INSTRUCTOR’S RESOURCE GUIDE
EDWARD B. SAFF
Vanderbilt University
A. DAVID SNIDER
University of South Florida
FUNDAMENTALS
OF DIFFERENTIAL EQUATIONS
SIXTH EDITION
FUNDAMENTALS OF
DIFFERENTIAL EQUATIONS
AND BOUNDARY VALUE PROBLEMS
FOURTH EDITION
R. Kent Nagle
Edward B. Saff
A. David Snider
Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors.
Copyright © 2004 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston MA 02116
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior
written permission of the publisher. Printed in the United States of America.
ISBN
0-321-17318-X
1 2 3 4 5 6 QEP 06 05 04 03
Contents
Notes to Instructor
Software Supplements
1
1
Instructional Utility Software
1
Interactive Differential Equations CD-ROM
2
Instructor’s Maple/Matlab/Mathematica Manual
2
Computer Labs
2
Group Projects
2
Technical Writing Exercises
3
Student Presentations
3
Homework Assignments
3
Syllabus Suggestions
3
Numerical, Graphical, and Qualitative
4
Engineering/Physics Applications
5
Biology/Ecology Applications
6
Supplemental Group Projects
Answers to Even-Numbered Problems
8
15
Chapter 1
15
Chapter 2
21
Chapter 3
25
Chapter 4
31
Chapter 5
43
Chapter 6
57
Chapter 7
61
Chapter 8
71
Chapter 9
81
Chapter 10
91
Chapter 11
97
Chapter 12
103
Chapter 13
117
Notes to the Instructor
One goal in our writing has been to create flexible texts that afford the instructor a variety of topics and make
available to the student an abundance of practice problems and projects. We recommend that the instructor read
the discussion given in the preface in order to gain an overview of the prerequisites, topics of emphasis, and
general philosophy of the texts.
An additional resource to accompany the texts is the on-line tool, MyMathLab. MyMathLab is ideal for lecturebased, lab-based, and on-line courses and provides students and instructors with a centralized point of access to
the multimedia resources available with the texts. The pages of the actual text are loaded into MyMathLab and
extensive course-management capabilities, including a host of communication tools for course participants, are
provided to create a user-friendly and interactive on-line learning environment. Instructors can also remove,
hide, or annotate Addison-Wesley preloaded content, add their own course documents, or change the order in
which material is presented. A link to the Instructional Utility Software package is also included. For more
information visit our Web site at www.mymathlab.com or contact your Addison-Wesley sales representative for
a live demonstration.
INSTRUCTIONAL UTILITY SOFTWARE
Written specifically for the texts and available via MyMathLab, this utility has the following items:
GRAPHICAL METHODS
Graph, tabulate or evaluate functions
Direction field
Phase plane diagram
MATRIX OPERATIONS
Solve a system of linear equations
Determinant and inverse of a matrix
Matrix multiplication
Eigenvalues and eigenvectors
NUMERICAL SOLUTIONS OF INITIAL VALUE PROBLEMS
Euler’s method subroutine
Euler’s method with tolerance
Improved Euler’s method subroutine
Improved Euler’s method with tolerance
Fourth order Runge-Kutta method subroutine
Fourth order Runge-Kutta method with tolerance
FOURTH ORDER RUNGE-KUTTA METHOD FOR SYSTEMS
System with 2 equations
System with 3 equations
System with 4 equations
OTHER COMPUTATIONAL METHODS
Roots of a polynomial equation
Newton’s method
Simpson’s Rule for definite integrals
Simpson’s Rule for linear differential equations
2
NOTES TO THE INSTRUCTOR
INTERACTIVE DIFFERENTIAL EQUATIONS CD-ROM
Written by Beverly West (Cornell University), Steven Strogatz (Cornell University), Jean Marie McDill
(California Polytechnic State University—San Luis Obispo), John Cantwell (St. Louis University), and Hubert
Hohn (Massachusetts College of Art), this CD-ROM is a revised version of a popular software directly tied to
the text. It focuses on helping students visualize concepts with applications drawn from engineering, physics,
chemistry, and biology. This software runs on supported Windows or Macintosh operating systems and is
bundled free with every book.
INSTRUCTOR’S MAPLE/MATLAB/MATHEMATICA MANUAL
The Instructor’s Maple/Matlab/Mathematica Manual, 0-321-17320-1, was written as an aid for anyone
interested in coordinating the use of computer algebra systems with their course. This supplement, including
sample worksheets, is available upon request from Addison Wesley and includes the following.
•
specific instruction in the use of Maple, Matlab and Mathematica to obtain graphic (direction fields,
solution curves, phase portraits), numeric (built-in and user defined), and symbolic information about
differential equations;
•
a sampling of techniques that can be used to ease the introduction of Maple into the differential equations
classroom (including sample worksheets directly related to the text); and
•
a collection of additional projects that are particularly amenable to solution using a computer algebra
system.
While this supplement is written for use with MAPLE, the general ideas can be adapted for use with
MATHEMATICA, MATLAB, or any other sophisticated numerical and/or computer algebra software.
Computer Labs: A computer lab in connection with a differential equations course can add a whole new
dimension to the teaching and learning of differential equations. As more and more colleges and universities set
up computer labs with software such as MATLAB, MAPLE, DERIVE, MATHEMATICA, PHASEPLANE,
and MACMATH, there will be more opportunities to include a lab as part of the differential equations course.
In our teaching and in our texts, we have tried to provide a variety of exercises, problems, and projects that
encourage the student to use the computer to explore. Even one or two hours at a computer generating phase
plane diagrams can provide the students with a feeling of how they will use technology together with the theory
to investigate real world problems. Furthermore, our experience is that they thoroughly enjoy these activities.
Of course the software provided free with the texts is especially convenient for such labs.
Group Projects: Although the projects that appear at the end of the chapters in the texts can be worked out by
the conscientious student working alone, making them group projects adds a social element that encourages
discussion and interactions that simulate a professional work place atmosphere. Group sizes of 3 or 4 seem to
be optimal. Moreover, requiring that each individual student separately write up the group’s solution as a
formal technical report for grading by the instructor also contributes to the professional flavor.
Typically our students each work on 3 or 4 projects per semester. If class time permits, oral presentations by the
groups can be scheduled and help to improve the communication skills of the students.
The role of the instructor is, of course, to help the students solve these elaborate problems on their own and to
recommend additional reference material when appropriate.
Some additional Group Projects are presented in this guide (see page 8).
INSTRUCTOR’S MAPLE/MATLAB/MATHEMATICA MANUAL
3
Technical Writing Exercises: The technical writing exercises at the end of most chapters invite students to
make documented responses to questions dealing with the concepts in the chapter. This not only gives students
an opportunity to improve their writing skills, but it helps them organize their thoughts and better understand
the new concepts. Moreover, many questions deal with critical thinking skills that will be useful in their careers
as engineers, scientists, or mathematicians.
Since most students have little experience with technical writing, it may be necessary to return ungraded the
first few technical writing assignments with comments and have the students redo the exercise. This has
worked well in our classes and is much appreciated by the students. Handing out a “model” technical writing
response is also helpful for the students.
Student Presentations: It is not uncommon for an instructor to have students go to the board and present a
solution to a problem. Differential equations is so rich in theory and applications that it is an excellent course to
allow (require) a student to give a presentation on a special application (e.g. almost any topic from Chapters 3
and 5), on a new technique not covered in class (e.g. material from Section 2.6, Projects A, B or C in Chapter
4), or on additional theory (e.g. material from Chapter 6 which generalizes the results in Chapter 4). In addition
to improving students’ communication skills, these “special” topics are long remembered by the students. Here,
too, working in groups of 3 or 4 and sharing the presentation responsibilities can add substantially to the
interest and quality of the presentation. Students should also be encouraged to enliven their communication by
building physical models, preparing part of their lectures on video cassette, etc.
Homework Assignments: We would like to share with you an obvious, nonoriginal, but effective method to
encourage students to do homework problems.
An essential feature is that it requires little extra work on the part of the instructor or grader. We assign
homework problems (about 10 of them) after each lecture. At the end of the week (Fridays), students are asked
to turn in their homeworks (typically 3 sets) for that week. We then choose at random one problem from each
assignment (typically a total of 3) that will be graded. (The point is that the student does not know in advance
which problems will be chosen.) Full credit is given for any of the chosen problems for which there is evidence
that the student has made an honest attempt at solving. The homework problem sets are returned to the students
at the next meeting (Mondays) with grades like 0/3, 1/3, 2/3 or 3/3 indicating the proportion of problems for
which the student received credit. The homework grades are tallied at the end of the semester and count as one
test grade. Certainly there are variations on this theme. The point is that students are motivated to do their
homework with little additional cost (= time) to the instructor.
