1
Chapter 5
Linear Second-Order Equations I
In this chapter we undertake a study of linear second-order differential equations
with constant coefficients. We will learn that such equations arise as models
of simple spring-mass systems and RLC electrical circults. Then, in Sections
5.2–5.7, we will discuss methods for visualizing computing solutions. Finally, in
Sections 5.8 and 5.9, we return to spring-mass systems and study their behavior.
5.1 Introduction: Modeling Vibrations
Many mechanical systems are capable of oscillatory motion, or vibrations. The
most basic model of such a system is a mass attached to a spring. As you might
imagine, whether—and in what manner—such a system vibrates depends upon
a combination of factors: the stiffness of the spring, the size of the mass, the
amount of friction present, and the nature of any external forces present.
Simple Spring-Mass Systems
Consider a mass m attached to the end of a spring that is fiyed at its opposite
end as in Figure 1. Assume that y = 0 is the position of the mass when it is
at rest and that y(t) > 0 when the spring is stretched and y(t) > 0 when the
spring is compressed.
k
m
Figure 1
y(t)
Assume further that Hooke’s law applies so that the restorative force exerted by
the spring on the mass at any time t is given by −ky(t), where k is a positive
constant representing the stiffness of the spring. Note that this force is negative
when the spring is stretched and positive when the spring is compressed. If no
other forces are acting upon the mass, and if m is constant, then Newton’s
second law tells us that, at any time t,
my �� (t) = −ky(t).
So the position of the mass satisfies the differential equation
my �� + ky = 0,
(1)
2
Chapter 5. Linear Second-Order Equations I
where each term
� in the equation has units of force. If we divide through by m
and set ω = k/m, then the equation becomes
y �� + ω 2 y = 0.
(2)
It is easy to see that cos ωt and sin ωt are particular solutions. Therefore, because
of the linearity and homogeneity of (2), it follows that
y(t) = c1 cos ωt + c2 sin ωt
is a solution for any constants c1 and c2 . The constants c1 and c2 can be determined from the initial displacement and the initial velocity of the mass. Thus
the mass experiences simple harmonic motion.
Damping
We now consider the effect of including a viscous damping force. Our assumption
will be that such a force is proportional to the velocity of the mass and in a
direction opposite that of the motion. Thus the force takes the form of −ry � ,
where r is a positive constant. This type of damping may be thought of as being
due to a shock absorber-like device called a dashpot and is indicated in Figure
3. When the velocities at its ends differ (in the present arrangement one end is
fixed), the dashpot absorbs energy as it compresses or extends.
y(t)
k
m
r
Figure 2
Newton’s second law applied to the system results in
my �� = −ky − ry � ,
which after rearrangement is
my �� + ry � + ky = 0.
(3)
External Forces
When an external force F acts on the mass, it is simply added to the right side
of the differential equation:
my �� + ry � + ky = F.
Depending upon the nature of the force, F may be a function of t, y, or both
t and y. Be careful to notice here that the nonhomogeneous term on the right
side of the equation is precisely the applied force, but only if the terms on the
left side each have units of force as well.
5.1. Introduction: Modeling Vibrations
3
One simple way that an external force may be applied to the mass in a
spring-mass system is by moving the opposite end of the spring. So suppose that
a function f (t) describes the motion of the left end of the spring as in Figure 3.
(The left end of the dashpot is still fixed.) Assume that f (t) = y(t) = 0 when the
system is at rest. (f (t) and y(t) are measured from different reference points.)
f(t)
k
y(t)
m
r
Figure 3
The only difference that this makes in the derivation of equation (3) is that at
time t the spring is stretched by an amount y(t) − f (t), and so the Hooke’s law
force becomes −k(y(t) − f (t)). Newton’s second law now gives
my �� (t) = mg − k (y(t) − f (t)) − ry � ,
which we rearrange to obtain
my �� + ry � + ky = kf (t).
(4)
Therefore, the external force that results is kf (t), where k is the spring stiffness
constant. (Note that the term kf (t) has units of force, just as ky does.)
Gravity Doesn’t Matter (Usually)
Consider a mass m attached to the end
of a spring that hangs vertically from a
fixed support as in Figure 4. The situation is slightly different here because the
weight of the mass is an addition force.
For our present purposes we will assume
that there is no damping and no external
-s
force. The first part of the figure shows
m
0
the spring hanging under its own weight.
m
y(t)
The middle part shows the spring with
Figure 4
the attached mass at rest (or equilibrium).
The elongation of the spring at this equilibrium position is denoted by s. The
last part of the figure is a snapshot of the system in motion. The displacement
of the mass from equilibrium is denoted by y(t), with the downward direction
taken to be positive.
First consider the system at equilibrium. The force on the mass due to gravity is its weight mg, where g is the usual gravitational acceleration. This force
4
Chapter 5. Linear Second-Order Equations I
is positive because we’re assuming that downward is the positive direction. Precisely balancing this downward force is the upward force that the spring exerts
on the mass, which is −ks. Now, because the mass is at equilibrium, the sum
of these two forces must be zero; that is,
ks = mg.
(5)
Now consider the mass at time t, when its displacement from equilibrium is
y(t). The spring is stretched an amount y(t) + s, and so the restorative force
of the spring is −k (y(t) + s). The sum of the forces on the mass is therefore
mg − k (y(t) + s). Thus Newton’s second law gives
my �� (t) = mg − k (y(t) + s).
The relationship ks = mg and a little rearrangement result in
my �� + ky = 0,
the same differential equation as for the system in Figure 1. So we’re seeing
that the differential equation for vertical motion is independent of gravitational
force. This is also true in the presence of damping and external forces. However,
it is not true if the spring force is a nonlinear function of the position of the
mass. See Problem 11.
For an arrangement like Figure 4, an external force
kf (t) results if f (t) describes the position of the top
end of the spring. (See Figure 5.) One can think of
applying this force in a simple, natural way. Imagine
that you are holding the top end of the spring. Moving your hand up and down in some way will cause
the mass to respond (i.e., to move as well), but in
a way that doesn’t necessarily follow closely the motion of your hand. Much of what we will study in
this chapter will lead to an understanding of how the
mass responds to such a force.
f(t)
m
y(t)
Figure 5
The Series RLC Circuit
In Section 2.4 we learned that the current in an RC or an RL circuit obeys a linear first-order differential equation. The reader may wish to review that section
for descriptions of the basic circuit components and statement of Kirchhoff’s
voltage law.
The current in a series RLC electrical circuit exhibits behavior that is completely analogous to that of the mechanical spring-mass model discussed above.
Here, we will look at an RLC circuit with a voltage source, as shown schemati-
5.1. Introduction: Modeling Vibrations
5
cally in Figure 6. Application of Kirchhoff’s voltage law results in
VL + VR + VC − vs (t) = 0.
Using VR = IR and VL = LI � , this becomes
LI � + RI + VC = vs (t).
vs
Now, since I = C VC� , this becomes
R
L
1
1
V = vs (t),
(6)
C
C
C
where V = VC and we have divided through
by C. This is a linear second-order differential equation with constant coefficients for
Figure 6
the voltage across the capacitor. If we diff�
erentiate with respect to t, multiply by C, and use CVC = I, then we get
LV �� + RV � +
1
I = vs� (t),
(7)
C
a linear second-order equation with constant coefficients for the current I(t).
LI �� + RI � +
Notice that equations (6) and (7) are exactly the same sort of equation as
(4), which governs the motion of a damped spring-mass system. Thus both
problems will have the same type of solutions. In fact, direct analogies can be
made between
• the inductance L and the mass m;
• the resistance R and the damping coefficient r;
• the capacitance C and the spring “compliance” 1/k.
As in the spring-mass problem, what we have here is an idealized model of a
circuit, resulting from the assumption of linear device characteristics.
Problems
1. In the British engineering system of units, the basic units of length, mass, and time
are the foot, the slug, and the second, respectively. Force has units of pounds, and
the acceleration due to gravity is approximately g = 32.1 ft/s2 .
(a)
(b)
(c)
(d)
What is the mass of an object that weighs 1 pound?
How much does a mass of 1 slug weigh?
What is the mass of a person who weighs 128 lbs?
What are the units of the spring stiffness k and the damping coefficient r? (Hint:
ky and ry � have units of force.)
2. In the mks (or SI) system of units, the basic units of length, mass, and time are the
meter, the kilogram, and the second, respectively. Force has units of Newtons, and
the acceleration due to gravity is approximately g = 9.8 m/s2 .
6
Chapter 5. Linear Second-Order Equations I
(a)
(b)
(c)
(d)
What
What
What
What
is the mass of an object that weighs 1 Newton?
is the weight (in Newtons) of a 1-kilogram mass?
is the mass of an object that weighs 49 Newtons?
are the units of the spring stiffness k and the damping coefficient r?
3. (a) A 10-lb weight stretches a spring 1 inch. What is the spring stiffness k?
(b) A 0.3-kg mass stretches a spring 0.06 meters. What is the spring stiffness k?
(c) An object is attached to a spring with known stiffness k = 147 dynes/cm, stretching it 4 cm. What are the weight and mass of the object (in cgs units)?
4. Suppose that in Figure 4, the mass is 2 kg, and it stretches the spring 0.1 m.
(a) Write down the differential equation governing the undamped, unforced motion.
(b) If the mass is pulled down an additional 0.2 meters and gently released, what is
the appropriate initial value problem for the subsequent motion?
5. An unknown mass stretches the spring in Figure 4 by an amount s = 0.04 m.
(a) Write down the differential equation governing the undamped, unforced motion.
(b) If the mass is pushed up 0.1 meters above equilibrium and projected downward
with an initial velocity of 1 m/s, what is the appropriate initial value problem for
the subsequent motion of the mass?
6. Suppose that in Figure 4, the mass is 2 kg, and it stretches the spring 0.1 m. In addition, a damping force of 4 N per unit velocity is present. Write down the differential
equation governing the damped, unforced motion.
7. Suppose that for the same system as in Problem 6, the top end of the spring moves
according to f (t) = 0.1 sin π t for t ≥ 0.
(a) Write down the differential equation governing the mass’s damped, forced motion.
(b) If the mass is at rest at time t = 0, what is the appropriate initial value problem
for the subsequent motion of the mass?
8. Suppose that the spring-mass system of Figure 4 is upside down, so that the weight
of the mass compresses the spring rather than stretching it.
(a) Taking the upward direction to be positive, set up the differential equation for
the undamped motion of the mass.
(b) Include into the equation a damping term and a force term caused by motion of
the bottom end of the spring.
9. A simple spring-mass system can be thought of as a crude model of your car’s
suspension system. Suppose that the car has mass m, the stiffness of the suspension
is k, and the shock absorbers provide a damping force with coefficient r. Now suppose
that you’re driving on a “washboarded” road with a sinusoidal surface of amplitude
a ft and high spots that are δ feet apart. If your speed is S ft/s, what is an appropriate
differential equation for the forced vertical motion of the body of the car?
10. Consider an undamped spring-mass system as in Figure 2 with mass m = 10 kg
and spring stiffness k = 50 N/m. The mass is composed of iron, and the other end
of the spring is attached to a fixed magnet that exerts a force on the mass that is
5.2. The Structure of Solutions
7
inversely proportional to the square of the distance between it and the mass. The
natural length of the spring is 1.2 m, and the magnetic force on the mass compresses
the spring by 0.2 m. Letting y(t) denote the displacement from equilibrium, find the
differential equation governing the undamped, forced motion of the spring.
11. (a) Set up the differential equation for the undamped motion of the mass in Figure 1,
given that the force exerted by the spring is proportional to the cube of the amount
that the spring is stretched. (b) Repeat for the hanging mass in Figure 4.
5.2 The Structure of Solutions
Motivated by the applications described in the preceding section, let us consider
linear second-order differential equations written in the form
y �� + py � + qy = f,
(1)
where p and q are constants and f is a continuous function on some interval
I. The focus of this section is a theorem that states the structure of general
solutions to (1). While we are restricting out attention to the constant-coefficient
case, the theorem is true also for equations of the form (1) when p and q, like
f , are (not necessarily constant) continuous functions on I. In the next chapter
we will study the non-constant coefficient case and provide a proof the theorem
on the structure of general solutions.
Let’s begin here with some observations concerning solutions of (1) that
are basic consequences of linearity. Since (1) is a linear differential equation,
superposition tells us the following important facts:
i) If u and v are any two solutions of
y �� + py � + qy = 0,
(2)
then c1 u + c2 v is also a solution for any pair of constants c1 and c2 .
ii) If u and v are any two solutions of (1), then u−v satisfies the corresponding
homogeneous equation (2).
iii) If ỹ is a solution of (1) and y is a solution of (2), then y + ỹ is a solution
of (1).
