61 
Substitute this value in (8), then 
/2ap 
+ 2 mp* = 0 
“ + (2Q + P ~) P f + - Q>> -Q/ + ^y-+ 2 f 
From (7) we obtain 
.(9) 
9 r* 3 a P l 
py = Qy--§~-y 
Then (9) becomes 
Op 1 
a o + ^ 
\ 3 p 
4a pQ 2pcd _ 
3 + 3 
2Q 2 - Q ^jy + Imp? 
( 10 ) 
But, 
2 ntf - *** + M. = | (3?re p _ 2aQ + « 1} _ o 
3 3 
Then (10) may be written 
V n - (3Q + ~ y + + 2Q 2 - Q 1 )? = 0 . . .(11) 
Eestore the values of p and Q, then 
<Vy 
dx* 
( d Sma 1 - 2am 1 \dy ( aim 11 - m 1 ^ 11 
I dx & */4 a 3 4 - 27m 2 Jdo? 1 2am 1 - Sma 1 
}y=o (12) 
W 4 a 3 + 27m 2 
6 am (a 1 ) 3 - 6a(m 1 ) 3 - 6a 2 m 1 (a 1 ) 2 
(2am 1 - Sma 1 ) (4a 3 + 27m 2 
Equation (12) is, therefore, the second differential resol- 
vent of the cubic (1). The value of y which satisfies (12) 
is a root of the cubic (1), and, either of the three roots of 
the cubic (1) is a particular solution of the differential 
equation (12). 
3. When ( a ) is constant, then, the first and second 
differential resolvents respectively become 
4a 3 + 27m 2 dy „ 3m 2a 
dx 
dx 2 
- {IM 
6am 1 
- 2am 1 
+ r -Ta-y + ^ =0 
(13) 
Jia 3 + 27m 2 / j dx 4a 3 + 27 m*' V ~ 
4. If (m) be such as to satisfy 
= 1 
m 1 
4 a 3 
27 
4- m 2 
( 15 ) 