Syllabus Suggestions: To serve as a guide in constructing a syllabus for a one-semester or two-semester
course, the prefaces to the texts list sample outlines that emphasize methods, applications, theory, partial
differential equations, phase plane analysis, or computation or combinations of these. As a further guide in
making a choice of subject matter, we provide below a listing of text material dealing with some common areas
of emphasis.
References to material in Chapters 11, 12, or 13 refer to the expanded text Fundamentals of Differential
Equations and Boundary Value Problems, 4th edition.
4
NOTES TO THE INSTRUCTOR
Numerical, Graphical, and Qualitative Methods
The sections and projects dealing with numerical, graphical, and qualitative techniques for solving differential
equations include:
Section 1.3, Direction Fields
Section 1.4, The Approximation Method of Euler
Project A for Chapter 1, Taylor Series Method
Project B for Chapter 1, Picard’s Method
Project D for Chapter 1, The Phase Line
Section 3.6, Improved Euler’s Method, which includes step-by-step outlines of the improved Euler’s method
subroutine and improved Euler’s method with tolerance. These outlines are easy for the student to translate into
a computer program (cf. pages 129, 130).
Section 3.7, Higher-Order Numerical Methods: Taylor and Runge-Kutta, which includes outlines for the
Fourth Order Runge-Kutta subroutine and algorithm with tolerance (cf. pages 138, 139).
Project G for Chapter 3, Stability of Numerical Methods
Project H for Chapter 3, Period Doubling and Chaos
Section 4.7, Qualitative Considerations for Variable-Coefficient and Nonlinear Equations, which discusses the
energy integral lemma, as well as the Airy, Bessel, Duffing, and van der Pol equations.
Section 5.3, Solving Systems and Higher-Order Equations Numerically, which describes the vectorized forms
of Euler’s Method and the Fourth-Order Runge-Kutta method, and discusses an application to population
dynamics.
Section 5.4, Introduction to the Phase Plane, which introduces the study of trajectories of autonomous systems,
critical points, and stability.
Section 5.7, Dynamical Systems, Poincaré Maps, and Chaos, which discusses the use of numerical methods to
approximate the Poincaré map and how to interpret the results.
Project B for Chapter 5, Designing a Landing System for Interplanetary Travel
Project C for Chapter 5, Things That Bob
Project F for Chapter 5, Strange Behavior of Competing Species-Part I
Project D for Chapter 9, Strange Behavior of Competing Species-Part II
Project D for Chapter 10, Numerical Method for ∆u = f on a Rectangle
Project D for Chapter 11, Shooting Method
Project E for Chapter 11, Finite-Difference Method for Boundary Value Problems
Project C for Chapter 12, Computing Phase Plane Diagrams
Project D for Chapter 12, Ecosystem of Planet GLIA-2
ENGINEERING/PHYSICS APPLICATIONS
5
Appendix A, Newton’s Method
Appendix B, Simpson’s Rule
Appendix D, Method of Least Squares
Appendix E, Runge-Kutta Procedure for n Equations
The instructor who wishes to emphasize numerical methods should also note that the text contains an extensive
chapter on series solutions of differential equations (Chapter 8).
Engineering/Physics Applications
Since Laplace Transforms is a subject vital to engineering, we have included a detailed chapter on this topic—
see Chapter 7. Stability is also an important subject for engineers, so we have included an introduction to the
subject in Section 5.4 along with an entire chapter addressing this topic—see Chapter 12. Further materials
dealing with engineering/physics applications include:
Project C for Chapter 1, Magnetic “Dipole”
Project A for Chapter 2, Torricelli’s Law of Fluid Flow
Section 3.1, Mathematical Modeling
Section 3.2, Compartmental Analysis, which contains a discussion of mixing problems and of population
models.
Section 3.3, Heating and Cooling of Buildings, which discusses temperature variations in the presence of air
conditioning or furnace heating.
Section 3.4, Newtonian Mechanics
Section 3.5, Electrical Circuits
Project B for Chapter 3, Curve of Pursuit
Project C for Chapter 3, Aircraft Guidance in a Crosswind
Project D for Chapter 3, Feedback and the Op Amp
Project E for Chapter 3, Bang-Bang Controls
Section 4.1, Introduction: The Mass-Spring Oscillator
Section 4.7, Qualitative Considerations for Variable-Coefficient and Nonlinear Equations
Section 4.8, A Closer Look at Free Mechanical Vibrations
Section 4.9, A Closer Look at Forced Mechanical Vibrations
Project F for Chapter 4, Apollo Reentry
Project G for Chapter 4, Simple Pendulum
6
NOTES TO THE INSTRUCTOR
Chapter 5, Introduction to Systems and Phase Plane Analysis, which includes sections on coupled mass-spring
systems, electrical circuits, and phase plane analysis.
Project B for Chapter 5, Designing a Landing System for Interplanetary Travel
Project C for Chapter 5, Things that Bob
Project E for Chapter 5, Hamiltonian Systems
Project B for Chapter 6, Transverse Vibrations of a Beam
Chapter 7, Laplace Transforms, which in addition to basic material includes discussions of transfer functions,
the Dirac delta function, and frequency response modeling.
Projects for Chapter 8, which deal with Schrödinger’s equation, buckling of a tower, and aging springs.
Project B for Chapter 9, Matrix Laplace Transform Method
Project C for Chapter 9, Undamped Second-Order Systems
Chapter 10, Partial Differential Equations, which includes sections on Fourier series, the heat equation, wave
equation, and Laplace’s equation.
Project A for Chapter 10, Steady-State Temperature Distribution in a Circular Cylinder
Project B for Chapter 10, A Laplace Transform Solution of the Wave Equation
Project A for Chapter 11, Hermite Polynomials and the Harmonic Oscillator
Section 12.4, Energy Methods, which addresses both conservative and nonconservative autonomous
mechanical systems.
Project A for Chapter 12, Solitons and Korteweg-de Vries Equation
Project B for Chapter 12, Burger’s Equation
Students of engineering and physics would also find Chapter 8 on series solutions particularly useful, especially
Section 8.8 on Special Functions.
Biology/Ecology Applications
Project D for Chapter 1, The Phase Line, which discusses the logistic population model and bifurcation
diagrams for population control.
Problem 32 in Exercises 2.2, which concerns radioisotopes and cancer detection.
Problem 37 in Exercises 2.3, which involves a simple model for the amount of a hormone in the blood during a
24-hour cycle.
Section 3.1, Mathematical Modeling
Section 3.2, Compartmental Analysis, which contains a discussion of mixing problems and population models.
Problem 19 in Exercises 3.6 which considers a generalization of the logistic model.
Problem 20 in Exercises 3.7, which involves chemical reaction rates.
Project A for Chapter 3, Aquaculture, which deals with a model for raising and harvesting catfish.
BIOLOGY/ECOLOGY APPLICATIONS
7
Section 5.1, Interconnected Fluid Tanks, which introduces systems of differential equations.
Section 5.3, Solving Systems and Higher-Order Equations Numerically, which contains an application to
population dynamics.
Section 5.4, Introduction to the Phase Plane, which contains exercises dealing with the spread of a disease
through a population (epidemic model).
Project A for Chapter 5, The Growth of a Tumor
Project D for Chapter 5, Periodic Solutions to Volterra-Lotka Systems
Project F for Chapter 5, Strange Behavior of Competing Species-Part I
Project G for Chapter 5, Cleaning Up the Great Lakes
Project D for Chapter 9, Strange Behavior of Competing Species-Part II
Problem 19 in Exercises 10.5, which involves chemical diffusion through a thin layer.
Project D for Chapter 12, Ecosystem on Planet GLIA-2
The basic content of the remainder of this instructor’s manual consists of supplemental Group Projects along
with the answers to the even numbered problems. These answers are for the most part not available any place
else since the text only provides answers to odd numbered problems, and the Student Solution Manual contains
only a handful of worked solutions to even numbered problems.
We would appreciate any comments you may have concerning the answers in this manual. These comments can
be sent to the authors’ email addresses given below. We also would encourage sharing with us (= the authors
and users of the texts) any of your favorite Group Projects.
E. B. Saff
esaff@math.vanderbilt.edu
A. D. Snider
snider@eng.usf.edu
Supplemental Group Projects
Group Project for Chapter 2
Designing a Solar Collector
You want to design a solar collector that will concentrate the sun’s rays at a point. By symmetry this surface will
have a shape that is a surface of revolution obtained by revolving a curve about an axis. Without loss of
generality, you can assume that this axis is the x-axis and the rays parallel to this axis are focused at the origin
(see Figure GP-1). To derive the equation for the curve, proceed as follows:
a. The law of reflection says that the angles γ and δ are equal. Use this and results from geometry to show
that β = 2α.
