For 2xp+2yq+4zp^2-z=0 : find the complete integral (the…

For $2xp+2yq+4zp^2-z=0$: find the complete integral (the general and singular integrals turn out not to exist as real surfaces), and find the solution satisfying $z^2=x$ on $y=x$.

Step 1 · Charpit's equations and two first integrals

With $F_x=2p$, $F_y=2q$, $F_z=4p^2-1$, $F_p=2x+8zp$, $F_q=2y$: \[ \dot p=-p(1+4p^2),\quad \dot q=-q(1+4p^2),\quad \dot z=z(1+4p^2). \] From $\dot p/\dot z=-p/z$: $pz=a$ (constant). From $\dot q/\dot p=q/p$: $q=bp$.

Step 2 · Complete integral

With $p=a/z$, $q=ab/z$, substituting into $F=0$ and multiplying by $z$: \[ 2a(x+by)+4a^2-z^2=0 \implies z^2 = 2a(x+by+2a). \]

Step 3 · General and singular integrals do not exist as real surfaces

Eliminating $a$ from (I) and $\partial_aI=0$ (for fixed $b$) gives $z^2=-(x+by)^2/4\le0$, so no real general integral of this reduced form exists. Similarly $F_p=F_q=0$ forces $y=0$ and then $z^2=-x^2/4$, so there is no real singular integral either.

Step 4 · Cauchy problem: $z^2=x$ on $y=x$

On $x=y=x_0$, $z=\sqrt{x_0}$: the strip condition (with $dy=dx$) gives $p_0+q_0=1/(2\sqrt{x_0})$, i.e. $a(1+b)=1/2$. Substituting $x=y=x_0$ into the complete integral and using this relation gives $x_0=x_0+4a^2$, forcing $a\to0$ with $ab\to1/2$ held fixed. Taking this limit in $z^2=2a(x+by+2a)$: \[ z^2 = 2\cdot\tfrac12\cdot y = y \implies z=\sqrt y. \]

Verification

Complete integral: substituting $p=-a(x+by+2a)^{-1/2}\cdot(\cdots)$ … more directly, differentiating $z^2=2a(x+by+2a)$ gives $2zp=2a$, $2zq=2ab$, so $p=a/z$, $q=ab/z$, and $2xp+2yq+4zp^2-z = 2a(x+by)/z+4a^2/z-z = [2a(x+by+2a)-z^2]/z=0$. \checkmark
Final solution: $z=\sqrt y$ gives $p=0$, $q=1/(2\sqrt y)$, and $2xp+2yq+4zp^2-z = 2y\cdot\tfrac{1}{2\sqrt y}-\sqrt y = \sqrt y-\sqrt y=0$. \checkmark At $y=x$: $z^2=y=x$. \checkmark