Chapter 23
Linear, Second-Order Differential Equations
Autonomous Differential Equation
Linear, second-order differential equation has
following form:
ÿ + a1 ẏ + a2 y = b.
(23.1)
As in Chapter 21, we divide the above
problem in two parts: Steady State Part and
Homogeneous Part. Steady state part is
given by
a2 y = b
(23.2)
b
y= .
a2
(23.3)
and solution is
1
Homogeneous part is given by
ÿ + a1 ẏ + a2 y = 0
(23.4)
Let the solution of homogeneous part be denoted by yh , then the complete solution of
(23.1) is
y = yh + y.
(23.5)
Solution of Homogeneous Part
In order to solve (23.4), we will use guess
and verify method. Let the solution be
y(t) = A exprt , A 6= 0.
(23.6)
We want to know for what values of r, (23.6)
satisfies (23.4). Putting (23.6) in (23.4), we
have
ÿ +a1 ẏ +a2 y = A exprt (r2 +a1 r +a2 ) (23.7)
2
Then we choose r such that
(r2 + a1 r + a2 ) = 0
(23.8)
(23.8) is known as characteristic equation.
It is a quadratic equation and will have two
roots given by
p
−a1 ± a21 − 4a2
r1 , r2 =
.
(23.9)
2
These roots are also as eigen values or characteristic roots.
Three Cases
(I) Real Roots & r1 6= r2 if a21 > 4a2
(II) Complex Roots & r1 6= r2 if a21 < 4a2
(III) Repeated Roots & r1 = r2 if a21 = 4a2
3
In the case of distinct roots, r1 6= r2 , we
have two solutions to the homogeneous part
y1 = A1 expr1 t & y2 = A2 expr2 t . (23.10)
The general solution to the homogeneous part in case of distinct roots is given
by
yh = c1 expr1 t +c2 expr2 t
(23.11)
where c1 and c2 are two arbitrary constants
(Theorem 23.2).
In the case of repeated roots, r1 =
r2 = r, the solutions to the homogeneous part
is given by
y1 = A1 exprt & y2 = tA2 exprt
where r = − a21 .
4
(23.12)
The general solution to the homogeneous part in case of repeated roots is given
by
yh = c1 exprt +c2 t exprt
(23.13)
where c1 and c2 are two arbitrary constants
(Theorem 23.2).
Complex Numbers
Let i be an imaginary number defined as
i=
√
−1.
(23.14)
A complex number z can be expressed as
z = h + iv
(23.15)
where h and v are two real numbers. Two
complex numbers z1 and z2 are called conjugate complex numbers if
z1 = h + iv & z2 = h − iv.
5
(23.16)
Two Properties of conjugate complex numbers
1. The sum of two conjugate complex numbers is a real-valued number, z1 + z2 =
2h.
2. The difference between two conjugate complex numbers multiplied by i yields a
real valued number (z1 − z2 )i = −2v.
6
Trigonometric Functions
Complex numbers can also be expressed
as trigonometric or circular functions (sines and cosines). Some commonly used values of the sine and cosine functions are:
Values of Sine and Cosine Functions
Degree
00
900
1800
2700
3600
Radians
0
sin(θ)
0
1
0
-1
0
π
2
π
3π
2
2π
7
cos(θ)
1
0
-1
0
1
Some Important Properties of
Sine and Cosine Functions
1. Both sine and cosine functions are bounded between 1 and -1. They are said
to have amplitude of one.
2. Both sine and cosine functions repeat their values every 2π radian
sin(θ) = sin(θ+2nπ) & cos(θ) = cos(θ+2nπ)
(23.17)
for n = 0, 1, 2, .... They are said to have
period of 2π radians.
3. Properties 1. and 2. imply that these
functions generate cyclical behavior.
4. Sine and cosine functions are symmetrical between positive and negative values
of the domain.
sin(−θ) = −sin(θ) & cos(−θ) = cos(θ).