Figure GP-1: Curve that generates a solar collector
y
dy
= tan α . Use this, the fact that x = tan β , and the double angle formula to
dx
b. From calculus recall that
show that
y
=
x
dy
2 dx
1−
( )
dy 2
dx
.
c. Now show that the curve satisfies the differential equation.
(1)
dy −x + x 2 + y 2
=
dx
y
d. Solve equation (1).
e. Describe the solutions and identify the type of collector obtained.
SUPPLEMENTAL GROUP PROJECTS
9
Group Projects for Chapter 3
Delay Differential Equations
In our discussion of mixing problems in Section 3.2,
we encountered the initial value problem
(1)
x′ ( t ) = 6 −
3
x ( t − t0 ) ,
500
x (t ) = 0
for t ∈ −t0 ,0
where t0 is a positive constant. The equation in (1) is an
example of a delay differential equation. These
equations differ from the usual differential equations
by the presence of the shift (t − t0 ) in the argument of
the unknown function x(t). In general, these equations
are more difficult to work with than are regular
differential equations, but quite a bit is known about
them.†
a. Show that the simple linear delay differential
equation
Use the method of steps to show that the
solution to the initial value problem
x′ (t ) = − x (t − 1) ,
x (t ) = 1 on [−1, 0]
is given by
x (t ) =
k
k t − ( k − 1)
∑ ( −1) k ! , for n − 1 ≤ t ≤ n,
k =0
n
where n is a nonnegative integer. (This problem
can also be solved using the Laplace transform
method of Chapter 7.)
c. Use the method of steps to compute the solution
to the initial value problem given in (1) on the
interval 0 ≤ t ≤ 15 for t0 = 3.
x′ (t ) = a x (t − b ) ,
(2)
where a and b are constants, has a solution of
the form x (t ) = Ce st , for any constant C,
provided s satisfies the transcendental equation
s = ae−bs .
b. A solution to (2) for t > 0 can also be found
using the method of steps. Assume that
x (t ) = f (t ) for − b ≤ t ≤ 0. For 0 ≤ t ≤ b,
equation (2) becomes
x′ (t ) = a x (t − b ) = a f (t − b ) ,
and so
1
x (t ) = ∫ a f (v − b ) dv + x (0 ) .
0
Now that we know x(t) on [0, b], we can repeat
this procedure to obtain
1
x (t ) = ∫ a x (v − b ) dv + x (b )
b
for b ≤ t ≤ 2b. This process can be continued
indefinitely.
_____________
†See, for example, Differential-Difference Equations, by R. Bellman and K. L. Cooke, Academic Press, New York, 1963, or Ordinary and
Delay Differential Equations, by R. D. Driver, Springer-Verlag, New York, 1977.
10
SUPPLEMENTAL GROUP PROJECTS
Extrapolation
When precise information about the form of the error
in an approximation is known, a technique called
extrapolation can be used to improve the rate of
convergence.
Suppose the approximation method converges
with rate O (h p ) as h → 0 (cf. Section 3.6). From
theoretical considerations assume we know, more
precisely, that
y(x; h) = φ(x) + h p a p (x) + O(h p +1),
(1)
where y(x; h) is the approximation to φ(x) using step
size h and a p (x) is some function that is independent
of h (typically we do not know a formula for a p (x),
only that it exists). Our goal is to obtain
approximations that converge at the faster rate
O( h p+1).
h
We start by replacing h by
in (1) to get
2
hp
h
y x; = φ(x) + p a p(x) + O(h p+1).
2
2
p
If we multiply both sides by 2 and subtract equation
(1), we find
h
p
p
p+1
).
2 y x; − y(x; h) = (2 − 1)φ(x) + O(h
2
Solving for φ(x) yields
(
)
2 p y x; h2 − y ( x; h)
φ ( x) =
+ O(h p +1 ).
p
2 −1
Hence
(
)
p
h
h 2 y x; 2 − y ( x; h)
y * x; :=
2
2 p −1
has a rate of convergence of O(h p
+1
).
a. Assuming
h
y * x; = φ(x) + h p +1a p +1(x) + O(h p+2 ),
2
show that
(
)
(
)
2 p +1 y * x; h − y * x; h
h
4
2
y ** x; :=
has a
p
+
1
4
−1
2
p +2
rate of convergence of O(h
).
b. Assuming
h
y * * x; = φ (x) + h p+2 a p +2 (x) + O(h p +3 ),
4
show that
(
)
(
2 p + 2 y ** x; h − y ** x; h
h
8
4
y *** x; :=
p+2
8
−
2
1
+3
has a rate of convergence of O(h p ).
)
c. The results of using Euler’s method
1 1 1
with h = 1, , , to approximate the
2 4 8
solution to the initial value problem y ′ = y,
y(0) = 1, at x = 1 are given in Table 1.2,
page 27. For Euler’s method, the extrapolation
procedure applies with p = 1. Use the results in
Table 1.2 to find an approximation to e = y(1)
1
by computing y * * * 1; . [Hint: Compute
8
1
1
1
y * 1; , y * 1; , and y * 1; ; then
2
4
8
1
1
compute y * *1; , and y * *1; .
4
8
d. Table 1.2 also contains Euler’s approximation
1
for y(1) when h = . Use this additional
16
information to compute the next step in the
extrapolation procedure; that is, compute
1
y * * * *1; .
16
SUPPLEMENTAL GROUP PROJECTS
11
Group Projects for Chapter 5
Effects of Hunting on Predator~Prey Systems
As discussed in Section 5.3 (page 257), cyclic
variations in the populations of predators and their prey
have been studied using the Volterra-Lotka predatorprey model given by the system
(1)
dx
= Ax − Bxy,
dt
(2)
dy
= −Cy + Dxy,
dt
where A, B, C, and D are positive constants, x(t) is the
population of prey at time t, and y(t) is the population
of predators. It can be shown that such a system has a
periodic solution (see Project D). That is, there exists
some constant T such that x(t) = x(t + T) and
y(t) = y(t + T) for all t. This periodic or cyclic variation
in the populations has been observed in various
systems such as sharks-food fish, lynx-rabbits, and
ladybird beetles-cottony cushion scale. Because of this
periodic behavior, it is useful to consider the average
populations x and y defined by
x :=
1 T
1 T
x (t ) dt , y := ∫ y (t ) dt
∫
T 0
T 0
a. Show that x = C/D and y = A/B. [Hint: Use
equation (1) and the fact that x(0) = x(T) to show that
x′ ( t )
∫0 ( A − By (t )) dt = ∫0 x (t ) dt = 0.]
T
T
b. To determine the effect of indiscriminate
hunting on the populations, assume hunting
reduces the rate of change in a population by a
constant times the population. Then, the
predator-prey system satisfies the new set of
equations
(3)
dx
= Ax − Bxy − εx = ( A − ε ) x − Bxy,
dt
(4)
dy
= −Cy + Dxy − δ y = − (C + δ ) y + Dxy,
dt
where ε and δ are positive constants with ε < A.
What effect does this have on the average
population of prey? On the average population
of predators?
c. Assume the hunting is done selectively, as in
shooting only rabbits (or shooting only lynx).
Then we have ε > 0and δ = 0 (or ε = 0 and
δ > 0) in (3)-(4). What effect does this have on
the average populations of predator and prey?
d. In a rural county, foxes prey mainly on rabbits
but occasionally include a chicken in their diet.
The farmers decide to put a stop to the chicken
killing by hunting the foxes. What do you
predict will happen? What will happen to the
farmers’ gardens?
_________________________
† The derivation of these equations is found in Attitude Stabilization and Control of Earth Satellites, by O.H. Gerlach, Space Science
Reviews, #4 (1965), 541–566
12
SUPPLEMENTAL GROUP PROJECTS
Limit Cycles
In the study of triode vacuum tubes, one encounters the van der Pol equation:†
(
)
y″ − µ 1 − y 2 y′ + y = 0,
where the constant µ is regarded as a parameter. In Section 4.7 (page 205), we used the mass-spring oscillator
analogy to argue that the nonzero solutions to the van der Pol equation with µ = 1 should approach a periodic limit
cycle. The same argument applies for any positive value of µ .
a. Recast the van der Pol equation as a system in normal form and use software to plot some typical trajectories
for µ = 0.1, 1, and 10. Rescale the plots if necessary until you can discern the limit cycle trajectory; find
trajectories that spiral in, and ones that spiral out, to the limit cycle.
b. Now let µ = −0.1, −1, and −10. Try to predict the nature of the solutions using the mass-spring analogy.
Then use the software to check your predictions. Are there limit cycles? Do the neighboring trajectories
spiral into, or spiral out from, the limit cycles?
c. Repeat parts (a) and (b) for the Rayleigh equation
2
y″ − µ 1 − ( y′ ) y′ + y = 0.
