# Calculus - Differential Equations (II)

Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Calculus
Differential Equations (II)
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Outline
1
Second-Order Equations with Constant Coefficients
Mass Spring System and it’s Formulation
2nd Order ODE with Constant Coefficients
2
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations
Method of Undetermined Coefficients
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Outline
1
Second-Order Equations with Constant Coefficients
Mass Spring System and it’s Formulation
2nd Order ODE with Constant Coefficients
2
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations
Method of Undetermined Coefficients
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (I)
In this section and section 6.5, we explore the mathematics
describing mechanical vibrations.
In Figure 6.20, we show a mass
hanging from a spring that is
suspended from the ceiling.
Figure: [6.20] Spring-mass
system.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (II)
We call the natural length of the
spring (the length before the spring
is hung from the ceiling) l.
Observe that hanging the mass from
the spring will stretch the spring a
distance ∆l from its natural length.
We now measure the displacement
u(t) of the mass from this equilibrium
position.
Figure: [6.20] Spring-mass
system.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (III)
We consider downward (beyond the
equilibrium position) to be a positive
displacement, while upward (below
the equilibrium position) is
considered to be the negative
direction.
The mass in Figure 6.20 has been
displaced from its natural length by a
total of u(t) + ∆l.
Figure: [6.20] Spring-mass
system.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (IV)
The motion of a spring-mass system
is described by Newton’s second law
of motion:
F = ma.
Primary forces acting on the mass:
Gravity force: mg (downward).
Spring restoring force: −k(u + ∆l)
for some positive constant k (called
the spring constant).
Damping force: −cv, where v = u0 is
the velocity of the mass and c is a
constant.
Figure: [6.20] Spring-mass
system.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (V)
Combining these forces, Newton’s 2nd law gives the following:
mu00 (t) = mg − k[u(t) + ∆l] − cu0 (t)
or
mu00 (t) + cu0 (t) + ku(t) = mg − k∆l
If the mass is not in motion, i.e., u(t) = 0, then we also have
u0 (t) = u00 (t) = 0, and the last equation reduces to
0 = mg − k∆l.
While this provides a means of solving for the spring constant k
in terms of the mass and ∆l, this also simplifies the equation to
mu00 (t) + cu0 (t) + ku(t) = 0.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Mass Spring System and it’s Formulation (VI)
The equation
mu00 (t) + cu0 (t) + ku(t) = 0.
is an example of a second-order differential equation, since
there is a second derivative in the equation. In fact, it is an
especially simple type of differential equation, because the
coefficients of u00 , u0 and u are all constants. By solving the
equation we learn how to analyze spring motion.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Outline
1
Second-Order Equations with Constant Coefficients
Mass Spring System and it’s Formulation
2nd Order ODE with Constant Coefficients
2
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations
Method of Undetermined Coefficients
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (I)
A second order constant coefficient ode takes the following
form:
ay00 (t) + by0 (t) + cy(t) = 0,
where a, b and c are constants.
Notice that equation asks us to find a function whose first
and second derivatives are similar enough that the
combination ay00 (t) + by0 (t) + cy(t) adds up to zero.
One candidate for such a function is the exponential
function ert . So, we assume that
y = ert ,
for some constant r.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (II)
Observe that if y(t) = ert , then y0 (t) = rert and y00 (t) = r2 ert .
Substituting all of this into the differential equation, we get
ar2 ert + brert + cert = 0
or (ar2 + br + c)ert = 0.
Since ert &gt; 0, this can only happen if
ar2 + br + c = 0.
The last equation is called the characteristic equation.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (III)
The solutions of the characteristic equation
ar2 + br + c = 0.
are given by the quadratic formula as:
√
√
−b + b2 − 4ac
−b − b2 − 4ac
r1 =
and r2 =
.
2a
2a
So, there are three possibilities for solutions of the differential
equation:
1
r1 and r2 are distinct real solutions (if b2 − 4ac &gt; 0),
2
r1 = r2 are (repeated) real solutions (if b2 − 4ac = 0) or,
3
r1 and r2 are complex solutions (if b2 − 4ac &lt; 0).
All three of these cases lead to different solutions of the
differential equation, which we must deal with separately.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (IV)
Case 1: If r1 and r2 are distinct real solutions of
ar2 + br + c = 0,
then
y1 = er1 t
and y2 = er2 t
are two solutions of
ay00 (t) + by0 (t) + cy(t) = 0,
and
y(t) = c1 er1 t + c2 er2 t
is the general solution.
We illustrate this in example 4.1.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Finding General Solutions
Example (4.1)
Find the general solution of
1
y00 − y0 − 6y = 0 and
2
y00 + 4y0 − 2y = 0.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (V)
Case 2: If r1 = r2 are repeated roots of the equation
ar2 + br + c = 0,
then we have found only one solution of:
y1 = er1 t .
