Find the integral surface of (y-z) p + (z-x) q = x-y, p = (∂…

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

Step 1 · Characteristic equations

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

Step 2 · First integral

Multipliers $(1,1,1)$: denominator $(y-z)+(z-x)+(x-y)=0$, so $dx+dy+dz=0$ identically: \[ x+y+z = c_1. \tag{I} \]

Step 3 · Second integral

Multipliers $(x,y,z)$: denominator $x(y-z)+y(z-x)+z(x-y)=0$, so $x\,dx+y\,dy+z\,dz=0$ identically: \[ x^2+y^2+z^2 = c_2. \tag{II} \]

Step 4 · General solution

\[ x^2+y^2+z^2 = \Phi(x+y+z). \tag{$*$} \]

Step 5 · Apply the initial condition

Parametrise $z=y=-x$ by $t$: $x=-t$, $y=t$, $z=t$. Then $x^2+y^2+z^2=3t^2$ and $x+y+z=t$, so $\Phi(t)=3t^2$. Hence \[ x^2+y^2+z^2 = 3(x+y+z)^2. \]

Verification

On the curve: $3t^2=3(-t+t+t)^2=3t^2$. $\checkmark$ Implicit differentiation with $S=x+y+z$ gives $p=\dfrac{3S-x}{z-3S}$, $q=\dfrac{3S-y}{z-3S}$, and \[ (y-z)p+(z-x)q = \frac{(y-x)(3S-z)}{z-3S} = -(y-x)=x-y. \checkmark \]