_________________
†Historical Footnote: Experimental research by E. V. Appleton and B. van der Pol in 1921 on the oscillations of an electrical circuit
containing a triode generator (vacuum tube) led to the nonlinear equation now called van der Pol’s equation. Methods of solution were
developed by van der Pol in 1926-1927. Mary L. Cartwright continued research into nonlinear oscillation theory and together with J. E.
Littlewood obtained existence results for forced oscillations in nonlinear systems in 1945.
SUPPLEMENTAL GROUP PROJECTS
13
Group Project for Chapter 13
David Stapleton, University of Central Oklahoma
Satellite Attitude Stability
In this problem, we determine the orientation at which
a satellite in a circular orbit of radius r can maintain a
relatively constant facing with respect to a spherical
primary (e.g. a planet) of mass M. The torque of
gravity on the asymmetric satellite maintains the
orientation.
Suppose (x, y, z) and (x , y , z ) refer to coordinates in
two systems that have a common origin at the
satellite’s center of mass. Fix the xyz-axes in the
satellite as principal axes; then let the z -axis point
toward the primary and let the x -axis point in the
direction of the satellite’s velocity. The xyz-axes may
be rotated to coincide with the x y z -axes by a rotation
φ about the x-axis (roll), followed by a rotation θ about
the resulting y-axis (pitch), and a rotation ψ about the
final z-axis (yaw). Euler’s equations from physics (with
high order terms omitted† to obtain approximate
solutions valid near (φ, θ, ψ) = (0, 0, 0)) show that the
equations for the rotational motion due to gravity
acting on the satellite are
..
.
I x φ = −4ω 20 ( I z − I y )φ − ω 0 ( I y – I z − I x )ψ
..
I y θ = −3ω 20 ( I x − I z )θ
..
I z ψ = −ω 20 ( I y − I x )ψ + ω0 ( I y − I z − I x )φ,
(where ω0 =
GM
is the angular frequency of the
r3
orbit and the positive constants I x , Iy , I z are the
moments of inertia of the satellite about the x, y, and
z-axes).
a. Find constants c1, …, c5 such that these
equations can be written as two systems
φ 0
d ψ 0
=
dt φ c1
θ 0
0
0
0
c3
1
0
0
c4
0 φ
1 ψ
c2 φ
0 ψ
and
d
dt
θ 0
θ = c
5
1 θ
0 θ
b. Show that the origin is asymptotically stable for
the first system in (a) if
(c2 c4 + c3 + c1 )2 − 4c1c3 > 0, c1c3 > 0, and
c2 c4 + c3 + c1 > 0 and hence deduce that
I y > I x > I z yields an asymptotically stable
origin. Are there other conditions on the
moments of inertia by which the origin is
stable?
c. Show that for the asymptotically stable
configuration in (b) the second system in (a)
becomes a harmonic oscillator problem, and
find the frequency of oscillation in terms of
I x , Iy , I z and ω 0 . Phobos maintains
I y > I x > I z in its orientation with respect to
Mars, and has angular frequency of orbit
(I x − I z )
= .23, show that
ω0 = .82 rad/hr. If
Iy
the period of the libration for Phobos (the period
with which the side of Phobos facing Mars
shakes back and forth) is about 9 hours.
_________________________
† The derivation of these equations is found in Attitude Stabilization and Control of Earth Satellites, by O.H. Gerlach, Space Science
Reviews, #4 (1965), 541–566
ANSWERS TO EVEN-NUMBERED PROBLEMS
CHAPTER 1
Exercises 1.1 (page 5)
2. ODE, 2nd order, ind. var. x, dep. var. y, linear
4. PDE, 2nd order, ind. var. x, y; dep. var. u
6. ODE, 1st order, ind. var. t, dep. var. x, nonlinear
8. ODE, 2nd order, ind. var. x, dep. var. y, nonlinear
10. ODE, 4th order, ind. var. x, dep. var. y, linear
12. ODE, 2nd order, ind. var. x, dep. var. y, nonlinear
14.
dx
= kx 4 , where k is the proportionality constant
dt
16.
dA
= kA2 , where k is the proportionality constant
dt
Exercises 1.2 (page 14)
4. Yes
6. No
8. Yes
10. Yes
12. Yes
20. a.
b.
22. a.
b.
24. Yes
26. No
28. Yes
–1, –5
0, –1, –2
5
1
c1 = , c2 =
3
3
c1 =
e2
2 e −1
, c2 =
3
3
16
CHAPTER 1 INTRODUCTION
Exercises 1.3 (page 22)
2. a.
y = –2 – 2x
Figure 1
Solution to Problem 2(a)
c.
Figure 2
Solution to Problem 2(b)
As x → ∞ , solution becomes infinite. As x → − ∞ , solution again becomes infinite and has
y = –2 – 2x as an asymptote.
4. Terminal velocity is v = 2.
Figure 3
Solutions to Problem 4 satisfying v(0) = 0,
v(0) = 1, v(0) = 2, v(0) = 3
6. a.
b.
c.
2
For x > 1 we have x + sin y > 0, so y′ > 0 and hence y is increasing.
d2 y
dx
d.
2
=
d
( x + sin y ) = 1 + y ′ cos y
dx
= 1 + ( x + sin y ) cos y
1
= 1 + x cos y + sin(2 y )
2
For x = y = 0 we have y ′ = 0 and, from the formula in part (c), it follows that
d2 y
= 1 > 0 when
dx 2
x = y = 0. Thus x = 0 is a critical point and by the 2nd derivative test, y has a relative minimum at (0, 0).
EXERCISES 1.3
8. a.
7
b.
d2x
dt
c.
2
=
d 3
dx
(t − x3 ) = 3t 2 − 3 x 2
dt
dt
= 3t 2 − 3 x 2 (t 3 − x3 )
= 3t 2 − 3 x 2 t 3 + 3 x5
No, because if 1 < x < 2 and t > 2.5, then
away from x = 1.
10.
12.
Figure 5
dx 3
= t − x3 > 0, and so the particle moves to the right,
dt
17
18
CHAPTER 1 INTRODUCTION
14.
Figure 6
16.
18. Every solution approaches zero.
EXERCISES 1.4
Exercises 1.4 (page 28)
2.
xn
0.1
0.2
0.3
0.4
0.5
yn
4.000
3.998
3.992
3.985
3.975
(rounded to three decimal places)
4.
6.
8.
10.
xn
0.1
0.2
0.3
0.4
0.5
yn
1.00
1.220
1.362
1.528
1.721
xn
1.1
yn
0.1
1.2
1.3
1.4
1.5
0.209 0.32463 0.44409 0.56437
N
1
2
4
8
φ (π )
π
π
2
1.20739
1.14763
xn
yn
actual
0
0
0
0.1
0
0.00484
0.2
0.01
0.01873
0.3
0.029
0.04082
0.4
0.0561
0.07032
0.5
0.09049
0.10653
0.6
0.13144
0.14881
0.7
0.17830
0.19658
0.8
0.23047
0.24933
0.9
0.28742
0.30657
1.0
0.34868
0.36788
19
20
CHAPTER 1 INTRODUCTION
14. h = 0.5
xn
0.5
1.0
1.5
2.0
yn
1.000
1.500
3.750
24.844
xn
0.5
1.0
yn
1.231
h = 0.1
1.5
2.0
3.819 1,023,041.61 overflow
h = 0.05
xn
0.5
1.0
1.5
2.0
yn
1.278
5.936
overflow
overflow
h = 0.01
xn
0.5
1.0
1.5
2.0
yn
1.321
18.299
overflow
overflow
16. T(1) ≈ 82.694°; T(2) ≈ 76.446°
CHAPTER 2
Exercises 2.2 (page 46)
Exercises 2.3 (page 54)
2. No
2. Neither
4. Yes
4. Linear
6. Yes
6. Linear
8. y 4 = 4 ln x + C
8. y = 2 x 2 + x ln x + Cx
10. x = Cet
10. y = − x −3 + Cx −2
3
12. 4v 2 = 1 + Cx −8 / 3
12. y =
x 3 e −4 x
+ Ce −4 x
3
14. y =
x3 2 x 2
−
+ x + Cx −3
6
5
14. y = tan( x3 + C )
16. ln(1 + y 2 ) = e − x + C
2
π
18. y = tan − ln(cos x)
3
1
20. y = −1 − 1 + 4 x3 + 8 x 2 + 8 x
2
22. y = 4 −
x3
3
24. y = ln 4 x 4 − 3
(
26. y = 1 − ln 1 + x
)
2
2
32. 79.95 mCi
b.
4
+ x2 + x 2 − x + C
( x 2 + 1) 2
18. y =
e− x
+ e −4 x
3
20. y =
3 x 2 x 9 x −3
− +
5
2
10
22. y = 1 – x cot x + csc x
28. y = 3e 2t −t ; maximum value is y(1) = 3e.
