Solution: The given partial differential equation is \begin{equation}\label{eq:18Apr2025-1} (2xy - 1)p + (z - 2x^2)q = 2(x - yz). \end{equation} So, the auxiliary equations are \begin{equation}\label{eq:18Apr2025-2} \frac{\mathrm{d} x}{2xy - 1} = \frac{\mathrm{d} y}{z - 2x^2} = \frac{\mathrm{d} z}{2(x - yz)}. \end{equation} We recall that for a given partial differential equation $Pp + Qq = R$, we have \[ \frac{\mathrm{d} x}{P} = \frac{\mathrm{d} y}{Q} = \frac{\mathrm{d} z}{R} = \frac{S \mathrm{d} x + T \mathrm{d} y + U \mathrm{d} z}{SP + T Q + UR}, \] where $S,T,U$ are functions of $x,y,z$. We call $(S, T, U)$ a multiplier and we choose this so that $SP + T Q + UR = 0$. Then we have $S \mathrm{d} x + T \mathrm{d} y + U \mathrm{d} z = 0$.

Consider the multipliers, $(2x, 2y, 1)$, then we have \begin{align*} \frac{2x \mathrm{d} x + 2y \mathrm{d} y + \mathrm{d} z}{2x(2xy - 1) + 2y(z - 2x^2) + 2(x - yz)} = \frac{2x \mathrm{d} x + 2y \mathrm{d} y + \mathrm{d} z}{0}. \end{align*} This implies \[ 2x \mathrm{d} x + 2y \mathrm{d} y + \mathrm{d} z = 0 \implies x^2 + y^2 + z = c_1, \] where $c_1$ is a constant.

Again choose a multiplier $(z, 1, x)$, we have \[ \frac{z \mathrm{d} x + \mathrm{d} y + x \mathrm{d} z}{z(2xy - 1) + (z - 2x^2) + 2x(x - yz)} = \frac{z \mathrm{d} x + \mathrm{d} y + x \mathrm{d} z}{0}. \] Thus, we have \begin{align*} z \mathrm{d} x + \mathrm{d} y + x \mathrm{d} z = 0 & \implies d(xz + y) = 0 \\ & \implies xz + y = c_2, \end{align*} where $c_2$ is a constant. Thus, a general solution will be \begin{equation}\label{eq:18Apr2025-3} f\left( x^2 + y^2 + z, xz + y \right) = 0, \end{equation} where $f$ is any arbitrary function.

Now to find a particular integral, we have $x = 1, y = 0$ which implies \begin{align*} c_1 = z + 1 \quad \text{ and } \quad c_2 = z \implies c_1 = c_2 + 1. \end{align*} Thus, a particular integral will be \[ x^2 + y^2 + z - xz - y - 1 = 0 . \]