Find the general integral of the equation 2yt - xs + 2q =…

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

Step 1 · Reduce to a first-order PDE in $q$

Since $\dfrac{\partial}{\partial y}(2yq) = 2q + 2y\,\dfrac{\partial q}{\partial y} = 2q+2yt$, the equation becomes \[ \frac{\partial}{\partial y}(2yq) - x\frac{\partial q}{\partial x} = x^2 y, \] i.e. a first-order linear PDE in $q(x,y)$: \[ -x\,\frac{\partial q}{\partial x} + 2y\,\frac{\partial q}{\partial y} + 2q = x^2 y. \tag{1} \]

Step 2 · Solve (1) via characteristics

The characteristic equations are $\dfrac{dx}{-x} = \dfrac{dy}{2y} = \dfrac{dq}{x^2y-2q}$. From the first pair, $\dfrac{dy}{dx} = -\dfrac{2y}{x}$, giving the first integral \[ u = x^2 y = c_1. \tag{I} \] Along a characteristic $y = c_1/x^2$, the equation for $q$ becomes the linear ODE \[ \frac{dq}{dx} - \frac{2q}{x} = -\frac{c_1}{x}. \tag{2} \] With integrating factor $x^{-2}$: \[ \frac{d}{dx}\!\left(\frac{q}{x^2}\right) = -\frac{c_1}{x^3} \;\Longrightarrow\; \frac{q}{x^2} = \frac{c_1}{2x^2} + \varphi(c_1), \] where $\varphi$ is an arbitrary function of $c_1=x^2y$. Hence: \[ \frac{\partial z}{\partial y} = \frac{x^2 y}{2} + x^2\varphi(x^2 y). \tag{3} \]

Step 3 · Integrate for $z$

Integrate (3) with respect to $y$; for the second term substitute $\xi=x^2y$, $d\xi = x^2\,dy$, so $\int x^2\varphi(x^2y)\,dy = \int\varphi(\xi)\,d\xi = \Phi(x^2y)$ with $\Phi'=\varphi$: \[ z = \frac{x^2 y^2}{4} + \Phi(x^2 y) + \psi(x), \] where $\psi(x)$ is an arbitrary differentiable function of $x$.

Verification

Let $\xi = x^2 y$. Compute \[ q = \frac{x^2 y}{2} + x^2\Phi'(\xi), \qquad t = \frac{x^2}{2} + x^4\Phi''(\xi), \qquad s = xy + 2x\Phi'(\xi) + 2x^3 y\,\Phi''(\xi). \] \begin{align*} 2yt &= x^2 y + 2x^4 y\,\Phi'',\\ xs &= x^2 y + 2x^2\Phi' + 2x^4 y\,\Phi'',\\ 2q &= x^2 y + 2x^2\Phi'. \end{align*} Therefore $2yt - xs + 2q = x^2 y + 2x^4y\Phi'' - x^2y - 2x^2\Phi' - 2x^4y\Phi'' + x^2y + 2x^2\Phi' = x^2 y$. $\checkmark$