Find the general solution of x^2 p + y^2 q = z^2, p = (∂…

Find the general solution of \[ x^2 p + y^2 q = z^2, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z^2}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{x^2}=\dfrac{dy}{y^2}$: $\dfrac{dx}{x^2}-\dfrac{dy}{y^2}=0$. Integrating: \[ \frac{1}{y} - \frac{1}{x} = c_1. \tag{I} \]

Step 3 · Second integral

From $\dfrac{dy}{y^2}=\dfrac{dz}{z^2}$: $\dfrac{dy}{y^2}-\dfrac{dz}{z^2}=0$. Integrating: \[ \frac{1}{z} - \frac{1}{y} = c_2. \tag{II} \]

Step 4 · General solution

The general solution is $\Phi(c_1,c_2)=0$, or equivalently, expressing $c_2$ as an arbitrary function of $c_1$: \[ \frac{1}{z} - \frac{1}{y} = \Phi\!\left(\frac{1}{y} - \frac{1}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

With $u=1/y-1/x$, implicit differentiation of $F=1/z-1/y-\Phi(u)=0$ gives \[ p = -\frac{z^2\,\Phi'(u)}{x^2}, \qquad q = \frac{z^2}{y^2}\!\left(1+\Phi'(u)\right). \] Check $x^2p+y^2q$: \begin{align*} x^2 p + y^2 q &= -z^2\Phi' + z^2(1+\Phi') = z^2. \checkmark \end{align*}