member of the new set Salso occurs in the complementary solution, then members
of the set Sare modified again. This is analogous to what one does in the study
of ordinary differential equations. Here one can assume a particular solution of the
given difference equation which is some linear combination of the functions in S.
This requires that an assumed particular solution have the form
yp(n) = An +B
where A and B are undetermined coefficients. Substitute this assumed particular
solution into the difference equation and obtain
A(n+ 1) + B+ 2An + 2 B= 3n
which simplifies to
(A+ 3B) + 3An = 3n.
Comparing like terms produces the equations 3 A= 3 and A+ 3 B= 0. Solving for A
and Bproduces A= 1 and B=− 1 / 3 .Hence, the particular solution becomes
yp(n) = n−^1
3
.
The general solution can be written as the sum of the complementary and particular
solutions.
yn=yc(n) + yp(n) = c 1 (−2)n+n−^13.
Example 10-43. (Variation of parameters)
Determine a particular solution to the difference equation
yn+2 +a 1 (n)yn+1 +a 2 (n)yn=fn, (10 .60)
where a 1 (n), a 2 (n), fn are given functions of n.
Solution: Assume that two independent solutions to the linear homogeneous equa-
tion
L(yn) = yn+2 +a 1 (n)yn+1 +a 2 (n)yn= 0
are known. Denote these solutions by unand vnso that by hypothesis L(un) = 0 and
L(vn) = 0 .Assume a particular solution to the nonhomogeneous equation (10.60) of
the form
yn=αnun+βnvn, (10 .61)
