I have the following equation:
$$ \begin{cases} H_{xx} + H_{yy} = xy \\ H(x,0) = 0 \\ H(x,1) = x \\ H(0,y) = 0 \\ H(1,y) = 0 \end{cases} $$
We want to solve this, so from inspection of eigenvalues and eigenfunctions we see the solution takes the form of: $ H(x,y) = \sum\limits_{n=1}^{\infty} \! Y_{n}(y)\sin(n\pi x) $.
Using undertermined coefficients we see:
$$ \sum_{n=1}^{\infty}\left[ Y_{n}''-n^2\pi^2 Y_{n}(y) \right]\sin(n\pi x) = xy $$
$$ E_{n}(y) = 2\int_0^1 \! xy\sin(n\pi x)\,\mathrm{d}x = \frac{2y\Bigl(-n\pi\cos(n\pi )+\sin(n\pi)\Bigr)}{n^2 \pi ^2} = -\frac{2(-1)^n y}{n\pi} $$
Thus complementary solution being,
$$ Y_{n}(y) = a_{n}\cosh(n\pi y) + b_{n}\sinh(n\pi y) + c + dy $$
$c=0,\quad d = \dfrac{2}{\pi^3}\dfrac{\cos(n\pi)}{n^3}$
Applying conditions:
$ H(x,0) = 0 = \displaystyle\sum_{n=1}^{\infty} a_{n}\sin(n\pi x),\hspace{0.2cm} \rightarrow \hspace{0.1cm} a_{n}=0 $
$ H(x,1) = x = \displaystyle\sum_{n=1}^{\infty} \Bigl[ b_{n}\sinh(n\pi) + d \Bigr] \sin(n\pi x) $
How would we continue from here to determine the $b_{n}$ coefficient? Is there a more elegant way to solve the non-homogeneous Poisson problem in general besides method of undetermined coefficients?