Find the general integral of the equation yt - q = 2x^2 y, t…

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

Step 1 · Recognise the exact derivative

Divide the equation by $y^2$: \[ \frac{t}{y} - \frac{q}{y^2} = \frac{2x^2}{y}. \] Since $\dfrac{\partial}{\partial y}\!\left(\dfrac{q}{y}\right) = \dfrac{ty-q}{y^2} = \dfrac{yt-q}{y^2}$, this is \[ \frac{\partial}{\partial y}\!\left(\frac{q}{y}\right) = \frac{2x^2}{y}. \]

Step 2 · Integrate with respect to $y$

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

Step 3 · Integrate again with respect to $y$

Using $\int y\ln y\,dy = \tfrac{y^2}{2}\ln y - \tfrac{y^2}{4}$ (by parts): \[ z = 2x^2\!\left(\frac{y^2}{2}\ln y - \frac{y^2}{4}\right) + \frac{y^2}{2}\,\varphi(x) + \psi(x) = x^2 y^2\ln y - \frac{x^2 y^2}{2} + \frac{y^2}{2}\,\varphi(x) + \psi(x). \] Since $\varphi(x)$ is arbitrary, absorb $-x^2/2$ into it by writing $\Phi(x) = \varphi(x) - x^2$ (still arbitrary): \[ z = x^2 y^2 \ln y + \frac{y^2}{2}\,\Phi(x) + \psi(x). \]

Verification

\[ q = 2x^2 y\ln y + x^2 y + y\,\Phi(x), \qquad t = 2x^2\ln y + 3x^2 + \Phi(x). \] \begin{align*} yt - q &= y\!\left(2x^2\ln y + 3x^2 + \Phi(x)\right) - \left(2x^2 y\ln y + x^2 y + y\,\Phi(x)\right)\\ &= 2x^2 y\ln y + 3x^2 y + y\,\Phi(x) - 2x^2 y\ln y - x^2 y - y\,\Phi(x) = 2x^2 y. \checkmark \end{align*}