16 Mr. W. H. L. Russell on Linear Differential Equations. [June 16, 

 To determine A r , A*"- 1 , . . . , let 



1 _L 1 



z— , or a?=a+ -. 



x — a z 



Then the solution of the resulting linear differential equation in (z) will be 

 of the form 



y = It 1 (*)€ A 0+ A l* +A 2* 2 + . • . Ar* r ; 



where R^z) can he expanded in descending powers of (z), and therefore 

 A , A 1} A r , . . . be determined as before. 



In the same way, by putting x=b + ~ y we may determine B . B x , . . . 



In order to exhibit the accuracy of this reasoning, I will form some dif- 

 ferential equations from given primitives, and then see if the above process 

 will enable us to reproduce these primitives as solutions. 



Let us take the primitive 



From this we may deduce the differential equation 



(x*-x 2 -x+l)^-(x* + 2x 3 --'6x 2 -3x+4) C ^ 

 x dx* v ' dx 



— (2x 5 + x* + 4x 3 — w 2 —5x—4)y = 0. 



Let 



y = 'R(x)ef d *(f-+ v *K 

 If we use higher powers of (x) in the exponential, they will not give us a 

 result. 



dy d if 



Substituting in the differential equation the values of — > — ^ given in 



ctx tix 



the earlier part of this paper, and equating the highest terms to zero, we 

 have 



v 2 —v-2=0 and 2^v — r 2 —^ — 2p—\=0, 



whence 



v = 2 and yu=3, or v= — 1, and yu = 0, 



and therefore 



x 2 



fdx(ii-\-t>x)=x 2 + 3x, or — — • 



The divisors of the first term are x— 1 and x+ 1. 



Let x — - + 1, and the differential equation becomes 



(2z 7 + . . . ) § -(2*< + . . .) ^ + (3z 5 + . . . )y=0, 

 az az 



which gives a solution of the form y=R 1 (z)e z , when z— -. If we put 



x — l 



x= i — 1, the differential equation will be of the form 



(A* 8 + ...)^| + (Bz 7 + ...)^+(Cs 6 + ...)y=0, 



