PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 279 



\ _i B -- 

 Hence y=—e " +^e ^ \s the complete solution of the given equation. 



In my second Memoir on this subject, I 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 

 P 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., p. 608). 



,^ f 8 aE (a + R) ^I^ -8 (a2 + 3 a K + R'^) ^5^ 



+ 24(a + R)^-24f^}=^RV(«) 

 dz~* dz"' JO 



i where y=/ (3 + a), s = 2R, a = a — R. 

 m This becomes, by substitution, 



B / z\d--^ u ^ / , 3az 2^ '/-3 



I 



/ z\d-'-^t/ / ., 3a z z\d-Sy 

 Aaz[a+ -r] ^ — 8 ( a- + — -- + -r \ # 



Dividing by s«, we get 



4a^ d--i/ 2a rf^V _ So^ rf-3y _ ]_2a d-^ y _ 2 d'^ j/ 



24« d-*y 12 </-*> / _ 24 rf-°y _ 2 a^/(a) 



Writing e^ for z, and (_i)-»"_£=Jf „ for ~ <Lll, there results the symbo- 



lical form 



. 2 _2# -D-2 „ _^,-D-2 - ., _2./-D-3 



i„ _^/-I>-3 /-D-3 „, ,_D-4 



.i2/:^^\.24/=^,= |«VW- 



or, collecting the terms. 



roW-D-5 ,„,-D-4 /-D-31 



