114 
A MEMOIR ON DIFFERENTIAL EQUATIONS. 
[655 
or, since the first two terms on the right-hand vanish, the equation (a, 
becomes 
, ,\d0 , \d6 , dd dQ . r .d0 
( a> h) db +(a ’ C do + ^ a ’ ^ dd + ( a ’ ^ de + ( a ’ ^ df~°' 
dc 
dd 
de 
6) — 0 thus 
But by the Poisson-Jacobi theorem (a, b), «fee., are each of them a solution of 
(H, 6) = 0, viz. they are each of them a function of H, a, b, c, d, e, f This is 
then a linear partial differential equation wherein the variables are H, a, b, c, d, e, /; 
,, , . dd dd 
or, since there are no terms m yyy, y, we 
dH da 
may regard a, H as constants, and 
treat it as a linear partial differential equation in b, c, d, e, f, the solutions of the 
equation being in fact the integrals, or any functions of the integrals, of 
dJb _ de _ dd _ de _ df 
(a, b) ~ (a, c) ~ (a, d)~(a, e)~ (a, /)' 
48. Suppose any four integrals are V, c', d!, e', so that a general integral is 
a, b', c', d', e'), then b', c', d!, d qua functions of H, a, b, c, d, e, f are integrals 
of the original equation (H, 6) — 0 ; hence changing the notation and writing b, c, d, e 
in place of these accented letters we have (H, a, b, c, d, e) as solutions of the two 
equations (H, 0) = 0, (a, 0) = 0 ; viz. a being any integral of the first of these equations, 
we see how to find four other integrals (b, c, d, e) which are such that 
(H, a) = 0, (H, b) = 0, (H,c)= 0, (H,d) = 0, (H, e) = 0, 
(a , b) = 0, (a , c) = 0, (a , d) = 0, (a , e) = 0. 
49. We proceed in the same course and endeavour to find 0, so that not only 
(.H, 0) = 0, (a, 0) = 0, but also (b, 0) = 0. Assuming here 0 — 0 (H, a, b, c, d, e) an 
arbitrary function of the integrals, the first and second equations are satisfied; for the 
third equation, we have 
.7 .. T7 . d0 /7 . d0 ,, T.d0 s d0 /7 7X d0 . d0 
(b, 0) = (b, + h ^db + ( b ’ G ^dc + ( b ’ d ^dd + ( ' b ’ e ^de’ 
viz. the first three terms here vanish, and the equation (b, 0) = 0 becomes 
d0 
d0 
d0 
C ' ) dc + ^ d ^ dd + 6 ^ de ~ °’ 
where, b, c, d, e being solutions as well of (H, 0) = 0 as of (a, 0) = 0, we have (6, c) a 
solution of these two equations, and as such a function of H, a, b, c, d, e ; and so 
(b, d) and (6, e) are each of them a function of the same variables. The above is 
therefore a linear partial differential equation wherein the variables are H, a, b, c, d, e, 
but as the equation does not contain 
d0 d0 
d0 
da ’ ° r db’ 
we may regard H, a, b as con 
stants ; and the solutions of the equation are, in fact, the integrals of 
de dd de 
(6, c) “ (0, d) ~ (0, e) ’