
PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 279 
a 
Hence — e %+ = e ® is the complete solution of the given equation. 
In my second Memoir on this subject, | exemplified the use of a theorem in 
general differentiation, by solving the problem of determining the law of force by 
which the particles of a sphere must act on a point, so that the whole attraction 
may be the same as if the sphere were collected at its centre of gravity. The 
solution of this problem led to a differential equation which was shewn, by an 
indirect process, to be satisfied by the law of force varying as the distance, or in- 
versely as its square. I propose, at present, to solve this differential equation. 
Ex. 4. The equation is (vol. xiv., ae 608). 
2+3aR+ 

be 7 [8 aR (a+R) 



d-> 4 Aq 
att =FR LS 
where y= f (e+ a), 2=2R, a=a—R. 
This becomes, by substitution, 
2 \d Ay Bond, & gy da? y 
MR a) pee ga 
d~*y tray ay? 2 
+24(a4 +5) aes g4Q 4 = 5a ue 2. 



Dividing by z*, we get 
4a? d-*y 2a d-Py Seda gititads y 2d- ry 













z+ dz-2 if dz-2 28 ae Ze de 8 2 d2-8 
Q4ad-*y 12d-*y 24d 2 a? f (a) 
Te Ce eS ee 55 2 
Writing < for z, and (—1)-« === : 
| ead 2 
lical form 
— D=2 Deo D8 
ON ad i el 1p Gen Eas ye 8 ate —26/ 
= y a =p y <a eeaa ay —=— y 
J Sis pa a 
+i 7 — +2! — OL rp ee 
ae = y =D yt ae ED y 
D—4 D 2 . 
+12 Sy 424 a ya Fe f(a) mle 
or, collecting the terms, 



