I keep getting stuck on this problem, so if someone could point out where my method is flawed and how I should approach this problem, that would be extremely useful.
We're considering the square region $0\leq x \leq \pi, 0\leq y \leq \pi$ with the boundary conditions $u(\pi,y)=u(0,0)=0, u(x,\pi)=f(x)=\cos(\dfrac{3}{2}x)$, and the left border ($u(0,y$) is "insulated", which I interpret to mean $\partial_xu=0$.
I start off with the condition $u(x,y)=X(x)Y(y)\rightarrow\dfrac{X''}{X}+\dfrac{Y''}{Y}=0$ and I equate the X fraction with $+\lambda$ and the Y fraction with $-\lambda$. The error I run into is when I solve for X: I get $X''-\lambda X=0$ and $\lambda=n^2\rightarrow X(x)=c_1e^{nx}+c_2e^{-nx}, X(\pi)=0\rightarrow c_2=-c_1e^{2\pi n}$. $X(x)=c_1(e^{nx}-e^{n(2\pi-x)}\rightarrow X'(x)=c_1(ne^{xn}+ne^{n(2\pi-x)})=0\rightarrow c_1=0$.
So now I get the trivial solution... my professor mentioned that we get the trivial solution because the assumption $u(x,y)=X(x)Y(y)$, but he never told us what to do instead and he doesn't give any details in the lecture notes, so I'm pretty lost. Any help would be greatly appreciated.
From $$ \frac{X''}{X}+\frac{Y''}{Y}=\lambda $$ and the boundary conditions we get for $X$ $$ X''-\lambda\,X=0,\quad X'(0)=0,\quad X(\pi)=0. $$ This has the nontrivial solution $$ X(x)=\cos\Bigl(\frac{2\,k+1}{2}\,x\Bigr),\quad k=0,1,2,\dots $$ corresponding to $$ \lambda=-\Bigl(\frac{2\,k+1}{2}\Bigr)^2. $$ Can you go on from here?