Find the general solution of 2xp + (y-x)q = x^2, p = (∂…

Find the general solution of \[ 2xp + (y-x)q = x^2, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{2x} = \frac{dy}{y-x} = \frac{dz}{x^2}. \tag{1} \]

Step 2 · First integral

From the first two ratios, $\dfrac{dy}{dx}=\dfrac{y-x}{2x}=\dfrac{y}{2x}-\dfrac12$, i.e. \[ \frac{dy}{dx} - \frac{y}{2x} = -\frac{1}{2}. \] Integrating factor $\mu=x^{-1/2}$: \[ \frac{d}{dx}\!\left(x^{-1/2}y\right) = -\tfrac12 x^{-1/2} \;\Longrightarrow\; x^{-1/2}y = -\sqrt{x}+c_1. \] Hence: \[ \frac{x+y}{\sqrt{x}} = c_1. \tag{I} \]

Step 3 · Second integral

From $\dfrac{dx}{2x}=\dfrac{dz}{x^2}$: $x^2\,dx=2x\,dz \Rightarrow dz=\tfrac{x}{2}\,dx$. Integrating: \[ z = \frac{x^2}{4} + c_2 \;\Longrightarrow\; z - \frac{x^2}{4} = c_2. \tag{II} \]

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$: \[ z = \frac{x^2}{4} + \Phi\!\left(\frac{x+y}{\sqrt{x}}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

Let $u=(x+y)/\sqrt{x}$, so $u_x=(x-y)/(2x^{3/2})$, $u_y=1/\sqrt{x}$. Then $p=\tfrac{x}{2}+\Phi'\cdot\tfrac{x-y}{2x^{3/2}}$, $q=\Phi'/\sqrt{x}$. \begin{align*} 2xp+(y-x)q &= x^2 + \frac{(x-y)\Phi'}{\sqrt{x}} + \frac{(y-x)\Phi'}{\sqrt{x}} = x^2. \checkmark \end{align*}