IN THE DIFFERENTIAL CALCULUS. 55 
y=d-* (dt—a)-? {x} (aa) x X}, 
which is precisely the solution given by Professor Booun’s process. 
We shall conclude the present Memoir by comparing the methods here exhi- 
bited in two particular cases of the second example. 
Case 1. Let X=0; then the first method gives y=A xe”: whilst the second 
and third methods give 
y=d-* (d'—a)~? . 0. 
Now, in a former paper (Z7’ransactions of the Royal Society of Edinburgh, 
vol. xiv., p. 252), I have shewn, in Cor. 1. to Example 3, that 
(dt—a)-* . O=Ace™ 74 Bae™* 

Hence y=d-* fA eV 24 Bre} 
_A et” Bz a2 x B ae 
ae ai hag ° 
The condition B=2 A a? reduces this solution to the former. 
1 ah 
Case 2. Let x=1(4+% :) : 
a J rv £ 
then, by the first method, 
nial eee 1 ett 
NZI MRE. Tiel ite i Kew Ale 
Beto art 
xt J a 
dP pees 31, @ 
x dx Gite) 
y=Ane* + ore®* [eae (ae +) 
hs a? x b 
=Aze ways 
By the second and third methods, 
ayy le a ke eg VON 
7@ jini (<3 Fe == 
Also at @_ aya Ft 2a+07t 
d—a@ 
Micelle at ST 
Va 3a Var ) 
—2 3 a at 
=(d—a? a 
( - ( dai 2 a) 
an ordinary linear differential equation, of which the solution is 
y=Axe” "+B eee e., 

y=(d—a®)~* (db 42a4+d7*) o( 
This agrees with the former solution by making B=0. 
VOL. XX. PART I. 
