For pq=z^mxy (m an integer): find the complete and general…

For $pq=z^mxy$ ($m$ an integer): find the complete and general integrals; determine for which $m$ the complete integral is a quadric; and for $m=0$, find the Cauchy integral through $x=1$, $z=\sqrt{1+y^2}$.

Step 1 · Complete integral

With $u=z^{(2-m)/2}$ ($m\neq2$), the equation reduces to $u_xu_y=\tfrac{(2-m)^2}{4}xy$; separating $u_x=ax$, $u_y=\tfrac{(2-m)^2}{4a}y$ and integrating: \[ z^{(2-m)/2} = \frac{ax^2}{2}+\frac{(2-m)^2y^2}{8a}+b. \] (For $m=2$, use $u=\ln z$ instead: $\ln z=\dfrac{ax^2}{2}+\dfrac{y^2}{2a}+b$.)

Step 2 · General integral (envelope in $a$)

Setting $\partial_a\bigl[\tfrac{ax^2}{2}+\tfrac{(2-m)^2y^2}{8a}\bigr]=0$ gives $a^2=(2-m)^2y^2/(4x^2)$, i.e. $a=(2-m)y/(2x)$; substituting back: \[ z^{(2-m)/2} = \frac{(2-m)xy}{2}+b. \]

Step 3 · Quadric classification

The complete integral is a polynomial of degree $\le2$ in $z$ exactly when the exponent $(2-m)/2$ equals $1$ or $2$: $m=0$ gives $z=\tfrac{ax^2}{2}+\tfrac{y^2}{2a}+b$ (an elliptic paraboloid), and $m=-2$ gives $z^2=\tfrac{ax^2}{2}+\tfrac{2y^2}{a}+b$ (an ellipsoid/hyperboloid type).

Step 4 · Cauchy integral for $m=0$

Here $z=\tfrac{ax^2}{2}+\tfrac{y^2}{2a}+b$. On $x=1$, $z=\sqrt{1+y_0^2}$: the strip condition gives $q_0=y_0/a$, and matching $q=y/a$ to $q_0=y_0/\sqrt{1+y_0^2}$ gives $a=\sqrt{1+y_0^2}$. Substituting into the complete integral and simplifying gives $b=1/(2a)$, i.e. $2ab=1$ for every member of the family. Substituting $b=1/(2a)$ back and taking the envelope in $a$ (setting $\partial_a=0$) gives $a=\sqrt{1+y^2}/x$, and substituting back: \[ z = x\sqrt{1+y^2}. \]

Verification

The corrected general integral $z^{(2-m)/2}=(2-m)xy/2+b$ was checked directly against $pq=z^mxy$ and confirmed to satisfy it identically for representative $m$ — note the source's draft introduced a spurious factor of $\sqrt2$ in this step (from an algebra slip solving $x^2/2=(2-m)^2y^2/(8a^2)$ for $a$); the corrected coefficient is $(2-m)/2$, not $(2-m)/(2\sqrt2)$.
Cauchy integral: $z=x\sqrt{1+y^2}$ gives $p=\sqrt{1+y^2}$, $q=xy/\sqrt{1+y^2}$, so $pq=xy=z^0xy$. \checkmark At $x=1$: $z=\sqrt{1+y^2}$. \checkmark