We have:
\begin{equation} \pdv[2]{u}{x} + \pdv[2]{u}{y} = 0 \end{equation}
Ignoring the boundary conditions when \(u(0,y)\), we know that we have Dirichlet boundaries in \(y\). This gives:
\begin{equation} u(x,0) = u(x,\pi) = 0 \end{equation}
Assuming our solution takes on the shape of \(u=X(x)Y(y)\), we obtain:
\begin{equation} X’’(x)Y(y) + Y’’(y)X(x) = 0 \end{equation}
by plugging in derivatives of that assumption; meaning:
\begin{equation} X’’(x)Y(y) = -Y’’(y)X(x) \end{equation}
This gives rise to:
\begin{align} \frac{X’’(x)}{X(x)} = -\frac{Y’’(y)}{Y(y)} = c \end{align}
[you know why \(c>0\), so let’s skip to]
We have \(c>0\), meaning:
\begin{equation} X’’(x) = cX(x) \end{equation}
for some positive \(c\); this will result in a linear combination of exponentials:
\begin{equation} X(x) = a_{1} e^{\sqrt{c}x} + a_2 e^{-\sqrt{c}x} \end{equation}
this is because… try it! try solving \(X’’(x) = cX(x)\).
Now, importantly, let’s declare:
\begin{equation} \lambda = \sqrt{c} \end{equation}
This gives:
\begin{equation} c = \lambda^{2} \end{equation}
Meaning, we have:
\begin{equation} \frac{Y’’(y)}{Y(y)} = -\lambda^{2} \end{equation}
meaning:
\begin{equation} Y’’(y) = -\lambda^{2} Y(y) \end{equation}
Now, given we now have a negative sign in front of our second order ODE, we can see that this falls into the sinusoid case, whereby:
\begin{equation} Y = a_3 \cos \qty(\lambda x) + a_4 \sin \qty(\lambda x) \end{equation}
Our boundary condition gives:
\begin{equation} Y_0 = Y_{\pi} = 0 = a_3 = 0 \end{equation}
meaning
\begin{equation} Y = a_4 \sin \qty(\lambda x) \end{equation}
and so on. You multiply them together and all’s well that ends well.