The notion of linear independence is important for this discussion. Two functions are said to be linearly dependent on an interval I if and only if one of
them is a constant multiple of the other.∗ So we can say that two functions are
linearly independent on an interval I if neither is a constant multiple of the
other. (Note that the zero function is a constant multiple of any other function.)
It is often to convenient to this of this in terms of the ratio of the two functions.
* For more than two functions, a more careful definition is needed.
8
Chapter 5. Linear Second-Order Equations I
Two functions u and v are linearly dependent on an interval I if and only the
ratio u/v is constant wherever v �= 0 on I.
• Example 1 Consider the functions u = eat and v = ebt on I = (−∞, ∞), where
a and b are constants. If a = b, then u and v are identical and therefore clearly
linearly dependent on I. On the other hand, if a �= b, then u and v are linearly
independent. To see why, note that u/v = e(a−b)t is not constant on I unless
a = b. Thus nether u nor v is a constant multiple of the other on I.
• Example 2 Consider the functions u = cos ωt and v = sin ωt on I = (−∞, ∞),
where ω is a positive constant. Neither u nor v is a constant multiple of the
other on I; for otherwise the functions tan ωt and cot ωt would be constant on
their domains.
• Example 3 Let f be any non-zero continuous function on an interval I, and let
g be any non-constant continuous function on I. Then f and the product f g are
linearly independent on I. For instance, each of the following pairs are linearly
independent on I = (−∞, ∞):
{2, 2t}, {t, t2 }, {et , t et }, {cos t, t2 cos t}, {e−t , e−t tan−1 t}.
�
We are now ready to state our theorem, whose proof in the more general case
of non-constant coefficients will be the subject of a section in Chapter 6.
THEOREM 1 There exists a pair of linearly independent solutions of (2) on I.
Given any such pair u and v, every solution of (2) on I is given by y = c1 u + c2 v
for some pair of constants c1 and c2 . If, in addition, ỹ is any particular solution
of (1), then every solution of (1) on I is given by
y = c1 u + c2 v + ỹ
for some pair of constants c1 and c2 . Furthermore, c1 and c2 can be chosen so
that y satisfies arbitrary initial values y(t0 ) = y0 and y � (t0 ) = m0 with t0 any
point in I.
Theorem 1 is the basis for the following procedure for finding a general solution of (1):
i) Find a pair of linearly independent solutions u and v of the corresponding
homogeneous equation (2).
ii) Find a particular solution ỹ of (1).
iii) Write y = c1 u + c2 v + ỹ.
Now, of course, everything hinges upon learning how to find such u, v, and ỹ,
which will occupy us for much of this chapter and the following one.
5.2. The Structure of Solutions
9
Let us also note that, according to Theorem 1, if u and v are linearly independent solutions of (2), then the general solution of (2) given by c1 u+c2 v must
contain all solutions of (2). Thus all such general solutions of (2) are equivalent.
Similarly, all general solutions of (1) constructed according to the above procedure are equivalent. So henceforth we will refer to any such general solution as
the general solution.
• Example 4 Consider the homogeneous equation
y �� − y = 0.
This is easily seen to have linearly independent solutions et and e−t on (−∞, ∞).
Thus, every solution is of the form
y = c1 et + c2 e−t .
Another pair of linearly independent solution consists of cosh t and sinh t, and
so every solution is also of the form
y = c1 cosh t + c2 sinh t.
These two descriptions are equivalent since they describe precisely the same collection of functions,∗ and so either may be used to decribe the general solution.
• Example 5 Consider the equation
y �� + y = 10e−2t .
The homogeneous equation y �� +y = 0 is easily seen to have linearly independent
solutions cos t and sin t on (−∞, ∞). Thus, every solution of y �� + y = 0 is of
the form
c1 cos t + c2 sin t.
To find a particular solution of y �� + y = 10e−2t , we look for it in the form of
y = we−2t , for which y �� = 4we−2t . Thus we want
4we−2t + we−2t = 10e−2t .
That equation is true for all t if a = 2; so we have the particular solution
ỹ = 2e−2t . Therefore, the general solution of y �� + y = 5e−2t is
y = c1 cos t + c2 sin t + 2e−2t .
Furthermore, the last assertion in Theorem 1 guarantees that c1 and c2 can be
found so that any initial conditions y(t0 ) = y0 and y � (t0 ) = m0 are satisfied.
* Recall that 2 cosh t = et + e−t and 2 sinh t = et − e−t .
10
Chapter 5. Linear Second-Order Equations I
Problems
Given that et and e−t are solutions of y �� − y = 0, solve each initial-value problem in
Problems 1–6.
1. y �� − y = 0, y(0) = 1, y � (0) = 0
2. y �� − y = 0, y(0) = 1, y � (0) = −1
3. y �� − y = 1, y(0) = y � (0) = 0
4. y �� − y = 1 − t, y(0) = 1, y � (0) = 0
5. y �� − y = e−t/2 , y(0) = y � (0) = 0
6. y �� − y = e−2t , y(0) = 0, y � (0) = 0
Given that cos 2t and sin 2t are solutions of y �� +4y = 0, solve each initial-value problem
in Problems 7–12.
7. y �� + 4y = 0, y(0) = 1, y � (0) = 0
9. y �� + 4y = 1, y(0) = y � (0) = 0
11. y �� + 4y = 4e−t/2 , y(0) = y � (0) = 0
8. y �� + 4y = 0, y(0) = 1, y � (0) = −1
10. y �� + 4y = 1 − t, y(0) = 1, y � (0) = 0
12. y �� + 4y = e−t , y(0) = 0, y � (0) = 0
13. a) Verify that e−t and te−t are linearly independent solutions of y �� + 2y � + y = 0.
b) Find a particular solution of y �� + 2y � + y = cos t in the form of a cos t + b sin t.
c) Write down the general solution of y �� + 2y � + y = cos t.
14. a) Verify that e−t cos t and e−t sin t are linearly independent solutions of
y �� + 2y � + 2y = 0.
b) Find a particular solution of y �� + 2y � + 2y = 2 sin t in the form of a cos t + b sin t.
c) Write down the general solution of y �� + 2y � + 2y = 2 sin t.
15. a) For what value(s) of k does the equation y �� + 5y � + 6y = ek t not have a solution
of the form aek t where a is a constant?
b) For what value(s) of q does the equation y �� + q y = cos 4t not have a solution of
the form a cos 4t + b sin 4t where a and b are constants?
16. Show that the equation y �� + 3y � + 2y = e−t
a) does not have a solution of the form ae−t where a is a constant;
b) does have a solution of the form ate−t where a is a constant.
17. Show that the equation y �� + 2y � + y = e−t
a) does not have a solution of the form ae−t where a is a constant;
b) does not have a solution of the form ate−t where a is constant;
c) does have a solution of the form at2 e−t where a is a constant.
18. Show that the equation y �� + 2y � + y = te−t
a) does not have a solution of the form (at + b)e−t where a and b are constants;
b) does not have a solution of the form t (at + b)e−t where a and b are constants;
c) does have a solution of the form t2 (at + b)e−t where a and b are constants.
5.3. The Characteristic Equation
11
5.3 The Characteristic Equation
We first consider homogeneous Linear Second-Order equations in the form
y �� + py � + qy = 0,
(1)
where p and q are constants, and we assume that q �= 0. (If q = 0, the equation
is easily reduced to one of first order.) It is sensible to look for solutions of the
form eλt , where λ is a constant, by virtue of the fact that differentiation of eλt
produces constant multiples of itself. Putting y = eλt into the equation yields
λ2 eλt + pλeλt + q eλt = 0.
Now dividing through by eλt produces a quadratic equation for λ:
λ2 + p λ + q = 0.
(2)
This quadratic equation is called the characteristic equation corresponding
to the differential equation (1).
The point is that if λ satisfies the characteristic equation (2), then eλt satisfies
the differential equation (1). The characteristic equation will, of course, have
different types of solutions depending on the coefficients p and q. In fact, the
quadratic formula gives the roots:
�
�
1�
λ1 , λ2 =
−p ± p2 − 4q .
2
Thus there are three cases:
i) If p2 − 4q > 0, then there are two distinct real roots λ1 , λ2 .
ii) If p2 − 4q = 0, then there is only one (repeated) root: λ1 = λ2 = −p/2.
iii) If p2 − 4q < 0, then there is a pair of (nonreal) complex conjugate roots.
Distinct Real Roots
In the first case, in which p2 − 4q > 0, we have completely solved our problem.
That is, since λ1 and λ2 are two distinct real roots of the characteristic equation,
eλ1 t and eλ2 t are two linearly independent solutions of the differential equation,
and so Theorem 1 of Section 5.4 tells us that the general solution of (1) is
y = c 1 eλ1 t + c 2 eλ2 t ,
(3)
where c1 , c2 are arbitrary constants.
So let us suppose for the time being that p2 − 4q > 0, so that λ1 and λ2 are
real and distinct. Since we assume that q �= 0, we are guaranteed that zero is
not a root and therefore guaranteed that the roots are either both positive, both
negative, or of opposite sign. The general behavior of solutions is determined
by which of these is true. Examples 1 through 3 illustrate this.
12
Chapter 5. Linear Second-Order Equations I
• Example 1 Consider the equation
2y �� + y � − 6y = 0.
While we could divide through by 2 to put the equation into the precise form
of (1), there really is no need to. We may as well use
2λ2 + λ − 6 = 0
as the characteristic equation. Factoring gives (2λ − 3)(λ + 2) = 0 and hence
solutions are λ1 = 3/2 and λ2 = −2. Therefore, e 3t/2 and e−2t are solutions
of the differential equation. Since these are linearly independent, the general
solution is given by
y = c1 e 3t/2 + c2 e−2t .
• Example 2 Consider the equation
y �� + 5y � + 6y = 0.
The corresponding characteristic equation is
λ2 + 5λ + 6 = 0,
which upon factoring becomes (λ+2)(λ+3) = 0 and hence has solutions λ1 = −2
and λ2 = −3. Therefore, e−2t and e−3t are solutions of the differential equation.
Since these are linearly independent, the general solution is given by
y = c1 e−2t + c2 e−3t .
• Example 3 Consider the equation
y �� + y � − 6y = 0.
The corresponding characteristic equation is
λ2 + λ − 6 = 0,
which upon factoring becomes (λ − 2)(λ + 3) = 0 and hence has solutions λ1 = 2
and λ2 = −3. Therefore, e2t and e−3t are solutions of the differential equation.
Since these are linearly independent, the general solution is given by
y = c1 e 2t + c2 e−3t .
A Repeated Real Root
In the second case, in which p2 − 4q = 0 and the characteristic polynomial has
only one (repeated) root λ1 = λ2 = −p/2, we find only a single exponential solution u = e−pt/2 and thus only a one-parameter family of exponential solutions
y = c1 e−pt/2 . Since two linearly independent solutions are necessary to describe
the general solution, we need to find a second linearly independent solution. One
5.3. The Characteristic Equation
13
approach to finding one is to look for it in the form of v(t) = w(t)e−pt/2 , which
is the known solution c1 e−pt/2 with the parameter c1 replace with an unknown
function w(t) to be determined. (This kind of approach is called “variation of
parameters” or “variation of constants.”)
Let’s first recall that this case arises when p2 − 4q = 0, i.e., q = p2 /4. So the
differential equation may be written as
y �� + py + p2 y/4 = 0.
Now from v(t) = we−pt/2 , we compute
�
p
p �
v � = w� e−pt/2 − we−pt/2 = w� − w e−pt/2
2
2
and
�
p �
p � � p � −pt/2
v �� (t) = w�� − w� e−pt/2 −
w − w e
2
2
2
�
2 �
p
= w�� − p w� +
w e−pt/2 .
4
It follows that
�
�
�
p2
p �
v �� + pv � + p2 v/4 = w�� − p w� +
w + p w� − w + p2 w/4 e−pt/2
4
2
�� −pt/2
=w e
.
Therefore v = we−pt/2 is a solution if w is any function for which w�� = 0. The
simplest such function is w(t) = t. So we’ve found that a second, linearly independent solution is v = t e−p/2 . Consequently, when the characteristic equation
has a single repeated real root, the general solution of y �� + py + qy = 0 is
y = c1 e−pt/2 + c2 t e−pt/2
= (c1 + c2 t)e−pt/2 .
(4)
• Example 4 Consider the equation
y �� + 4y � + 4y = 0.
The characteristic equation is λ2 + 4λ + 4 = 0, which is the same as (λ + 2)2 = 0.
Thus there is only the repeated root λ1 = −2, and so e−2t and t e−2t are
linearly independent solutions of the differential equation. The general solution
is therefore
y = (c1 + c2 t)e−2t .
