To solve the partial differential equation $z = p^2 x + q^2 x$ using Charpit's method, let's follow these steps:

1. Rewrite the PDE in terms of the variables $p$ and $q$:

   • The given PDE is: $z = p^2 x + q^2 x$
   • In Charpit's method, we have $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$.

2. Compute the Charpit Equations:

   • Charpit's equations for a first-order PDE $F(x, y, z, p, q) = 0$ are: $\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{p F_p + q F_q} = \frac{dp}{-F_x - p F_z} = \frac{dq}{-F_y - q F_z}$
   • For our PDE, define $F(x, y, z, p, q) = z - p^2 x - q^2 x = 0$.

3. Calculate Partial Derivatives of $F$:

   • $F_p = -2px$
   • $F_q = -2qx$
   • $F_x = -p^2 - q^2$
   • $F_y = 0$
   • $F_z = 1$

4. Substitute into Charpit’s Equations:

   • The system becomes: $\frac{dx}{-2px} = \frac{dy}{-2qx} = \frac{dz}{p(-2px) + q(-2qx)} = \frac{dp}{p^2 + q^2} = \frac{dq}{0}$

5. Solve the System of Equations:

   • From $\frac{dq}{0}$, we find that $q$ is constant.
   • From $\frac{dx}{-2px} = \frac{dy}{-2qx}$,