
PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 255 


or y= (1-myai 2) =i OE 
Suppose J=2a.e8. G 
3a, foe —m/=T th) =0 by (A); 

which can be satisfied only by making 1—m/—1 [n = =0; giving, consequently, 

only one value of n - 
AS 3 
y= —n 1S the complete solution. 




Hence 
2 2 
Cor. If m= Bae Fa 
oe ay Oe =|f= pase 
|n YT 2 Al ? 
n=1 and y= — 
—dty ‘ad 
Ex. 2. SSN Tee ae 
The equation in @ is ff e~ 
0 (1-my aa Beh) liewest 
ea +36. (1—-mv=1 244) ! (Ex. 1 and A) 




Cor. If =n; this expression becomes infinite. We must, in this case, write 
n+c in place of 7, expand in terms of c, and finally put c=0. 











ae —"4(1c 6+ &e. 
We have, thus, ~— ie are a S = == ‘ roe 
A 7 ay lm 1-mV=1("2? 4 2 fF c+ a.) 
|r In dn 
—né —né 
i. e ahs ce = an ae 
Lie ee a RE 
ne e-n4 i e-”46 


