A MEMOIR ON DIFFERENTIAL EQUATIONS. 
122 
[655 
a solution of the equation H = const, involving the arbitrary constants x 0 , y 0 , z 0 
(one more than required for a complete solution). 
The theorem is here stated in the form proper for the solution of the partial 
differential equation H = const.; a more general statement will be given further on. 
66. I take first n = 2, or the independent variables to be x, y\ here p, q are 
determined by the equations a = a 0 , b = b 0 , c = c 0 , H = const., and it is to be shown 
that p dx + q dy = dV. 
Considering p, q, p 0 , q 0 as functions of the independent variables x, y, then 
differentiating in regard to x, and eliminating ^, C ^°, we find 
viz. this is 
But in the same way 
da 
+ 
da 
dq 
da 
da 0 
da 0 
dx 
dq 
dx ’ 
dp ’ 
dp 0 ’ 
dq 0 
db 
1 
db 
dq 
db 
db 0 
db 0 
dx 
dq 
dx ’ 
dp ’ 
dp 0 ’ 
dq 0 
dc 
1 
do 
dq 
dc 
dc 0 
dc 0 
dx 
+ 
dq 
dx’ 
dj)’ 
dp 0 ’ 
dq n 
dH 
dx 
+ 
dHdq 
dq dx ’ 
dH 
dp ’ 
o, 
0 
= 0, 
d(ao, b 0 ) id (H, c) d(H, c) dq\ „ 
d(p 0 , <lo)\d(p, x) d (p, q)dx) + 
d(a 0 , b 0 ) jd(H, c) d (H, c) dpi ,o 0 
d(p 0 , q 0 ) (d (q, y) d(q, p) dy) 
adding these two equations we have 
d(a 0 , b 0 ) | c \ + d (H, c) /dq 
d (p 0) q 0 ) d(p, q) [dx 
+ &c. = 0, 
the terms denoted by the &c. being the like terms with b, c, a and c, a, b in 
place of a, b, c. We have (.H, a) = 0, (H, b) = 0, (.H, c) = 0, and the equation, in 
fact, is 
(v d(ao, b 0 ) d(H, c) ) fdq _ dp\ = Q . 
1 d(p 0 , q 0 ) d(p, q)) [dx dy) 
viz. we have = 0, the condition for an exact differential. 
dx dy 
67. Coming now to the case where the independent variables are x, y, z, we 
proceed in the same way with the equations H — const., a = a 0 , b=b 0 , c = c 0 , d = d 0 , 
e=e 0 . Differentiating in regard to x, and eliminating 
dp dq dp,, dq 0 dr 0 
dx ’ dx ’ dx ’ dx ’ dx ’