Introduction
Problems encountered so far have mostly been static in that the quantities
and equations involved are for a particular period of time. For instance, the
current price of a good depends on the current demand of consumers.
However, it may be that this year’s price for a good, such as a car, depends
on last year’s demand; or a manufacturing quota for a particular month depends
on the demand in that month in the previous year (or years). This is the concept
of a lagged response. These examples are not static but dynamic situations in
which economic models are viewed as a sequence of discrete periods. The value
of an economic quantity in one period may depend on data from the previous
period, or previous n periods for some integer n > 0.
Difference equations are used to analyse dynamic models. An nth order
difference equation, for instance, might express the price of a good as a function
of the demands in the previous n years.
12.2 Difference Equations
Consider a sequence X0, X1, . . . of quantities, which we denote simply by
Xt. By a sequence we mean a list in a specific order. In the economic situations
that will concern us, t denotes time measured in discrete units 0, 1, 2, . . ..
261
262 Elements of Mathematics for Economics and Finance
In this case, the period before time t + 1 starting at time t is referred to as
period t.
Thus, period 0 is the initial period. The sequence Xt can be regarded
as the values of a ‘step’ function X of a continuous time variable i, where
X(i) = Xt for all i in period t. Thus, X has constant value Xt in period t (see,
for example, Fig. 12.1).
One way of generating a sequence Xt is by using an nth order difference
equation, which relates the general term of the sequence to the previous n
terms. We shall only be concerned with the linear difference equations
(LDEs). They are of the form:
Xt + at−1Xt−1 + at−2Xt−2 + . . . + at−nXt−n = b,
where the coefficients ai and b are constants (independent of t). If b = 0, the
difference equation is said to be homogeneous; otherwise it is inhomogeneous.
By ‘solving’ a difference equation, we mean expressing the function X mentioned
earlier explicitly as a function of t. An nth order difference equation
determines the sequence uniquely if the first n terms of the sequence are specified.
For example, consider the first order difference equation
Xt − 3Xt−1 = −1,
(or Xt = 3Xt−1 − 1) where X0 = 2. Using the difference equation, we can
generate successively the terms of the corresponding sequence Xt:
X1 = 3X0 − 1 = 3 × 2 − 1 = 5,
X2 = 3X1 − 1 = 3 × 5 − 1 = 14,
and so on. To compute the term X100 in this way would mean explicitly evaluating
X1, X2, . . . , X99 successively. This is laborious, but if we solve the linear
difference equation, the computation is simple. Accept for the moment (for reasons
that will be explained later) that the solution of this difference equation
is
Xt =
1
2
(1 + 3t+1).
Substituting t = 0 gives X0 = 1
2 (1 + 31) = 2 (as it should be); t = 1 gives
X1 = 1
2 (1 + 32) = 5, and t = 2 gives X2 = 1
2 (1 + 33) = 14, agreeing with our
previous computations. The term X100 is simply 1
2 (1 + 3101), which is a huge
number and best left expressed in this form.
Difference equations are sometimes known as time series (if t denotes time)
or recurrence relations. The sequence Xt is also known as the time path of
the function X being measured, giving the successive values in time of X.