34. a.
16. y =
x5
5
T = Ce kt + M
70.77
24. a.
y (t ) = 5e −10t + 35e −20t ; the term 5e−10t
eventually dominates.
b.
y (t ) = 50te −10t + 40e−10t
26. 0.860
1/ 3
30. b.
x 1
y = − + Ce−6 x
2 12
36. 20°
38. v(t ) = 196 − 186e−t / 20 m/sec; limiting velocity
is 196 m/sec.
21
22
CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS
1 − e −2 x
32. y =
−2 x
6
−1 + (2e − 1)e
0≤ x≤3
3< x
Figure 8
36. b.
yh ( x) = x −3
x6
6
d.
v( x) =
e.
y ( x ) = Cx −3 +
x3
6
Exercises 2.4 (page 65)
2. Linear with y as dep. var.
20. 2 arcsin x + sin( xy) −
4. Separable
6. Separable and linear with y as dep. var.
22. e xy −
8. Exact
10. x 2 + xy − y 2 = C
12. e x sin y − x3 + y1/ 3 = C
14. y =
C + et (t − 1)
16. e xy −
1 + et
x
=C
y
18. x 2 + xy 2 − sin( x + y ) − e y = C
24. x =
3 y2 / 3
=C
2
x
= e −1
y
e−t
et − 1
26. x tan y + ln y – 2x = 0
28. a.
x cos(xy) + g(y)
b. xe xy − x 4 + g ( y )
where g is a function of y only.
30. b.
c.
x2 y
x5 y 2 + x 6 y 3 + x 4 y 3 = C
CHAPTER 2 REVIEW
34. µ = x 2 y −2 ; x 2 + y 2 = Cy
18. y = –x – 2 + tan(x + C)
20. tan(x – y) + sec(x – y) – x = C
22. y −2 = Ce−2 x −
e2 x
and y ≡ 0.
2
24. y = [( x − 2)2 + C ( x − 2)−1/ 2 ]2 and y ≡ 0
Figure 9
Orthogonal trajectures for Problem 34
Exercises 2.5 (page 71)
2. Integrating factor depending on x alone.
4. Exact
6. Linear, integrating factor depends on x alone.
8. µ = y −4 , x 2 y −3 − y −1 = C and y ≡ 0.
1/ 3
3e x
+ Ce −3 x
26. y =
4
28. y −2 = − x −
1
+ Ce 2 x and y ≡ 0.
2
y −3
30. ln[( y − 3)2 + ( x + 2) 2 ] − 2 arctan
=C
x+2
2
2
6
6
8
8
32. x + + x + y + − y + = C
5
5
5
5
x4
10. µ = x −2 (also linear); y =
− x ln x + Cx
3
34. t 2 + x 2 = Ct
12. Exact; x 2 y 3 + x − ln y = C
36. y =
14. µ = x3 y 2 , 3 x 4 y 2 + x5 y 3 = C , but not y ≡ 0.
16. y = Cx
Exercises 2.6 (page 0)
2. Homogeneous
3
Cx − x 4
and y ≡ 0.
θ ( ln θ + C )
2
38. y =
4
40. cos( x + y ) + sin( x + y ) = Ce x − y
y
42. sec
x2
y
+ tan 2
x
= Cx
4. Bernoulli
6. Homogeneous
46. b.
y = x+
5x
C − x5
8. y ′ = G (ax + by )
10. y =
x
and y ≡ 0.
C − ln x
Chapter 2 Review (page 81)
2. y = −8 x 2 − 4 x − 1 + Ce 4 x
12. x3 + 3 xy 2 = C
4.
y
14. sin − ln θ = C
θ
16. y = xeCx
x3 4 x 2 3 x
−
+ + Cx −3
6
5
4
6. y −2 = 2 ln 1 − x 2 + C and y ≡ 0.
23
CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS
24
8. y = (Cx 2 − 2 x3 )−1 and y ≡ 0.
10. x + y + 2 y1/ 2 + arctan( x + y ) = C
12. 2 ye 2 x + y 3 e x = C
14. x =
t 2 (t − 1)
+ t (t − 1) + 3(t − 1) ln t − 1 + C (t − 1)
2
16. y = (cos x ) ln cos x + C cos x
18. y = 1 − 2 x + 2 tan
(
12θ 2
20. y = Cθ −3 −
5
2x + C
)
1/ 3
22. (3 y − 2 x + 9)( y + x − 2)4 = C
24. 2 xy + sin x − cos y = C
2
26. y = Ce − x / 2
28. ( y + 3)2 + 2( y + 3)( x + 2) − ( x + 2)2 = C
30. y = Ce 4 x − x −
1
4
32. y 2 = x 2 ln( x 2 ) + 16 x 2
2
2x
34. y = x 2 sin x +
π2
36. sin(2 x + y ) −
x3
2
+ e y = sin 2 +
3
3
1
x
38. y = 2 − arctan
4
2
40.
8
1 − 3e
−4 x
− 4x
2
CHAPTER 3
Exercises 3.2 (page 98)
2. 2.5 − 2e−3t / 25 , 5.78 min
(t + 100) 108 (t + 100)−3
−
4.
, 18.92 min
5
5
6. 43.8 min
8. (0.2)(1 − e−3t /125 ) g/cm3 ; 28.88 sec
14. 187,500
p (t + h) − p (t )
= p ′(t ) and
h
p (t − h ) − p (t )
= p ′(t ), adding these two
lim
−h
h →0
equations and dividing by 2 yields the given
formula.
16. Since lim
h →0
24. 41.94 years
26. a.
31,323 yr
b.
28,333 yr
c.
percent of mass remaining
Exercises 3.3 (page 107)
2. 57.5°F
4. 19.7 min.
6. 2:26 P.M.; 1:37 P.M.
8. 3:51 A.M.; 3:51 P.M.
10. 59.5°F; never
a
18. a. P(t ) = exp Ce−bt +
b
a
a
b. P(t ) = exp ln( P0 ) − e−bt +
b
b
c.
b > 0: P(t ) → ea / b ;
a
If b < 0 and ln( P0 ) > , then P(t) →
b
a
If b < 0 and ln( P0 ) < , then P(t) → 0;
b
a
If b < 0 and ln( P0 ) = , then P(t ) ≡ ea / b .
b
12. The impatient friend
14. 127.5°F
Exercises 3.4 (page 115)
2. 40t + 50e−(0.8)t − 50 ft; 13.75 sec
4. 1.67 sec
6. 12.45e −2t + 4.91t − 12.45 m, 22.9 sec
20. 6.9315 light years
8. 206.8 sec, 86.8 sec
22. 90 min., vanishes at t =
25
26 CHAPTER 3 MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS
1.96t + 0.39(e−5t − 1)
0 ≤ t ≤ 15.5
10. x(t ) =
−50(t −15.5)
]
15.5 ≤ t
30 + 0.098(t − 15.5) + 0.035[1 − e
0.098 m/sec
12. 16.48 sec; 1535 m
1/ n
mg
14.
k
(
)
(
)
T / k2
T / k2
ω /k
= Ce −t / 2 I , where C = e 0
16. e ω / k k ω − T
k ω0 − T
18. 1.61t 2 m, 5.65 m/sec
20. α 0 = arctan µ
5
5
1
+ 6t −
t ≤ ln 5 ≈ 0.805
2
t
2
2
2e
22. x(t ) =
13 / 5
37
1
10t + 5
− 2 ln 5 −
t > ln 5
6t / 5
6
2
6e
10 m/sec
m0
m0
1 2 β
24. v(t ) = β ln
− gt , x(t ) = β t − gt − (m0 − α t ) ln
2
α
m0 − α t
m0 − α t
Exercises 3.5 (page 122)
2. capacitor voltage = −
resistor voltage =
current =
10, 000
100, 000, 000
10, 000
cos100t +
sin 100t +
e−1,000,000t V
100, 000, 001
100, 000, 001
100, 000, 001
10, 000
1
10, 000
cos100t +
sin 100t −
e−1,000,000t V
100, 000, 001
100, 000, 001
100, 000, 001
10, 000
1
10, 000
cos100t +
sin 100t −
e−1,000,000t A
100, 000, 001
100, 000, 001
100, 000, 001
4. From (2), I =
1
dE
.
E (t ) dt. From the derivative of (4), I = C
∫
L
dt
d 1 2
LI + RI 2 = EI (power generated by the voltage source equals the power
dt 2
inserted into the inductor plus the power dissipated by the resistor). Multiply the equation above (4) by I,
dq
d 1
and then replace q by CEC in the capacitor term, and derive RI 2 + CEC2 = EI (power
replace I by
dt
dt 2
generated by the voltage source equals the power inserted into the capacitor plus the power dissipated by the
resistor).
