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

Find the integral surface of \[ 2xz\,p + 2yz\,q = z^2-x^2-y^2, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $x+y+z=0$, $x^2+y^2+z^2=a^2$.

Step 1 · Characteristic equations and first integral

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

Step 2 · Second integral

Multipliers $(2x,2y,2z)$ give denominator $2z(x^2+y^2+z^2)$ and numerator $d(x^2+y^2+z^2)$. Equating to the first ratio and simplifying: \[ \frac{d(x^2+y^2+z^2)}{x^2+y^2+z^2} = \frac{dx}{x} \;\Longrightarrow\; \frac{x^2+y^2+z^2}{x} = c_2. \tag{II} \]

Step 3 · General solution

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

Step 4 · Apply the initial condition

Parametrise $x+y+z=0$, $x^2+y^2+z^2=a^2$ by $y=t$, $z=-x-t$, giving $x^2+xt+t^2=a^2/2$. On the curve $c_1=t/x$, $c_2=a^2/x$, so $x=a^2/c_2$, $t=a^2c_1/c_2$. Substituting into $x^2+xt+t^2=a^2/2$ gives $c_2^2=2a^2(1+c_1+c_1^2)$, so $\Phi(s)=\sqrt{2a^2(1+s+s^2)}$. Substituting $s=y/x$ into ($*$) and squaring: \[ (x^2+y^2+z^2)^2 = 2a^2(x^2+xy+y^2). \]

Verification

On the initial curve, $z=-(x+y)$ gives $2(x^2+xy+y^2)=a^2$, so both sides of the boxed relation equal $a^4$. $\checkmark$
Implicit differentiation of $S^2=2a^2(x^2+xy+y^2)$ ($S=x^2+y^2+z^2$) gives $zp=\tfrac{a^2(2x+y)}{2S}-x$, $zq=\tfrac{a^2(x+2y)}{2S}-y$, and using $a^2(x^2+xy+y^2)=S^2/2$: \[ 2xzp+2yzq = \frac{2a^2(x^2+xy+y^2)}{S}-2(x^2+y^2) = S-2(x^2+y^2)=z^2-x^2-y^2. \checkmark \]