Find the general integral of the equation y^2 s + 2yp = 1, s…

Find the general integral of the equation \[ y^2 s + 2yp = 1, \qquad s = \frac{\partial^2 z}{\partial x\,\partial y},\quad p = \frac{\partial z}{\partial x}. \]

Step 1 · Recognise the exact derivative

Since $\dfrac{\partial}{\partial y}\!\left(y^2\,\dfrac{\partial z}{\partial x}\right) = 2y\,\dfrac{\partial z}{\partial x} + y^2\,\dfrac{\partial^2 z}{\partial x\,\partial y}$, the left-hand side is an exact derivative with respect to $y$: \[ \frac{\partial}{\partial y}\!\left(y^2\,p\right) = 1. \]

Step 2 · Integrate with respect to $y$

\[ y^2\,p = y + \varphi(x), \] where $\varphi(x)$ is an arbitrary differentiable function of $x$. Hence: \[ \frac{\partial z}{\partial x} = \frac{1}{y} + \frac{\varphi(x)}{y^2}. \tag{1} \]

Step 3 · Integrate with respect to $x$

\[ z = \int\!\left(\frac{1}{y} + \frac{\varphi(x)}{y^2}\right)dx = \frac{x}{y} + \frac{\Phi(x)}{y^2} + \psi(y), \] where $\Phi(x) = \int \varphi(x)\,dx$ is an arbitrary differentiable function of $x$, and $\psi(y)$ is an arbitrary differentiable function of $y$.

Verification

\[ p = \frac{1}{y} + \frac{\Phi'(x)}{y^2}, \qquad s = \frac{\partial p}{\partial y} = -\frac{1}{y^2} - \frac{2\Phi'(x)}{y^3}. \] \begin{align*} y^2 s + 2yp &= y^2\!\left(-\frac{1}{y^2} - \frac{2\Phi'}{y^3}\right) + 2y\!\left(\frac{1}{y} + \frac{\Phi'}{y^2}\right) \\ &= -1 - \frac{2\Phi'}{y} + 2 + \frac{2\Phi'}{y} = 1. \checkmark \end{align*}