Find the general integral of the equation ps - qr = 0, p =…
Find the general integral of the equation \[ ps - qr = 0, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y},\quad r = \frac{\partial^2 z}{\partial x^2},\quad s = \frac{\partial^2 z}{\partial x\,\partial y}. \]
Explanation of Method
Writing $r=\partial_x p$ and $s=\partial_y p$, the equation $p\,\partial_y p - q\,\partial_x p = 0$ says that $\nabla p$ is parallel to $\nabla z=(p,q)$, which forces $p$ to be a function of $z$ alone; separating variables in $p=f(z)$ and integrating gives the general integral.
Step 1 · Factor the equation
Rewrite $ps - qr = 0$ as \[ p\,\frac{\partial p}{\partial y} - q\,\frac{\partial p}{\partial x} = 0, \tag{2} \] since $s = \partial p/\partial y$ and $r = \partial p/\partial x$. This says $\nabla p$ is parallel to the vector $(q,p)=(\partial z/\partial y,\partial z/\partial x)$, i.e. $\nabla p$ is parallel to $\nabla z$. On any level curve $z=\text{const}$ we have $dz=p\,dx+q\,dy=0$, and along the same curve $dp=r\,dx+s\,dy=0$ by (2). Hence $p$ is constant on the level curves of $z$, so \[ p = \frac{\partial z}{\partial x} = f(z), \tag{3} \] for some arbitrary differentiable function $f$.
Step 2 · Solve the ODE (3)
Equation (3) is a first-order ODE in $x$ with $y$ as parameter. Separating variables: \[ \frac{dz}{f(z)} = dx. \] Integrating, $F(z) = x + \varphi(y)$ where $F(z)=\int dz/f(z)$ is an arbitrary function of $z$ and $\varphi(y)$ is an arbitrary function of $y$. Writing this implicitly as $\Phi(z) = x+\varphi(y)$ and solving for $z$ (with $G=\Phi^{-1}$, again arbitrary): \[ z = G\!\left(x + \varphi(y)\right). \]
Verification
Let $\xi = x + \varphi(y)$. Then $p=G'(\xi)$, $q=G'(\xi)\varphi'(y)$, $r=G''(\xi)$, $s=G''(\xi)\varphi'(y)$, so \[ ps - qr = G'(\xi)G''(\xi)\varphi'(y) - G'(\xi)\varphi'(y)G''(\xi) = 0. \checkmark \]
Result
The general integral is: