AND PARTIAL DIFFERENTIAL RESOLVENTS. 43 



2. To find the second differential resolvent of the cubic 

 (i) it will be convenient to write (6) as follows : — 



y J +py % -Qy + ^y = o, .... (7) 



where 



(4a 3 -f 27m 2 ) p = 6am 1 — qma 1 . 



Differentiate (7) with respect to x, then 



y"+2.pyy* 4 . i ,y_Qj,'_QV+^ + ^ = o. . (8) 



From (7) there results 



1 r\ x ? 2a PV 



yy =&y -py* — 5^ 



= Qy z + — +mp. 

 Substitute this value in (8), then 



2/" +^ 2 Q+^y^ + (^-Q.^y- Qf + 2S.' + 



T ( 2ap l ( 2a l p 

 3 

 + 2mp z = o (9) 



From (7) we obtain 



py z =Qy~Y~y'- 



Then (9) becomes 



AapQ, , 2pa T . 



— ^-^ \-- JL - =0 (10) 



3 3 



But 



2mp z — ^- 1- -J-— = - (^mp — 2«Q -f a 1 ) = o. 



