Find the complete and singular integrals of 4xyz =…

Find the complete and singular integrals of \[ 4xyz = pq+2px^2y+2qxy^2. \]

Step 1 · Introduce $w=z+x^2y^2$

Since $(p+2xy^2)(q+2x^2y)=pq+2px^2y+2qxy^2+4x^3y^3$, the equation becomes \[ 4xy(z+x^2y^2) = (p+2xy^2)(q+2x^2y), \qquad\text{i.e.}\qquad PQ=4xy\,w. \]

Step 2 · Solve $PQ=4xyw$ via $w=X(x)Y(y)$

With $P=X'Y$, $Q=XY'$: $X'Y'=4xy$, separated as $X'=2ax$, $Y'=2y/a$, i.e. $X=ax^2+b$, $Y=y^2/a+e$ for arbitrary constants $a,b,e$. Then \[ w = XY = (ax^2+b)\!\left(\frac{y^2}{a}+e\right) = x^2y^2+eax^2+\frac{b}{a}y^2+be. \] Since rescaling $X\to\lambda X$, $Y\to Y/\lambda$ leaves $P,Q,w$ unchanged, only two of $\{a,b,e\}$ are essential; writing $A=ea$, $B=b/a$ (so the constant term $be=AB$), the genuinely two-parameter result is \[ w = x^2y^2+Ax^2+By^2+AB. \]

Step 3 · Write the complete integral

Since $z=w-x^2y^2$: \[ z = Ax^2+By^2+AB. \]

Step 4 · Find the singular integral

With $F=pq+2px^2y+2qxy^2-4xyz$: $F_p=q+2x^2y=0 \Rightarrow q=-2x^2y$; $F_q=p+2xy^2=0 \Rightarrow p=-2xy^2$. Substituting into the original equation: \[ 4xyz = (-2xy^2)(-2x^2y)+2(-2xy^2)x^2y+2(-2x^2y)xy^2 = -4x^3y^3 \implies z=-x^2y^2. \]

Verification (complete integral)

With $z=Ax^2+By^2+AB$: $p=2Ax$, $q=2By$: \[ pq+2px^2y+2qxy^2 = 4ABxy+4Ax^3y+4Bxy^3 = 4xy(Ax^2+By^2+AB)=4xyz. \checkmark \]

Verification (singular integral)

$z=-x^2y^2$: $p=-2xy^2$, $q=-2x^2y$: \[ pq+2px^2y+2qxy^2 = 4x^3y^3-4x^3y^3-4x^3y^3=-4x^3y^3 = 4xy(-x^2y^2)=4xyz. \checkmark \]
Note: the source's draft dropped one of the two integration constants as "WLOG zero," which loses the $x^2$ term and gives only a one-parameter family; the corrected two-parameter complete integral above was verified directly against the PDE.