7 5—2 
[626 
ticular case 
where e is 
nv., further 
as variable, 
differential 
an solution 
/) - 2Mde) ; 
comes 
yY 
xif - y% 
and also a 
626] 
dx dy 
ON THE GENERAL DIFFERENTIAL EQUATION — 0. 
595 
Multiplying the numerator and the denominator by 
d (x -y)-v2e (x 2 — y 2 ) + 2 (dX — v'F) de, 
the denominator becomes 
= (x — y) 3 
\d + 2e(x + y)J 2 — 4>e 
\ X ~V 
which, introducing herein the C of Euler’s equation, is 
= (x — y) 3 (dr — 4eC). 
We have therefore 
z(x — y) 3 (d 2 — 4eC) = {2a + b(x + y) +c. 2xy + dxy (x + y) + e. 2xy (x- — xy +- y 2 ) 
+ 2 *Je (x - y) (x \JY- y dX) -2 d XY} x [d (x - y) + 2e (x 2 -y 2 ) + 2 de (dX - dY)}. 
Using (& to denote the same value as before, the function on the right-hand is, in 
fact, 
= (x — y) 3 {2be — cd + dC + 2 de V®}; 
and, this being so, the required relation between 2", G is 
^ (d 3 — 4e(7) = {2be — cd + dC + 2 de V®}. 
To prove this, we have first, from the equation 
= C + d(x + y) + e(x + yf, 
to express ® as a function of x, y. This equation, regarding therein C as a variable, 
gives 
and we have therefore 
dx dy dC 
dX^dYVS ’ 
-Ve = Vxf = VF|, 
dC 
viz. dX j will be a symmetrical function of x, y. Putting, as before 
M = 
dX-dY 
we have 
and thence 
We have 
x-y 
C = M 2 - d (x + y) - e (x + y) 2 , 
dC cTijdM . 
= 2M — d — 2e (x + y). 
dx 
dM 
dx 
1 X' dX-dY 
dx x — y 2 dX (x — y) 2 *