If y(t) is a function of a variable t, an equation of the form
dy
dt
= ay + b (13.1)
where a, b are constants, not both 0, is known as a first order linear differential
equation. The equation is homogeneous if b = 0; otherwise it is
inhomogeneous.
The associated linear homogeneous differential equation of (13.1) is
dy
dt
= ay. (13.2)
If y1(t), y2(t) are solutions of (13.1), then y
1 = ay1 +b and y
2 = ay2 +b, which
implies y
1
− y
2 = a(y1 − y2). Let y = y1 − y2. Then y = y
1
− y
2 and therefore
y = ay, which means y is a solution of (13.2).
It follows that any two solutions of equation (13.1) differ only in a solution
of (13.2), the associated homogeneous equation. Therefore, the general solution
of y = ay+b is obtained by adding a particular solution of that equation to
the general solution of the homogeneous equation y = ay. The general solution
of (13.2) is known as the complementary solution of (13.1). The analogy
with the theory of difference equations is clear (see Section 12.6).
To find a particular solution of (13.1), try one that does not change with
t (i.e., one that is time invariant if t denotes time). Try y = k, where k is a
constant to be determined. In this case dy
dt = 0, since k is constant, and so for
(13.5) to hold, we require 0 = ak + b or k = −b
a (if a = 0). Therefore,
y = −b
a
- Differential Equations 289
is a particular solution if a = 0. Next, we solve the homogeneous equation
dy
dt
= ay.
Since
dt
dy
1
dy
dt
,
(see (6.10) then
dt
dy
1
ay
and so
t =
1
ay
dy =
1
a
1
y
dy =
1
a
ln y + k,
where k is a constant. Therefore, putting c = ka, this simplifies to
ln y = at − c
y = eat−c = e
−ceat
(using the product rule for indices (1.11)). Therefore,
y = Aeat
where A(= e−c) is a constant.
Therefore, the general solution of
dy
dt
= ay + b,
if a = 0, is
y = Aeat − b
a
where A is a constant.
In the case a = 0, equation (13.1) reduces to
dy
dt
= b
whose solution is y =
bdt = b
dt = bt+K (remember b is a constant), where
K is a constant of integration. To sum up:
The general solution of the equation dy
dt
= ay + b is:
y =
⎧⎨
⎩
Aeat − b
a
if a = 0,
bt + K if a = 0,
where A and K are constants.
290 Elements of Mathematics for Economics and Finance
Example 13.1
Solve the differential equation
dy
dt
= (3.4)y + 17
where y = 3 when t = 0.
Solution. Using the general solution in the box above,
y = Ae3.4t − 17
3.4
= Ae3.4t − 5.
When t = 0, 3 = y = Ae0 −5 = A − 5 so that A = 8 and so the solution is
y = 8e3.4t − 5.
Example 13.2
A model for the population y(t), in millions, of some country at time t states
that the rate of change of the population is given by
dy
dt
= −0.05y + 4.5.
The population at time t = 0 is 100 million.
- Evaluate y(10), correct to 2 decimal places.
- Find the value of t for which y(t) = 91, correct to 1 decimal place.
Solution. We are given that
dy
dt
= −0.05y + 4.5.
The solution is
y = Ae
−0.05t − 4.5
(−0.05)
= Ae
−0.05t + 90. - Since 100 = y(0) = Ae0 + 90 = A + 90, then A = 10, so
y(t) = 10e
−0.05t + 90.
Therefore,
y(10) = 10e
−0.5 + 90 = 96.07 (correct to 2 decimal places). - Differential Equations 291
- If 91 = 10e−0.05t + 90, it follows that 1 = 10e−0.05t, which gives
e
−0.05t = 0.1.
Take the natural logarithm of each side:
ln 0.1 = lne
−0.05t = −0.05t.
Therefore,
t = − 1
0.05
ln 0.1 = 46.1 (correct to 1 decimal place).
A sketch of this solution is shown in Fig. 13.1. The dashed line corresponds to
the line y = 90.
0 10 20 30 40
80
85
90
95
100
y
t
Figure 13.1 The graph of the solution of Example 13.2 on population
decline. The population converges to 90 million.
