The story so far...

We have found the general solutions of Laplace's equation, $\frac{\partial^2T(x,y)}{\partial x^2}+\frac{\partial^2T(x,y)}{\partial y^2}=0$.
The general solutions are: $$T(x,y)=X(x)\cdot Y(y)= \begin{cases} {\rm e}^{kx}\sin{ky}&\\ {\rm e}^{ky}\sin{kx}& \end{cases} \begin{eqnarray*} {\rm e}^{-kx}\sin{ky}\quad& {\rm e}^{kx}\cos{ky}\quad& {\rm e}^{-kx}\cos{ky}\\ {\rm e}^{-ky}\sin{kx}\quad& {\rm e}^{ky}\cos{kx}\quad& {\rm e}^{-ky}\cos{kx} \end{eqnarray*}$$
Applying the boundary conditions of the hot plate problem leaves $T(x,y)=e^{-\frac{n\pi}{10}y}\sin{\frac{n\pi}{10}x}$
as physically sensible solutions.
A linear combination of all of these with coefficients $b_n$ is $T(x,y)=\sum_{n=1}^{\infty}b_ne^{-\frac{n\pi}{10}y}\sin{\frac{n\pi}{10}x}$.

Applying the final boundary condition

Compare the linear combination above with the Fourier sine series: $f(x)=\sum_{n=1}^{\infty}b_n\sin{\frac{n\pi x}{l}}$.
They are identical for $l=10$ and $y=0$ (because then the e-term is 1).
The BC for the heated edge hasn't been used yet: $T(x,0)=100$.
Therefore, we can substitute: $f(x)=T(x,0)=100=\sum_{n=1}^{\infty}b_n\sin{\frac{n\pi x}{10}}$.
The Fourier coefficients are, generally: $b_n=\frac{2}{l}\int_0^lf(x){\sin{\frac{n\pi x}{l}}}{\rm d}x$.
So, in this case: $b_n=\frac{2}{10}\int_0^{10}100\sin{\frac{n\pi x}{10}}{\rm d}x$.
Integrate: $=20(-\frac{10}{n\pi})\left[\cos{\frac{n\pi x}{10}}\right]_0^{10}$,
insert limits: $=-\frac{200}{n\pi}\left(\cos{n\pi}-\cos{0}\right)$,
and tidy up: $=\begin{cases}\frac{400}{n\pi}&(\text{for $n$ odd})\\0&(\text{for $n$ even})\end{cases}$.
Finally, insert the $b_n$ into $T(x,y)$: $$T(x,y)=\frac{400}{\pi}\sum_{\text{odd}~n=1}^{\infty}\frac{1}{n}{\rm e}^{-\frac{n\pi}{10}y}\sin{\frac{n\pi}{10}x}$$

And __that's it__ !

Summary - Laplace's equation

The next on the list of PDE in physics is the diffusion equation.