Complex Roots
In the third case, in which the characteristic polynomial has a pair of complex
conjugate roots, the preceding method still produces a pair of solutions, but
14
Chapter 5. Linear Second-Order Equations I
they are complex solutions:
e(α±β i)t = eαt e±β i t , where α = −p/2 and β =
�
4q − p2 /2.
It is not difficult to verify this if we simply assume that the differentiation
d ct
formula dt
e = ce ct is valid when c is a complex constant. However, we have
not yet given a meaning to imaginary powers of e, and, moreover, our goal is to
find real solutions. In the following section we will address these issues and find
out how to produce a pair of linearly independent, real solutions from a single
complex solution.
Problems
In Problems 1 through 12, solve the appropriate characteristic equation; then write
down the general solution of the differential equation.
1. y �� + 5y � + 4y = 0
4. y �� − 4y = 0
7. y �� − 6y � + 9y = 0
10. y �� + 2y � + y = 0
2. y �� + 4y � − 5y = 0
3. y �� − 2y � − 8y = 0
8. y �� + y � + y/4 = 0
9. y �� − 2y � + y = 0
5. y �� + 5y � + 6y = 0
11. y �� + 6y � + 9y = 0
6. y �� − 4y � = 0
12. y �� − 4y � + 4y = 0
13. Suppose that p is a constant and consider the first-order equation y � + py = 0.
(a) Find the characteristic equation by substituting y = ert into the equation.
(b) Solve the characteristic equation and write down the general solution of the differential equation.
14. (a) Find three linearly independent solutions of y ��� + y �� − 4y � − 4y = 0.
(b) Find four linearly independent solutions of y (iv) − 5y �� + 4y = 0.
5.4 Complex Roots
In this section we will look at the problem of finding a pair of linearly independent, real solutions of
y �� + p q � + qy = 0
(1)
when p2 − 4q < 0 and the roots of the characteristic equation λ2 + p λ + q = 0
are the complex conjugates
�
�
1�
λ1 , λ2 =
−p ± i 4q − p2 ,
2
2
where i = −1. Let us first simplify our notation by setting
�
α = −p/2 and β = 4q − p2 /2
5.4. Complex Roots
15
so that these roots are α ± β i. If we accept that the differentiation rule
d ct
e = c e ct
dt
is valid for any complex constant c (and we do), then it is easy to check that
e(α+β i)t and e(α−β i)t are indeed solutions of the differential equation. In addition, these functions can be shown to be linearly independent. However, we are
interested only in real solutions. So our goal here is to find out how to obtain
a pair of linearly independent real solutions of (1) from the complex solutions
e(α±β i)t .
Complex Numbers
The imaginary
unit i is defined by the property i2 = −1. Often this is expressed
√
as i = −1. The set of imaginary numbers is defined to be the set of all “numbers” of the form b i, where b is real. The set of complex numbers C is defined
to be the set of all “numbers” having the form a + b i, where a and b are real
numbers and the symbol i has the property that i2 = −1. The real numbers a
and b are called the real and imaginary parts, respectively, of a + b i.
The absolute value (a.k.a. modulus or magnitude) of a + b i is defined as
�
|a + b i| = a2 + b2 .
Addition, subtraction, and multiplication are defined in the natural way:
(a + b i) ± (c + d i) = (a ± c) + (b ± d) i
(a + b i)(c + d i) = ac + adi + bc i + bdi2
= (ac − bd) + (ad + bc)i
The complex number a − b i is called the complex conjugate of a + b i (and
vice versa). Note that the product of complex conjugates is the square of the
absolute value:
(a + b i)(a − b i) = a2 − b2 i2 = a2 + b2 = |a ± b i|2 .
Complex division amounts to using the conjugate multiplication property to
make the denominator real:
For instance,
a + bi
a + bi c − di
(ac + bd) + (bc − ad)i
=
·
=
.
c + di
c + di c − di
c 2 + d2
2+i
2 + i 1 + 3i
(2 − 3) + (6 + 1)i
1
7
=
·
=
i.
=− +
1 − 3i
1 − 3i 1 + 3i
12 + 3 2
10 10
16
Chapter 5. Linear Second-Order Equations I
The Euler-DeMoivre formula defines imaginary powers of e and may
be justified by a variety of arguments, including one using Taylor series. (See
Exercises 14 and 19.) It states that, for any real number θ,
eiθ = cos θ + i sin θ.
(2)
Thus we have, for instance,
√
1
3
iπ/3
e
= +
i, eiπ/2 = i, e±iπ = −1, e3iπ/2 = −i, and e2iπ = 1.
2
2
�
It follows from this definition that |eiθ | = sin2 θ + cos2 θ = 1. Therefore, for
any real θ, the complex number eiθ lies on the unit circle in the complex plane.
(See Figure 1.) The number θ corresponds to the usual polar angle θ.
Imaginary
i
eiθ
-1
sinθ
θ
cos θ
1
Real
-i
Figure 1
Any complex number z = a + bi may be expressed in polar form as
z = r (cos θ + i sin θ) = reiθ ,
√
where r = |z| = a2 + b2 , cos θ = a/r, and sin θ = b/r. The angle θ is called the
argument of z, sometimes denoted by arg(z). This angle is not unique, but is
usually understood to be the principal value, which is taken from the interval
(−π, π]. For instance,
• arg(i) = π/2, because i = 0 + 1 i = cos(π/2) + i sin(π/2) = e iπ/2 ;
√
√
• arg(1 + i) = π/4, because 1 + i = 2 (cos(π/4) + i sin(π/4)) = 2 e iπ/4 ;
• arg(−1) = π, because − 1 = −1 + 0 i = cos π + i sin π = eiπ .
By the addition formulas for sine and cosine,
and division of polar forms behave as expected
z1 = r1 eiθ1 and z2 = r2 eiθ2 , then
z1
z1 z2 = r1 r2 ei (θ1 +θ2 ) and
=
z2
it follows that multiplication
(see Problem 23); that is, if
r1 i (θ1 −θ2 )
e
.
r2
5.4. Complex Roots
17
The Complex-valued Exponential Function
A complex-valued function of a real variable t may be expressed in the form
f (t) = u(t) + i v(t),
where u and v have real values. The definition of the derivative can be extended
in a straightforward manner so that
f � (t) = u� (t) + i v � (t).
The Euler-DeMoivre formula (2) defines the complex-valued function eit in
terms of its real and imaginary parts:
eit = cos t + i sin t.
With this definition we can easily verify the differentiation rule
for any imaginary constant bi as follows:
d
dt
e bit = bie bit
d
(cos bt + i sin bt) = −b sin bt + bi cos bt = bi (cos bt + i sin bt).
dt
Since cosine is an even function and sine is an odd function, the EulerDeMoivre formula also gives the identity
e−it = cos t − i sin t.
Therefore, the identity e(α±β i)t = e αt e±β it is equivalent to
e(α±β i)t = eαt (cos β t ± i sin β t) .
(3)
Thus we can write e(α±β i)t in terms of its real and imaginary parts as
e(α±β i)t = u(t) ± i v(t),
where
u(t) = eαt cos β t and v(t) = eαt sin β t.
(4)
This is the key to “extracting” a pair of linearly independent real solutions of
(1) from the complex solutions e(α±β i)t .
Real Solutions from Complex Roots
�
Our claim is that u and v from (4), with α = −p/2 and β = 4q − p2 /2, are
both real solutions of (1) when p2 − 4q < 0. Note that each of u and v can be
expressed as a linear combination of u ± i v:
1
((u + i v) + (u − i v)),
2
i
v = ((u − i v) − (u + i v)).
2
u=
18
Chapter 5. Linear Second-Order Equations I
Therefore, since u ± i v are solutions, and since the differential equation is homogeneous and linear, it follows that u and v are solutions as well. A straightforward argument shows that u and v are also linearly independent.
We sum up the result of the preceding discussion as a theorem:
THEOREM 1 If the characteristic equation λ2 + p λ + q = 0 has a pair of
complex conjugate roots α ± β i with β �= 0, then e αt cos β t and e αt sin β t are
linearly independent, real solutions of (1). Thus the general solution of (1) is
y = e αt (c1 cos β t + c2 sin β t) .
(5) �
• Example 1 Consider the equation
y �� + 2y � + 10y = 0.
The characteristic equation is λ2 + 2λ + 10 = 0, whose roots are
√
�
1�
−2 ± 4 − 40 = −1 ± 3i.
2
These roots have real part α = −1 and imaginary part ±β where β = 3. Thus,
e−t cos 3t and e−t sin 3t are solutions, and the general solution of the equation
is given by
y = e−t (c1 cos 3t + c2 sin 3t) .
• Example 2 Consider the equation y �� + 4y = 0. The characteristic equation
λ2 + 4 = 0 has purely imaginary roots ±2i. So the roots have real part α = 0
and imaginary part ±2. Thus, cos 2t and sin 2t are solutions, and the general
solution of the equation (which we could have written down by inspection) is
y = c1 cos 2t + c2 sin 2t.
• Example 3 Consider the equation y �� −y � +17y/4 = 0.�The √
associated
� characteristic equation is λ2 − λ + 17/4 = 0, whose roots are 12 1 ± 1 − 17 = 1/2 ± 2i.
These roots have real part α = 1/2 and imaginary part ±β where β = 2. Thus,
e t/2 cos 2t and e t/2 sin 2t are solutions, and the equation has the general solution
y = e t/2 (c1 cos 2t + c2 sin 2t) .
Summary
According to Theorem 1, when the characteristic equation has complex roots
α ± β i with β �= 0, the solution of (1) will exhibit oscillatory behavior with
amplitude proportional to eαt . Thus the amplitude of the oscillations decays
exponentially if the roots have negative real part, is constant if the real part of
the roots is zero, and grows exponentially if the roots have positive real part.
This was illustrated in the preceding Examples 1–3.
5.4. Complex Roots
19
We have now completely solved the homogeneous, linear, second-order differential equation with constant coefficients:
y �� + py � + qy = 0.
The following summarizes the results in terms of the roots λ1 and λ2 of the
characteristic polynomial λ2 + p λ + q.
λ1 , λ2 real and distinct:
λ 1 = λ2 :
λ1 , λ2 = α ± β i, β �= 0 :
y = c 1 eλ1 t + c 2 eλ2 t
y = (c1 + c2 t)eλ1 t
y = eαt (c1 cos β t + c2 sin β t)
Problems
Find the general solution of each equation in Problems 1 through 6.
1. y �� + 9y = 0
2. y �� − 4y � + 5y = 0
3. y �� + 6y � + 18y = 0
4. y �� + 6y � + 10y = 0
5. y �� + 4y � + 8y = 0
6. y �� + 3y � + 52 y = 0
Solve each initial-value problem
�
y �� + 4y = 0
7.
8.
y(0) = 1, y � (0) = −1

 y �� + y � + 1 y = 0
2
10.
11.

y(0) = 0, y � (0) = 1
in Problems 7 through 12.
�
y �� + 2y � + 2y = 0
y(0) = 1, y � (0) = 0

 y �� + 1 y � + 1 y = 0
2
8

�
y(0) = 1, y (0) = 0
9.
�
y �� + 4y � + 5y = 0
y(0) = 0, y � (0) = 1

 y �� + 3 y � + 5 y = 0
2
8
12.

�
y(0) = 2, y (0) = −1
αt
αt
13. Verify directly that
� u = e cos β t and v = e sin β t each satisfy equation (1) if
α = −p/2, β = 4q − p2 /2, and p2 − 4q < 0.
14. In the general solution y = eαt (c1 cos β t + c2 sin β t), express the constants c1 and c2
in terms of initial values y(0) = y0 and y � (0) = v0 .
15. Show that if p > 0 and q �= 0, then every solution of y �� + py � + q y = 0 approaches
zero as t → ∞. What happens in the case where q = 0?