The second solution is
y2 = ter1 t .
The general solution of
ay00 (t) + by0 (t) + cy(t) = 0,
is then
y(t) = c1 er1 t + c2 ter1 t .
We illustrate this case in example 4.2.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Finding General Solutions (Repeated Roots)
Example (4.2)
Find the general solution of y00 − 6y0 + 9y = 0.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
2nd Order ODE with Constant Coefficients (VI)
Case 3: If r1 and r2 are complex roots of the equation
ar2 + br + c = 0,
we can
√ write these as r1 = u + vi and r2 = u − vi, where
i = −1. To interpret a complex exponential like e(u+vi)t we rely
on Euler’s formula:
eiθ = cos θ + i sin θ.
The solution corresponding to r = u + vi is then
e(u+vi)t = eut+vti = eut evti = eut (cos vt + i sin vt).
It can be shown that
y1 = eut cos vt
and y2 = eut sin vt
are solutions of the differential equation. So, the general
solution of the differential equation is
y1 (t) = c1 eut cos vt + c2 eut sin vt.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Finding General Solutions (Complex Roots)
Example (4.3)
Find the general solution of the equations
1
y00 + 2y0 + 5y = 0 and
2
y00 + 4y = 0.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Initial Value Problem
We can now find the general solution of any equation of the
form
ay00 (t) + by0 (t) + cy(t) = 0.
Notice that the general solution of a second-order differential
equation always involves two arbitrary constants.
To determine the value of these constants, we specify two initial
conditions, most often
y(0) and y0 (0)
A second-order differential equation, plus two initial conditions
is called an initial value problem.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Solving an Initial Value Problem (I)
Example (4.4)
Find the solution of the initial
value problem
y00 + 4y0 + 3y = 0, y(0) = 2,
y0 (0) = 0.
Figure: [6.21] y = −e−3t + 3e−t .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Solving an Initial Value Problem (II)
Example (4.5)
Find the solution of the initial
value problem y00 + 9y = 0,
y(0) = 4, y0 (0) = −6.
Figure: [6.22]
y = 4 cos 3t − 2 sin 3t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Spring-Mass System with No Damping (I)
Example (4.6)
A spring is stretched 6 inches when an 8-pound weight is
attached. The mass is then pulled down an additional 4 inches
and released. Neglect damping. Find an equation for the
position of the mass at any time t and graph the position
function.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Spring-Mass System with No Damping (II)
Figure: [6.23] u(t) =
1
3
cos 8t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Spring-Mass System with Damping (I)
For a real spring-mass system, there is always some damping,
so that the idealized perpetual motion of example 4.6 must be
modified somewhat. In example 4.7, again take note of the
steps required to obtain the equation of motion.
Example (4.7)
A spring is stretched 5 cm when a 2-kg mass is attached. The
mass is set in motion from its equilibrium position with an
upward velocity of 2 m/s. The damping constant equals c = 4.
Find an equation for the position of the mass at any time t and
graph the position function.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Spring-Mass System with Damping (II)
√
2
Figure: [6.24] u(t) = − √195
e−t sin 195t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
RLC Electric Circuit Model (I)
The charge in simple electrical circuits
can be modeled with the same
equation we just used for the motion
of a spring-mass system.
An RLC-circuit illustrated in the figure,
consists of
a resistor R,
a capacitor C,
an inductor L and
a voltage source V.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
RLC Electric Circuit Model (II)
Assume that there is no impressed
voltage. Let
Q(t) (coulombs) be the total
charge on the capacitor at time t
and
I(t) be the current.
Then I = Q0 (t).
It is known that
The voltage drop across the resistor is IR.
Q
The voltage drop across the capacitor is .
C
The voltage drop across the inductor is LI 0 (t).
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
RLC Electric Circuit Model (III)
These voltage drops must sum to the
impressed voltage. If there is none,
then
LI 0 (t) + RI(t) +
1
Q(t) = 0
C
or
LQ00 (t) + RQ0 (t) +
1
Q(t) = 0.
C
Observe that this is the same as the
equation for describing the motion of
a spring-mass system, except for the
names of the constants.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Finding the Charge in an Electrical Circuit
Example (4.8)
A series circuit has an inductor
of 0.2 henry, a resistor of 300
ohms and a capacitor of 10−5
farad. The initial charge on the
capacitor is 10−6 coulomb and
there is no initial current. Find
the charge on the capacitor
and the current at any time t.
Figure: [6.25]
Q(t) = 10−6 2e−500t − e−1000t .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Outline
1
Second-Order Equations with Constant Coefficients
Mass Spring System and it’s Formulation
2nd Order ODE with Constant Coefficients
2
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations
Method of Undetermined Coefficients
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations (I)
Imagine yourself trying to tape an important event on a video
camera. You might be more concerned with keeping a steady
hand than with understanding the mathematics of motion
control, but mathematics plays a vital role in helping you
produce a professional-looking tape.
