave equation using separation of variables method
$$
u_{tt}=4(u_{xx}+u_{yy}) \quad (0<x<1, \quad 0<y<1, \quad 0<t)
$$
And boundary conditions are
$$
u(0,y,t)=sin(\pi y), \quad u_x(1,y,t)=0, \quad u(x,0,t)=0, \quad u(x,1,t)=0, \quad u(x,y,0)=cos(\pi x)sin(\pi y), \quad u_t(x,y,0)=1
$$
Because the solution must be in separation of variables method the following approach would be used
$$
u(x,y,t)=X(x)Y(y)T(t)
$$
But in order to use separation of variables method, the equation and the boundary conditions must be homogeneous. In my case $u(0,y,t)=sin(\pi y)$ boundary condition breaks the homogeneity. So I need to make the boundary conditions homogeneous. I tried to separate $u$ into two different functions;
$$
u(x,y,t)=v(x,y,t)+w(x,y)
$$ 
$w$ function here would be used to make the boundary conditions homogeneous. The $w(x,y)$ function I am looking for needs to satisfy following conditions;
$$
w_{xx}=-w_{yy} 
$$
So that the equations remains homogeneous.
$$
w(0,y)=sin(\pi y)
$$
$$
w_x(1,y)=0
$$
$$
w(x,0)=0
$$
$$
w(x,1)=0
$$
So that boundary conditions along x and y axis become homogenous.
When I satisfy homogeneity for $u(0,y,t)$, boundary condition along $y$ direction becomes nonhomogeneous. I need help.
 A: After going through this problem, I can't imagine why one would ever want to assign this problem, there's a lot to keep track of. This is only an outline. Also, please for the love of god check my work.
After letting $u(x, t) = v(x, t) + \sin{\pi y}$, we get the new problem
$\begin{cases}
v_{tt} - 4 \Delta v= -\pi^2 \sin{\pi y}\\
BC: \begin{cases}
v(x, t) = 0, x=0, y=0, 1\\
v_{x}(x,t) = 0, x=1
\end{cases}\\
IC: \begin{cases}
v(x,0) = \cos{\pi x} \sin{\pi y} - \sin{\pi y}\\
v_{t}(x,0)=0
\end{cases}
\end{cases}
$
The idea, because we're dealing with a forced problem now after shifting the BC, is to use an eigenfunction expansion to solve the problem. In order to do this, we have to somewhat some the homogeneous version of the above problem via SepVar.
Doing so, Assuming $v(x, t) = \phi(t)X(x)Y(y)$, we see that $X(x) = \sin{\left(\frac{\pi}{2}(2n-1)x\right)}$, and $Y(y) = \sin{\left( m \pi y \right)}$.
Let $\alpha_n^2 = \frac{\pi}{2}(2n-1)$ and $\beta_m^2 = m \pi$. Thus, as the moment, we've got that 
$$v(x, t) = \sum_{n, m = 1}^{\infty} \phi_{n,m}(t)\sin{(\alpha_n^2 x)} \sin{(\beta_m^2 y)}$$
We can find that the expansion of the forcing function in this way will be
$$-\pi^2 \sin{(\pi y)} = -\sin{(\pi y)} \sum_{n=1}^{\infty}\frac{4\pi}{2n-1}\sin{(\alpha_n^2 x)}$$
via the coefficients 
$$A_{n,m} = 4\int_0^1 \int_0^1 f(x,y)\sin{(\alpha_n^2 x)}\sin{(\beta_m^2y)}$$
You'll find that $m=1$ is the only possibility.
Now we do the typical thing of plugging in our solution into the original to solve for $\phi_{n,1}(t)$. This results in a 2nd order nonhomogeneous ODE. Let $\lambda_{n,1}^2 = \alpha_n^2 + \beta_1^2$. You should get
$$\phi_{n,1}''(t)+4\lambda_{n,1}^2 \phi_{n,1}(t)=-\frac{4\pi}{2n-1}$$
So that $$\phi_{n,1}(t) = A_n\cos{(2\lambda_n x)}+B_n \sin{(2 \lambda_n x)} - \frac{\pi}{(2n-1)\lambda_n^2} $$
Applying the IC's, you should get that $B_n = 0$, and to get A_n, we have to computer another inner product with $v(x,0)$. We get 
$$A_n = \frac{4(20n^2+12n+17)}{\pi (2n-3)(2n-1)(2n+1)(4n^2+4n+5)}$$
So we're done, we've found all of the pieces to put together our solution for $v(x,t)$. Remembering our expression for $u$, we see that
$$u(x,t) = \sin{(\pi y)}\left[1 + \frac{4}{\pi} \sum_{n=1}^{\infty} \left[ \frac{20n^2+12n+17}{(2n-3)(2n-1)(2n+1)(4n^2+4n+5)} \right]\cos{(2 \lambda_n t)}\sin{(\alpha_n x)} \right]$$