16. Show that substitution of x = it into the Taylor series for ex suggests the EulerDeMoivre formula.
17. With eit defined by the Euler-DeMoivre formula, show that, for real a and b,
d (a+bi)t
e
= (a + b i) e(a+bi)t .
dt
18. Assuming that ddt eit = i eit , justify the Euler-DeMoivre formula by showing that
�
�
d
eit
eit
= 0 for all t, and
= 1 when t = 0.
dt cos t + i sin t
cos t + i sin t
20
Chapter 5. Linear Second-Order Equations I
19. The conjugate of a complex number z = a + b i is defined as√
z̄ = a − b i. The absolute
value of a complex number z = a + b i is defined as |z| = a2 + b2 . Show that the
following are true for any complex number z = a + b i.
z + z̄
z − z̄
(a)
=a
(b)
=b
(c) z z̄ = |z|2
(d) ez = ez̄
(e) |ez | = ea
2
2i
z1
1
20. Show that
=
z1 z̄2 for any complex numbers z1 and z2 with z2 �= 0.
z2
|z2 |2
i
1+i
1
21. Using the result of Problem 20, compute and simplify: (a)
(b)
(c)
1+i
2−i
i
22. Prove the following:
eiθ1
= ei (θ1 −θ2 )
eiθ2
z1
r1 i (θ1 −θ2 )
(d) If z1 = r1 eiθ1 and z2 = r2 eiθ2 , then z1 z2 = r1 r2 ei (θ1 +θ2 ) and
=
e
.
z2
r2
ez1
(e) If z1 = a1 + b1 i and z2 = a2 + b2 i, then ez1 ez2 = e z1 +z2 and z2 = e z1 −z2 .
e
(a) eiθ1 eiθ2 = ei (θ1 +θ2 )
(b)
1
= e−iθ
eiθ
(c)
23. Write each number in polar form reiθ :
(a) z = 1 − i,
(b) z = 2i
(c) z = −3
(d) z = 1 −
24. Write each number in rectangular form a + bi:
(a) z = e−iπ/3
(b) z = e3iπ
(c) z = 2e 5iπ/6
(d) z =
√
√
3i
2 e−3iπ/4
25. Given constants a and b, the (second-order) Euler-Cauchy equation is
t2 y �� + aty � + b y = 0.
(a) Show that the change of independent variable s = ln t results in an equation with
constant coefficients, namely
d2 y
dy
+ (a − 1) + b y = 0.
ds2
ds
(Hint:
dy
dt
=
ds dy
dt ds .)
(b) Let λ1 and λ2 be the roots of the quadratic polynomial λ2 + (a − 1)λ + b.
Describe the general solution of the homogeneous Euler-Cauchy equation in each
of the three cases: (i) λ1 , λ2 real and distinct, (ii) λ1 = λ2 , and (iii) λ1 , λ2 = α±β i.
Find the general solution of the Euler-Cauchy equations in Problems 26 through 31.
26. t2 y �� − 3ty � + 3y = 0
27. t2 y �� + 5ty � + 4y = 0
28. t2 y �� + ty � + y = 0
29. t2 y �� + 5ty � + 3y = 0
30. t2 y �� + 3ty � + 2y = 0
31. t2 y �� + ty � − y = 0
5.5. Nonhomogeneous Equations
21
5.5 Nonhomogeneous Equations
In this section we will begin to address the problem of finding a particular
solution of a nonhomogeneous equation
y �� + py � + qy = f.
We begin with a discussion of the operator form of the equation, which we will
exploit with a technique called exponential shift in the case where f has an
exponential factor.
Operator Form and the Characteristic Polynomial
It is convenient to define the notation
d
d2
d3
D = , D2 = 2 , D3 = 3 , . . .
dt
dt
dt
In other words, we are defining an operator D that acts on differentiable functions y according to
Dy = y � ,
with “powers” of D representing repeated application of D:
D2 y = D(Dy) = Dy � = y �� ,
D3 y = D(D2 y) = Dy �� = y ��� ,
and so on. We also define the identity operator I —or D0 —by
I y = D0 y = y.
It also makes sense to left-multiply any of these operators by a constant (or
a function) and think of the result as another operator. For instance, 5D2 is an
operator that acts on a function y(t) as follows:
5D2 y = 5y �� .
We can also add two or more operators to obtain another operator. For instance,
D2 + 3D + 2 I is an operator that works like this:
(D2 + 3D + 2 I )y = D2 y + 3Dy + 2 I y = y �� + 3y � + 2y.
Now notice that a linear second-order differential equation,
y �� + py � + qy = f,
can be expressed as
or even more succinctly as
(D2 + pD + q I ) y = f,
P (D)y = f,
22
Chapter 5. Linear Second-Order Equations I
where P is the quadratic polynomial
P (λ) = λ2 + p λ + q.
This is precisely the quadratic polynomial that appears in the characteristic
equation for the associated homogeneous differential equation. This polynomial
is called the characteristic polynomial of the operator P (D).
Given the zeros of P (λ),
�
�
1�
λ1 , λ2 =
−p ± p2 − 4q ,
2
we can write P (λ) in the factored form
P (λ) = (λ − λ1 )(λ − λ2 )
and in the expanded form
P (λ) = λ2 − (λ1 + λ2 )λ + λ1 λ2 .
So it is natural to wonder whether the “factored” operator
(D − λ1 I )(D − λ2 I )
is the same as the “expanded” operator
D2 − (λ1 + λ2 )D + λ1 λ2 I .
In other words, is it true that
�
�
(D − λ1 I )(D − λ2 I )y = D2 − (λ1 + λ2 )D + λ1 λ2 I y
for any given y? We can check as follows:
(D − λ1 I )(D − λ2 I )y = (D − λ1 I )(y � − λ2 y)
= D(y � − λ2 y) − λ1 (y � − λ2 y)
= y �� − (λ1 + λ2 )y � + λ1 λ2 y
�
�
= D2 − (λ1 + λ2 )D + λ1 λ2 I y.
So, indeed, the two operators are identical. Thus it makes sense to factor
an operator P (D) = D2 + pD + q I or expand a factored operator such as
(D − λ1 I )(D − λ2 I ) in the natural way, exactly as we would factor the polynomial P (λ) = λ2 + pλ + qλ or expand the product (λ − λ1 I )(λ − λ2 I ).∗
* This relies on p and q being constants; it is not true in general when p or q is a non-constant
function of t.
5.5. Nonhomogeneous Equations
23
Exponential Shift
Consider the operator D applied to a product of the form w(t)ek t . By the
product rule,
D(wek t ) = w� ek t + w kek t = ek t (w� + kw).
So the result is ek t times the operator D + k I applied to w; that is,
D(wek t ) = ek t (D + k I )w.
(1)
This is the most basic form of exponential shift, which may be viewed as a
shortcut for differentiating w(t)ek t without using the product rule. It essentially
says that we can bring the ek t factor outside the operator by replacing D with
D + kI.
• Example 1 We can differentiate the product e−2t cos 3t using exponential shift
(as a shortcut for the product rule) as follows:
D(e−2t cos 3t) = e−2t (D − 2 I ) cos 3t = e−2t (−3 sin 3t − 2 cos 3t).
�
• Example 2 Suppose that we want to find an antiderivative of t2 e−3t (which
is typically found using integration by parts). This is equivalent to finding a
particular solution of the differential equation
Dy = t2 e−3t .
It makes sense to look for y in the form of w(t)e−3t , where w is to be determined.
Substituting y = w(t)e−3t into the differential equation gives us
D (we−3t ) = t2 e−3t ,
which by exponential shift becomes
e−3t (D − 3 I )w = t2 e−3t .
So the w we want satisfies
w� − 3w = t2 .
Now, substituting w = at2 + bt + c gives us
2at + b − 3 (at2 + bt + c) = t2 ,
which simplifies to
−3at2 + (2a − 3b)t + b − 3c = t2 .
By equating coefficients we find a = −1/3, b = −2/9, and c = −2/27. So an
antiderivative of t2 e−3t is
�
1 � 2
y=−
9t + 6t + 2 e−3t .
�
27
24
Chapter 5. Linear Second-Order Equations I
Now consider an operator P (D) = D2 + pD + q I applied to a product of
the form w(t)ek t . First observe that two applications of the basic rule (1) for
D(wek t ) produce an analogous rule for D2 (wek t ) :
D2 (wek t ) = D(D(wek t )) = D(ek t (D + k I )w)
= ek t (D + k I )(D + k I )w
= ek t (D + k I )2 w.
Consequently,
P (D) (wek t ) = D2 (wek t ) + p D(wek t ) + q wek t
= ek t (D + k I )2 w + p ek t (D + k I )w + q wek t
�
�
= ek t (D + k I )2 + p (D + k I ) + q I w
= ek t P (D + k I )w.
So again, the ek t factor may be brought outside of the operator P (D) by replacing D with D + k I . This result, which is stated in Theorem 1 for general
linear operators, can greatly simply many tedious calculations. The technique
it provides is called exponential shift.
THEOREM 1 Let P (D) = pn Dn +pn−1 Dn−1 +· · ·+p1 D +p0 I , where pn , . . . , p0
may be constants or functions of the independent variable. Then for any n-times
differentiable function w and any constant k,
P (D)(wek t ) = ek t P (D + k I )w.
(2) �
Particular Solutions of Nonhomogeneous Equations
Consider the problem of finding a particular solution of
y �� + py + qy = φ(t)ek t ,
(3)
where φ(t) is a given continuous function (possibly constant). Note that the
operator form of the equation is
P (D)y = φ(t)ek t ,
where P (λ) = λ2 + pλ + q. We will use exponential shift to prove the following
theorem.
THEOREM 2 The function y = w(t)ek t is a solution of
P (D)y = φ(t)ek t
if and only if w satisfies
� 2
�
D + P � (k)D + P (k) I w(t) = φ(t).
(4)
(5) �
5.5. Nonhomogeneous Equations
25
Proof. We begin by substituting y = w(t)ek t into (4):
P (D)(w(t)ekt ) = φ(t)ekt .
Now by exponential shift, this becomes
and division by ekt gives
ekt P (D + k I )w(t) = φ(t)ekt ,
P (D + k I )w(t) = φ(t).
(6)
The operator P (D + k I ) can be expanded as follows:
P (D + k I ) = (D + k I )2 + p (D + k I ) + q I
= D2 + (2k + p)D + (k 2 + pk + q) I .
Now, since P � (λ) = 2λ + p, we can write
P (D + k I ) = D2 + P � (k)D + P (k) I .
Thus equation (6) can be written as
� 2
�
D + P � (k)D + P (k) I w(t) = φ(t),
which is equation (5).
Now consider the situation in which the function φ is constant; i.e.,
y �� + py + qy = c0 ek t .
(7)
By Theorem 2, there will be a solution y = w(t)ekt , provided that w satisfies
� 2
�
D + P � (k)D + P (k) I w(t) = c0 .
(8)
If P (k) �= 0, this equation has the constant solution
c0
w(t) =
.
P (k)
However, if P (k) = 0 and P � (k) �= 0, then equation (8) becomes
�
�
D + P � (k)I Dw = c0 ,
which has a solution satisfying
Dw =
c0
,
P � (k)
w(t) =
c0
t.
P � (k)
and so we can take
If P (k) = P � (k) = 0 (i.e., k is a double root of P (λ)), then equation (8) becomes
D 2 w = c0 ,
26
Chapter 5. Linear Second-Order Equations I
which gives
1
c 0 t2 .
2
These results are summarized in the following theorem.
w(t) =
THEOREM 3 Let P (λ) = λ2 + pλ + q. Then:
(i) If P (k) �= 0, then equation (7) has the particular solution
c0 kt
y=
e .
P (k)
(ii) If P (k) = 0 and P � (k) �= 0, then equation (7) has the particular solution
c0
y= �
t ekt .
P (k)
(iii) If P (k) = P � (k) = 0, then equation (7) has the particular solution
c0 2 kt
y=
t e .
2
�
• Example 3 Consider the problem of finding a particular solution of
y �� + 2y � + 2y = 2e−3t .
First notice that the characteristic polynomial associated with the left side of
the equation is P (λ) = λ2 + 2λ + 2. So P (−3) = 5 �= 0. By part (ii) of Theorem
2, a particular solution of the differential equation is therefore y = 25 e−3t .
• Example 4 Consider the problem of finding a particular solution of
y �� − 9y = 9e−3t .
The characteristic polynomial associated with the left side of the equation is
P (λ) = λ2 − 9. So P (−3) = 0 and P � (−3) = −6 �= 0. By part (ii) of Theorem 2,
a particular solution of the differential equation is therefore
y=
9
3
t e−3t = − t e−3t .
−6
2
• Example 5 Consider the problem of finding a particular solution of
y �� + 4y � + 4y = 4e−2t .
The characteristic polynomial associated with the left side of the equation is
P (λ) = λ2 + 4λ + 4 = (λ + 2)2 . So P (−2) = 0 and P � (−2) = 0. So by part (iii)
of Theorem 2, a particular solution of the differential equation is
y=
4 2 −2t
t e
= 2t2 e−3t .
2
�
5.5. Nonhomogeneous Equations
27
We now consider the situation where the function φ in (3)–(6) is a polynomial
with degree n. By Theorem 2, y = w(t)ekt will be a solution of (3) if
� 2
�
D + P � (k)D + P (k) I w(t) = φ(t).
(9)
Here, a specific formula for a particular solution w(t) is not practical. However,
we can describe the expected form fairly simply. If P (k) �= 0, there will be a
polynomial solution w(t) of degree n. If P (k) = 0 and P � (k) �= 0, then (9)
becomes
�
�
D + P � (k)I Dw = φ(t),
and so there will be a solution for which Dw is a polynomial with degree n.
