We will solve the Laplace equation: $$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0$$ With boundary conditions $$V(0,y)=V(l,y)=V(x,0)=0 \textrm{ and } V(x,l)=V_0 \textrm{ for some } l \in \mathbb{R}$$

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Assume a seperable solution of the form: $$V(x,y)=X(x)Y(y)$$ Substituting and choosing a negative and real separation constant, we get $$\frac{-X''}{X}=\frac{Y''}{Y}=-k^2$$

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For the X equation: $$\textrm{let } A_i \in \mathbb{R} \textrm{ be constants}$$ $$X'' - k^2 X=0$$ $$\implies X=A_1 cos(k x) + A_2 sin(k x)$$ Applying the boundary conditions $$V(0,y)=0=V(l,y)$$ We get $$A_1=0 \textrm{ and } kl=n \pi \textrm{ for } n \in \mathbb{Z}$$

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For the Y equation: $$Y'' + k^2 Y=0$$ $$\implies Y=A_3 e^{ky} + A_4 e^{-ky}$$ Apply the boundary condition $$V(x,0)=0$$ $$\implies A_3=-A_4$$ $$\implies Y=A_3 e^{ky} - A_3 e^{-ky}=2 A_3 sinh(ky)$$

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$$\textrm{let } B_n = 2 A_2 A_3$$ $$\textrm{So } V_n(x,y)=B_n sin \left ( \frac{n \pi x}{l} \right ) sinh \left ( \frac{n \pi y}{l} \right )$$ But this cannot satisfy $$V(x,l)=V_0$$ So we must consider a linear combination of these solutions, ie $$V(x,y)=\sum_{n=1}^{\infty} V_n(x,y)$$

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So the boundary condition becomes $$V_0=\sum_{n=1}^{\infty} B_n sinh(n \pi) sin \left ( \frac{n \pi x}{l} \right )$$ Use the Fourier trick (orthogonality of sines) $$\int_{0}^{l} V_0 sin \left ( \frac{m \pi x}{l} \right ) dx = \int_{0}^{l} \sum_{n=1}^{\infty} \left [ B_n sinh(n \pi) sin \left ( \frac{n \pi x}{l} \right ) sin \left ( \frac{m \pi x}{l} \right ) \right ] dx$$ $$V_0 \left [ \frac{-l}{m \pi} cos \left ( \frac{m \pi x}{l} \right ) \right ]_{0}^{l} = \int_{0}^{l} B_m sinh(m \pi) sin \left ( \frac{m \pi x}{l} \right ) dx$$ $$V_0 \left [ \frac{(-1)^{m+1} l}{m \pi} + \frac{l}{m \pi} \right ] = B_m sinh(m \pi) \frac{l}{2}$$ $$2 V_0 [ 1 - (-1)^m ] = m \pi B_m sinh(m \pi)$$ $$B_m = \frac{2 V_0 [ 1 - (-1)^m ]}{m \pi sinh(m \pi)} \implies B_m=0 \textrm{ for even } m$$ Thus, $$V(x,y) = \sum_{m=1}^{\infty} \frac{2 V_0 [ 1 - (-1)^m ]}{m \pi sinh(m \pi)} sinh \left ( \frac{m \pi y}{l} \right ) sin \left ( \frac{m \pi x}{l} \right )$$