1. **State the problem:** Solve the partial differential equation (PDE) $$x^2 r + 2 x y s + y^2 t = 0$$ where $r = \frac{\partial^2 z}{\partial x^2}$, $s = \frac{\partial^2 z}{\partial x \partial y}$, and $t = \frac{\partial^2 z}{\partial y^2}$. 2. **Write the Monge's subsidiary equations:** From the PDE, the subsidiary equations are given by $$x^2 dp \, dy + y^2 dq \, dx = 0$$ and the quadratic form in differentials $$x^2 (dy)^2 - 2 x y dx \, dy + y^2 (dx)^2 = 0.$$ 3. **Simplify the quadratic form:** The quadratic form can be rewritten as $$ (x \, dy - y \, dx)^2 = 0,$$ which implies $$x \, dy - y \, dx = 0.$$ 4. **Integrate the differential equation:** Integrating $$x \, dy - y \, dx = 0$$ gives $$\frac{y}{x} = a_1$$ or equivalently $$\frac{x}{y} = a_1,$$ where $a_1$ is an arbitrary constant. 5. **Use the subsidiary equation:** From the first subsidiary equation, we have $$x \, dp + y \, dq = 0.$$ 6. **Rewrite in total differential form:** Note that $$x \, dp + p \, dx + q \, dy + y \, dq = d(x p) + d(y q) = dz,$$ where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. 7. **Interpretation:** The equation $$d(x p) + d(y q) = dz$$ relates the differentials of $x p$, $y q$, and $z$. This can be used to find the general solution of the PDE by integrating accordingly. **Final answer:** The characteristic equation is $$\frac{y}{x} = a_1,$$ and the subsidiary equation $$x dp + y dq = 0$$ leads to the relation $$d(x p) + d(y q) = dz,$$ which can be integrated to find the general solution of the PDE.