Antidifferentiation then gives a polynomial of degree n + 1, in which we can
assume the constant term is zero. If P (k) = P � (k) = 0, then (9) becomes simply
D2 w = φ(t).
Antidifferentiation then gives a polynomial of degree n + 2, in which we can
assume that the lowest degree term has degree 2. These observations lead to the
following theorem.
THEOREM 4 Let P (λ) = λ2 + pλ + q. If φ(t) is a polynomial of degree n with
leading coefficient cn , then equation (3) has a particular solution y = w(t)ekt ,
where w(t) is a polynomial for which the following are true:
(i) If P (k) �= 0, then w(t) has degree n.
(ii) If P (k) = 0 and P � (k) �= 0, then w(t) = t ψ(t), where ψ is a polynomial of
degree n.
(iii) If P (k) = P � (k) = 0, then w�� (t) = φ(t).
�
• Example 6 Consider the problem of finding a particular solution of
y �� + 3y � + 10y = 10t2 e−3t .
The operator form of the equation is
(D2 + 3D + 10 I )y = 10t2 e−3t .
The characteristic polynomial is P (λ) = λ2 + 3λ + 10; so P (−3) = 10. So
part (ii) of Theorem 4 tells us to look for a particular solution of the form
y = (at2 + bt + c)e−3t . So we substitute this into the operator form of the
equation and use exponential shift:
e−3t ((D − 3I )2 + 3 (D − 3I ) + 10 I )(at2 + bt + c) = 10t2 e−3t .
Thus
(D2 − 3D + 10I )(at2 + bt + c) = 10t2 .
28
Chapter 5. Linear Second-Order Equations I
Computation of the left side gives
2a − 3(2at + b) + 10(at2 + bt + c) = 10t2 .
Equating coefficients gives a = 1, b = 3/5, and c = −1/50. Therefore, we have
the particular solution
y=
1
(50t2 + 30t − 1)e−3t .
50
�
• Example 7 Consider the problem of finding a particular solution of
y �� + 3y � + 2y = 3t e−t .
The characteristic polynomial is P (λ) = λ2 +3λ+2; so P (−1) = 0 and P � (−1) =
1 �= 0. The operator form of the equation is
(D2 + 3D + 2 I )y = 3t e−t .
Part (ii) of Theorem 4 tells us to look for a particular solution of the form
y = t (at + b)e−t . So we put this into the equation and use exponential shift:
et ((D − I )2 + 3 (D − I ) + 2 I ) (t (at + b)) = 3t e−t .
Thus
(D2 + D)(at2 + bt) = 3t.
Computation of the left side gives
2a + 2at + b = 3t.
Equating coefficients produces a = 3/2 and then b = −3. Therefore, we have
the particular solution
y = t (3t/2 − 3)e−t =
3
t (t − 2)e−t .
2
�
• Example 8 Let’s look now at the problem of finding a particular solution of
y �� + 4y � + 4y = 4t (1 − t) e−2t .
The characteristic polynomial is P (λ) = λ2 + 4λ + 4 = (λ + 2)2 . Thus P (−2) =
P � (−2) = 0. By part (iii) of Theorem 4, there will be a particular solution
y = we−2t , where
D2 w = 4t (1 − t).
Antidifferentiating twice gives
w=
2 3 1 4 1 3
t − t = t (2 − t).
3
3
3
5.6. Real Solutions from Complex Solutions
29
So we have the particular solution
1
y = t3 (2 − t)e−2t .
3
Problems
1. Let P (D) = D2 + 3D + 4 I . Compute:
(a) P (D) t2
(b) P (D) sin t
(d) P (D) t−1
(c) P (D) cos 2t
2. Use exponential shift to compute y �� for (a) y = t−1 e 2t and (b) y = e−t ln t.
3. Let P (D) = D2 + D + 2 I . Use exponential shift to compute:
(a) P (D)(t2 et )
(b) P (D)(t e−3t )
(c) P (D)(e−t sin t)
(d) P (D) t−1 et
4. Use exponential shift to find a solution of y �� = 4t2 e−2t in the form y = w(t)e−2t .
5. Use exponential shift to find a solution of y �� = (t2 − t)e−t in the form y = w(t)e−t .
In Problems 6 through 8, use Theorem 3 to find a particular solution.
6. y �� + 5y � + y = 7e−t/2
7. y �� + 4y � + 3y = 6e−3t
8. y �� + 6y � + 9y = e−3t
In Problems 9 through 17, write the equation in operator form and use exponential shift
along with Theorem 3 or 4 to find a particular solution.
9. y �� + y = te−t
12. y �� + y � + 2y = t2 e−3t
15. y �� + 3y � + 2y = t2 e−t
10. y �� − y = te−t
11. y �� + 2y � + y = te−t
13. y �� + y � + y = te−t
14. y �� + 2y � + y = te−2t
16. y �� + y � + y = t2 e−2t
17. y �� + 43 y � + 13 y = te−t
Find the general solution of each equation in Problems 18 through 20.
18. y �� + 2y � + 2y = te−2t
19. y �� + y � + y/4 = te−t/2
20. y �� + 4y � + 3y = 3e−3t
Solve the given initial-value problem in Problems 21 through 24.
21. y �� − 4y = 9te t , y(0) = 1, y � (0) = 0
23. y �� + 5y + 6y = te−2t , y(0) = y � (0) = 0
22. y �� + 4y = te−t , y(0) = 0, y � (0) = 1
24. y �� + 2y � + y = e−t , y(0) = y � (0) = 0
25. (a) Show that the operator (D + t I )(D − tI ) is not the same operator as D2 − t2 I .
(b) Show that the operator (D + t I )(D − tI ) is not the same as (D − t I )(D + tI ).
(c) What are the correct expanded forms of (D + t I )(D − tI ) and (D − t I )(D + tI )?
26. Find the general solution of y �� + 3y � + 2y = 0 as follows:
i) Write the equation in operator form and factor the operator to obtain
(D + I )(D + 2 I )y = 0.
ii) Let w = (D + 2 I )y and solve the first-order equation (D + I )w = 0 for w.
iii) Solve the first-order equation (D + 2 I )y = w for y.
30
Chapter 5. Linear Second-Order Equations I
5.6 Real Solutions from Complex Solutions
In the Section 5.4 we saw how the complex solution e(α+iβ)t gives rise to two
real solutions through its real and imaginary parts, eαt cos βt and eαt sin βt.
In this section we will use a similar idea to help us find particular solutions of
nonhomogeneous equations when the nonhomogeneous term has a trigonometric
form. The basis of this is the following theorem.
THEOREM 1 Let f and g be continuous and real valued on an interval I. Then
the complex-valued function z = x + iy satisfies the complex equation
z �� + pz � + qz = f + i g
if and only if
x�� + px� + qx = f and y �� + py � + qy = g.
�
Let’s think for a moment about why this is true. Suppose that z = x + iy.
Then, by definition,
z � = x� + iy � and z �� = x�� + iy �� .
Thus, a little rearranging shows that
z �� + pz � + qz = (x�� + px� + qx) + i (y �� + py � + qy).
Thus z �� + p z � + q z = f + i g if and only if
x�� + px� + qx = f and y �� + py � + qy = g,
because two complex numbers are equal if and only if their real parts are equal
and their imaginary parts are equal.
The following corollary is a special case of Theorem 1, since
e(ρ+iω)t = e ρt cos ωt + i e ρt sin ωt.
COROLLARY 1 Let c0 be a real constant. Then x and y satisfy
x�� + px� + qx = c0 e ρt cos ωt and y �� + py � + qy = c0 e ρt sin ωt
if and only if the complex-valued function z = x + iy satisfies the complex
equation
z �� + pz � + qz = c0 e(ρ+ω i)t .
(1) �
The upshot of this is that we can simultaneously find particular solutions of the
real equations
x�� + px� + qx = c0 e ρt cos ωt and y �� + py � + qy = c0 e ρt sin ωt
5.6. Real Solutions from Complex Solutions
31
by finding a particular solution of the complex equation (1) and then extracting
its real part x and its imaginary part y. While doing this demands that we use
complex arithmetic, the exponential form of the nonhomogeneous term in the
complex equation simplifies the computation considerably.
An added advantage of this approach is that Theorem 3 of the previous
section remains true for equation (1). Note that part (iii) of that theorem never
applies, because complex roots of P (λ) are always distinct. We particularize the
result for our purposes here as
THEOREM 2 Suppose that P (λ) = λ2 +pλ+q has complex roots α±β i. Then:
(i) If ρ + ωi equals neither of α ± β i, then (1) has the particular solution
c0
z=
e(ρ+ω i)t .
P (ρ + ωi)
(ii) If ρ + ωi equals one of α ± β i, then (1) has the particular solution
c0
z= �
t e(ρ+ω i)t .
P (ρ + ωi)
�
• Example 1 Consider the problem of finding particular solutions of
x�� + x� + 2x = 2 cos t and y �� + y � + 2y = 2 sin t.
Since the nonhomogeneous terms are the real and imaginary parts of 2e it , respectively, we can obtain both solutions at once by finding a particular solution
of the complex equation
z �� + z � + 2z = 2e it
and then extracting real and imaginary parts. The characteristic polynomial is
P (λ) = λ2 + λ + 2; thus P (i) = i2 + i + 2 = 1 + i. Since that is not zero, it
follows that a particular complex solution is
z=
Now, since
2
1+i
=
2 (1−i)
2
2
2
e it =
e it .
P (i)
1+i
= 1 − i, we have
z = (1 − i)e it .
We finish the job by finding the real and imaginary parts of z:
z = (1 − i)e it = (1 − i)(cos t + i sin t) = cos t + sin t + i (sin t − cos t).
Therefore,
x = cos t + sin t and y = sin t − cos t
are particular solutions of the original real equations.
32
Chapter 5. Linear Second-Order Equations I
• Example 2 Consider the problem of finding a particular solution of
y �� + y = e−t cos t.
Since e−t cos t is the real part of the complex function e−t e it = e(−1+i)t , the
particular solution we seek will be the real part of a particular solution of
z �� + z = e(−1+i)t .
The characteristic polynomial is P (λ) = λ2 + 1; so
P (−1 + i) = (−1 + i)2 + 1 = 1 − 2i + i2 + 1 = 1 − 2i.
Since that is not zero, a particular solution is
z=
Since
1
1−2i
=
z=
1+2i
5 ,
1
e(−1+i)t .
1 − 2i
we have
1 + 2i (−1+i)t
1
e
= (1 + 2i) e−t (cos t + i sin t)
5
5
�
�
1
= e−t cos t − 2 sin t + i (2 cos t + sin t) ,
5
from which we can see that the real part is
y=
1 −t
e (cos t − 2 sin t).
5
• Example 3 Let’s find a particular solution of the equation
y �� + 2y � + 2y = e−t sin t.
Since e−t sin t is the imaginary part of e−t e it = e(−1+i)t , the particular solution
we seek will be the imaginary part of some particular solution of
z �� + 2z � + 2z = e(−1+i)t .
The characteristic polynomial is P (λ) = λ2 + 2λ + 2; thus
P (−1 + i) = (−1 + i)2 + 2(−1 + i) + 2 = 1 − 2i + i2 − 2 + 2i + 2 = 0.
So we use conclusion (ii) of Theorem 2. Since P � (λ) = 2λ + 2, we have
P � (−1 + i) = 2(−1 + i) + 2 = 2i.
So a particular complex solution is
z=
1 (−1+i)t
i
te
= − te(−1+i)t .
2i
2
5.6. Real Solutions from Complex Solutions
33
Now, using Euler’s formula,
z==−
i −t
te (cos t + i sin t)
2
1 −t
te (sin t − i cos t).
2
Finally, we extract the imaginary part:
1
y = − te−t cos t.
2
=
Using Polar Form to Do the Division
The results of Theorem 2 involve division by a complex number: either P (ρ+ωi)
or P � (ρ + ωi). By writing this denominator in polar form Reφi , the division may
be done by dividing e(ρ+ω i)t by eφi . This gives the exponential factor
e(ρ+ω i)t
= e(ρ+ω i)t−φi = e ρt+(ω t−φ)i = e ρt e(ω t−φ)i ,
eφi
which gives real and imaginary parts in a nice form that reveals the angle φ as
a phase angle for the trigonometric terms:
e ρt e(ω t−φ)i = e ρt ( cos(ωt − φ) + i sin(ωt − φ))
• Example 4 Let us illustrate the method by finding the real and imaginary parts
of the function
1
z=
e(−1+2i)t .
