I have found all solutions on the form $u = F(x) G(y)$ that fufills $u(x,0)=u(0,y)=u(2,y) =0$
And that is: $$u(x,y) = \sum_{n=1}^\infty A_n\sin\left(\frac{n\pi x}{2}\right)\sinh\left(\frac{n\pi y}{2}\right)$$
I'm now supposed to find an other solution that also fulfills $u(x,1) = \sin(\pi x)\cos(2\pi x).$ And I am quite stuck on where to start. I thought I might plug inn y = 0 and then get
$$u(x,1) = \sum_{n=1}^\infty A_n\sin\left(\frac{n\pi x}{2}\right)\sinh\left(\frac{n\pi}{2}\right) =\sin(\pi x)\cos(2\pi x)$$ But I do not how to move forward form here. Can someone please help? :)
Let us denote $B_n = A_n \sinh\left(\dfrac{n\pi}{2}\right)$ and rewrite $$ u(x,1) = \sum_{n=1}^\infty A_n\sin\left(\frac{n\pi x}{2}\right)\sinh\left(\frac{n\pi}{2}\right) =\sin(\pi x)\cos(2\pi x) $$ as $$ \sum_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)=\sin(\pi x)\cos(2\pi x) $$ This is the Fourier series of the function $f(x) = \sin(\pi x)\cos(2\pi x)$ over the system $\left\{X_n(x) = \sin\left(\frac{n\pi x}{2}\right)\right\}_{n=1}^{\infty}$ on the range $[0,2]$. So, we have $$ A_n \sinh\left(\dfrac{n\pi}{2}\right) = B_n = \frac{\langle f \mid X_n \rangle}{\langle X_n \mid X_n \rangle}, $$ where $\langle q_1 \mid q_2 \rangle = \int\limits_0^2 q_1(x) \overline{q_2(x)} dx$ (dot product).
After calculating we obtain $$ B_n = \begin{cases} -\dfrac{1}{2}, & n = 2,\\ \dfrac{1}{2}, & n = 6,\\ 0, & n \neq 2 \land n \neq 6, \\ \end{cases} $$ and $$ A_n = \begin{cases} -\dfrac{\mathrm{csch}(\pi )}{2}, & n = 2,\\ \dfrac{\mathrm{csch}(3 \pi )}{2}, & n = 6,\\ 0, & n \neq 2 \land n \neq 6. \\ \end{cases} $$
This leads us to the answer $$ u(x,y) = \dfrac{\mathrm{csch}(3 \pi )}{2} \sin(3 \pi x)\sinh(3 \pi y) - \dfrac{\mathrm{csch}(\pi )}{2} \sin(\pi x)\sinh(\pi y). $$