120 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 
[655 
Adding these together, we have 
(V u\ , d(K t H) ¡dr dg\ d(K, H) fdp dr 
^ y ' /7 (n \ /7/j/ /7-r / /7 (nr Hip. 
d (q, r) \dy dz) d (r, p) \dz 
fdr_dq\, 
\dy dz) 
d (r, p) 
fdp eZr\ d (K, H) fdq _ cZp\ _ ^ 
Ve?z dx) ~d(p, q) \dx dy) 
viz. if p dx + q dy + r dz be an exact differential, then (H, K) = 0, which is the theorem 
in question. 
62. In the case where the variables are (x, y, p, q), we have simply 
viz. p dx + qdy being a complete differential, (AT, H) = 0. Conversely, if (K, H) = 0, 
dx dy 
= 0 5 this equation would imply that K, H considered as functions of p, q, 
are functions one of the other: and, supposing it to hold good, we could not from 
the equations H = 0, K = 0 determine p, q as functions of x, y, for, eliminating one 
of the variables p, q, the other would disappear of itself. We hence obtain the 
complete statement of the converse theorem, viz. the functions H, K being such that 
it is possible from the equations H= 0, K = 0 to express p, q as functions of x, y, 
then, if (H, K) = 0, we have p dx + q dy an exact differential. 
63. Returning to the case of the variables (x, y, z, p, q, r), if p, q, r are 
determined as functions of x, y, z by the three equations H = 0, K = 0, L — 0, then, 
by what precedes, in order that p dx+ qdy + r dz may be a complete differential, we 
must have (H, K) = 0, (H, L) = 0, (K, L) = 0; and it further appears that if these 
equations are satisfied, then we have, conversely, 
dr dq _ dp dr _ dq dp _ _ 
dy dz ’ dz dx ’ dx dy ’ 
that is, pdx + qdy +rdz is an exact differential; viz. this is the case unless we have 
between H, K, L the relation 
d(H, K) d(H, K) d(H, K) _ A 
7/ \> 7/ 7/ \ '“'l 
~d (q, r) ’ d (r, p) ’ d (p, q) 
H, L , H, L , H, L 
K, L , K, L , K, L 
where in the determinant the second and third lines are the same functions of 
H, L and K, L respectively that the first line is of H, L. 
The determinant is, in fact, equal to the square of 
d(H, K, L) 
d (p, q, r) ’