Find the complete integral of pq=x^my^nz^l (l≠2)
Find the complete integral of $pq=x^my^nz^l$ ($l\neq2$).
Explanation of Method
Substitute $u=z^\alpha$ with $\alpha=(2-l)/2$ to remove the explicit $z$-power, reducing the equation to a separable form in $u_x,u_y$. Treat $l=2$ separately via $u=\ln z$.
Step 1 · Substitute $u=z^{(2-l)/2}$
With $p=\dfrac{z^{l/2}}{\alpha}u_x$, $q=\dfrac{z^{l/2}}{\alpha}u_y$, the equation $pq=x^my^nz^l$ becomes \[ u_xu_y = \alpha^2x^my^n = \frac{(2-l)^2}{4}x^my^n. \]
Step 2 · Separate and integrate
Set $u_x=ax^m$, $u_y=\dfrac{(2-l)^2}{4a}y^n$; integrating (for $m,n\neq-1$): \[ u = \frac{ax^{m+1}}{m+1}+\frac{(2-l)^2y^{n+1}}{4a(n+1)}+b. \]
Step 3 · Special case $l=2$
Here $\alpha=0$ is degenerate; use $u=\ln z$ instead, giving $u_xu_y=x^my^n$, separated as $u_x=ax^m$, $u_y=y^n/a$: \[ \ln z = \frac{ax^{m+1}}{m+1}+\frac{y^{n+1}}{a(n+1)}+b. \]
Verification (numeric, $l\neq2$)
Direct substitution of $p,q$ recovered from $u=z^{(2-l)/2}$ back into $pq$ confirms $pq=x^my^nz^l$ identically (checked symbolically for representative $m,n,l$). \checkmark
Verification ($l=2$)
$p/z=ax^m$, $q/z=y^n/a$, so $pq=x^my^nz^2$. \checkmark
Result
The complete integral is: