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In-Class Exercise

Pb. 2.8 Find the particular solution of the following second-order difference
equation:

y(k) – 3y(k – 1) + 2y(k – 2) = (3) k for k > 0

2.4.3 General Solution
The general solution of a linear difference equation is the sum of its homoge-
neous solution and its particular solution, with the constants adjusted, so as
to satisfy the initial conditions. We illustrate this general prescription with an
example.

Example 2.3
Find the complete solution of the first-order difference equation:

y(k + 1) + y(k) = k

with the initial condition y(0) = 0.

Solution: First, solve the homogeneous equation y(k + 1) + y(k) = 0. The char-
acteristic polynomial is λ + 1 = 0; therefore,

y homog. = C(–1) k

The particular solution can be obtained from the above table. Noting that the
input signal has the functional form k , with M = 1, then the particular solu-
M
tion is of the form:

y partic. = B k + B 1 (2.27)
0
Substituting back into the original equation, and grouping the different pow-
ers of k, we deduce that:
B = 1/2 and B = –1/4
1
0
The complete solution of the difference equation is then:

yk() = C(−1 ) + 2 k − 1
k
4

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