292 Elements of Mathematics for Economics and Finance
13.2.1 Stability
If α is any positive number, e−αt will tend to 0 and eαt will tend to ∞ (increase
without bound) as t tends to ∞. It follows that the solution
y = Aeat − b
a
of the differential equation
dy
dt
= ay + b (a = 0),
will converge on the equilibrium value −b
a as t tends to ∞ when a < 0. In this case, we say the solution of the equation is stable. Solutions that diverge are said to be unstable; for instance, when a > 0.
In the case a = 0, the differential equation is dy
dt = b, where b = 0. The
general solution y = bt+K, where K is a constant, is evidently divergent since
bt will tend to ±∞ with t. To sum up:
The solution of the differential equation dy
dt
= ay + b is - unstable if a ≥ 0;
- stable if a < 0. In this case the solution y converges on the particular solution −b/a as equilibrium value. The solution in Example 13.1 is unstable. The solution in Example 13.2 is stable and the equilibrium value is 90. This means the population converges towards this value as t increases (see Fig 13.1). It takes almost 46 years to reach 91 million (see the second part of Example 13.2). However, by decreasing one of the parameters, namely changing 4.5 to 3.85, it takes less than 10 years to fall from 100 to 91 million. Note the equilibrium value −b a is never realized by y = Aeat − b a if A = 0. This is because eat never takes the value 0, since ex > 0 for any number x.
13.3 Nonlinear First Order Differential
Equations
Nonlinear differential equations are more difficult to analyse. For these equations,
there are specialised techniques depending on the type of equation. - Differential Equations 293
One such equation that occurs in economics is the Bernoulli equation
dy
dt
= ay + byn (13.3)
where a, b, n are constants and n > 1.
This can be solved by linearizing it in the following way. Let z = y1−n.
Then by the chain rule (6.8)
dz
dt
= dz
dy
× dy
dt
= d(y1−n)
dy
× dy
dt
= (1 − n)y
−n dy
dt
.
Therefore,
y
−n dy
dt
1
(1 − n)
dz
dt
.
Multiplying equation (13.3) throughout by y−n gives
y
−n dy
dt
= ayy
−n + byny
−n.
That is
1
(1 − n)
dz
dt
= ay1−n + b = az + b.
Therefore,
dz
dt
= (1 − n)az + (1 − n)b,
which is a linear first order differential equation. This can be solved by the
method discussed earlier in Section 13.2.
Example 13.3
Solve
dy
dt
= y − 2y2,
given that y(0) = 1
5 .
Solution. This is the Bernoulli equation (13.3) with a = 1, b = −2 and n = 2.
Let z = y1−n = y1−2 = y−1. Then following through the above technique, the
given differential equation transforms to
dz
dt
= −z + 2.
The general solution is
z = Ae
−t − 2
(−1)
= Ae
−t + 2.
294 Elements of Mathematics for Economics and Finance
Since z = y−1, then
y
−1 = Ae
−t + 2.
We are given that y = 1
5 when t = 0. Therefore,
1
5
−1
= Ae0 + 2 = A + 2.
It follows that A = 3, y−1 = 3e−t + 2, and therefore
y =
1
3e−t + 2.
13.3.1 Separation of Variables
If a first degree differential equation can be expressed in the form
f(y)dy
dt
= g(t),
where f is a function only of y and g a function only of t, we can sometimes
solve the equation by the method of separation of variables.
The technique is more easily understood if we treat dy and dt as individual
quantities (called differentials – see Appendix A) whose ratio is the derivative
dy
dt . For the purposes of our current discussion, it is enough just to accept dy
and dt can be regarded as separate entities. The usefulness of this idea will
become apparent from the following examples.
Example 13.4
Solve the differential equation
y2 dy
dt
= 8t + 1,
given that y(0) = 6.
Solution. Write the equation as
y2dy = (8t + 1)dt.
Integrate both sides:
y2dy =
(8t + 1)dt.
- Differential Equations 295
Then
1
3y3 = 4t2 + t + K,
where K is a constant.
When t = 0, y = 6 and so 1
3 (63) = 0+0+K, which gives K = 72. It follows
that
1
3y3 = 4t2 + t + 72
which can also be written as
y3 = 3(4t2 + t + 72).
Example 13.5
Solve the equation
dy
dt
1
2y3t2,
given that y(0) = 1.