−2 + 3i
The polar form of the denominator is
�
√
−2 + 3i = 22 + 32 eφi = 13 eφi ,
√
√
where cos φ = −2/ 13 and sin φ = 3/ 13. Since cos φ < 0 and sin φ√> 0, φ is a
second-quadrant angle, and so we may take it to be φ = cos−1 (−2/ 13). Now
z=√
1
1
1
e(−1+2i)t = √ e(−1+2i)t−φi = √ e−t+(2t−φ)i ;
φi
13 e
13
13
so
1
z = √ e−t ( cos(2t − φ) + sin(2t − φ)).
13
Therefore its real and imaginary parts are
1
1
x = √ e−t cos(2t − φ) and y = √ e−t sin(2t − φ),
13
13
√
where φ = cos−1 (−2/ 13) ≈ 2.159.
34
Chapter 5. Linear Second-Order Equations I
• Example 5 Say we want to find a particular solution of
y �� + y � + 7y = 7 cos 5t.
The associated complex equation is
z �� + z � + 7z = 7e5it .
The characteristic polynomial is P (λ) = λ2 + λ + 7, and its value at 5i is
P (5i) = −25 + 5i + 7 = 18 + 5i. So a particular complex solution is
7
e5it .
18 + 5i
Now we write 18 + 5i in polar form:
�
√
18 + 5i = 182 + 52 eφi = 349 eφi ,
√
√
where cos φ = 18/ 349 and sin φ = 5/ 349. This means that φ is a firstquadrant angle; so we may as well write it as φ = tan−1 (5/18). Now
z=
7
7
e5it = √
e(5t−φ)i .
φi
349 e
349
Since our real differential equation has cosine on its right side, we want y to be
the real part of z, which is
7
cos(5t − φ),
y=√
349
z=√
where φ = tan−1 (5/18) ≈ 0.271.
Problems
In Problems 1 through 6, use an appropriate complex equation to find a pair of particular
solutions.
� ��
� ��
x + x = cos t
x + x = e−t cos t
1.
2.
y �� + y = sin t
y �� + y = e−t sin t
� ��
�
x + x� + 3x = 5 cos t
x�� + x� + 5x = cos 2t
3.
4.
y �� + y � + 3y = 5 sin t
y �� + y � + 5y = sin 2t
� ��
�
x − x = 8e−t cos 2t
x�� + x� + x = e−2t cos t
5.
6.
��
−t
y − y = 8e sin 2t
y �� + y � + y = e−2t sin t
7. Use the results of Problem 1 to write down a particular solution of
(a) y �� + y = cos t + sin t
(b) y �� + y = 3 sin t − cos t
8. Use the results of Problem 2 to write down a particular solution of
(a) y �� + y = e−t (cos t + sin t)
(b) y �� + y = 3e−t (5 sin t − cos t)
5.7. Unforced Vibrations
35
9. Use the results of Problem 3 to write down a particular solution of
(a) y �� + y � + 3y = 10 cos t + sin t
(b) y �� + y � + 3y = 15 sin t − 5 cos t
10. Use the results of Problem 4 to write down a particular solution of
(a) y �� + y � + 5y = cos 2t + sin 2t
(b) y �� + y � + 5y = 3 sin 2t − cos 2t
11. Use the results of Problem 5 to write down a particular solution of
(a) y �� − y = 8e−t (cos 2t + 2 sin 2t)
(b) y �� − y = 8e−t (3 sin 2t − 2 cos 2t)
12. Use the results of Problem 6 to write down a particular solution of
(a) y �� + y � + y = 13e−2t (cos t + 2 sin t)
(b) y �� + y � + y = 13e−2t (5 cos t − sin t)
In Problems 13 through 15, find the general solution.
13. y �� + 2y � + 2y = 5 cos t
14. y �� + 3y � + 2y = e−t sin t
15. y �� + y = sin t
In Problems 16 through 18, solve the initial-value problem.
16. y �� + 3y � + 2y = 10 cos t, y(0) = 1, y � (0) = 0
17. y �� + 2y � + 5y = 17 cos 2t, y(0) = 0, y � (0) = 0
18. y �� + 2y � + 2y = 5 cos t, y(0) = 1, y � (0) = 0
The purpose of Problems 19 through 21 is to look at the method of undetermined
coefficients, which is an approach to finding particular solutions that is an alternative
to using a complex equation.
19. Find a solution of y �� + y � + 10y = 10 sin 3t in the form y = A cos 3t + B sin 3t.
20. Find a solution of y �� +y � +10y = 10e−t sin 3t in the form y = e−t (A cos 3t+B sin 3t).
(Exponential shift will help here.)
21. Find a solution of y �� + 2y � + 2y = e−t sin t in the form y = t e−t (A cos t + B sin t).
(Again, exponential shift will make the job a little easier.) Why do you think that
the factor of t is needed in this trial solution?
5.7 Unforced Vibrations
In Section 5.1 we learned that models of simple spring-mass systems and electrical circuits give rise to Linear Second-Order differential equations with constant
coefficients. The equations arising from unforced systems are homogeneous. In
this section we will look at these equations and their solutions. A brief review of
Section 5.1 might be appropriate before proceeding with the rest of this section.
Recall the equation of motion for an damped, unforced spring-mass system:
my �� + ry � + ky = 0,
where m, r, and k are the mass, damping coefficient, and spring stiffness, respectively. The equation for the current in an unforced RLC circuit is
L I �� + R I � + I/C = 0
36
Chapter 5. Linear Second-Order Equations I
with L, R, and C being the inductance, resistance, and capacitance, respectively.
By dividing through by the leading coefficient in each of these equations, we
arrive at an equation of the form
y �� + py � + qy = 0.
(1)
Since p ≥ 0 and q > 0, this a special case of the homogeneous equations we
have seen already in this chapter. Note that the coefficient p will be zero only
if there is no damping in the system.
Undamped, Unforced Vibrations
When no damping is present, we have p = 0, and so the equation takes the form
y �� + qy = 0 with q > 0. To simplify the notation in what follows, let q = ω02 , so
that the equation becomes
y �� + ω02 y = 0.
The general solution of this equation is easily seen to be
y = c1 cos ω0 t + c2 sin ω0 t.
Thus, ω0 is the angular frequency (in radians per unit time) of the unforced
vibration. The frequency of the vibration is ω2π0 cycles per unit time. We call
these quantities the natural angular frequency and natural frequency of the
system. An undamped system can vibrate at a different frequency, but only if
an external force is applied.
In the case of an undamped spring-mass system, we have q = k/m, and so
the natural angular frequency is
�
ω0 = k/m .
In the case of an LC circuit, we have q = 1/(LC), and so the natural angular
frequency is
√
ω0 = 1/ LC.
We have essential determined that undamped, unforced vibrations are purely
sinusoidal in nature with a natural frequency determined solely by system parameters. All of this is independent of the initial state of the system. The amplitude of the vibration, however, is determined by initial conditions.
Consider a solution y = c1 cos ω0 t + c2 sin ω0 t. We would like to find the
amplitude of this sinusoidal solution. To do so we need to express the solution
as a single sine or cosine term. Recall the trigonometric identity
cos(α − β) = cos α cos β + sin α sin β.
5.7. Unforced Vibrations
37
Because of this, we introduce numbers A > 0 and φ, related to c1 and c2 by
c1 = A cos φ and c2 = A sin φ,
and then express y as
y = A cos(ω0 t − φ),
where
A=
�
c21 + c22 and tan φ =
c2
.
c1
The quadrant in which the angle φ lies is determined by the signs of c1 and c2 .
A is the amplitude of the solution, and φ is called the phase angle or angular
phase shift. The actual phase shift (with units of time) is φ/ω0 ; i.e., the
graph of cos(ω0 t − φ) is that of cos ω0 t shifted horizontally by φ/ω0 .
Let us now look at the initial-value problem
y �� + ω02 y = 0, y(0) = y0 , y � (0) = v0 .
It is easy to determine that the solution is
y = y0 cos ω0 t +
v0
sin ω0 t.
ω0
From this and the preceding discussion, we can write the solution as
�
v0
y = A cos(ω0 t − φ), where A = y02 + v02 /ω02 and tan φ =
.
y0 ω 0
• Example 1 Consider the initial-value problem
√
y �� + 9y = 0, y(0) = −2 3, y � (0) = 6.
The general solution of the differential equation is described by
y = c1 cos 3t + c2 sin 3t.
√
The initial conditions require c1 = −2 3 and c2 = 2. Thus the amplitude and
phase angle are given by
√
2
1
√ = −√ .
A = 12 + 4 = 4, tan φ =
−2 3
3
Since c1 < 0 and c2 > 0, φ should be a second quadrant angle. Therefore, we
take φ = 5π/6, and so the solution may be written as
y = 4 cos(3t − 5π/6).
38
Chapter 5. Linear Second-Order Equations I
Damped, Unforced Vibrations
The motion of a damped, unforced spring-mass system and the current in an
unforced RLC circuit are each governed by a differential equation of the form (1)
where p and q are both positive. The roots of the corresponding characteristic
equation are, as usual,
�
�
1�
λ 1 , λ2 =
−p ± p2 − 4q .
2
So to examine the solutions of (1), we need to consider the following three cases.
Case 1: p2 − 4q > 0. Here, we
� have two distinct real roots λ1 and λ2 , which are
negative because p > 0 and p2 − 4q < p. Thus the general solution is
y = c 1 eλ1 t + c 2 eλ2 t ,
and we can see that in this case there is no vibration and every solution must
tend to zero as t → ∞. This is called the overdamped case.
Case 2: p2 − 4q = 0. Here, we have the repeated real root λ1 = λ2 = −p/2,
which is necessarily negative because p > 0. Thus the general solution is
y = (c1 + c2 t)e−pt/2 ,
and we can see that again there is no vibration and every solution must tend to
zero as t → ∞. This is called the critically damped case.
Case 3: p2 − 4q < 0. In this case the roots are the complex conjugates
�
�
1�
α ± iβ =
−p ± i 4q − p2 .
2
Thus the general solution is
where β =
�
y = e−pt/2 (c1 cos β t + c2 sin β t) ,
4q − p2 /2. Just as in the undamped case we can write this as
y = Ae−pt/2 cos(β t − φ),
�
where A = c21 + c22 and tan φ = c2 /c1 . Here we have damped vibration in which
we can think of the quantity Ae−pt/2 as the exponentially decaying amplitude.
This is called the underdamped case and is the only one of the three cases in
which the system actually vibrates. It is important to observe the distinct roles
played by the real and imaginary parts of the roots of the characteristic equation.
The imaginary part determines the angular frequency of the vibrations and the
real part determines the rate of damping. Just as in the first two cases, every
solution must tend to zero as t → ∞, because the roots have negative real part.
We have learned that the behavior of a damped, unforced system can be
characterized in terms of the roots of the corresponding characteristic equation.
The system experiences (damped) vibrations precisely when these roots have
5.7. Unforced Vibrations
39
nonzero imaginary parts. The undamped, unforced case also fits nicely into this
picture as the case in which the roots of the characteristic equation are purely
imaginary. Note that case 3 reduces to the undamped case when p = 0.
Figures 1a–1d show the solution of the initial-value problem
1
y �� + py � + 4y = 0, y(0) = , y � (0) = −2
2
with p = 5, 4, 2, and 0, respectively. Figures 1a–c illustrate cases 1–3, while Figure 1d illustrates the undamped case. These four pictures essentially illustrate
the possible behaviors of an unforced system.
Figure 1a
Figure 1b
Figure 1c
Figure 1d
Problems
1. An undamped spring-mass system has mass m = 2 slugs and stiffness k = 8 lb/ft.
(a) Find the natural angular frequency and the natural frequency of the system.
(b) Find the resulting motion if the mass is gently released from an initial position
of y0 = .75 ft. What is the amplitude of the motion?
(c) Find the resulting motion if the mass is set in motion from its equilibrium position
with an initial velocity of v0 = 2 ft/s. What is the amplitude of the motion?
(d) Find the resulting motion if the mass is set in motion from an initial position of
y0 = .75 ft with an initial velocity of v0 = 2 ft/s. What is the amplitude?
2. An LC circuit has inductance L = 1 mH and capacitance C = 10 µF. Find the
natural angular frequency and the natural frequency of the circuit.
3. A damped spring-mass system has mass m, damping coefficient r, and stiffness k.
Give conditions on m, r, and k which lead to the (a) overdamped, (b) critically
damped, and (c) underdamped cases.
4. Suppose we use a spring-mass system as a simple model of a car’s suspension system.
We determine that the car has mass m = 1000 kg and that the stiffness of the
suspension is k = 16, 000 N/m. What amount of damping from shock absorbers
would critically damp any unforced motion?
5. Consider an RLC circuit with resistance R, inductance L, and capacitance C. Give
conditions on R, L, and C which lead to the (a) overdamped, (b) critically damped,
and (c) underdamped cases.
