( 1246 ) 
which are impossible unless 
(5a+26'+-c) r, + 2br, + (a4 20'+-c) r, — 2br, = 0 
or 
b fa Jo) (BD Balor Sti, ee a MI) 
This condition breaks up into three conditions which will be considered 
separately. 
Supposing in the first place 
ao = and y= 0 
the differential equation may be solved. For putting 
5 
— == Ae 
5 zig 1 by 
we obtain the linear differential equation 
dt a 1 2b' He — 2(b' He) z + ez? 
-— —t{t = - 
dz b'z 85? z 
A particular integral of this equation being 
t=a-+t get ye? 
where 
2b' He 2 (b'+-¢) c 
OR On (oneeays 
the general integral of the original differential equation takes the 
form 
a 
ot — 2 (a+-B+7) HAD (B42y) y — 86" (1-25) ” = const. 
which for small values of 2 and y may be expanded in the form 
et+yt Fy -+ F,...= const. 
In this ease therefore the origin is a centrum. 
7. If, in the second place 
a =f) and a e=0 
the corresponding differential equation 
dy — w+ ba? + 2b'xy — by’ 
dert y + av® + 2bey — ay” 
has three particular integrals of the form 
iS Aat B 
for substituting this value and equalling the coefficients of the different 