Solution. Rearrange the equation to obtain
2y
−3dy = t2dt.
Integrating both sides:
2
y
−3dy =
t2dt,
gives
2
−3 + 1y
−3+1 =
1
3t3 + K,
where K is a constant. Then
−y
−2 =
1
3t3 + K.
Since y = 1 when t = 0, then K = −1. Therefore
−y
−2 =
1
3t3 − 1.
Multiply throughout by −3 to get
3y
−2 = −t3 + 3
or
3
y2 = 3− t3.
We can also write this as
y2 =
3
3 − t3 .
296 Elements of Mathematics for Economics and Finance
Example 13.6
Solve the equation
t
dy
dt
= y2,
where y(1) = −1
2 .
Solution. Write the equation in the form
y
−2dy = t
−1dt
and integrate both sides to obtain
y
−2dy =
t
−1dt.
That is,
−y
−1 = lnt + K (13.4)
where K is a constant. When t = 1, y = −1
2; so we have
−
−1
2
−1
= ln1+K = 0+K.
Therefore 2 = K, and substituting this in (13.4) gives
−y
−1 = lnt + 2
which can be rearranged as
y = − 1
2 + lnt
.
13.4 Second Order Linear Differential Equations
The general second order differential equation is of the form
d2y
dt2 + a
dy
dt
- by = c, (13.5)
which can also be written as
y
+ ay
+ by = c.
- Differential Equations 297
Here a, b, c are constants,
y
= dy
dt
and y
= d2y
dt2 .
The equation is homogeneous if c = 0; otherwise it is inhomogeneous.
The associated homogeneous differential equation to (13.5) is
d2y
dt2 + a
dy
dt
- by = 0. (13.6)
As in the first order case, it can easily be shown that any two solutions of
equation (13.5) differ by a solution of (13.6). Thus, the general solution of
(13.5) is any particular solution of (13.5) plus the general solution of (13.6).
As before, the general solution of the associated homogeneous differential
equation is known as the complementary solution of (13.5).
13.4.1 The Homogeneous Case
Consider the homogeneous linear differential equation (13.6). The general second
order homogeneous linear difference equation had solutions of the form
Xt = Aαt, where A, α are constants. So we might try solutions of this form for
the differential equations case.
However, as the function ex is easier to differentiate than the general exponential
αx (recall that dex
dx = ex), we shall try solutions of the form y = Aeαt,
with A, α constants. This is not a major change because αx can be expressed
as a power of e, noting that et ln α = αt (since t ln α = ln(αt) = loge(αt)).
If y = Aeαt (A = 0), then y = A d
dt (eαt) = Aαeαt and y = Aα d
dt (eαt) =
Aα2eαt. Therefore, y = Aeαt is a solution of the homogeneous equation if and
only if
Aα2eαt + aAαeαt + bAeαt = 0.
Dividing throughout by Aeαt gives
α2 + aα + b = 0.
This is the condition for α to be a root of the quadratic equation
x2 + ax + b = 0
which we will call the characteristic equation of the differential equation. Its
roots are the characteristic roots. The similarity with difference equations
is clear (see Section 12.6).
If we allow A = 0, then y = 0, which is still a solution of the homogeneous
differential equation. It follows that y = Aeαt is a solution for any constant
298 Elements of Mathematics for Economics and Finance
A, where α is any one of the two characteristic roots. If α, β are the two
characteristic roots, there are two combinations of this basic type solution that
give the general solution of y +ay +b = 0 depending on where α, β are equal
or not. They are as follows:
y =
Aeαt + Beβt if α = β,
(A + tB)eαt if α = β,
where A, B are constants. The values of A, B can be determined from boundary
conditions; for instance the values of y(0) and y(0) are given, or the
values of y(0) and y(1).
Example 13.7
Solve the differential equations
- d2y
dt2 + 5dy
dt
- 6y = 0; y(0) = 0 and y
(0) = 4,
- d2y
dt2
− dy
dt
− 6y = 0; y(0) = 1 and y
(0) = 5, - d2y
dt2
− 6dy
dt
- 9y = 0; y(0) = 1 and y
(0) = 1.
Solution.
- The characteristic equation is
x2 + 5x + 6 = (x + 2)(x + 3) = 0.
