Solution: The given equation is \begin{equation}\label{eq:28Feb2025-1} z = pq. \end{equation} We recall that for $f(x,y,z,p,q) = 0$, the Charpit's auxiliary equations are \[ \frac{\mathrm{d} x}{-f_p} = \frac{\mathrm{d} y}{-f_q} = \frac{\mathrm{d} z}{-p f_p - q f_q} = \frac{\mathrm{d} p}{f_x + p \mathrm{d} z} = \frac{\mathrm{d} q}{f_y + q \mathrm{d} z}. \] In this problem, we have $f = z - pq$. Thus, the above equations become \begin{align*} \frac{\mathrm{d} x}{q} = \frac{\mathrm{d} y}{p} = \frac{\mathrm{d} z}{2pq} = \frac{\mathrm{d} p}{p} = \frac{\mathrm{d} q}{q}. \end{align*} From the last two equations, we get \begin{align*} \frac{\mathrm{d} p}{p} = \frac{\mathrm{d} q}{q} & \implies \ln p = \ln q + \ln a \\ & \implies p = aq, \end{align*} where $a$ is a constant.

Substituting this into \eqref{eq:28Feb2025-1}, we get \begin{align*} z = aq^2 & \implies q = \sqrt{\frac{z}{a}}\\ & \implies p = a \sqrt{\frac{z}{a}} = \sqrt{az}. \end{align*} So, we have \begin{align*} \mathrm{d} z = p \mathrm{d} x + q \mathrm{d} y & \implies \mathrm{d} z = \sqrt{az} \mathrm{d} x + \sqrt{\frac{z}{a}} \mathrm{d} y \\ & \implies \int \frac{\mathrm{d} z}{\sqrt{z}} = \int \sqrt{a} \mathrm{d} x + \int \frac{1}{\sqrt{a}} \mathrm{d} y \\ & \implies 2\sqrt{z} = \sqrt{a} x + \frac{1}{\sqrt{a}} y + c \\ & \implies z = \frac{1}{4} \left( \sqrt{a} x + \frac{1}{\sqrt{a}} y + c \right)^2, \end{align*} where $a,b$ are constants. Thus, the complete integral of the given PDE is: \[ z = \frac{1}{4} \left( \sqrt{a} x + \frac{1}{\sqrt{a}} y + c \right)^2. \]