Linear Difference Equations 263
t
1 2 3 4 5
0
2
4
6
8
X 10
Figure 12.1 Graph representing the solution of the difference equation
Xt = 2Xt−1 − 1 with X0 = 2.
A sequence Xt may be visualised by plotting the graph of the corresponding
step function X. For instance, the sequence Xt given by the linear
difference equation
Xt = 2Xt−1 − 1,
where X0 = 2, has first four terms: 2, 3, 5, 9. This can be represented graphically
as in Fig. 12.1. (On a line segment, the heavy dot represents the end point
that belongs to the graph.)
The values of X change only at discrete values of t. So for instance,
Xt = X0 = 2 for 0 ≤ t < 1,
Xt = X1 = 3 for 1 ≤ t < 2,
Xt = X2 = 5 for 2 ≤ t < 3,
Xt = X3 = 9 for 3 ≤ t < 4,
and so on.
264 Elements of Mathematics for Economics and Finance
12.3 First Order Linear Difference Equations
First order linear difference equations can be dealt with as a special case of
second order linear difference equations. However, it is instructive to see them
analysed from basics.
Consider a general first order linear difference equation written in the form
Xt = aXt−1 + b. (12.1)
This means that each term is a times the previous term and then b is added to
this product. Therefore, Xt−1 = aXt−2 + b, Xt−2 = aXt−3 + b, and so on. It
follows that
Xt = a(aXt−2 + b) + b
= a2Xt−2 + (a + 1)b
= a2(aXt−3 + b) + (a + 1)b
= a3Xt−3 + (a2 + a + 1)b.
Eventually, we have
Xt = atXt−t + (at−1 + at−2 + . . . + a + 1)b.
The term
at−1 + at−2 + . . . + a + 1
is the sum of a geometric series and is equal to
(1 − at)
(1 − a)
if a = 1;
otherwise the sum is equal to
1 + 1 + . . . + 1 + 1(t terms ) = t.
Therefore,
Xt =
⎧⎨
⎩
atX0 +

1 − at
1 − a

b if a = 1,
X0 + bt if a = 1.
If a = 1, we can rewrite the solution (collecting together the terms involving
at) in the form:
Xt = Aat + b
1 − a
, (12.2)
where A is a constant that can be determined from X0. (The example given
earlier in Section 12.2 had a = 3 and b = −1.)
Linear Difference Equations 265
Example 12.1
A bank customer borrows $15,000. Interest is 9.6% per annum on the outstanding
balance. The customer can afford to repay at most $400 each month.
How long will it take to repay the loan?
How much does the customer owe after 1 year?
Solution.
Let Xt be the amount owed after t months. Then X0 = 15,000. At the
end of t months, interest of 9.6
12% = 0.8% of the current balance of Xt−1 is
added and a repayment made of $400, therefore
Xt =

1 +
0.096
12

Xt−1 − 400
= 1.008Xt−1 − 400.
From (12.2) we have (with a = 1.008 and b = −400):
Xt = A(1.008)t − 400
1 − 1.008
= A(1.008)t + 50,000,
where A is a constant.
Since X0 = 15,000, then putting t = 0 in this equation gives (as 1.0080 = 1)
15,000 = A + 50,000.
Therefore A = −35,000 and so
Xt = −35,000(1.008)t + 50,000.
This is explicitly the balance owing after t months. The balance is 0 at
time t if
35,000(1.008)t = 50,000,
that is
(1.008)t =
50,000

35,000

10
7 .
Taking natural logarithms of both sides:
t ln(1.008) = ln

10
7

.
Therefore
t =
ln
10
7

ln(1.008) ,
which is approximately 44.76 months. Therefore, the loan will be paid off
at the end of the 45th month.
266 Elements of Mathematics for Economics and Finance

