I was trying to solve this and got stuck right before the final solution, so that's the problem:
$$ \begin{matrix} &u_{xx}+u_{yy}=0 & x\in (0,2), y\in (0,1) \\ &u(x,0)=0 & x\in[0,2]\\ &u(x,1)=0 &x\in[0,2] \\ &u(0,y)=sin(\pi y) &y\in[0,1] \\ &u(2,y)=sin(3\pi y) &y\in[0,1] \\ \end{matrix}$$
I tried looking for solutions of the form $$ u(x,y)=X(x)Y(y)$$
by solving this PDE using the method of separation of variables I got to the following general solution
$$u(x,y)=\sum_{n=1}^{\infty}[C_{n}e^{n\pi x}+D_{n}e^{-n\pi x}]sin(n\pi y )$$ an alternative : $$ u(x,y)=\sum_{n=1}^{\infty}[A_{n}sinh(n\pi x)+B_{n}sinh(n\pi(2-x))]sin(n\pi y )$$ In my textbook the alternative general solution is presented first and I have the feeling that it might be better in order to express an exact solution to the initial conditions. if that's true, what's the reason and how can the alternative solution be derived?
in order to solve for the initial conditions by using the first method I get to the conclusion that $C_{1}+D_{1}=1$ and $C_{3}e^{6n\pi}+D_{3}e^{-6n\pi}=1$ although how can this system of solutions (which is also from the textbook) for the previous two equations can be achieved? $$\\C_{1}=\frac{1}{1-e^{4\pi}},D_{1}=\frac{e^{4\pi}}{1-e^{-4\pi}},C_{3}=\frac{1}{e^{6\pi}-e^{-6\pi}},D_{3}=\frac{1}{e^{-6\pi}-e^{6\pi}}$$
We've already went over ODE's, sturm-liouville and Fourier series in my course but I'm missing something and can't understand how a specific general equation can be achieved with what I got so far. I'd appreciate any help or insight