6. An LC circuit with C = 1 µF has a natural angular frequency of 105 radians per second. What amount of resistance added to the circuit would critically damp unforced
oscillations in the current?
40
Chapter 5. Linear Second-Order Equations I
7. (a) Determine the constants c1 , c2 in terms of initial values y(0) = y0 , y � (0) = v0 for
the general solution in the underdamped case.
(b) Write the solution in the form y = Ae−pt/2 cos(ω t − φ) with A, φ expressed in
terms of initial values y(0) = y0 , y � (0) = v0 .
8. (a) Determine the constants c1 , c2 in terms of initial values y(0) = y0 , y � (0) = v0 for
the general solution in the overdamped case.
(b) Repeat part (a) for the critically damped case.
9. A damped, unforced spring-mass system is observed to vibrate at an angular frequency of 6 radians per second with an amplitude that decreases by a factor of 1/2
every 3 seconds.
(a) Find the differential equation that governs the motion.
(b) If the spring stiffness is known to be k = 1 lb/ft, find the mass m and the damping
coefficient r.
10. Consider a damped spring-mass system with stiffness k = 8 dynes/cm and damping coefficient r = 16 dynes/(cm/sec). Find the general solution of the differential
equation that governs unforced motion of the mass if
(a) m = 6 grams
(b) m = 8 grams
(c) m = 10 grams
11. (a–c) For the spring mass systems in Problem 10, find the motion of the mass if it
is gently released from an initial position of y0 = 10 cm.
12. Consider a spring-mass system with mass m = 2 kg and stiffness k = 18 N/m. Find
the general solution of the differential equation that governs unforced motion of the
mass if the damping coefficient (in units of N/(m/s)) is
√
(a) r = 0
(b) r = 4 5
(c) r = 12
(d) r = 20
13. (a–d) For the spring-mass systems in Problem 12, find the motion of the mass if it
is set in motion from its equilibrium position with an initial velocity of v0 = 1 m/s.
14. Solve for the solutions shown in Figures 1a–1d.
5.8 Periodic Force and Response
In the preceding section we investigated the behavior of unforced spring-mass
systems and electrical circuits governed by second-order, linear differential equations with constant coefficients. From a physical point of view, an unforced system that is initially at rest should remain at rest for all time. Since the governing
differential equation for an unforced system is homogeneous, this principle is reflected in the fact that the “rest” solution (i.e., with zero initial conditions) is in
fact the zero solution. Moreover, any nontrivial behavior of an unforced system
results solely from being initially perturbed from rest. Such unforced motion is
often referred to as the free response of the system. One of the lessons of the
5.8. Periodic Force and Response
41
last section was that the type of free response depends only on the coefficients
of the differential equation (i.e., system characteristics) and not on the initial
conditions.
In this section we will investigate forced response, i.e., response to an
external force, specifically when that force is periodic. Because of the form of
equations (5) and (6) in Section 5.1, we will write our equation in the form
y �� + py � + ω02 y = ω02 f (t).
(1)
Recall
that, for the spring-mass system shown in Figure 3 of Section 5.1, ω0 =
�
k/m and f (t) may be thought of as the position of the support to which the
spring is attached. (The resulting external force is actually kf (t).) On the other
1
hand, if y represents the voltage in an RLC circuit, then ω0 = √LC
and f (t) is
the source voltage that drives the circuit. In any case, we refer to the function
f (t) as a forcing function or input to the system. Our goal is to understand how
the system “responds” to a given periodic input.
For this purpose, there is no need for generality; we could study any particular
solution of (1). However, it is sensible to concentrate on either (a) the rest
solution of the equation, which represents behavior that is caused solely by
the input, or (b) the steady-state solution, which contains no terms that
satisfy the homogeneous equation. Since p and ω02 are both positive, the roots of
the characteristic polynomial have negative real part, and so every solution of
the homogeneous equation approaches 0 as t → ∞. Thus every solution of (1)
approaches the steady-state solution as t → ∞. So in that sense the steady-state
solution is what remains after any free-response has died away.
Gain and Phase Angle
A periodic input f may be written as (or at least approximated by) a linear
combination of sinusoidal terms with various angular frequencies. When f has
that form, any particular solution of (1) will be a linear combination of particular
solutions corresponding to the individual terms. So we will concentrate on simple
sinusoidal inputs f in the form
f (t) = sin ωt or cos ωt.
We assume throughout that ω > 0 and refer to ω as the angular input frequency.
ω
(The input frequency is 2π
.)
We will see that in most cases, the steady-state solution will be sinusoidal
in nature with the same angular frequency ω, and so the response may be
characterized by the gain and phase angle of the system at that angular
input frequency. The gain is defined as
G=
amplitude of the steady-state solution
.
amplitude of the input
42
Chapter 5. Linear Second-Order Equations I
Since we are using inputs with amplitude 1, the gain will be precisely the amplitude of the steady-state solution. The phase angle is the difference in phase
between the steady state solution and the input. For instance, if the input cos ωt
results in the steady-state solution
G cos(ωt − φ),
then the phase angle is φ. For a given system, the gain and phase angle are
functions of ω and together describe the frequency response of the system.
To help us find particular solutions we will use the complex equation
z �� + p z � + ω02 z = ω02 eiω t .
Once we find a particular complex solution of this equation, we can extract its
real and imaginary parts to obtain particular real solutions of (1) corresponding
to f (t) = cos ωt and f (t) = sin ωt, respectively. However, the gain and phase
angle at angular input frequency ω may be obtained directly from the complex
steady-state solution. If the complex steady-state solution is given by
z = A e iω t ,
where A is a complex constant with polar form A = |A| e−iφ , then
z = |A| e−iφ e iω t = |A| e i(ω t−φ) .
Thus the gain is G = |A| and φ is the phase angle.
Forced Response, No Damping
When p = 0, ω0 is the natural angular frequency of the system. Our present
purpose is to study the steady-state solution of the complex equation
z �� + ω02 z = ω02 e iω t .
Case 1: ω �= ω0 ; i.e., the input frequency is different from the natural frequency
of the system. The characteristic polynomial is P (λ) = λ2 + ω02 , so P (iω) =
ω02 − ω 2 . Consequently, the complex steady-state solution is
z=
ω02
e iω t .
ω02 − ω 2
(2)
Note that real and imaginary parts are, respectively,
x=
Therefore the gain is
ω02
ω02
cos
ωt
and
y
=
sin ωt .
ω02 − ω 2
ω02 − ω 2
G = |A| =
ω02
.
|ω02 − ω 2 |
(3)
5.8. Periodic Force and Response
43
So notice that inputs with frequencies close to the natural frequency are greatly
amplified; in fact, it is easy to see that G → ∞ as ω0 − ω → 0. Also note that
if ω0 is fixed, then G → 0 as ω → ∞, and if ω is fixed, then G → 1 as ω0 → ∞.
Therefore, if ω is very large relative to ω0 , then the amplitude of the response is
much smaller than that of the input, and if ω0 is very large relative to ω, then
the response is nearly identical to the input.
Now to determine the phase angle, we use the fact that −1 = e−iπ to write
the polar form of the constant A as
�
G
if ω < ω0 ,
A=
G e−iπ if ω > ω0 .
Consequently
z=
�
G e iω t
G ei (ω t−π)
if ω < ω0 ,
if ω > ω0 .
So the phase angle is φ = 0 if ω < ω0 and φ = π if ω > ω0 . Thus the response
is in phase with the input if ω < ω0 and 180◦ out of phase if ω > ω0 .
• Example 1 Consider the equation
y �� + 16y = 16 cos ωt.
For any ω �= 4 a particular solution is
y=
16
cos ωt.
16 − ω 2
For 0 < ω < 4, the solution is a positive multiple of the input. For ω > 4, the
solution is a negative multiple of the input. In either case, the amplitude is very
large when ω ≈ 4.
• Example 2 Consider the equation
y �� + ω02 y = ω02 (cos t + cos 3t + cos 6t).
By superposition and (2), a particular solution is
�
�
cos t
cos 3t
cos 6t
2
y = ω0
+
+
,
ω02 − 1 ω02 − 9 ω02 − 36
provided that ω0 �= 1, 3, 6. If ω0 is close to one of these exceptional values, then
the corresponding term in the solution will dominate the others. In that sense,
the system “selects” any component of the input whose frequency is close to
its natural frequency and amplifies it more than the others. So by changing the
natural frequency of the system, one can “tune” the system to respond to a
particular component frequency of the input.
44
Chapter 5. Linear Second-Order Equations I
• Example 3: Beats The phenomenon of beats occurs when an undamped system
is forced at a frequency near its natural frequency. A low-frequency oscillation is
produced that has a frequency equal to half the difference between the natural
frequency and the frequency of the input. Audible beats are often used to tune
two guitar or piano strings to the same pitch. When the strings vibrate at close
to the same pitch, beats can be heard. As the frequency of one string is tuned
to the other, the beats slow down and disappear.
The existence of beats can be seen in the rest solution of
y �� + ω02 y = ω02 cos ωt.
We have just found that a particular solution of this equation is
y=
Thus the general solution is
ω02
ω02
cos ωt.
− ω2
y = c1 cos ω0 t + c2 sin ω0 t +
ω02
cos ωt.
ω02 − ω 2
The rest solution (see Problem 12) turns out to be
y=
ω02
(cos ωt − cos ω0 t),
ω02 − ω 2
as is easy to check. By one of the “sum to product” identities from trigonometry,
this may be written in the form
�
� �
�
2ω02
ω + ω0
ω − ω0
y= 2
t sin
t .
sin
2
2
ω0 − ω 2
0
When ω ≈ ω0 , the lower-frequency factor 2 sin( ω−ω
2 t) may be thought of as the
0
slowly oscillating amplitude of the higher-frequency factor sin( ω+ω
Untitled-1
2 t). This
slowly oscillating amplitude is called the envelope of the higher-frequency oscillation. Figure 3 shows a typical plot.
�
Figure 3
5.8. Periodic Force and Response
45
Case 2: ω = ω0 . This is the situation in which the system is forced at its own
natural frequency. To find a particular solution of the complex equation
z �� + ω02 z = ω02 e iω0 t ,
note that P (iω0 ) = 0 and P � (iω0 ) = 2iω0 . So we have the complex solution
z=
1
i
t e iω0 t = − ω0 t e iω0 t .
2iω0
2
Since −i = e−π/2 , this becomes
1
ω0 t e i (ω0 t−π/2) ,
2
whose real and imaginary parts are
�
1
π� 1
x = ω0 t cos ω0 t −
= ω0 t sin ω0 t,
2
2
2
�
�
1
π
1
y = ω0 t sin ω0 t −
= − ω0 t cos ω0 t.
2
2
2
◦
So we see that the response oscillates 90 out
of phase with the input. More important, we
see that the amplitude of the response grows
linearly and without bound as t → ∞. This
phenomenon is called resonance. It is what
occurs when an undamped system is forced
at precisely its own natural frequency. For
this reason, the system’s natural frequency ω0
is sometimes called its resonant frequency. A
resonant response is shown in Figure 4.
z=
(6)
Figure 4
Obviously, resonance is undesirable in a
mechanical system (such as a car’s suspension
or an airplane wing) and can result in mechanical failure due to extreme stress.
Thus, knowledge of the natural frequency of a system and the frequencies at
which a system is likely to be forced are important in engineering design.
The undamped model is a highly idealized one; any real physical system will
possess at least a small amount of damping. We will see in what follows that
there is no such thing as true resonance if any amount of damping is present.
However, the gain of a damped system can be extremely large at input frequencies close to the natural frequency of the corresponding undamped system.
46
Chapter 5. Linear Second-Order Equations I
Forced Response with Damping
Our goal here is to find the steady-state solution of
z �� + pz � + ω02 z = ω02 e iωt ,
where p > 0, in order to determine the gain and phase angle of the system at
angular input frequency ω.
• Example 4 Consider the equation
z �� + z � + 10z = 10e3it .
The angular input frequency here is 3, and the amplitude of the input is 1. The
characteristic polynomial evaluated at 3i is P (3i) = −9 + 3i + 10 = 1 + 3i. Thus
a particular solution is
9
z=
e3it .
1 + 3i
√
√
√
Now we write 1+3i = Reiφ where R = 10,
cos φ = 1/ 10, and sin φ = 3/ 10.
√
Since sin φ > 0, we can take φ = cos−1 (1/ 10). So
√
√
10
z=√
e3it = 10 e(3t−φ)i = 10 (cos(3t − φ) + i sin(3t − φ)).
10 eiφ
From this we observe that the gain and phase angle of the system at angular
input frequency 3 are
√
√
G = 10 ≈ 3.16 and φ = cos−1 (1/ 10) ≈ 1.25 ≈ 71.6◦ .
