Find the complete integral of p^2+q^2 = z^2(x+y)

Find the complete integral of \[ p^2+q^2 = z^2(x+y). \]

Explanation of Method

The equation is homogeneous of degree 2 in $z,p,q$, so the substitution $u=\ln z$ removes $z$ entirely, reducing it to a separable equation $u_x^2+u_y^2=x+y$ in $F(x,p)=F(y,q)$ form.

Step 1 · Substitute $u=\ln z$

With $z=e^u$, $p=e^uu_x$, $q=e^uu_y$: \[ u_x^2+u_y^2 = x+y. \]

Step 2 · Separate variables

Write $u_x^2-x=y-u_y^2$ and set both sides equal to $a$: \[ u_x=\sqrt{x+a},\qquad u_y=\sqrt{y-a}. \]

Step 3 · Integrate

\[ u = \tfrac23(x+a)^{3/2}+\tfrac23(y-a)^{3/2}+b. \]

Verification

$u_x^2+u_y^2=(x+a)+(y-a)=x+y$. \checkmark Recovering $p=zu_x$, $q=zu_y$: $p^2+q^2=z^2(x+y)$. \checkmark

Result

The complete integral is:

\[ \boxed{ \ln z = \tfrac23(x+a)^{3/2}+\tfrac23(y-a)^{3/2}+b } \]