Laplace's equation in 2 dimension

Since you are looking for separable solutions assume it has the form.

$u(x,y)=X(x)Y(y)$

This gives

$u_{yy}=XY'' \quad u_{xx}=X''Y$

This gives

$X''Y+XY''=0 \iff \frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$

With

$\lambda^2 > 0$ (why?)

This gives the system

$X''+\lambda^2X=0 \quad Y''-\lambda^2 Y=0$

The boundary conditions gives

$X(0)=X(\pi)=0$

The solutions to the X equation are

$X =c_1\cos(\lambda x)+c_2\sin(\lambda x)$

This gives

$X(0)=c_1=0$

$X =c_2\sin(\lambda x)$

$X(\pi)=0 =c_2\sin(\lambda \pi) \iff n\pi =\lambda \pi \implies \lambda = n$

with

$n \in \mathbb{Z}^+$

$X(x)=\sin(n x)$

Now if you solve the Y equation you get

$Y(y)=d_1\cosh(ny)+d_2\cosh(ny)$

This gives

$u(x,y)=X(x)Y(y)=\sin(n x)( d_1\cosh(ny)+d_2\cosh(ny))$

For C just use this formula

For D just mimic the process done in b)