124 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 
[655 
and each of these, as containing the same letter twice in the denominator, that is, 
as having two identical columns, is = 0; the theorem is thus proved. And in the 
same way ^ ^ ^ are each = 0; that is, pdx+ qdy + rdz = dV. 
69. The proof would fail if the factors multiplying ^ — —, &c., or any one of 
these factors, were =0. I have not particularly examined this, bat the meaning must 
be that here the equations a = a 0 , &c., H = const., fail to give for p, q, r expressions 
as functions of x, y, z, ¿r 0 , y 0 , z 0 , H; whenever such expressions are obtainable, we have 
pdx + qdy + rdz = dV. 
The proof in the case of a greater number of variables, say in the next case 
where the independent variables are x, y, z, w, would probably present greater difficulty, 
but I have not examined this. 
70. Taking the independent variables to be x and y, we may from the equations 
a = ct 0 , b = b 0 , c = c 0 , H = const, (which last equation may also be written H = H 0 = const.) 
find p, q, p 0 , q 0 as functions of x, y, x 0 , y 0 , H; and we have then the theorem that, 
considering only II as a constant, 
p dx + q dy — p 0 dx 0 — q 0 dy 0 — d V. 
we have 
to prove the further equations 
dx 0 
dp 0 _ 
dx 
\d(b 0 , 
c 0 ) d(a, H)] 
! dH 
d (a 0 , 
bo, C 0 ) 
— 0 
doo 0 i 
[d (Po, 
^o) d{p, q) J 
I dq 
d (x 0 , 
Po, <7o) 
dp 0 v 1 
\d(b, i 
c) d (a 0 , H 0 )} 
i dH 0 
d (a, 
b, c) 
— 0 
dx | 
[cl(p, 
q) d(p 0 , q 0 )J 
1 dq 0 
d (x, 
v, q) 
and it is to be shown that the coefficients of ~, are equal and of opposite 
signs, and that the other two terms are equal ; viz. this being so, subtracting the 
two equations, we have the required relation ^+^ = 0. Now H, are the same 
functions of a, b, c and of a 0 , b 0 , c 0 ; and there is no real loss of generality in assuming 
c = H, c 0 = H 0 ; but this being so, the first coefficient is 
and the second is 
d(b 0 , H 0 ) d(a, H) d (H 0 , a 0 ) d (6, H) 
d (p 0 , q 0 ) d (p, q) d(p 0 , q 0 j cl (p, q) 5 
d{b, H) d(a 0 , H 0 ) + d(H, a) d(b 0 , H 0 ) 
d (p, q) d(p 0 , q 0 ) d (p } q) d(p 0 , q 0 ) ’ 
which only differ by their signs. As regards the other two terms, we have identically 
da, 7 T T\ db, TT x dH, d(a, b, H) 
a)+ <k (a ’ b)= i<ktYy