We want to find a general solution of this equation. The general solution will need to
have two arbitrary constants so that a specific solution can be found once
y
0
and
y
1
are
given (just as the general solution of a firstorder equation contains one arbitrary
constant).
As the equation is linear and homogeneous, two very useful properties apply (compare
this with homogeneous systems of linear equations,
A
x
=
0
):
a constant multiple of a solution is a solution,
the sum of two solutions is a solution.
164
11
11.6. Homogeneous secondorder difference equations
Activity 11.5
Show this. Given two sequences
x
t
and
z
t
which satisfy the
homogeneous difference equation
y
t
+2
+
a
1
y
t
+1
+
a
2
y
t
= 0, show that
x
t
+
z
t
and
cx
t
,
where
c
is a constant, also satisfy this equation
Therefore, it follows that if we know two solutions
x
t
and
z
t
of the difference equation,
then
y
t
=
Ax
t
+
Bz
t
is also a solution for any constants
A, B
∈
R
.
We next set about finding solutions. Knowing what we do about the general solution of
the homogeneous firstorder equation (a geometric progression), let’s try a solution of
the form
y
t
=
m
t
where
m
is some constant to be determined.
Substituting
y
t
=
m
t
into the difference equation we obtain
y
t
+2
+
a
1
y
t
+1
+
a
2
y
t
=
m
t
+2
+
a
1
m
t
+1
+
a
2
m
t
=
m
t
(
m
2
+
a
1
m
+
a
2
) = 0
.
If
m
= 0, we get
y
t
= 0, so we ignore this possibility. Then
y
t
=
m
t
is a solution of the
difference equation if and only if
m
satisfies
m
2
+
a
1
m
+
a
2
= 0. This equation,
z
2
+
a
1
z
+
a
2
= 0. is known as the
auxiliary equation
.
Let’s look at an example.
Example 11.6
We find the general solution of the difference equation
y
t
+2

5
y
t
+1
+ 6
y
t
= 0
.
The auxiliary equation is
z
2

5
z
+ 6 = 0
.
This equation factors,
z
2

5
z
+ 6 = (
z

2)(
z

3) = 0, with solutions
z
= 2 and
z
= 3. Therefore both
x
t
= 2
t
and
z
t
= 3
t
are solutions and the general solution is
y
t
=
A
(2
t
) +
B
(3
t
)
,
A, B
∈
R
.
Now suppose we are given initial conditions
y
0
= 1 and
y
1
= 4. Then the difference
equation determines all remaining numbers in the sequence. We have
y
2
= 5
y
1

6
y
0
= 5(4)

6(1) = 14,
y
3
= 5
y
2

6
y
1
, and so on.
The specific solution of the difference equation with these initial conditions is found
by substituting
t
= 0 and
t
= 1 into the general solution. We have
y
0
=
A
+
B
= 1
and
y
1
= 2
A
+ 3
B
= 4
.
Solving these equations for
A
and
B
, we find
B
= 2 and
A
=

1. Therefore the
solution of
y
t
+2

5
y
t
+1
= 6
y
t
= 0 with initial conditions
y
0
= 1 and
y
1
= 4 is
y
t
=

(2
t
) + 2(3
t
)
.
We can immediately check that
y
0
=

2
0
+ 2(3
0
) =

1 + 2(1) = 1 and
y
1
=

2 + 2(3) = 4 as required. A further check is to find that
y
2
=

(2
2
) + 2(3
2
) =

4 + 18 = 14 as we found earlier.
165
11
11. Difference equations
The auxiliary equation
z
2
+
a
1
z
+
a
2
= 0 of a secondorder difference equation is a
quadratic equation and the general solution to the difference equation depends on
whether the auxiliary equation has two distinct solutions, or just one solution, or no
(real) solutions. Thus, the form of general solution depends on the value of the
discriminant,
a
2
1

4
a
2
. We consider each case in turn.