
PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 301 
ae 
—qQ 
=A (a?+1)* + x 

xX 
=A (a? +1)? +(a* + a) (a? +1)" 3 pet 
Cor. 1. Let X=6z; then the solution of the equation a*x,—au,=62 is 
tip =A (a? +1)" — (ab +a). (3 ah =) 
auta+l1 
then the valueof (a—a@%)-!X is — 2 
therefore the solution of the equation a? u, — a ees ont 
a (z+1) 
is u, = A(a? + 1)?— (at + a)* 
a ee © Mitral 1 ee Ee (by Ex. 6, Art. 24.) 
Ex. 7. paren 
This equation may be written 
(A+ aat+b)u, =0. 
Let a, @ be the roots of the equation 2?+az+6=0, then 
(a*—a) (a*—f) uw, =0 
u,=A (A?—a)-1.0+B(a?—8)-!.0 
=A (1+a’+B(1+ 2)" (Ex. 4.) 
Cor. If a=, we obtain, by the usual process, u,=(A+B2) (1+a?)*. 
Ex. 8. AU,+@ A? U,+buU,=¢ (1 +e)" 
The solution is 
maz plet— a0 + el Fe ays B10 + ole) 
else) ete) 
(a—8) (e—a) (a—f) (e—8) 
me 2\a 200 ce (1+e?)” 
=A (1+a’)”+B (1+) + aaah 
Cor. If e=a, we must proceed as in Cor. 3, Ex. 4, Class 1, of Differential 
Equations, and we shall obtain 
=A (1+a?)"+B(1+?)"+ 
ex(1+a’)-* 
2a+a 
VOL. XVI. PART III. 4G 
u,= A (1+)? + B(1+ 6)" 4 
