Find the complete and singular integrals of z^2(p^2+q^2+1) = 1

Find the complete and singular integrals of \[ z^2(p^2+q^2+1) = 1. \]

Step 1 · Substitute $w=z^2$

With $p=w_x/(2\sqrt w)$, $q=w_y/(2\sqrt w)$, the equation becomes \[ w_x^2+w_y^2 = 4(1-w). \]

Step 2 · Travelling-wave reduction

With $w=f(ax+by)+c$, $a^2+b^2=1$ (normalising), separating and integrating the resulting ODE in $f$ gives \[ w = 1-(ax+by+c)^2. \]

Step 3 · Find the singular integral

With $F=z^2(p^2+q^2+1)-1$: $F_p=2z^2p=0$, $F_q=2z^2q=0$ give $p=q=0$ ($z\neq0$); substituting into $F=0$ gives $z^2=1$: \[ z = \pm1. \]

Verification

From $z^2=1-(ax+by+c)^2$: $p=-a(ax+by+c)/z$, $q=-b(ax+by+c)/z$, so \[ p^2+q^2+1 = \frac{(ax+by+c)^2}{z^2}+1 = \frac{1}{z^2} \implies z^2(p^2+q^2+1)=1. \checkmark \] Singular: $p=q=0$, $z=\pm1$: $1\cdot(0+0+1)=1$. \checkmark