the calculus of functions. 251 
whose solutions are 
<P i x —y) — X {x—y* a+b—x+y } 
and (p (a?-] -y) = (a — b — 2X 
X 
— *y)x [&+y,a- 
.b — "x—y j 
hence the general solution of the equation 
dtp (a— x, y ) 
d-p ( x , b—y ) 
dy 
dx 
is 4 / (x,y) —J ( dx + dy) x + y , a + 6 — 
x— y} ”f“ 
55 
1 
*© 
1 
f 
+ ay) X {m—y,a 
X 
jf 
** 
i 
1 
Ex. 2. Given the equatior 
d x, — ] 
y I 
l 
1 
f -^1 
l x 1 
dx 
dy 
the partial differential equation to be solved is in this case 
(x,y) f d-\, (x, y) 
dy z ~~ y z dx % 
and its solution is 
i'( x ,y)-— (j) +<p(xy) 
determining <p and <p so as to fulfil the conditions of the equa- 
tion, we have 
4 - (x,y) = c (x+y) +fd (ay) • ( ~) 2 x { xy, j- y } 
Problem XLI. 
Given the equation 
F {x,y,$(x,y),ty(ex,y), &c. &c.|=o 
and let a? x = x and ( 3 ?y =y, 
then there may be pq different forms of the function con- 
tained in the general expression \|/ (cfx, jQ*y), r varying from o 
to p— 1, and 5 varying from o to q — 1. 
In the first place it may be observed, that if we substitute 
ocx for x in such a quantity as 
d n y (a 3 x , (2y) 
dx n 
LI 
MDCCCXVI. 