(23.18)
8
5.
dsin(θ)
dcos(θ)
= cos(θ) &
= −sin(θ).
dθ
dθ
(23.19)
Using the above properties, the complex
number can be expressed in terms of trigonometric functions as follows
h + vi = R[cos(θ) + i sin(θ)]
(23.20)
√
where R = h2 + v 2 , h = R cos(θ), and v =
R sin(θ).
Euler’s Formula
expiθ = cos(θ) + i sin(θ)
(23.21a)
exp−iθ = cos(θ) − i sin(θ)
(23.21b)
9
De Moivre’s Theorem
The conjugate complex numbers h ± vi,
when raised to the nth power can be expressed as
(h ± vi)n = Rn [cos(nθ) ± i sin(nθ)]. (23.22)
Analysis of Linear Second-Order
Differential Equation With Complex Roots
The roots of characteristic equation are
given by
r1 , r2 =
−a1 ±
p
a21 − 4a2
.
2
(23.23)
We will have complex roots if a21 < 4a2 .
10
(23.23) can be written as
p
−a1 ± −1 4a2 − a21
r1 , r2 =
. (23.24)
2
√
4a2 −a21
−a1
Now letting h = 2 and v =
,
2
(23.24) can be written as
√
r1 , r2 = h ± vi.
(23.25)
We can now express the general solution of
the homogeneous part as
ht
yh = exp
¡
vit
c1 exp
−vit
+c2 exp
¢
. (23.26)
Using the Euler’s formula, (23.26) can be written as
yh = B1 expht cos vt + B2 expht sin vt
(23.27)
where B1 and B2 are two arbitrary constants.
11
Theorem 23.4: The complete solution
to the linear, autonomous, second-order
differential equation (23.1) is
y(t) = c1 exp
r1 t
+c2 exp
r2 t
b
+ , if r1 6= r2
a2
b
y(t) = (c1 + tc2 ) exp + , if r1 = r2 = r
a2
rt
b
y(t) = B1 exp cos vt + B2 exp sin vt + ,
a2
ht
ht
if roots are complex.
12
Constants of Integration
In solutions arbitrary constants like c1 ,
c2 , B1 , B2 figure. In order to pin down their
values, we need some initial conditions. Since, solutions involve two constants, we need
two initial conditions. Usually, initial conditions specify values of y and ẏ at time t = 0.
Using these two conditions, we can find the
values of constants.
Convergence to Steady State
We would like to derive conditions under
which limt→∞ y(t) = ab2 . If limt→∞ y(t) =
b
a2 , then the steady state is stable equilibrium, otherwise it is unstable.
13
Theorem 23.5: The solution to the autonomous linear, second-order differential equation converges to steady state
equilibrium if and only if the real parts
of the characteristic roots are negative.
Conditions of Convergence: Three Cases
(I) Real Roots : r1 6= r2 then r1 & r2 < 0
a1
(II) Complex : r1 6= r2 then h ≡ − < 0
2
(III) Repeated Roots : r1 = r2 = r then r < 0
14
An Economic Example
Inflation and Unemployment: Phillips Curve
Let, p, π, m and U denote actual inflation rate, expected inflation rate, growth rate
in money supply, and unemployment rate respectively. If P is the price level then p =
Ṗ
P . Similarly, if M is the money supply then
m = Ṁ
M . Suppose that growth rate of money
supply, m, is constant.
Phillips curve posits negative association
between inflation rate and unemployment rate. The basic Phillips curve assumed that
there is negative association between actual
inflation rate and unemployment rate:
p = f (U ), f 0 (U ) < 0.
(23.27)
However, later economists modified this
relationship. They posited that there is negative relationship between unexpected in15
flation and unemployment rate, where unexpected inflation is given by p − π:
p − π = f (U ), f 0 (U ) < 0.
(23.28)
This later relationship is known as Expectation-Augmented Phillips Curve. Suppose that expected-augmented curve is linear
p − π = −βU, β > 0.
(23.29)
Now the question is how expectation regarding inflation, π, is formed. Assume that agents use adaptive expectation i.e.
π̇ = j(p − π), 0 < j < 1.
(23.30)
Finally, suppose that unemployment rate, U ,
evolves as follows:
16
U̇ = −k(m − p), k > 0.
