Find the general integral of the equation s + xy + yz = 0, s…

Find the general integral of the equation \[ s + xy + yz = 0, \qquad s = \frac{\partial^2 z}{\partial x\,\partial y}. \]

Step 1 · Solve the ODE for $w=z_y$

\[ \frac{\partial w}{\partial x} + yw = -xy. \tag{1} \] The integrating factor is $e^{xy}$. Multiply (1) by $e^{xy}$: \[ \frac{\partial}{\partial x}\!\left(w e^{xy}\right) = -xy\,e^{xy}. \] Integrate with respect to $x$ (by parts, $u=x$, $dv=e^{xy}dx$, so $\int xe^{xy}dx = \tfrac{xe^{xy}}{y}-\tfrac{e^{xy}}{y^2}$): \[ w e^{xy} = -x e^{xy} + \frac{e^{xy}}{y} + \varphi(y). \] Divide by $e^{xy}$: \[ w = \frac{\partial z}{\partial y} = -x + \frac{1}{y} + \varphi(y)\,e^{-xy}, \tag{2} \] where $\varphi(y)$ is an arbitrary differentiable function of $y$.

Step 2 · Integrate for $z$

Integrate (2) with respect to $y$, treating $x$ as a parameter: \[ z = \int\!\left(-x + \frac{1}{y} + \varphi(y)\,e^{-xy}\right)dy = -xy + \ln y + \int \varphi(y)\,e^{-xy}\,dy + \psi(x), \] where $\psi(x)$ is an arbitrary differentiable function of $x$.

Verification

From (2), $q = -x + \tfrac1y + \varphi(y)e^{-xy}$, so $s = \partial q/\partial x = -1 - y\varphi(y)e^{-xy}$. Check $\partial w/\partial x + yw = -xy$: \begin{align*} \frac{\partial w}{\partial x} + yw &= \left(-1 - y\varphi\, e^{-xy}\right) + y\!\left(-x + \frac{1}{y} + \varphi\, e^{-xy}\right) \\ &= -1 - y\varphi\, e^{-xy} - xy + 1 + y\varphi\, e^{-xy} = -xy. \checkmark \end{align*} Since $\partial w/\partial x + yw = -xy$ holds identically, and $s=\partial w/\partial x$, the original equation $s + xy + yz = 0$ is satisfied for all choices of $\varphi(y)$ and $\psi(x)$. $\checkmark$