X12 = −35,000(1.008)12 + 50,000 ≈ $11,488.15.
Example 12.2
A bank savings account pays 5% per annum interest. Initially, a saver deposits
£1,000. After 10 years, what will be the value of this customer’s savings account
if
no further deposits are made;
£100 is deposited at the end of each year?
Solution. Let Xt be the value of the savings account after t years.
In this case
Xt = (1+0.05)Xt−1 = 1.05Xt−1.
Then from (12.2), with a = 1.05 and b = 0, we have
Xt = A(1.05)t,
where A is a constant.
Since X0 = 1,000, then 1,000 = X0 = A(1.05)0 = A. Therefore,
Xt = 1,000(1.05)t.
We want
X10 = 1,000(1.05)10 ≈ £1,628.89.
The difference equation is now
Xt = (1.05)Xt−1 + 100.
Then from (12.2), with a = 1.05 and b = 100, we have
Xt = A(1.05)t +
100
1 − 1.05
= A(1.05)t − 2,000.
Since 1,000 = X0, then
1,000 = X0 = A(1.05)0 − 2,000 = A − 2,000.
Consequently, A = 3,000 and
Xt = 3,000(1.05)t − 2,000.
The value after 10 years is
X10 = 3,000(1.05)10 − 2,000 ≈ £2,886.68.
Linear Difference Equations 267
12.4 Stability
Suppose an economic model has associated with it the first order linear difference
equation
Xt = aXt−1 + b (a = 1).
We showed that this has solution:
Xt = Aat + b
1 − a
,
where A is a constant (independent of time t).
If −1 < a < 1, then at tends to 0 as t tends to infinity (i.e., t increases
indefinitely). Then Xt converges to the value b
1−a , called the equilibrium
value.
The convergence is oscillatory if a is negative and is uniform if a is
positive. The model or difference equation in this case is said to be stable.
See Figs. 12.2 and 12.3, respectively.
If a < −1 or a > 1, then Xt diverges in that, numerically, Xt increases
without bound. If a is negative, the divergence is oscillatory, whereas if a
is positive, the divergence is uniform. See Figs. 12.4 and 12.5, respectively.
The model or difference equation is unstable in this case.
Example 12.3
Solve the following linear difference equations:
Xt = −0.5Xt−1 + 0.25, where X0 = 0.5,
Xt = 1
3Xt−1 + 1, where X0 = 1
2 ,
Xt = −2Xt−1 + 3, where X0 = 3,
Xt = 3Xt−1 + 5, where X0 = 3.5.
In each case, comment on stability and display the solution graphically for
0 ≤ t < 5.
Solution.
1.
Xt = A(−0.5)t +
0.25
1 − (−0.5)
= A(−0.5)t +
0.25
1.5
= A(−0.5)t +
1
6.
268 Elements of Mathematics for Economics and Finance
t
0 1 2 3 4 5
0
0.2
0.4
X 0.6
Figure 12.2 Graph of the solution of Example 12.3.1 showing oscillatory
convergence to the equilibrium value.
Since 0.5 = X0 = A + 1
6, then A = 1
3 . Therefore
Xt =
1
3
(−0.5)t +
1
6.
This is a stable linear difference equation, with oscillatory convergence to
the equilibrium value 1
6 as shown in Fig. 12.2.
2.
Xt = A

1
3
t
+
1
1 − 1
3
= A

1
3
t
+
3
2.
Since 1
2 = X0 = A + 3
2, then A = −1. Therefore,
Xt =
3
2
−

1
3
t
.
This is a stable linear difference equation, uniformly converging to the
equilibrium value 3
2 as shown in Fig. 12.3.
3.
Xt = A(−2)t +
3
1 − (−2)
= A(−2)t + 1.
Linear Difference Equations 269
t
1 2 3 4 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
X
Figure 12.3 Graph of the solution of Example 12.3.2 showing uniform
convergence to the equilibrium value.
Since 3 = X0 = A + 1, then A = 2 and
Xt = 2(−2)t + 1.
This is an unstable linear difference equation. The divergence is oscillatory
(see Fig. 12.4).
4.
Xt = A(3)t +
5
1 − 3
= A(3)t − 2.5.
Since 3.5 = X0 = A − 2.5, then A = 6. Therefore,
Xt = 6(3)t − 2.5.
This is an unstable linear difference equation. The divergence is uniform
(see Fig. 12.5).
270 Elements of Mathematics for Economics and Finance
t
1 2 3 4 5
-10
0
10
20
30
X
Figure 12.4 Graph of the solution of Example 12.3.4 showing oscillatory
divergence.
12.5 The Cobweb Model
The Cobweb model is an economic model for analysing periodic fluctuations
in price, supply, and demand that oscillate towards equilibrium. It is assumed
that the quantities involved change only at discrete time intervals and that
there is a time lag in the response of suppliers to price changes.
For instance, the supply this year of a particular agricultural product depends
on the price obtained from the previous year’s harvest. The demand for
the produce will depend of course on this year’s price. Another example is that
of package holidays. The holiday company’s supply of holidays for this season
will depend on the prices obtained for last season’s.
In general, we assume that the supply function at time t for a single good
is
QS,t = aPt−1 + b.
Here QS,t is the supply at time t and Pt−1 the price at time t−1 (the previous
period). The demand equation is
QD,t = cPt + d.
Linear Difference Equations 271
t
1 2 3 4 5
0
100
200
300
400
X 500
Figure 12.5 Graph of the solution of Example 12.3.3 showing uniform
divergence.
Here a, b, c, d are constants, with a > 0 and c < 0. Initially t = 0, and then t
increases one unit at a time.
Assuming equilibrium in period t, we have QD,t = QS,t. That is,
aPt−1 + b = cPt + d
or
Pt = a
c
Pt−1 + b − d
c
.
This is a first order linear difference equation, with a
c < 0, since a is positive
and c is negative. The sequence Pt generated gives the equilibrium prices
for each period. Since a/c = 0, as a/c is negative, we can solve the difference
equation using (12.2) to obtain an expression for Pt in the form
Pt = A
a
c
t
+

