655] 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 133 
having the integrals 
a = p, 
b =q, 
c = px — qy, 
, 1 X 
d = —1—, 
r q 
_ z 1 
pq r 2 5 
(whence H = — abe). We have H, a, b, a system of conjugate integrals and, in terms 
of these, 
hence, writing X for the constant value of V, we have 
that is, 
X = Jja dx+bdy + ^dzj, 
X = ax + by + 2 \J{ab (z + H)}, 
or say, 
4ab (z + H) — {ax + by — X) 2 , 
a solution containing really the two constants X and and thus a complete solution 
of the given equation pq — z = H. We have, in fact, 
that is, 
2ab p = a {ax + by — X), 
2ab q = b {ax + by — X); 
4a 2 6 2 pq = ab {ax + by — X) 2 , = 4a 2 & 2 {z + H), 
or 
- 
as it should be. 
pq = z + H,