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C =− 4 y 1 +() y 2() and C = y 1 +( ) y 2()
1 2
5 5 20
NOTE If the characteristic polynomial has roots of multiplicity m, then the
portion of the homogeneous solution corresponding to that root can be writ-
ten, instead of C λ , as:
k
1
C λ + C k λ +…+ C ( m) k m 1− λ k
2 ( )
k
1 ()
k
1 1 1
In-Class Exercises
Pb. 2.6 Find the homogeneous solution of the following second-order dif-
ference equation:
y(k) = 3y(k – 1) – 2y(k – 2)

with the initial conditions: y(0) = 1 and y(1) = 2. Then check your results
numerically.
Pb. 2.7 Find the homogeneous solution of the following second-order dif-
ference equation:

y(k) = [2 cos(θ)]y(k – 1) – y(k – 2)

with the initial conditions: y(–2) = 0 and y(–1) = 1. Check your results
numerically.

2.4.2 Particular Solution
The particular solution depends on the form of the input signal. The follow-
ing table summarizes the form of the particular solution of a linear equation
for some simple input functions:

Input Signal Particular Solution
A (constant) B (constant)
AM k BM k
Ak M B 0 k + B 1 k M–1 + … + B M
M
{A cos(ω 0 k), A sin(ω 0 k)} B 1 cos(ω 0 k) + B 2 sin(ω 0 k)

For more complicated input signals, the z-transform technique provides the
simplest solution method. This technique is discussed in great detail in
courses on linear systems.

© 2001 by CRC Press LLC