b − d
c

/

1 − a
c

= A
a
c
t

b − d
c − a
where A is a constant.
272 Elements of Mathematics for Economics and Finance
If −1 < a c < 0, we have stability and oscillatory convergence to an equilibrium value Pe = b − d c − a . (12.3) Example 12.4 A car manufacturer’s supply and demand functions at time t for a particular car model are QS,t = 3Pt−1 − 12, QD,t = −4Pt + 28. Show that over time, the car’s price will converge and give the equilibrium value. Solution. For equilibrium in period t, 3Pt−1 − 12 = −4Pt + 28, which simplifies to Pt = −3 4 Pt−1 + 10. (12.4) Since 0 > −3
4 > −1, the prices Pt converge to the equilibrium value Pe, which
we can compute using the formula obtained earlier (equation (12.3)) or in §12.3.
Using the latter with a = −3
4 and b = 10, we have
Pe = b

1 − a

10
1 −

−3
4
=
40
7
= 5
5
7.
Another way to compute Pe, when we know there is convergence, is to note
that in the limit as t tends to infinity, Pt = Pt−1 = Pe. Therefore,
Pe =

−3
4

Pe + 10.
Then

1 + 3
4

Pe = 10, from which it follows that Pe = 55
7 , as before.
Observe that the equilibrium price in this problem is independent of the value
of P0. The general solution of linear difference equation (12.4) is
Pt = A

−3
4
t
+
10
1 −
3
4
= A

−3
4
t
+
40
7
where A is a constant.
From this equation it is again clear that the equilibrium price is 40
7 and that
this does not depend on the value of A (which depends on the value of P0).

Linear Difference Equations 273
12.6 Second Order Linear Difference Equations
A time path sequence is uniquely determined, given a second order linear difference
equation and the values of the first two terms.
Example 12.5
Find the first five terms of the sequence Xt given by
Xt − 5Xt−1 + 3Xt−2 = 1,
given that X0 = 1, X1 = 3.
Solution. Rearranging the linear difference equation gives
Xt = 5Xt−1 − 3Xt−2 + 1.
Therefore,
X2 = 5X1 − 3X0 + 1 = 5 × 3 − 3 × 1 + 1 = 13,
X3 = 5X2 − 3X1 + 1 = 5 × 13 − 3 × 3 + 1 = 57,
X4 = 5X3 − 3X2 + 1 = 5 × 57 − 3 × 13 + 1 = 247.
Therefore, the first five terms in order are 1, 3, 13, 57, 247.
The general second order linear difference equation is of the form
Xt + aXt−1 + bXt−2 = c (12.5)
where we assume a, b, c are constants (independent of t).
First order linear difference equations may be considered as the special case
b = 0. This explains the remark made in Section 12.3 when discussing first
order linear difference equations: that they are special cases of second order
linear difference equations. However, we gave a self-contained account of the
general solution for the first order case. The general solution for the second
order linear difference equations is a little more involved.
The associated homogeneous linear difference equation of (12.5) is
Xt + aXt−1 + bXt−2 = 0. (12.6)
If Ut, Vt are sequences satisfying the linear difference equation (12.5), then
Ut + aUt−1 + bUt−2 = c
274 Elements of Mathematics for Economics and Finance
and
Vt + aVt−1 + bVt−2 = c.
Subtracting we have:
Ut − Vt + a(Ut−1 − Vt−1) + b(Ut−2 − Vt−2) = 0
or
Wt + aWt−1 + bWt−2 = 0
where Wt = Ut − Vt. Thus Wt satisfies the homogeneous linear difference
equation (12.6).
It follows that any two solutions of (12.5) differ by a solution of the associated
linear difference equation (12.6). Thus if we manage to find a particular
solution of (12.5) by guesswork or from theory, then the general solution
of an inhomogeneous linear difference equation is obtained by adding the particular
solution to the general solution of the associated homogeneous linear
difference equation (known as the complementary solution). In brief:
General Solution=Particular Solution+Complementary Solution.
(12.7)
This is true generally for linear difference equations and, in particular, for first
order ones. We saw in Section 12.3 that the general solution of a first order
linear difference equation
Xt = aXt−1 + b
is of the form
Xt = Aat + b
1 − a
(if a = 1).
It can be verified that Xt = Aat is the complementary solution (the general
solution of the associated linear difference equation Xt = aXt−1) and that
the sequence Xt, where Xt = b
1−a (for all t), is a particular solution of
Xt = aXt−1 + b since
b
1 − a
= a