6. Multiply (2) by I to derive
8. In cold weather, 96.27 hours. In (extremely) humid weather, 0.0485 seconds.
EXERCISES 3.6
Exercises 3.6 (page 132)
6. For k = 0, 1, 2, …, n, let xk = kh and zk = f ( xk ) where h =
8.
10.
" + zn−1 )
a.
h( z0 + z1 + z2 +
b.
h
( z0 + 2 z1 + 2 z2 +
2
c.
Same as (b)
" + 2 zn−1 + zn )
xn
1.2
1.4
1.6
1.8
yn
1.48
2.24780
3.65173
6.88712
xn
yn
0.1
1.15845
0.2
1.23777
0.3
1.26029
0.4
1.24368
0.5
1.20046
0.6
1.13920
0.7
1.06568
0.8
0.98381
0.9
0.89623
1.0
0.80476
12. φ (π) ≈ y (π; π2−4 ) = 1.09589
14. 2.36 at x = 0.78
16. x = 1.26
1
.
n
27
28 CHAPTER 3 MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS
20.
t
r = 1.0
r = 1.5
r = 2.0
0
0.0
0.0
0.0
0.2
1.56960
1.41236
1.19211
0.4
2.63693
2.14989
1.53276
0.6
3.36271
2.51867
1.71926
0.8
3.85624
2.70281
1.84117
1.0
4.19185
2.79483
1.92743
1.2
4.42005
2.84084
1.99113
1.4
4.57524
2.86384
2.03940
1.6
4.68076
2.87535
2.07656
1.8
4.75252
2.88110
2.10548
2.0
4.80131
2.88398
2.12815
2.2
4.83449
2.88542
2.14599
2.4
4.85705
2.88614
2.16009
2.6
4.87240
2.88650
2.17126
2.8
4.88283
2.88668
2.18013
3.0
4.88992
2.88677
2.18717
3.2
4.89475
2.88682
2.19277
3.4
4.89803
2.88684
2.19723
3.6
4.90026
2.88685
2.20078
3.8
4.90178
2.88686
2.20360
4.0
4.90281
2.88686
2.20586
4.2
4.90351
2.88686
2.20765
4.4
4.90399
2.88686
2.20908
4.6
4.90431
2.88686
2.21022
4.8
4.90453
2.88686
2.21113
5.0
4.90468
2.88686
2.21186
Exercises 3.7 (page 142)
h2
2. yn +1 = yn + h( xn yn − yn2 ) +
2
( yn + ( xn − 2 yn )( xn yn − yn2 ))
h2
h3
h4
4. yn +1 = yn + h( xn2 + yn ) + (2 xn + xn2 + yn ) + (2 + 2 xn + xn2 + yn ) + (2 + 2 xn + xn2 + yn )
2!
3!
4!
6. Order 2: φ (1) § order 4: φ (1) § EXERCISES 3.7
8. 0.63211
10. 0.70139 with h = 0.25
12. –0.928 at x = 1.2
14. 1.00000 with h =
16.
π
16
xn
yn
0.5
1.17677
1.0
0.37628
1.5
1.35924
2.0
2.66750
2.5
2.00744
3.0
2.72286
3.5
4.11215
4.0
3.72111
20. x(10) ≈ 2.23597 ×10−4
29
CHAPTER 4
Exercises 4.1 (page 159)
4. Both solutions approach zero as t → ∞.
6. The oscillations get larger until Hooke’s law becomes invalid.
8. −
10. a.
30
25
cos 3t − sin 3t
61
61
A cos Ωt + B sin Ωt =
k − mΩ 2
2 2
2
( k − mΩ ) + b Ω
2
cos Ωt +
bΩ
2 2
( k − mΩ ) + b 2 Ω 2
sin Ωt
b.
c.
Figure 10
31
32
CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
Exercises 4.2 (page 167)
Exercises 4.3 (page 177)
2. c1e −t + c2 e2t
2. c1 cos t + c2 sin t
4. c1e−3t + c2 te −3t
4. c1e5t cos t + c2 e5t sin t
6. c1e2t + c2 e3t
6. c1e2t cos 3t + c2 e2t sin 3t
8. c1et / 2 + c2 e −2t / 3
5t
5t
−t / 2
8. c1e−t / 2 cos
sin
+ c2 e
2
2
10. c1et / 2 + c2 tet / 2
12. c1e
(
)
−11+ 205 t / 6
14. 3 − e
(
)
−t
t
16.
+ c2 e
−11− 205 t / 6
12. c1 cos 7t + c2 sin 7t
14. c1et cos 5t + c2 et sin 5t
3t
4e e
−
3
3
18. 2e3t +
10. c1e−2t cos 2t + c2 e−2t sin 2t
16. c1e
7te3t
3
(3+
)
53 t / 2
+ c2 e
(3−
)
53 t / 2
18. c1e−7t + c2 et / 2
20. c1e−t + c2 et cos t + c3 et sin t
20. (2 − t )e 2t − 2
22. e−t cos 4t
ce7t
22.
3
24. ce −11t / 3
1
24. sin 3t + cos 3t
3
26. a. 2 cos t
26. et − 3tet
b. 2 cos t + c2 sin t , where c2 is arbitrary.
28. linearly independent
28. b = 5: y =
4e −t − e −4t
3
b = 4: y = (1 + 2t )e −2t
30. linearly independent
b = 2: y = e −t cos 3t + 3−1/ 2 e−t sin 3t
32. linearly dependent
32. a.
c
38. If c1 ≠ 0, then y1 = − 2 y2 .
c1
42. c1e−t + c2 et + c3 e6t
46. c1e + c2
48. et + e 2t
5
2π
2
34. I (t ) = e−t sin 25t
5
44. c1e−t + c2 e3t + c3 e5t
t
b.
y(t) = 0.3 cos 5t – 0.02 sin 5t m
( −1+ 3 )t + c e( −1− 3 )t
e
3
3
3
36. a. y (t ) = 1 − i e( −1+i )t + 1 + i e( −1−i )t
2
2
40. c1 x + c2 x −7
EXERCISES 4.5
42. c1 x 2 sin 2 ln x + c1 x 2 cos 2 ln x
44. c1 x −1 sin [2 ln x ] + c1 x −1 cos [2 ln x ]
Exercises 4.5 (page 192)
2. a.
t 1 1
− + sin 2t
4 8 4
b.
t 1 3
− − sin 2t
2 4 4
c.
11t 11
− − 3 sin 2t
4
8
Exercises 4.4 (page 186)
2. Yes
4. Yes
6. Yes
4. c1 + c2 e −t + t
8. Yes
6. c1e−2 x + c2 e−3 x + e x + x 2
10. y p = −10
8. c1e 2 x + c2 e− 2 x + tan x
12. x p = 3t 2 − 12t + 24
2
14. y p = 2 x (ln 2 ) + 1
16. θ p = −
10. Yes
−1
12. No
14. Yes
t sin t + cos t
2
16. Yes
18. c1e3t + c2 e−t − t 2 +
18. y p = −2t cos 2t
20. y p = −2t 2 cos 2t + t sin 2t
20. c1 cos 2θ + c2 sin 2θ +
22. x p = 2t 4 et
24. t 3 − t + 3
26. y p = t 2 e −t sin t + te−t cos t
26. cos 3t + 2 sin 3t + 3
( At 4 + Bt3 + Ct 2 + Dt + E ) et
28.
30. Aet sin t + Bet cos t
(
)
32. t At 6 + Bt 5 + Ct 4 + Dt 3 + Et 2 + Ft e−3t
34. −
e −t
4
sin t
36. −
4
sin θ − cos θ
3
22. c1e−3 x cos x + c2 e −3 x sin x + x 4 − x 2 + 2
24. y p = x 2 sin x + x cos x
28.
4t 1
+
3 9
e−4t 1 et e2t 7e3t
+ − −
+
60 12 10 6
6
30. t 2 − 4t + 7 −
et 27e−t te−t
−
−
4
4
2
32. ( At 2 + Bt + C )e 2t
34. A cos t + B sin t + C cos 2t + D sin 2t
36. e2t ( At 4 + Bt 3 + Ct 2 )
33
34
CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
38. −2 sin t + cos t
40. −3t 2 + 2t
42. a. y (t ) = A sin β t + B cos β t + yh , where A =
4mk − b 2
yh = e −bt / 2 m c1 cos
2m
b.
k − mβ 2
2 2
2
(k − m β ) + b β
4mk − b 2
t + c2 sin
2m
2
, B=
−b β
(k − m β 2 ) 2 + b 2 β 2
and
t
In each case, yh → 0 as t → ∞. So as t → ∞, y(t) approaches the function y p (t ) = A sin β t + B cos β t.
