$$

\frac{1}{G_{zx}}\frac{\partial ^2\phi (x,y)}{\partial^2 y}+\frac{1}{G_{zy}}\frac{\partial ^2\phi (x,y)}{\partial^2 x}=-2 \theta

$$

where $G_{zy}$, $G_{zx}$, $\theta$, $a$, and $b$ are constants and with BCs $\phi (0,y)=\phi (a,y)=0$ and $\phi (x,-b)=\phi (x,b)=0$. The solution that I have sets $\phi (x,y)=\sum _{k=1,3,5\text{...}}^{\infty } Y(y) \sin \left(\frac{\pi k}{a}x\right)$ and expands $-2 \theta$ in a Fourier sine series in the interval between $0$ and $a$ so we end up with a second order differential equation,

$$

\frac{Y''(y)}{G_{zx}}-\frac{\pi ^2 k^2 Y(y)}{a^2 G_{zy}}=-\frac{8 \theta }{\pi k}

$$

I tried several times to solve it by hand but end up making mistakes. Therefore I resorted to Mathematica and it gives me the following solution,

$$

\phi_{mine}(x,y)=-\sum_{k=1,3,5,...}^{\infty}\frac{8 a^2 G_{zy} \theta \left(\text{sech}\left(\frac{\pi b k}{2 a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}} \right) \cosh \left(\frac{\pi k y}{a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}}\right)-1\right)}{\pi ^3 k^3}\sin \left(\frac{\pi k x}{a}\right)

$$

This is almost exactly what is written in the solution that I have, which is

$$

\phi_{sol}(x,y)=\frac{8}{\pi^3} G_{zy} a^2 \sum_{k=1,3,5,...}^{\infty}\frac{(-1)^{(k-1)/2}}{k^3}\left( 1-\frac{\cosh \left(\frac{\pi k \mu }{a}y\right)}{\cosh \left(\frac{b \pi k \mu}{2 a}\right)} \right)\cos \left(\frac{\pi k}{a}x\right)

$$

where $\mu=\sqrt{\frac{G_{zx}}{G_{zy}}}$.

$\phi_{sol}(x,y)$ is missing $\theta$ but I suspect that it might be a typo. I know that $\phi_{mine}(x,y)$ is wrong because the next step in the process involves working out a constant $\beta$ where

$$

\beta=\frac{2 \int_{-b/2}^{b/2} \left(\int_0^a \phi (x,y) \, dx\right) \, dy}{G_{zx} a b^3}

$$

I get that $\phi_{sol}(x,y)$ converges to a value while $\phi_{mine}(x,y)$ goes to infinity. Therefore I'm doing something wrong but I'm not sure what. I've been trying to figure out the difference between my answer and the solution and all I can find is that somehow

$$

\sin \left(\frac{\pi k x}{a}\right)=\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}

$$

for $k=1,3,5,...$ Is it possible to change $\sin \left(\frac{\pi k x}{a}\right)$ into $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$ when only odd values of $k$ are used? I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.

As can be seen, $a$ determines the amplitude of the periodic functions so that $

\sin \left(\frac{\pi k x}{a}\right)$ is always $0$ at -1 and 1. $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$ on the other hand flips from $1$ to $-1$ at $\pm a$.