b
1 − a

b.
12.6.1 Complementary Solutions
In analogy with the complementary solution for the homogeneous first order
case, suppose we try a similar solution of (12.6), of the form Xt = Aut, where
A is a constant and u a number to be determined. Then
Aut + aAut−1 + bAut−2 = 0.

Linear Difference Equations 275
Dividing throughout by Aut−2 gives
u2 + au + b = 0.
So u is a root of the quadratic equation
x2 + ax + b = 0, (12.8)
known as the characteristic equation of the linear difference equation. Let
its roots be u, v (sometimes called the characteristic roots).
It can be shown that the general solution of the homogeneous linear difference
equation (12.6) is of the form
Xt =

Aut + Bvt if u = v,
(A + tB)ut if u = v,
where A and B are constants.
Example 12.6
Solve the homogeneous linear difference equation
Xt − 7Xt−1 + 10Xt−2 = 0
where X0 = 2, X1 = 13. Determine X10.
Solution. The characteristic equation is
x2 − 7x + 10 = 0
whose roots are 2 and 5. The general solution is therefore
Xt = A2t + B5t
where A and B are constants.
Since X0 = 2, then
2 = X0 = A20 + B50 = A + B.
Similarly, since X1 = 13, then
13 = X1 = A21 + B51 = 2A + 5B.
Thus, we have the simultaneous equations
A + B = 2, (12.9)
2A + 5B = 13. (12.10)
276 Elements of Mathematics for Economics and Finance
Multiplying equation (12.9) by two and subtracting the result from equation
(12.10) gives
3B = 13 − 4 = 9.
Therefore B = 3 and so, from (12.9), A = −1. The general solution is therefore
Xt = −2t + 3(5t).
Then
X10 = −210 + 3(510) = 29,295,851.
Example 12.7
Solve the linear difference equation
Xt − 12Xt−1 + 36Xt−2 = 0
where X0 = 1 and X1 = 8. Determine X9.
Solution. The characteristic equation is x2 − 12x + 36 = 0 or (x − 6)2 = 0.
This equation has two equal roots 6, 6. So the general solution of the linear
difference equation is of the form
Xt = (A + tB)6t
where A and B are constants.
Since X0 = 1, then
1 = X0 = (A + 0)60 = A
and since X1 = 8, then
8 = X1 = (A + 1 × B)61 or 8 = 6(A + B).
Since A = 1, then 8 = 6+6B and therefore B = 1
3 .
The general solution of the linear difference equation is
Xt =

