Find the general integral of the equation t = -x^2sin xy, t…

Find the general integral of the equation \[ t = -x^2\sin xy, \qquad t = \frac{\partial^2 z}{\partial y^2}. \]

Step 1 · First integration with respect to $y$

Since $x$ is treated as a constant and $\int \sin xy\,dy = -\dfrac{\cos xy}{x}$: \[ \frac{\partial z}{\partial y} = -x^2\cdot\left(-\frac{\cos xy}{x}\right) + \varphi(x) = x\cos xy + \varphi(x), \] where $\varphi(x)$ is an arbitrary differentiable function of $x$.

Step 2 · Second integration with respect to $y$

Since $\int \cos xy\,dy = \dfrac{\sin xy}{x}$: \[ z = \int\!\left(x\cos xy + \varphi(x)\right)dy = x\cdot\frac{\sin xy}{x} + y\,\varphi(x) + \psi(x) = \sin xy + y\,\varphi(x) + \psi(x), \] where $\psi(x)$ is a second arbitrary differentiable function of $x$.

Verification

\[ \frac{\partial z}{\partial y} = x\cos xy + \varphi(x), \qquad \frac{\partial^2 z}{\partial y^2} = -x^2\sin xy. \checkmark \]