44. sin 3t − cos 3t + e−t sin 2t
46. If λ 2 = 1 the general solution is −
sin t
λ2 −1
t cos t
+ c1 sin t + c2 cos t ; if λ 2 = 4, 9, ... the general solution is
2
+ c1 sin λ t + c2 cos λ t. In neither case is there a solution satisfying the boundary conditions.
Exercises 4.6 (page 197)
2. c1 cos t + c2 sin t + t sin t + (cos t ) ln cos t
4. c1et + c2 e −t − 2t − 4
t 2 e −t
2
6. c1e−t + c2 te−t +
8. c1 cos 3t + c2 sin 3t +
(sin 3t ) ln sec 3t + tan 3t − 1
9
2 −2t
10. c1e−2t + c2 te −2t + t e
2
12. c1 cos t + c2 sin t +
3t 2 e −2t
ln t −
4
e3t
− 1 − (cos t ) ln sec t + tan t
10
14. c1 cos θ + c2 sin θ + sin θ tan θ −
16. c1e−2t + c2 e −3t + 3t 2 − 5t +
18. c1e3t + c2 te3t +
e3t
2t
22. c1t 2 + c2 t 3 + t 3 ln t +
24. c1et + c2 et ln t +
t 2 et
4
1
6
19
6
sec θ
2
EXERCISES 4.8
35
Exercises 4.7 (page 208)
2. Inertia = 1, damping = 0, stiffness = –6y, and is negative for y > 0 (which explains why solutions
1
> 0 run away).
y=
(c − t ) 2
4. Suppose
a1
(3 − t )
Therefore
2
a2
+
a1
2
(2 − t )
2
+
a2
+
2
a3
(1 − t )2
≡ 0 for –1 < t < 1. By taking limits as t → 1, we conclude a3 = 0.
≡ 0, so a1 (2 − t )2 + a2 (3 − t )2 = (a1 + a2 )t 2 + (−4a1 − 6a2 )t + 4a1 + 9a2 ≡ 0.
(3 − t )
(2 − t )
Thus a1 + a2 = 0 and −4a1 − 6a2 = 0 and 4a1 + 9a2 = 0, which implies a1 = a2 = 0. Hence a1 = a2 = a3 = 0
and they are linearly independent.
6. y ′′ = −
k
d k 2
k 2
1
y=−
y , so ( y ′) 2 +
y = constant, or m( y ′)2 + ky 2 = 2m ⋅ (constant).
m
dy 2m
2
2m
8. By eq. (21), θ ′′ = −
g
A
sin θ =
(θ ′)2 g
d g
cos θ . Therefore 2 − A cos θ = constant.
dθ A
10. At t = 0, θ ′(0) = 0 and θ (0 ) = α , so
θ ′(0)2 g
(θ ′)2 g
− cos θ =
− cos α
A
A
2
2
g
= − cos α .
A
Therefore, cos θ (t ) =
A θ ′(t )
g
2
2
+ cos α ≥ cos α , and since cos θ is a decreasing function for
0 ≤ θ ≤ π , θ (t ) ≤ α .
16. y ′′ = − y − y 3 =
Therefore,
( y ′)2 y 2 y 4
d y2 y4
+
+
= K.
−
−
, so
2
2
4
4
dy 2
y2
( y ′) 2 y 4
≤K−
−
≤ K , and hence y ≤ 2 K .
2
2
4
Exercises 4.8 (page 219)
1
1
2. y = − cos 5t − sin 5t
4
5
41
=
sin(5t + φ ),
20
5
where φ = arctan − π = −2.246 .
4
()
π − arctan 54
41
2π
5
≈ 0.449 sec.
amp.=
, period =
, freq. =
. Passes through equilibrium at time
2π
5
20
5
36
CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
4. b = 0: y(t) = cos 8t
Figure 11 (b = 0)
5 −5t
b = 10: y (t ) = e −5t cos 39t +
e sin 39t
39
8 −5t
=
e sin 39t + φ ,
39
39
where φ = arctan
≈ 0.896.
5
(
)
Figure 12 (b = 10)
EXERCISES 4.8
b = 16: y (t ) = (1 + 8t )e−8t
Figure 13 (b = 16)
4
1
b = 20: y (t ) = e −4t − e −16t
3
3
Figure 14 (b = 20)
37
38
CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
1 + 2 ( −2 + 2 )t 1 − 2 ( −2 − 2 )t
6. k = 2: y (t ) =
+
e
e
2
2
Figure 15 (k = 2)
k = 4: y (t ) = (1 + 2t )e−2t
Figure 16 (k = 4)
EXERCISES 4.9
k = 6: y (t ) = e −2t cos 2t + 2e −2t sin 2t
= 3 e −2t sin 2t + φ ,
(
)
2
where φ = arctan
≈ 0.615 .
2
Figure 17 (k = 6)
8. Never
10. e−π / 127 ≈ 0.755 m
12. 0.080 sec
Exercises 4.9 (page 227)
2. M (γ ) =
1
(3 − 2γ 2 ) 2 + 9γ 2
Figure 18
Response curve for Problem 2
5
4. y (t ) = 1 + t sin t
2
16.
18
kg
7
39
40
CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
3
9
8. y (t ) = e −3t − e −t + 3; lim y (t ) = 3, which is the displacement of a simple spring with spring constant k
t →∞
2
2
= 6 when a force F0 = 18 is applied.
Figure 19
759t
759t
3
−5t / 4
12. y (t ) = (0.047)e −5t / 4 cos
sin
+ (0.058)e
+
sin(t + θ ) m,
4
4 10 9241
367
96
≈ 1.078
where θ = arctan ≈ 1.519;
5
4 2π
Chapter 4 Review (page 230)
2. c1e−t / 7 + c2 te−t / 7
4. c1e5t / 3 + c2 te5t / 3
6. c1e
( −4 +
)
30 t
+ c2 e
( −4 −
)
30 t
8. c1e−2t / 5 + c2 te−2t / 5
10. c1 cos
(
)
11t + c2 sin
(
11t
)
12. c1e−5t / 2 + c2 e2t + c3 te2t
14. c1e2t cos
( 3t ) + c2 e2t sin ( 3t )
CHAPTER 4 REVIEW
16. e−t c1 + c2 cos
( 2t ) + c3 sin ( 2t )
18. c1t 3 + c2 t 2 + c3 t + c4 + t 5
4
1
20. c1et / 2 + c2 e −t / 2 − cos t − t sin t
9
3
1092
4641
22. c1e11t + c2 e−3t +
cos t −
sin t
305
305
1 1 1
24. c1et / 2 + c2 e−3t / 5 − t − − tet / 2
3 9 11
26. c1e−3t cos
( 6t ) + c2 e−3t sin ( 6t ) + 311 e2t + 5
28. c1 x3 cos(2 ln x) + c2 x3 sin(2 ln x)
30. 3e−θ + 2θ e−θ + sin θ
32. et / 2 cos t − 6et / 2 sin t
34. e−7t + 4e2t
36. −3e−2t / 3 + te−2t / 3
1
1
2π
5
38. y (t ) = − cos 5t + sin 5t , amp. ≈ 0.320 m, period =
, freq. =
,
4
5
5
2π
1
5
tequil = arctan ≈ 0.179 sec.
5
4
41
CHAPTER 5
Exercises 5.2 (page 250)
3
2. x = c1e2t − c2 e−3t ; y = c1e2t + c2 e−3t
2
1
1
4. x = − c1e3t + c2 e −t ; y = c1e3t + c2 e−t
2
2
c + c
c − c
7
1
6. x = 1 2 et cos 2t + 2 1 et sin 2t + cos t − sin t ;
10
10
2
2
11
7
y = c1et cos 2t + c2 et sin 2t + cos t + sin t
10
10
26
45
5c
4
1
8. x = − 1 e11t − t −
; y = c1e11t + t +
11 121
11 121
4
c − 3c1
c1 + 3c2
1 t 1 −t
10. x = c1 cos t + c2 sin t ; y = 2
cos t − 2 sin t + 2 e − 2 e
2
12. u = c1e2t + c2 e−2t + 1; v = −2c1e 2t + 2c2 e−2t + 2t + c3
14. x = −c1 sin t + c2 cos t + 2t − 1; y = c1 cos t + c2 sin t + t 2 − 2
2
1
1
16. x = c1et + c2 e −2t + e4t ; y = −2c1et − c2 e−2t + c3 − e 4t
9
2
36
1
18. x(t ) = −t 2 − 4t − 3 + c3 + c4 et − c1tet − c2 e−t ; y (t ) = −t 2 − 2t − c1et + c2 e −t + c3
2
20. x(t ) =
3 t 1 3t
3
1
e − e , y (t ) = − et − e3t + e 2t
2
2
2
2
22. x(t ) = 1 +
9 3t 5 −t
3
5
e − e , y (t ) = 1 + e3t − e−t
4
4
2
2
24. No solutions
1
26. x = et {(c1 − c2 ) cos t + (c1 + c2 ) sin t} + c3 e2t ; y = et (c1 cos t + c2 sin t );
2
3
z = et {(c1 − c2 ) cos t + (c1 + c2 ) sin t} + c3 e 2t
2
28. x(t ) = c1 + c2 e −4t + 2c3 e3 t , y (t ) = 6c1 − 2c2 e −4t + 3c3 e3t , z (t ) = −13c1 − c2 e −4t − 2c3 e3t
30. λ 43
44
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
32. x =
y=
20 − 10 19
19
−50
36.