1 + t
3

6t.
Therefore
X9 =

1 +
9
3

69 = 4× 69 = 40,310,784.
Linear Difference Equations 277
12.6.2 Particular Solutions
A particular solution for the general second order linear difference equation
Xt + aXt−1 + bXt−2 = c
where a, b, c are constants is as follows:
Xt =
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
c
1 + a + b
if 1 + a + b = 0;
ct
2 + a
if 1 + a + b = 0 anda = −2;
1
2ct2 if a = −2 andb = 1.
(12.11)
The conditions on a and b in the above are respectively that:
1 is not a characteristic root (i.e., not a root of x2 + ax + b = 0);
One characteristic root is 1 and the other is not;
Both characteristic roots are 1.
That these are particular solutions is easily verified, but we will illustrate this
by example only.
It is not really necessary to remember the formulae for particular solutions.
For a second order linear difference equation with constant coefficients, only
one of the forms Xt = K, Xt = Kt, Xt = Kt2 will work as a solution with K
a constant. So one need only test three possibilities.
For instance, Xt = K (K constant) cannot be a solution of
Xt − 3Xt−1 + 2Xt−2 = 21,
since this would require K−3K+2K = 21 or 0 = 21, which is absurd. However,
trying Xt = Kt (so Xt−1 = K(t − 1), Xt−2 = K(t − 2)) we have:
Kt − 3K(t − 1) + 2K(t − 2) = 21
which simplifies to −K = 21. Therefore, K = −21 and so Xt = −21t is a
solution of the difference equation.
Example 12.8
Find particular solutions of the following difference equations:
Xt + 7Xt−1 + 12Xt−2 = 4,
Xt − 5Xt−1 + 4Xt−2 = 9,
Xt − 2Xt−1 + Xt−2 = 4.
278 Elements of Mathematics for Economics and Finance
Solution.
In this difference equation 1 + a + b = 1 + 7 + 12 = 20 = 0. Therefore,
Xt = c

1 + a + b

20

1
5
is a particular solution. That is, the repeating sequence 1
5 , 1
5 , 1
5 , . . . satisfies
this difference equation. To check this, note that 1
5 + 7 × 1
5 + 12 × 1
5 = 4.

Here 1 + a + b = 1− 5 + 4 = 0 and a = −5 = −2. Then
Xt = ct

a + 2

9t
−5 + 2
= −3t
is a particular solution. Thus X0 = 0, X1 = −3, X2 = −6,X3 = −9 and
so on.

In this case, a = −2, b = 1, c = 4 so a particular solution is
Xt =
1
2
4t2 = 2t2.
The general second order linear difference equation is solved by adding a
particular solution to the complementary solution. We illustrate this in the
next example.
Example 12.9
Solve the following difference equations:
Xt + 7Xt−1 + 12Xt−2 = 4;X0 = 1.2, X1 = 2.2,
Xt − 5Xt−1 + 4Xt−2 = 9;X0 = 0, X1 = 5,
Xt − 2Xt−1 + Xt−2 = 4.X0 = 1, X1 = 2.
Solution. Particular solutions were found for these equations in the previous
example; so we only need the complementary solutions.
The characteristic equation is
x2 + 7x + 12 = 0
whose roots are −3, −4. The complementary solution is therefore
Xt = A(−3)t + B(−4)t
with A, B constants.
Linear Difference Equations 279
Therefore, the general solution (of the given inhomogeneous difference
equation) is obtained by adding the particular solution Xt = 1
5 = 0.2
found in the previous example:
Xt = 0.2 + A(−3)t + B(−4)t.
Since X0 = 1.2, then 1.2 = X0 = 0.2 + A + B, which gives
A + B = 1.
Since X1 = 2.2, then 2.2 = X1 = 0.2 + A(−3) + B(−4), which gives
3A + 4B = −2.
Solving these two simultaneous equations for A and B gives
A = 6, B= −5.
The solution is therefore
Xt = 0.2 + 6(−3)t − 5(−4)t.
The characteristic equation is
x2 − 5x + 4 = 0
with roots 1, 4. The complementary solution is
Xt = A1t + B4t = A + B4t
with A, B constants. A particular solution we found, in Example 12.8.2,
was Xt = −3t. Therefore the general solution is
Xt = A + B(4)t − 3t.
Since X0 = 0, then 0 = X0 = A + B40 −0 = A + B. Therefore, A = −B.
Next X1 = 5, gives 5 = X1 = A + B41 − 3 × 1. That is A + 4B = 8. Since
A = −B, then 3B = 8 and so B = 8
3 and A = −8
3 . The general solution is
therefore
Xt = −8
3
+
8
3
(4t) − 3t.
280 Elements of Mathematics for Economics and Finance
The characteristic equation is
x2 − 2x + 1 = 0
which has two equal roots: 1, 1.
The complementary solution is therefore Xt = A+tB. A particular solution
we found was Xt = 2t2. Therefore the general solution is
Xt = A + tB + 2t2.
Now X0 = 1 gives 1 = A; while X1 = 2 gives
2 = A + 1 × B + 2(1)2 = A + B + 2.
Therefore A+B = 0. Since A = 1, then B = −1. The solution is therefore
Xt = 1− t + 2t2.
Example 12.10
A simplified Samuelson model for a national economy is provided by the following
difference equation, where Xt is the total national income in year t:
Xt − c(1 + w)Xt−1 + cwXt−2 = k.
Here c, w, k are positive constants and c < 1.
Find the general solution for the case c = 0.9, w = 0.5, k = 1, where X0 = 1
and X1 = 1.3.
Calculate the national income in years 10 and 20.
Solution. The difference equation for the given parameters is
Xt − 0.9(1.5)Xt−1 + 0.9(0.5)Xt−2 = 1,
that is
Xt − 1.35Xt−1 + 0.45Xt−2 = 1.
The characteristic equation is
x2 − 1.35x + 0.45 = 0
which has roots 0.75, 0.6. The complementary solution is therefore
Xt = A(0.75t) + B(0.6t)
where A and B are constants.
Linear Difference Equations 281
A particular solution is (see 12.11)
Xt =
1

