220 
EXACT DIFFERENTIAL EQUATIONS. 
Hence tlien a first integral of the above equation will be 
The method of integrating an exact differential equation 
which we shall illustrate, and which contains an implicit solu 
tion of the question whether a proposed equation is exact or 
not, appears to be primarily due to M. Sarrus (.Liouville, Tom. 
xiv. p. 131, note). 
Ek. Given y + Zx d £ + îy (g) + + = 
Supposing the above an exact differential, we are by defini 
tion permitted to write 
dx...[ 34). 
How a first and obvious condition is that the highest differ 
ential coefficient in an exact differential equation, being the 
one introduced by differentiation, can only present itself in the 
first degree. This condition is seen to be satisfied. 
Representing the highest differential coefficient but one by 
p, we can express (34) in the form 
dU — (y + 3xp + 2yp') dx + (y? + 2y^p) dp. 
Now let U x represent what the integral of the term contain 
ing dp would be were p the only variable. Then 
U x = x*p + y 2 y> 2 . 
Assume, then, removing all restriction, 
whence dU 1 = 
dx. 
Subtracting this from (34) 
(35).