(23.31)
(23.29), (23.30), and (23.31) constitute a
complete macro-model in three variables, inflation rate, p, expected inflation rate, π, and
unemployment rate, U . Question is: what
type of time paths this model implies for these
three variables?
In order to find the time paths, we combine these three equations as follows. Putting
(23.29) in (23.30), we get
π̇ = −jβU
(23.32)
which implies that
π̈ = −jβ U̇
(23.33).
Now put (23.31) in (23.33), we have
π̈ = jβk(m − p)
17
(23.34).
We want to eliminate p from (23.34). In order
to do so, rewrite (23.30) as
π̇
p = + π.
j
(23.35)
Putting (23.35) in (23.34), we have
π̈ + βk π̇ + jβkπ = jβkm
(23.36)
which is linear, autonomous, second-order differential equation in expected inflation rate,
π. Using the method discussed earlier, we can
solve it. Here,
a1 = βk, a2 = jβk, b = jβkm.
(23.37)
The steady-state value, π, is given by
π = m.
(23.37)
The characteristic roots of the homogeneous
part is given by
18
p
−βk ± β 2 k 2 − 4jβk
r1 , r2 =
.
2
(23.38)
(23.38) implies that we will have real and distinct roots if βk > 4j, real and repeated roots
if βk = 4j, and complex and distinct roots if
βk < 4j.
From (23.38), we can also derive that
steady-state is a stable equilibrium. In the
case of repeated root, r = −βk
< 0. Simi2
larly, in the case of complex root h = −βk
2 <
0. Finally,
p in thepcase of real distinct roots,
βk = β 2 k 2 > β 2 k 2 − 4jβk. Thus, both
r1 , r2 < 0.
Once, we have derived the time path of
expected inflation rate, π, we can find out
time path of actual inflation rate, p, by using
equation (23.35). Finally, by using (23.29) we
can derive time path of unemployment rate,
U.
19
(23.35) implies that the steady state value of inflation rate
p = π = m.
(23.39)
Thus, in the long run actual inflation rate
is equal to the growth rate of money supply.
Finally, (23.32) implies that the steady state
value of unemployment rate is
U = 0.
(23.40)
The steady state value of unemployment rate
is known as natural rate of unemployment. In this particular example, it is equal
to zero, but it need not always be the case.
(23.39) and (23.40) suggest that long run actual inflation rate is independent of unemployment rate. This long-run relationship is
captured by the notion that long run Phillips curve is vertical at U .
20
Non-Autonomous Differential Equation
The general form of linear, second-order,
non-autonomous differential equation is
ÿ + a1 (t)ẏ + a2 (t)y = b(t).
(23.41)
We are going to concentrate on a special form
of non-autonomous equation in which a1 and
a2 are constants. We will consider two special
cases:
Case I: b(t) = A0 + A1 t + .... + An tn
Case II: b(t) = expαt [A0 + A1 t + .... +
An tn ].
In both cases, one can find solution using the method of undetermined coefficients.
21
We want to find solution of following differential equation
ÿ + a1 ẏ + a2 y = b(t)
(23.42)
where b(t) is as defined in above two cases.
The method works as follows. First we find
solution of homogeneous part
ÿ + a1 ẏ + a2 y = 0
(23.43)
using the methods discussed earlier. Denote
this solution by yh . Now we need to find
particular solution, yp . Since, b(t) is variable there does not exist steady state. So
we have to find particular solution using the
method of undetermined coefficients.
Case I: b(t) = A0 + A1 t + .... + An tn
In this case, assume that
yp = B0 + B1 t + .... + Bn tn .
22
Trick here is to find values of Bs which will
satisfy differential equation (23.42). In order
to find out what Bs will do, derive ẏp and ÿp
and put them in differential equation (23.42).
This will provide you with series of equations
in Bs. Solving these equations, you can derive particular solution, yp .
Case II: b(t) = expαt [A0 + A1 t + .... + An tn ]
In this case, assume that
yp = expαt [B0 + B1 t + .... + Bn tn ].
and do all the steps discussed for case I. The
general solution to the differential equation
then is
y = yh + yp .
23