1 − 1.35 + 0.45

1
0.1
= 10.
The general solution is therefore of the form
Xt = A(0.75t) + B(0.6t) + 10.
Since X0 = 1, then 1 = X0 = A + B + 10, which gives
A + B = −9. (12.12)
Similarly, since X1 = 1.3, then 1.3 = X1 = A(0.75) + B(0.6) + 10. Therefore
0.75A + 0.6B = −8.7. (12.13)
Multiplying equation (12.12) by 0.6 and subtracting the result from equation
(12.13) gives
0.15A = −8.7 + 0.6 ×9 = −3.3.
It follows that A = −22 and, from (12.12), B = −9 − A = −9 + 22 = 13. The
general solution is therefore
Xt = −22(0.75t) + 13(0.6t) + 10. (12.14)
The national income in year 10 is
X10 = −22(0.7510) + 13(0.610) + 10 = 8.840 (to 3 decimal places).
(This is almost 9 times the first year’s national income X0 = 1.)
In year 20, the national income is
X20 = −22(0.7520) + 13(0.620) + 10 = 9.931 (to 3 decimal places).
If in the above example, we compute Xt for larger and larger t, the values
approach a limiting value of 10. This can be seen from equation (12.14). As t
increases, i.e., tends to ∞, the terms involving 0.75t and 0.6t tend to 0, so that
Xt tends to the value 10; see Fig. 12.6.
282 Elements of Mathematics for Economics and Finance
t
5 10 15 20 25
0
2
4
6
8
10
X 12
Figure 12.6 Graph of the solution Xt of Example 12.10 in continuous
time, illustrating the simplified Samuelson model converging to the value 10.
12.6.3 Stability
A second order linear difference equation
Xt + aXt−1 + bXt−2 = c
where a, b, c are constants, is stable if both its characteristic roots are strictly
between −1 and 1. Otherwise it is divergent.
If the characteristic roots are α, β, this condition is the same as requiring
−1 < α < 1 and −1 < β < 1. The general solution of the difference equation
must then be one of the forms:
Xt =

Aαt + Bβt + C if α = β,
(A + tB)αt + C if α = β,
where A, B are constants determined by the initial conditions (the values X0
and X1) and where
C = c
1 + a + b
.
The particular solution is Xt = C and the complementary solution is either
Xt = Aαt + Bβt or Xt = (A + tB)αt.

