Here is another example of how to solve the Laplace equation for a simple rectangular domain using separation of variables, under the assumption of different boundary conditions:

1. Write the Laplace equation in the form:

∇^2 u = 0

This is the general form of the Laplace equation, which holds for any rectangular domain.

2. Define the dimensions of the rectangular domain. Let’s say the domain has dimensions $L_x$ and $L_y$ in the $x$ and $y$ directions, respectively.

3. Separate the variables by expressing the solution u as a product of functions that depend on x and y separately:

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

Substituting this expression into the Laplace equation and separating the variables,

we get:

 X''(x)Y(y) + X(x)Y''(y) = 0

4. Solve for $X(x)$ and $Y(y)$ by setting each term equal to a constant:

 X''(x)Y(y) = -X(x) Y''(y)\\       {X''(x) \over  X(x)} = -{Y''(y) \over Y(y) }=const=-k^2

where $k$ is a constant.

5. The solutions to the ODEs will be of the form:

X(x) = Acos(kx) + Bsin(kx) \\ Y(y) = Ce^{ky} + De^{-ky}

or

X(x) = Ce^{kx} + De^{-kx} \\ Y(y) = Acos(ky) + Bsin(ky) 

where $A, B, C,$ and $D$ are constants that can be determined from boundary conditions.

. Substituting the expressions for $X(x)$ and $Y(y)$ into the original separation of variables solution $u(x,y) = X(x)Y(y)$, we get:

u(x,y) = (Acos(kx) + Bsin(kx))(Ce^{ky} + De^{-ky}) 

or

u(x,y) = (Acos(ky) + Bsin(ky))(Ce^{kx} + De^{-kx}) 

This is the general solution to the Laplace equation for a rectangular domain, in terms of the constants $A, B, C,$ and $D$.