Finding $u_0(x,y)$ for Laplace's Equation I looked at this resource to help me solve the below problem, but I still have questions on some parts.
Problem

My Questions
To solve Laplace's equation, I assumed $u(x,y) = X(x)Y(y)$. 
For $\lambda = 0$, when solving $X'' = 0$, I get $X(x)=c_2$. However, for λ = 0, the solutions says $X = 1$- how?
For $\lambda = 0$, $Y(y)= k_1y + k_2$, but the solution says $u_0(x,y) = y$. What happened to $k_2$?
 A: To provide some context, the resource quoted uses separation of variables. Using an ansatz of the form $u(x,t) = X(x)Y(y)$, one obtains a pair of ODEs,
$$ X''(x) + \lambda X(x) = 0, \ \ \ \ \ \ Y''(y) - \lambda Y(y) = 0.$$
These ODEs have solutions for various values of $\lambda$, and the full solution to the PDE is a linear superposition of $X_\lambda (x)Y_\lambda (y)$ for the various values of $\lambda$. The OP's questions seek clarification specifically about the contribution with $\lambda = 0$.
First, let's deal with the $Y(y)$ equation. It is true that the general solution to $Y''(y)  = 0$ is $Y(y) = k_1  y + k_2$. However, since $u(x,y)$ satisfies the boundary condition $u(x,0) = 0$, we want $Y(y)$ correspondingly to satisfy the boundary condition $Y(0) = 0$. This implies that $k_2 = 0$. So we have $Y(y) = k_1 y$.
Next, the solution to the $X(x)$ equation satisfying the appropriate boundary conditions is indeed $X(x) = c_2$. So the full $\lambda = 0$ solution is
$ u_{\lambda = 0}(x,t) = c_2 k_1 y.$
However, since both $c_2$ and $k_1$ are constants, we may as well redefine them, setting
$ c_{2, {\rm \ new}} = 1$ and $k_{1, {\rm \ new }} = c_{2, {\rm \ old }} \times k_{1,{\rm \ old}}$. This redefinition makes no difference to the solution.
For completeness, the full solution, including contributions from other values of $\lambda$, looks like
$$ u(x,y) = a_0 y + \sum_{n = 1}^\infty a_n \cos \frac{n \pi x }{a} \sinh \frac{n \pi y}{b},$$
(so $a_0$ is the same thing as $c_{2, {\rm \ old}} \times k_{1, {\rm \ old}}$), and the coefficients $a_n$ and $b_n$ are to be determined by comparison with the Fourier expansion of the $g(x)$ function appearing in the boundary condition at $y = b$.
