998 PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 
EQUATIONS OF DIFFERENCES. 
26. As the method of solving equations of differences of the second and 
higher orders, by treating the symbols of operation as symbols of quantity, and 
reducing the resulting fraction by decomposing it into partial fractions, has been 
little, if at all, employed, we shall commence with an example or two in ordinary 
equations of differences. 
Ext. Un 4 3tE Uy o4+b Up +6 U,=X. 
This may be written 
{+4)?+a(1+4)?+6(1+a)+clu,=X 
If we write a, for 1+, and suppose a, 8, y the roots of the equation 
A3+aA2+6A,+c=03 we get (A,—a) (A,—) (4,—Yy)u,=X, or 
te 1 
" (4,—4@) (4,—8) (4,—) 
This equation is reduced, by the decomposition of the fraction of operation 
into its equivalent partial fractions, to 
. (X+0). 

2S eS ee </e (X +0) 
=) (a—y) (a —a) ‘ PAE ieee y) Oa = 
1 
(X +0). 
W-a a) (y—B) (4, —¥) = 
Now “a (X+0) is the solution of the equation v,,,; —av,=X+0; 
hence it is equal to ax (a+ =) ; and similarly of the others. 
Hence the complete solution of the given equation is 
1 xX i xX’ 
th) = eo A+ ——— —S——_—_————— 9 ——— 
“=Gopa@an® Ata) * ray Gan Bt gee) 
1 D4 
aa ees C+ 
(=a) (y—8) * Se? 1) 
Cor. If a=, we must, as in similar cases, put a+c for 6, and expand in 
terms of c. The result is 
1 x 
1 ae a* x X (#+1) 
at ae (fu xara} -<) (Bite Ssyrsat) 
igri koga se 
+(—ap (C+2 35) 