Linear Difference Equations 283
In either case; if −1 < α < 1 and −1 < β < 1, the complementary solution
will tend to 0 as t tends to ∞. Then the general solution will tend to the
particular solution.
The three difference equations in the second example of the previous section
are divergent. The difference equation in the simplified Samuelson model example
is stable and the solution converges on the particular solution Xt = 10,
as is illustrated in Fig. 12.6.
When a characteristic root is 1 or −1, convergence is still possible in certain
cases. One is when the sequence is constant: Xt = k for all t, where k is constant.
The other case occurs when one characteristic root is 1 and the other one α
satisfies −1 < α < 1.
In this case the homogeneous difference equation is
Xt − (1 + α)Xt−1 + αXt−2 = 0
and the general solution is of the form
Xt = A(1t) + B(αt) = A + Bαt
where A, B are constants.
Since −1 < α < 1, then αt tends to the value 0 as t tends to ∞ and
therefore Xt converges to the solution Xt = A. The convergence is oscillatory
if −1 < α < 0.
Example 12.11
Consider the difference equation
2Xt − Xt−1 − Xt−2 = 0
(So the value of Xt in period t is the average of its values in the previous two
periods, t ≥ 2.)
Find the general solution and comment on stability.
Solution. The characteristic equation is 2×2 − x − 1 = 0, whose roots are
1, −1
2 . The general solution is therefore
Xt = A + B

−1
2
t
where A, B are constants.
As t tends to ∞,

−1
2
t will tend to the value 0 and therefore Xt converges
to the solution Xt = A. The value of A can easily be expressed in terms of X0
and X1. Since X0 = A + B and X1 = A − 1
2B, then A = 1
3 (X0 + 2X1).
284 Elements of Mathematics for Economics and Finance
EXERCISES
12.1. Solve the following difference equations, commenting on stability.
a) Xt = −3Xt−1 + 5; X0 = 2
b) Xt = −1
3Xt−1 + 5; X0 = −1
4
c) Xt = 2Xt−1 − 8; X0 = 9
d) Xt = 3
2Xt−1
e) Xt = 2
3Xt−1 − 10
f) Xt = 0.95Xt−1.
Evaluate X10 in each case.
12.2. Calculate the equilibrium price of a single good in an isolated market
where the supply, QS,t, demand QD,t and price Pt in period t are
given by
QS,t = 4Pt−1 − 5,
QD,t = −5Pt + 10.
Show that the prices Pt converge and find the equilibrium price.
12.3. Solve the difference equations:
a) Xt = Xt−1 + 3; X0 = 0
b) Xt = 3− Xt−1; X0 = 1
In each case, sketch the graph of the function Xt for t = 0, 1, 2, 3, 4.
12.4. Solve the following difference equations, commenting on stability in
each case:
a) Xt − 6Xt−1 + 9Xt−2 = 2; X0 = 1.5, X1 = 2
b) Xt + 2Xt−1 − 3Xt−2 = 7; X0 = 0, X1 = 4
c) Xt − 2Xt−1 + Xt−2 = 6; X0 = 1, X1 = 3
d) Xt − 2Xt−1 − 15Xt−2 = 8; X0 = 0, X1 = 1
e) 10Xt − 3Xt−1 − 4Xt−2 = 18; X0 = 2, X1 = 0
f) 4Xt − Xt−2 = 9; X0 = 5, X1 = 1
g) 3Xt − 4Xt−1 + Xt−2 = 0; X0 = 4, X1 = 0
12.5. Solve the difference equations:
Linear Difference Equations 285
a) 2Xt − Xt−1 − Xt−2 = 0; X0 = 0, X1 = 1
b) 2Xt − Xt−1 − Xt−2 = 1; X0 = 0, X1 = 1
12.6. A simple model for total national income Xt in year t satisfies the
difference equation
Xt − 1.26Xt−1 + 0.36Xt−2 = 1.
Show that Xt converges and give the equilibrium value.
12.7. A population model for a population Xt in year t is given by the
difference equation
9Xt − 9Xt−1 + 2Xt−2 = 100.
Show that the population converges and give its equilibrium value.
Evaluate X10, given that X0 = 48 and X1 = 49.
12.8. A second order linear difference equation Xt + aXt−1 + bXt−2 = c
with b = 0 may be considered first order. Use this to deduce the
solution of a general first order linear difference equation as given in
Section 12.3, from the theory of solutions of second order equations
given in Section 12.6.

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35,000

1 − a

1 + a + b

20

a + 2

1 − 1.35 + 0.45

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