Find the complete and singular integrals of z = px+qy+pq

Find the complete and singular integrals of \[ z = px+qy+pq. \]

Explanation of Method

Clairaut equation: the complete integral comes from $p=a$, $q=b$; the singular integral is the envelope, found by eliminating $a,b$ from the plane family and its two partial derivatives.

Step 1 · Complete integral

\[ z = ax+by+ab. \tag{I} \]

Step 2 · Eliminate $a,b$ for the envelope

\[ \partial_a: \ x+b=0 \implies b=-x, \qquad \partial_b: \ y+a=0 \implies a=-y. \] Substituting into (I): \[ z = (-y)x+(-x)y+(-y)(-x) = -xy. \]

Verification (complete integral)

$p=a$, $q=b$: $px+qy+pq=ax+by+ab=z$. \checkmark

Verification (singular integral)

$z=-xy$: $p=-y$, $q=-x$: \[ px+qy+pq = -xy-xy+xy = -xy = z. \checkmark \]

Result

The complete and singular integrals are:

\[ \boxed{ \text{Complete: } z=ax+by+ab; \qquad \text{Singular: } z=-xy } \]