In section 6.4, we modelled mechanical vibrations when the
motion is started by an initial displacement or velocity. In this
section, we extend that model to cases where an external force
such as a shaky hand continues to affect the system.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations (II)
As usual, the starting place for our model is Newton’s second
law of motion:
F = ma.
We now add an external force to the spring force and damping
force considered in section 6.4. If the external force is F(t) and
u(t) gives the displacement from equilibrium, as defined before,
we have
mu00 (t) + cu0 (t) + ku(t) = F(t).
The only change from the spring model in section 6.4 is that the
right-hand side is no longer zero.
If F(t) = 0 then the equation is homogeneous.
For F(t) 6= 0, we call the equation nonhomogeneous.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations (III)
Consider the nonhomogeneous equation of the form
mu00 (t) + cu0 (t) + ku(t) = F(t).
Our goal is to find the general solution of such an equation
Let up (t) denote one particular solution of the nonhomogeneous
equation. Notice that if u(t) is any other solution, then we have
that
m(u − up )00 + c(u − up )0 + k(u − up )
= (mu00 + cu0 + ku) − (mu00p − cu0p + kup ) = F − F = 0.
That is, the function u − up is a solution of the homogeneous
equation
mu00 + cu0 + ku = 0.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations (IV)
So, if the general solution of the homogeneous equation
mu00 + cu0 + ku = 0.
is c1 u1 + c2 u2 , then
u − up = c1 u1 + c2 u2
for some constants c1 and c2 and
u = c1 u1 + c2 u2 + up .
is the solution of the nonhomogeneous equation.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations (V)
Theorem (5.1)
Let u = c1 u1 + c2 u2 be the general solution of mu00 + cu0 + ku = 0
and let up be any solution of mu00 + cu0 + ku = F. Then the
general solution of mu00 + cu0 + ku = F is given by
u = c1 u1 + c2 u2 + up .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Solving a Nonhomogeneous Equation
Example (5.1)
Find the general solution of u00 + 4u0 + 3u = 30e2t given that
up = 2e2t is a solution.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Outline
1
Second-Order Equations with Constant Coefficients
Mass Spring System and it’s Formulation
2nd Order ODE with Constant Coefficients
2
Nonhomogeneous Equations: Undetermined Coefficients
Nonhomogeneous Equations
Method of Undetermined Coefficients
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Method of Undetermined Coefficients (I)
While example 5.1 showed us how to piece together the
solution of a nonhomogeneous equation from a particular
solution and the general solution of the corresponding
homogeneous equation, we still do not know how to find a
particular solution.
To find a particular solution we present a method here, called
the method of undetermined coefficients.
It works for equations with constant coefficients and where
the nonhomogeneous term is not too complicated.
The method essentially relies on our ability to make an
educated guess about the form of a particular solution.
We illustrate this technique for example 5.1.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Method of Undetermined Coefficients (II)
Consider the equation
u00 + 4u0 + 3u = 30e2t ,
The most likely candidate for the form of u(t) is a constant
multiple of e2t . (How else would u00 , 4u0 and 3u all add up to
30e2t ?)
Let A be a constant. Our educated guess is
up (t) = Ae2t ,
We next substitute this into the differential equation and solve
for A. (If it turns out to be impossible to solve for A, then we
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Method of Undetermined Coefficients (III)
Here, we have
u0p = 2Ae2t
and u00p = 4Ae2t
and so, requiring up to be a solution of the nonhomogeneous
equation gives us
30e2t = u00p + 4u0p + 3up
= Ae2t + 4(2Ae2t ) + 3(Ae2t )30e2t
= 15Ae2t .
So,
15A = 30
or A = 2.
A particular solution is then up (t) = 2e2t , as desired.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Solving a Nonhomogeneous Equation
Example (5.2)
Find the general solution of u00 + 2u0 − 3u = −30 sin 3t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Rules for Making Good Guess (I)
for each derivative.
In example 5.2, we started with A sin 3t and then added its
derivative B cos 3t. We do not need to add other terms,
because all other derivatives involve either sin 3t or cos 3t.
However, suppose that F(t) = 7t5 . The initial guess would
include At5 and the derivatives Bt4 , Ct3 and so on. The
initial guess is then
A5 t5 + A4 t4 + A3 t3 + A2 t2 + A1 t + A0 .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Rules for Making Good Guess (II)
There is one exception to the preceding rule.
If any term in the initial guess is also a solution of the
homogeneous equation, we must multiply the initial guess by a
sufficiently high power of t so that nothing in the modified guess
is a solution of the homogeneous equation. (In the case of
second-order equations, this means multiplying the initial guess
by either t or t2 .)
To see why, consider
u00 + 2u0 − 3u = 4e−3t .