The characteristic roots are therefore −2, −3. The solution is therefore
y = Ae
−2t + Be
−3t
and so y = −2Ae−2t − 3Be−3t. Since
0 = y(0) = Ae0 + Be0 = A + B,
then A = −B. Since also
4 = y
(0) = −2A − 3B
then
4 = −2A − 3(−A) = −2A + 3A = A.
Therefore A = 4 = −B, and the solution is
y = 4e
−2t − 4e
−3t. - Differential Equations 299
- The characteristic equation is
x2 − x − 6 = (x + 2)(x − 3) = 0.
The characteristic roots are therefore −2, 3. The solution is therefore
y = Ae
−2t + Be3t.
Then
y
= −2Ae
−2t + 3Be3t.
Since y(0) = 1, then
A + B = 1.
Since y(0) = 5, then
−2A + 3B = 5.
Solving the simultaneous equations gives A = −0.4 and B = 1.4. The
solution is therefore
y = −0.4e
−2t + 1.4e3t. - The characteristic equation is
x2 − 6x + 9 = (x − 3)2 = 0.
Therefore, there are two equal characteristic roots 3, 3. The solution is
therefore
y = (A + tB)e3t.
We have
1 = y(0) = (A + 0)e0 = A.
Since
y
= Be3t + (A + tB)3e3t,
using the rule for differentiation of a product of functions (6.6), then
y
(0) = Be0 + (A + 0)3e0
= B + 3A
= B + 3 (since A = 1).
Therefore, since y(0) = 1, then B = −2. It follows that the solution is
y = (1 − 2t)e3t.
300 Elements of Mathematics for Economics and Finance
13.4.2 The General Case
We have shown how to solve homogeneous linear differential equations and
therefore we can find complementary solutions in the inhomogeneous case. All
we need now is to find particular solutions of equation (13.5):
y
+ ay
+ by = c.
There are three cases of particular solutions:
y =
⎧⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎩
c
b
if b = 0,
c
a
t if b = 0, a = 0,
1
2ct2 if a = b = 0.
It is a simple exercise to show that these are indeed particular solutions of
equation (13.5). Note that the solution y = c
b is that obtained by assuming y
is constant (i.e., time invariant if t represents time).
The cases for a particular solution correspond, in order, to the cases when:
0 is not a characteristic root; exactly one characteristic root is 0; both characteristic
roots are 0. Compare this with the corresponding case for difference
equations in Chapter 12.
Example 13.8
If y = y(t) is a function of t, solve the following differential equations for y: - y − y − 6y = 6; y(0) = 0 and y(0) = 5,
- y + 5y + 6y = −12; y(0) = 2 and y(0) = 3,
- y − 6y + 9y = 18; y(0) = 0 and y(0) = 1,
- y − 4y = 8; y(0) = 0 and y(1) = 3.
Solution. - From Example 13.7.2, we know that the complementary solution is of the
form
y = Ae
−2t + Be3t.
(We do not apply boundary conditions until we have the complete general
solution.) - Differential Equations 301
A particular solution is y = 6
−6 = −1, so the general solution is
y = Ae
−2t + Be3t − 1.
Then
y
= −2Ae
−2t + 3Be3t.
Since
0 = y(0) = A + B − 1,
then A + B = 1. We also have
5 = y
(0) = −2A + 3B.
Solving the simultaneous equations A + B = 1 and −2A + 3B = 5 gives
A = −0.4 and B = 1.4. The solution is therefore
y = −0.4e
−2t + 1.4e3t − 1. - From Example 13.7.1, the complementary solution is
y = Ae
−2t + Be
−3t.
A particular solution is y = −12
6 = −2. Therefore, the general solution is
y = Ae
−2t + Be
−3t − 2.
Then
y
= −2Ae
−2t − 3Be
−3t.
Since y(0) = 2, then 2 = A+B−2 and since y(0) = 3, then 3 = −2A−3B.
Solving the simultaneous equations A + B = 4 and 2A + 3B = −3 gives
A = 15 and B = −11. The solution is therefore
y = 15e
−2t − 11e
−3t − 2. - From Example 13.7.3, the complementary solution is
y = (A + tB)e3t.
A particular solution is y = 18
9 = 2. The general solution is therefore
y = (A + tB)e3t + 2.