Figure
√ 5 is a plot of the input cos 3t together with the resulting response, which
is 10 cos(3t − φ). (Note that the phase shift is actually φ/3 ≈ 0.42.)
�
3
2
response
input
1
1
2
3
4
5
6
1
2
3
Figure 5
In the general case, the characteristic polynomial is P (λ) = λ2 + pλ + ω02 , so
P (iω) = ω02 − ω 2 + pωi. Thus we have the complex steady-state solution
z=
ω02
e iω t .
ω02 − ω 2 + pωi
5.8. Periodic Force and Response
47
Now our job is to express this z in the form G(ω)e i (ω t−φ) , where G(ω) is real
and positive. To do this, we first express the denominator in polar form as
ω02 − ω 2 + p ωi = R eiφ ,
where
�
ω02 − ω 2
pω
, and sin φ =
.
R
R
Note that sin φ = p ω/R > 0; so we may assume that 0 < φ < π and write
� 2
�
�
�
2
ω02 − ω 2
−1 ω0 − ω
−1
�
φ = cos
= cos
.
(7)
R
(ω02 − ω 2 )2 + p2 ω 2
R=
(ω02 − ω 2 )2 + p2 ω 2 , cos φ =
With this done, we can write
ω02
ω02 i (ω t−φ)
iω t
e
=
e
,
(8)
R e iφ
R
from which we observe that φ in (7) is the phase angle and that the gain is
z=
G(ω) =
ω02
ω02
=� 2
,
R
(ω0 − ω 2 )2 + p2 ω 2
which may also be written as
�
� �2 � �4 �−1/2
ω
ω
p2 − 2ω02
G(ω) = 1 + γ
+
, where γ =
.
ω0
ω0
ω02
(9)
Also we point out that the real and imaginary parts of z are
x = G(ω) cos(ωt − φ) and y = G(ω) sin(ωt − φ).
We’ll now list a few observations concerning the gain G(ω). You will be asked
to verify the less obvious of these in the problem set at the end of this section.
i) As p → 0+ , G(ω) approaches ω02 /|ω02 − ω 2 |, the gain in the undamped case.
ii) When ω is large relative to p and ω0 , G(ω) ≈ ω02 /ω 2 ≈ 0. Thus high-frequency
input produces small amplitude response.
iii) For any fixed p and ω0 , G(ω) is a bounded function with limω→0+ G(ω) = 1
and limω→∞ G(ω) = 0. Moreover,
• when p2 − 2ω02 ≥ 0, G(ω) < 1 and decreasing for all ω > 0;
• when p2 − 2ω02 < 0, there is a maximum gain
�
� 2ω 2 − p2 �
2ω 2
0
Gmax = G
= � 20
,
2
p 4ω0 − p2
which is slightly greater than G(ω0 ) = ω0 /p. Moreover, G(ω) > 1 when
48
Chapter 5. Linear Second-Order Equations I
�
0 < ω < 2ω02 − p2 . This describes the input frequencies that are amplified by the system. Input frequencies for which ω 2 > 2ω02 − p2 are
attenuated ; that is, 0 < G(ω) < 1.
We also make the following observations concerning the phase angle
�
�
ω02 − ω 2
−1
�
φ(ω) = cos
.
(ω02 − ω 2 )2 + p2 ω 2
iv) If there is no damping (i.e., p = 0), then either φ(ω) = 0 or φ(ω) = π.
v) For fixed p and ω02 , φ(ω) is an increasing function, with φ(ω) → 0 as ω → 0−
and φ(ω) → π as ω → ∞. Thus the response is nearly in phase with low
frequency input and nearly 180◦ out of phase with high frequency input.
Moreover, since 0 ≤ φ(ω) ≤ π, the response lags the input. For that reason
φ(ω) is often called the phase lag.
vi) φ(ω) = π/2 when ω = ω0 . That is, the response is 90◦ (or halfway) out of
phase with input at the natural frequency of the undamped system.
Bode Plots
For a given system, i.e., given values of p and ω0 , frequency response may be
summarized by plots of the gain G(ω) and the phase angle φ as functions of
ω
the angular input frequency ω (or input frequency f = 2π
). Plots of the gain
in decibels (i.e., 20 log10 G(ω)) and the phase angle in degrees, with logarithmic
scales on the horizontal axes, together comprise what is known by engineers as
a Bode plot. Figure 7 is an illustration.
Bode Plot: p = 2, ω 0 = 10
21
180
150
9
ϕ in degrees
Gain in dB
15
120
3
-3
-9
60
30
-15
-21
90
1
2
3
ω
5
10
(log10 scale)
0
20
Figure 7
1
2
3
5
10
ω (log10 scale)
20
5.8. Periodic Force and Response
49
The Quality Factor and Bandwidth
Engineers are fond of defining a dimensionless parameter Q = ω0 /p and calling
it the quality factor of the system. Writing the coefficient p in the differential
equation (1) as ω/Q, we have
ω0 �
y �� +
y + ω02 y = ω02 f (t).
(10)
Q
For our spring-mass systems, this is equivalent to
√
km �
��
my +
y + ky = kf (t).
Q
In terms of Q, our gain formula (9) becomes
G(ω) = �
1+γ
1
� �2
ω
ω0
+
�
ω
ω0
�4 , where γ =
1 − 2Q2
.
Q2
(11)
The system’s bandwidth B is the width of the frequency interval over which
1 2
G .
2 max
These are the frequencies for which the�energy of the response is at least half
�
of its maximum. Provided that Q >
1 + 1/2 ≈ 1.31, there are angular
frequencies ω1 < ω2 such that G(ω)2 ≥ 12 G2max for ω1 ≤ ω ≤ ω2 . After solving
for those, the bandwidth turns out to be
G(ω)2 ≥
ω2 − ω1
2π ��
�
�
�
�
ω0
2
2
2
2
=
Q − 1/2 + Q − 1/4 − Q − 1/2 − Q − 1/4
2πQ
B=
The quantity in the parentheses approaches 1 as Q → ∞. So, for large Q,
ω0
B≈
.
2πQ
(This turns out to be a reasonably good approximation for moderately large Q
as well.) To summarize, for equation (10) when Q is large,
ω0
Gmax ≈ Q and B ≈
.
(12)
2πQ
The level at which G(ω)2 = 12 G2max is approximately 3 dB below Gmax , since
�
20 log10 1/2 ≈ −3. Thus, when ω1 ≤ ω ≤ ω2 , G(ω) will be within 3 dB of
Gmax . Figure 8 illustrates this 3 dB “band,” as well as the corresponding band
of angular frequencies with width 2πB—using precise calculations.
50
Chapter 5. Linear Second-Order Equations I
Q = 1.4
21
gain in dB
15
9
3
-3
-9
-15
-21
ω0/10
ω0/2
ω0
ω (log10 scale)
2ω0
Figure 8
Figure 9 shows the 3 dB band and an angular frequency band of width 2πB—
using the estimates in (12) and centering the band at ω0 . Notice that for Q = 1.4
these estimates differ somewhat from the computed values used in Figure 8 but
nevertheless still provide rough qualitative information. (The situation becomes
much worse for smaller Q.) The second graph indicates that, even for a relatively
small Q = 4, the estimates in (12)—and centering the frequency band at ω0 —
comprise a rather good description of the peak in the curve.
Q = 1.4
gain in dB
21
15
15
9
9
3
3
-3
-3
-9
-9
-15
-15
-21
-21
ω0/10
ω0/2
Q=4
21
ω0
ω (log10 scale)
2ω0
ω0/10
Figure 9
ω0/2
ω0
ω (log10 scale)
2ω0
It is also worth remarking that Figures 8 and 9 are exactly the same for any
ω0 > 0. Their shapes depend only upon Q.
Problems
In Problems 1 through 4, use a formula from (3) to construct a particular solution.
√
√
1. y �� + 4y = 4 (3 sin t − 5 cos 3t)
2. y �� + y = cos( 0.9 t) + sin( 1.01 t)
�π �
π2
π2
3. y �� + 9y = 9 (cos t + cos 5t)
4. y �� +
y=
cos
t
100
100
9
5. Find a particular solution of
(a) y �� + 4y = 12 sin 2t
(b) y �� + 4y � + 4y = 12 sin 2t
5.8. Periodic Force and Response
51
6. Find a particular solution of
(a) y �� + 16y = 16 cos 4t
(b) y �� + y � + 16y = 16 cos 4t
In Problems 7 through 10, find a particular solution expressed in the form A cos(ω t − φ)
or A sin(ω t − φ) by first finding a particular solution of the corresponding complex
equation. State the phase angle and the gain.
7. y �� + 4y � + 14y = 13 cos 3t
8. y �� + 3y � + 5y = 5 sin t
10. y �� + y � + 32 y = 5 cos t
9. y �� + 2y � + y = 5 cos 2t
11. (a) Use an appropriate complex equation to find particular real solutions of
x�� + x = 8t cos t, y �� + y = 8t sin t.
(b) Use the results of part (a) to help write down the general solution of
y �� + y = 8t (sin t + cos t).
12. This problem concerns the phenomenon of beats.
(a) Show that the rest solution of y �� + ω02 y = ω02 cos ω t is
y=
ω02
ω02
(cos ω t − cos ω0 t) .
− ω2
(b) Use the trigonometric identity cos 2α − cos 2β = 2 sin(β + α) sin(β − α) to show
that the rest solution in part (a) may be written
�
�
�
�
2ω02
ω + ω0
ω − ω0
y= 2
sin
t sin
t .
ω0 − ω 2
2
2
(c) Make a rough sketch of the graph of the rest solution with ω = 7π and ω0 = 5π
on the interval [0, 2].
13. (a) Check that the solution of the initial-value problem
y �� + py � + q y = q f (t), y(0) = y0 , y � (0) = v0
can be written as y = u + w where u and w satisfy, respectively,
u�� + pu� + q u = 0, u(0) = y0 , u� (0) = v0 ,
w�� + pw� + q w = q f (t), w(0) = w� (0) = 0.
(b) Argue that if p, q > 0, then u(t) → 0 as t → ∞, regardless of the initial values,
and so y(t) ≈ w(t) for large t. Consequently, the long-term behavior of y does not
depend on initial values. Such systems are said to be dissipative. For dissipative
systems the effect of initial conditions on a solution is transient.
14. (a) Verify observation iii) concerning the gain G of a damped system.
�
�
(b) Show also that Gmax = ω02 / ω04 − ω�4 , where ω� = ω02 − p2 /2, provided that
ω02 > p2 /2.
15. Express the gain G(ω) of a damped spring-mass system in terms of the mass m, the
damping coefficient r, and the spring stiffness k.
52
Chapter 5. Linear Second-Order Equations I
16. Express the gain G(ω) of an RLC circuit in terms of the resistance R, the inductance
L, and the capacitance C.
17. Find coefficients p, ω02 > 0 so that maximum gain occurs at an angular frequency of
ω = 10 with a phase angle of φ = π/4.
18. Show that a damped system experiences maximum gain at a given angular input
frequency ω with a given phase angle 0 < φ < π/2 if
p = 2ω cot φ and ω02 = ω 2 (1 + 2 cot2 φ).
19. Suppose we wish to have a given maximum gain Gmax > 1 occur at a given angular
input frequency ω. Show that this happens if
�
��
�
G2max
G2max
2
2
2
2
p = 2ω
−1
and ω0 = ω
.
2
2
Gmax − 1
Gmax − 1
�
20. Show that, �
for fixed p, ω > 0, the maximum gain is Gmax = 1 + ω 2 /p2 , occurring
when ω0 = ω 2 + p2 .
21. Show that an underdamped
system experiences maximum gain at an angular input
�
frequency ω = 2β 2 − ω02 , where β is the angular frequency of the system’s damped
unforced vibrations.
22. Suppose that the weight of your car compresses its suspension system by 8 inches.
Suppose further that when you jump off your car’s bumper, you observe damped,
free vibration with a frequency of about 1 cycle per second. Find the input frequency
that will result in forced vibration of maximum amplitude.
23. Suppose that you are driving the same car as in Problem 22 on a washboarded road
on which the surface is approximately sinusoidal with successive peaks about 30 feet
apart. Find the speed (in mph) at which you and the car would experience the most
vibration. Also find the gain by using the result of Problem 14b.
24. For the same car as in Problems 22 and 23, find the quality factor Q.
25. A spring-mass system has mass m = 1/4 and stiffness k = 100. Its maximum gain
is estimated to be 20. Estimate the quality factor Q, the damping coefficient r, and
the bandwidth B.
25. A series RLC circuit has R = 10 Ω, L = 1 mH, and C = 1 µF. Find
estimate the bandwidth B.
ω0
2π
and Q, and