DHARM

RECURRENCE RELATIONS WITH CONSTANT COEFFICIENTS 65

NOTE Here f(n) is a constant and accordingly if we assume the general form of the particular
solution to be P, then we obtain nonexistence form of the equation P – P = 5. Therefore the correct way to
assume the form of f(n) is 5.1n.
(h) Consider the difference equation an – 2 an–1 + an–2 = 7 ; Since 1 is the characteristic
root of multiplicity 2 of the difference equation. So, f(n) can be written as 7.1n. Thus the particular
solution is of the form
n^2 .P.1n or n^2 .P ;
Substitute n^2 .P into the difference equation and simplify it, thus we obtain
P = 7/2
Hence, the particular solution is
an(p) = n^2 .7/2
(i) For the difference equation an – 5 an–1 + 6 an–2 = 3n + n; since 2 and 3 are its charac-
teristic roots. Here, f(n) = 3n + n. So, the form of f(n) is the combination of βn (where β = 3) and
a polynomial of degree 1. Thus, corresponding form of particular solution is the combination of
P 1 .n.3n and (P 2 .n +P 3 ) respectively.
or P 1 .n.3n + (P 2 .n +P 3 ) ;
Substitute particular solution into the difference equation and after simplification we
obtain
P 1 = – 3, P 2 = 1/3 and P 3 = 7/9;
Hence, the particular solution is
an(p) = (– 3).n.3n + (1/3).n + 7/9 ;
Now we summarize that section 3.3 discussed the method of finding the solution of
LRRCC in terms of the homogeneous solution and the particular solution. The complete solu-
tion of the LRRCC is the combination of both homogenous solution and the particular solution
which is a discrete numeric function. In the homogenous solution the unknown coefficients
can be determined using base condition/s. For a kth order difference equation the k-unknown
coefficients C 1 , C 2 , ...... Ck in the homogenous solution can be determined by using base condi-
tions an 0 , an 1 , ........ank–1 (where anj = anj–1+1 for 0 < j ≤ nk–1).
Let complete solution is of the form
an = an(h) + an(p)

i.e. an =

i

k

ii

n n

=

∑ +

1

CPα () (3.20)

Substituting the values of base conditions in equation (3.20), we obtain k linear equa-
tions, i.e.

an 0 =

i

k

ii

n n

=

∑ +

1

0

CPα () 0 ;

an 1 =

i

k

ii

n n

=

∑ +

1

1

CPα () (^1) ;
.
.
ank–1 =
i
k
ii
nk
nk

−
∑ + −
1
1
CPα () (^1) ;