The initial guess Ae−3t won’t work: for u = Ae−3t we have
u00 + 2u0 − 3u = 0,
so there’s no choice of A to make
u00 + 2u0 − 3u = 4e−3t .
However, Ate−3t does work, as seen in example 5.3.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Modifying an Initial Guess
Example (5.3)
Show that u(t) = Ae−3t is not a solution of u00 + 2u0 − 3u = 4e−3t ,
but that there is a solution of the form u(t) = Ate−3t .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
The Form of up for au00 + bu00 + cu = F(t)
A summary of rules for making good guesses is given in the
table below.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Finding the Form of Particular Solutions
Example (5.4)
Determine the form for a particular solution of the following
equations:
1
y00 + 4y0 = t4 + 3t2 + 2e−4t sin t + 3e−4t and
2
y00 + 4y = 3t2 sin 2t + 3te2t .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
The Motion of a Spring Subject to an External Force (I)
Example (5.5)
A mass of 0.2 kg stretches a spring by 10 cm. The damping
constant is c = 0.4. External vibrations create a force of
F(t) = 0.2 sin 4t newtons, setting the spring in motion from its
equilibrium position. Find an equation for the position of spring
at any time t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
The Motion of a Spring Subject to an External Force (II)
Notice in Figure 6.26 that after a very
brief time the motion appears to be
simple harmonic motion. We can
verify this by a quick analysis
Figure: [6.26] Spring motion
with an external force.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
The Motion of a Spring Subject to an External Force (III)
Recall that the solution comes in two pieces, the particular
solution
up (t) =
41
1
sin 4t −
cos 4t
6756
1689
and the solution of the homogeneous equation
√
√
uh (t) = c1 e−t cos 97t + c2 e−t sin 97t.
As t increases,
e−t → 0 and uh (t) → 0
regardless of the value of the constants c1 and c2 .
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
The Motion of a Spring Subject to an External Force (IV)
This says that as t → ∞ u = up (t) + uh (t) → up which is a simple
oscillation. For this reason,
the solution of the homogeneous equation is called the
transient solution and
the particular solution is called the steady-state solution.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Example (5.6)
For u00 + 3u0 + 2u = 20 cos 2t, find the steady-state solution.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (I)
There are several possibilities for the steady-state motion of a
mechanical system. Two interesting cases, called
resonance and
beats,
are introduced here. In their pure forms, both occur only when
there is no damping and the external force is a sine or cosine.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (II)
In these cases, then, the equation of motion is
mu00 + ku = F(t).
The characteristic equation for the homogeneous equation is
mr2 + k = 0,
which has solutions
r
r=&plusmn;
k
i
m
and the solution of the homogeneous equation is
c1 cos ωt + c2 sin ωt,
r
where ω =
k
is called the natural frequency of the system.
m
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (III)
Resonance occurs in a mechanical system when the external
force is a sine or cosine whose frequency exactly matches the
natural frequency of the system.
For example, suppose that
F(t) = sin ωt.
Then our initial guess
up (t) = A sin ωt + B cos ωt
matches the homogeneous solution and must be modified to
the guess
up (t) = t(A sin ωt + B cos ωt).
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (IV)
The graph of
up (t) = t(A sin ωt + B cos ωt).
would oscillate, but the presence of
the t factor would cause the
oscillations to grow larger and larger
without bound.
The graph of
y = 2t sin 4t
in Figure 6.28 illustrates this behavior.
Figure: [6.28] y = 2t sin 4t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (V)
Physically, resonance can cause impressive disasters:
A singer hitting a note (thus producing an external force) at
exactly the natural frequency of a wineglass can shatter it.
Soldiers marching in step across a bridge at exactly the
natural frequency of the bridge can create large
oscillations in the bridge that can cause it to collapse.
These phenomena are well known and it is for this reason that
soldiers are ordered to break stride while marching across a
bridge.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (VI)
The phenomenon of beats occurs when the forcing frequency is
close to the natural frequency.
For example, consider
u00 + 4u = 2 sin 2.1t
with u(0) = u0 (0) = 0.
The homogeneous solution is
c1 sin 2t + c2 cos 2t
and so, the forcing frequency of 2.1 is close to the natural
frequency of 2. The solution is
u(t) = 5.1219 sin 2t − 4.878 sin 2.1t.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (VII)
The graph in Figure 6.29 illustrates
the beats phenomenon of periodically
increasing and decreasing
amplitudes.
This can be heard when tuning a
piano. If a note is slightly off, its
frequency is close to the frequency of
the external tuning fork and you will
hear the amplitude variation shown in
the figure.
Figure: [6.29] Beats.
Second-Order Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Resonance and Beats (VIII)
Example (5.7)
For the system u00 + 5u = 3 sin ωt, find the natural frequency, the
value of ω that produces resonance and a value of ω that
produces beats.