145 
(8a 8 + 27»?)»t-9^ = 0 
CIOC 
(8a 5 + 27xi)n + Zr/p = 0 
CLOG 
Substitute the above values of l, m, n iu 
d—^ly^ + my + n 
Then (8a 3 + 27af)^ + 3^(2a/ - 3*# + a 2 ) = 0 
( 6 ) 
An equation, which is called the first Resolvent of the 
quartic (1). 
Each of the roots of this quartic is, therefore, a particular 
solution of (6), and any value of y which satisfies (6) is, 
therefore, a root of the quartic (1). 
2. A general solution of (6) is obtained as follows : — 
Let y = y x + v be a general solution (7) 
where y\ is one of the roots of the quartic (1), and, v a new 
variable to be determined. 
Substitute the above value of y in (6) and it becomes 
(8a 3 + 27x\)^~ + ^ + 3^^2a(yi + vf - dx^i + v) + a 2 j = 0 
From which is derived 
(8a 3 + 27*?)^ + 3^ 1 2 av* + (4a^ - 3*,>j = 0 (8) 
This equation is a well known integrable form, and, by 
integrating it in the usual way, putting (8a 3 + 27#?)P = 6a^ 
then we obtain 
»(« + /|P exp. J (^-2yi)Pc&|^ = exp.y"i ) p<&...(9) 
using the convenient notation of Professor Cayley, viz., 
e s = exp. 2 . 
Hence the general integral, or complete primitive of (6) is 
exp./'(|^-2yi)p& 
V = y 1 + Y1 /-/^ — ~ r — =j— 
c + J | P exp. j (^ - 2y 1 JPc& J dx 
( 10 ) 
