PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 

257 
Again, let a + &e 
ean Ta V@ 1. a3 x3 
1 
then Tan NPY 2)= 1.a3 5 — &e 
oY 1 
ee aa? 
dYs 1 il “1 
or tea i + 2p =) ¥2=0 
CO 
and y=A (y,— 7 y,) 
By solving the equations for y, and y, we obtain finally 
i iL 
2 hee, ec enihy 
mre (ale tS} 
The equations from which y, and y, are determined differ only in the term 
which does not contain y; and it will be seen hereafter that similar equations 
serve to give the solution of the other differential equations of this class, when 7 
is an integer. If a\/—1=m, these equations are 
a eee Sly A 
dx Tome) a2 
ea ey calee | z 
dae (5, ~ mm) ¥2= 0 
The following method of solving this equation has the ad- 
vantage of not appearing to take for granted the form in which y is expressed 
in terms of 2. 

12. OTHERWISE. 
—av =1ae—“=0 gives y= ————__; 
g ; d xt 5 d 1—-a/_—1ed! 
_liaV=Ied , 
l+atad ad 
Now 

yg Is the solution of the equation 
l+a?ad’ xd 
v+a@ad?adiv=0, or of 
v dv 1 
2 =a 
aE RE SO —() 
qin ine 2 , 
f EY pe eo 
oe dx \2Qa =e) poe 
which is the equation for determining y, given above. 
VOL. XVI. PART III. 

