Find the integral surface of x p + y q = z - x^2 - y^2, p =…

Find the integral surface of \[ x\,p + y\,q = z - x^2 - y^2, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $y=-2$, $z=x-x^2$.

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z-x^2-y^2}. \tag{1} \] From the first two ratios: $y/x=c_1$. \tag{I}

Step 2 · Second integral via a parameter

Let $\tau=\ln x$ (so $x=e^\tau$) and $y=c_1x=c_1e^\tau$ along a characteristic. The ODE for $z$ becomes \[ \frac{dz}{d\tau} = z - e^{2\tau}(1+c_1^2). \] With integrating factor $e^{-\tau}$: $\dfrac{d(ze^{-\tau})}{d\tau}=-e^{\tau}(1+c_1^2)$, so \[ ze^{-\tau} = -(1+c_1^2)e^{\tau}+c_2 \;\Longrightarrow\; z = -(x^2+y^2)+c_2x. \] Hence \[ c_2 = \frac{z+x^2+y^2}{x}. \tag{II} \]

Step 3 · General solution

\[ z = x\,\Phi\!\left(\frac{y}{x}\right) - x^2 - y^2. \tag{$*$} \]

Step 4 · Apply the initial condition

Set $y=-2$, $z=x-x^2$ in ($*$): $x-x^2 = x\Phi(-2/x)-x^2-4$, so $\Phi(-2/x)=1+4/x$. With $t=-2/x$ (so $4/x=-2t$): $\Phi(t)=1-2t$. Substituting $t=y/x$: \[ z = x\!\left(1-\frac{2y}{x}\right)-x^2-y^2 = x-2y-x^2-y^2. \]

Verification

At $y=-2$: $z=x+4-x^2-4=x-x^2$. $\checkmark$ Compute $p=1-2x$, $q=-2-2y$: \[ xp+yq = x-2x^2-2y-2y^2, \qquad z-x^2-y^2 = x-2y-2x^2-2y^2, \] which are identical. $\checkmark$