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y homog. (k) = λ k (2.23)

Substituting in the homogeneous equation, we obtain the following algebraic
equation:

N
∑ a λ kj − = 0 (2.24)
j
j=0

or

λ kN− a ( λ + a λ N−1 + a λ N−2 +…+ a λ + a ) = 0 (2.25)
N
0 1 2 N−1 N
The polynomial in parentheses is called the characteristic polynomial of the
system. The roots can be obtained analytically for all polynomials up to order
4; otherwise, they are obtained numerically. In MATLAB, they can be
obtained graphically when they are all real, or through the roots command
in the most general case. We introduce this command in Chapter 5. In all the
following examples in this chapter, we restrict ourselves to cases for which
the roots can be obtained analytically.
If we assume that the roots are all distinct, the general solution to the homo-
geneous difference equation is:

y k () = C λ k + C λ k +…+ C λ k (2.26)
homog. 1 1 2 2 N N
where λ , λ , λ , …, λ are the roots of the characteristic polynomial.
N
3
1
2
Example 2.2
Find the homogeneous solution of the difference equation
y(k) – 3y(k – 1) – 4y(k – 2) = 0

Solution: The characteristic polynomial associated with this equation leads
to the quadratic equation:
λ – 3λ – 4 = 0
2
The roots of this equation are –1 and 4, respectively. Therefore, the solution of
the homogeneous equation is:
k
y homog. (k) = C (–1) + C (4) k
2
1
The constants C and C are determined from the initial conditions y(1) and
2
1
y(2). Substituting, we obtain:

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