Since 0 = y(0) = (A + 0)e0 + 2 = A + 2, then A = −2. Since
y
= Be3t + (A + tB)3e3t,
using the rule for differentiation of a product of functions (6.6), then
y
(0) = B + 3A.
Therefore, since y(0) = 1 and A = −2, then B = 7. The general solution
is therefore
y = (7t − 2)e3t + 2.
302 Elements of Mathematics for Economics and Finance - The characteristic equation is
x2 − 4x = x(x − 4) = 0
and the characteristic roots are therefore 4, 0. The complementary solution
is
y = Ae4t + Be0t = Ae4t + B.
A particular solution is y = 8
−4 t = −2t. The general solution is therefore
y = Ae4t + B − 2t.
Since 0 = y(0) = A + B and 3 = y(1) = Ae4 + B − 2, then B = −A and
5 = Ae4 + B. Then 5 = Ae4 − A = A(e4 − 1). Therefore, A = 5
e4−1 = −B
and the general solution is
y =
5
e4 − 1
(e4t − 1) − 2t.
13.4.3 Stability
To discuss the stability of the second order linear differential equation
d2y
dt2 + a
dy
dt
- by = c,
we shall assume b = 0 in order to simplify matters by avoiding degenerate cases.
The condition b = 0 is equivalent to the condition that the characteristic roots
α, β = 0.
The general solution of the differential equation is then
y =
⎧⎪⎨
⎪⎩
Aeαt + Beβt + c
b
if α = β,
(A + tB)eαt + c
b
if α = β,
where A, B are constants.
Since eγt tends to 0 or ∞ according as γ < 0 or γ > 0, the solution
y will diverge if either α or β is positive; while if α, β are both negative, the
complementary solution tends to 0 and so y converges on the particular solution
c
b , the equilibrium value.
In Examples 13.8.1 and 13.8.3, the solution diverges, while in Example
13.8.2 it converges to the equilibrium value −2, the particular solution in that
case.
- Differential Equations 303
EXERCISES
13.1. Solve the following differential equations for the function y = y(t) of
t.
a) dy
dt
= 5y + 6; y(0) = 1,
b) dy
dt
= −3y + 4; y(0) =
1
3,
c) dy
dt
= 0.8y + 12; y(0) = 5.
Comment on stability for each of these equations and sketch the
graph of y against t.
13.2. Solve the following differential equations for y = y(t) and sketch the
graph of y against t.
a) dy
dt
= 4; y(0) = 7,
b) dy
dt
= 4t; y(0) = 1.
13.3. In a population model, the population y(t) (thousands) at time t
(years) satisfies
y
= −0.05y + 2.
The initial population is 100,000. What is the equilibrium value of
the population? When does the population fall to within 1,000 of
this equilibrium value? Sketch the graph of population against time.
13.4. Solve the following differential equations for y = y(t).
a) y = 1.2y − y2; y(0) = 2,
b) y = −2y + 5y1.2; y(0) = 1.
13.5. Solve the following differential equations for y = y(t).
a) y2 dy
dt
= 4t; y(0) = 3,
b) yt
dy
dt
= 1; y(1) = 1,
c) dy
dt
= yt; y(0) = 3,
d) dy
dt
2t + 1
6y2 ; y(0) = 1,
e) et dy
dt
= y2; y(0) = 0.5.
304 Elements of Mathematics for Economics and Finance
13.6. Solve the following differential equations for y = y(t). In each case,
comment on stability.
a) d2y
dt2
− 2dy
dt
− 15y = 0; y(0) = 5, y
(0) = 1,
b) d2y
dt2 + 8dy
dt
- 15y = 30; y(0) = 2, y
(0) = 1,
c) d2y
dt2
− 8dy
dt - 16y = 4; y(0) = 4, y
(0) = 10.
13.7. Solve the following differential equations.
a) d2y
dt2 + 3dy
dt
= 0; y(0) = 1, y
(0) = 3,
b) d2y
dt2
− 4y = 12; y(0) = 0, y
(0) = 6,
c) d2y
dt2
− 10dy
dt
= 5; y(0) = 0, y
(0) =
1
2,
d) d2y
dt2 = 10; y(0) = 1, y
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