e
(
( −7+ 19 )t /100 + −20 − 10
19
19
)
−7 + 19 t /100
19
+
50
e
(
)
−7 − 19 t /100
19
e
(−7− 19 )t /100 + 20;
+ 20
1
3
3
1
V1 = c2 − c1 e−t cos 3t − c2 + c1 e −t sin 3t + 5 L; V2 = c1 e−t cos 3t + c2 e−t sin 3t + 18 L
2
2
2
2
34. b.
c.
e
As t → + V1 → 5 L and V2 → 18 L.
400
≈ 36.4°F
11
38. A runaway arms race.
40. a. 3x 2 = x3
b. 6 x + 3 x 2 − 2 x3
c. 3 x 2 + 2 x3
d. 6 x + 3 x 2 − 2 x3
e. D 2 + D − 2
f. 6 x + 3 x 2 − 2 x3
Exercises 5.3 (page 261)
2. x1′ = x2 , x2′ = x12 + cos(t − x1 ); x1 (0) = 1, x2 (0) = 0
4. x1′ = x2 , x2′ = x3 , x3′ = x4 , x4′ = x5 , x5′ = x6 , x6′ = ( x2 )2 − sin x1 + e2t ; x1 (0) =
" = x6 (0) = 0
5
2
3
1
6. Setting x1 = x, x2 = x ′, x3 = y , x4 = y ′, we obtain x1′ = x2 , x2′ = − x1 + x3 , x3′ = x4 , x4′ = x1 − x3 .
2
2
3
3
8. x1 (0) = a, x2 (0) = p (0)b
10.
12.
i
ti
y (ti )
1
0.250
0.96924
2
0.500
0.88251
3
0.750
0.75486
4
1.000
0.60656
i
ti
y (ti )
1
1.250
0.80761
2
1.500
0.71351
3
1.750
0.69724
4
2.000
0.74357
14. y (8) ≈ 24.01531
16. x(1) ≈ 127.773; y(1) ≈ –423.476
18. Conventional troops
EXERCISES 5.3
20. a. period ≈ 2(3.14)
b. period ≈ 2(3.20)
c. period ≈ 2(3.34)
24. Yes, yes
26. x(1) ≈ 0.80300; y(1) ≈ 0.59598; z(1) ≈ 0.82316
x1′ = x2
28. a.
x2′ =
x1 (0) = 1
− x1
( x12
30. a.
ti
x1 (ti ) ≈ x(ti )
x3 (ti ) ≈ y (ti )
10
0.628
0.80902
0.58778
20
1.257
0.30902
0.95106
x3 (0) = 0
30
1.885
–0.30902
0.95106
x4 (0) = 1
40
2.513
0.80902
0.58779
50
3.142
–1.00000
0.00000
60
3.770
–0.80902
–0.58778
70
4.398
–0.30903
–0.95106
80
5.027
0.30901
–0.95106
90
5.655
0.80901
–0.58780
100 6.283
1.00000
–0.00001
x2 (0) = 0
+ x32 )3 / 2
x3′ = x4
x4′ =
i
b.
− x3
( x12 + x32 )3 / 2
ti
li
θi
ti
li
θi
1.0
5.27015
0.0
1.0
5.13916
0.45454
2.0
4.79193
0.0
2.0
4.10579
0.28930
3.0
4.50500
0.0
3.0
2.89358
–0.10506
4.0
4.67318
0.0
4.0
2.11863
–0.83585
5.0
5.14183
0.0
5.0
2.13296
–1.51111
6.0
5.48008
0.0
6.0
3.18065
–1.64163
7.0
5.37695
0.0
7.0
5.10863
–1.49843
8.0
4.92725
0.0
8.0
6.94525
–1.29488
9.0
4.54444
0.0
9.0
7.76451
–1.04062
10.0
4.58046
0.0
10.0
7.68681
–0.69607
b.
45
46
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
Exercises 5.4 (page 274)
2.
Figure 20
4. x = –6, y = 1
6. The line y = 2 and the point (1, 1)
8. x3 − x 2 y − y −2 = c
10. Critical points are (1, 0) and (–1, 0).
Integral curves:
for y > 0, x > 1, y = c x 2 − 1;
for y > 0, x < 1, y = c 1 − x 2 ;
for y < 0, x > 1, y = −c x 2 − 1;
for y < 0, x < 1, y = −c 1 − x 2 ;
all with c ≥ 0.
If c = 1, y = ± 1 − x 2 are semicircles ending at
(1, 0) and (–1, 0).
EXERCISES 5.4
12. 9 x 2 + 4 y 2 = c
Figure 21
14. y = cx 2 / 3
Figure 22
47
48
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
16. (0, 0) is a stable node.
Figure 23
18. (0, 0) is an unstable node.
(0, 5) is a stable node.
(7, 0) is a stable node.
(3, 2) is a saddle point.
Figure 24
EXERCISES 5.4
20.
{vy′′ == −v y ; (0, 0) is a center.
Figure 25
y′ = v
22.
3 ; (0, 0) is a center.
v ′ = − y
Figure 26
49
50
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
y′ = v
24.
3;
v ′ = − y + y
(0, 0) is a center.
(–1, 0) is a saddle point.
(1, 0) is a saddle point.
Figure 27
26.
x2 x4 y 2
+
+
= c; all solutions are bounded.
2
4
2
Figure 28
28. (0, 0) is a center. (1, 0) is a saddle.
EXERCISES 5.4
30. a.
x(t) → x*, y(t) → y*, f and g are continuous implies x ′(t ) ≡ f ( x(t ), y (t )) → f ( x*, y*) and
y ′(t ) ≡ g ( x(t ), y (t )) → g ( x*, y*).
t
b.
x(t ) = ∫ x ′(τ )d τ + x(T )
T
f ( x*, y*)
>
(t − T ) + x(T )
2
t
≡ f ( x*, y*) + C
2
c.
If f(x*, y*) > 0, f(x*, y*)t → LPSO\LQJ x(t) →
d, e. similar
32.
( y ′)2 y 4
+
= c by Problem 30.
2
4
Thus,
34. a.
y4
( y ′) 2
=c−
≤ c, so y ≤ 4 4c .
4
2
dI
b
b
= 0, that is when aSI – bI = 0 or S =
provided S (0) >
dt
a
a
b
so that S can in fact attain the value .
a
Peak will occur when
b
b
, while I ′(t ) < 0 if I > 0 and S < .
a
a
b.
I ′(t ) > 0 if I > 0 and S >
c.
It steadily decreases to zero.
36.
Figure 29
51
52
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
38. a.
b.
d 2
[ x + y 2 + z 2 ] = 0; the magnitude of the angular velocity is constant.
dt
All points on the axes are critical points: (x, 0, 0), (0, y, 0), (0, 0, z).
d.
y2
dy −2 x
= c (cylinder).
=
, so x 2 +
From (a), x 2 + y 2 + z 2 = K (sphere). Also
2
dx
y
The solutions are periodic.
e.
The critical point on the y-axis is unstable. The other two are stable.
c.
Exercises 5.5 (page 284)
2. x(t ) =
(
)
(
)
10
3 10
1 − 10 cos r1t − 1 + 10 cos r2 t , y (t ) =
[cos r1t − cos r2 t ], where r1 = 4 + 10 ,
20
20
r2 = 4 − 10 .
4. m1 x ′′ = −k1 x + k2 ( y − x), m2 y ′′ = −k2 ( y − x) − by ′
6. b.
37
x(t ) = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t + cos 3t
40
c.
111
y (t ) = 2c1 cos t + 2c2 sin t − c3 cos 2t − c4 sin 2t −
cos 3t
20
d.
23
9
37
23
9
111
x(t ) = cos t − cos 2t + cos 3t , y (t ) = cos t + cos 2t −
cos 3t
8
5
40
4
5
20
8. θ1 (t ) =
π
π 10
9.8 π
9.8
9.8 π 10
9.8
cos
t + cos
t ; θ 2 (t ) =
cos
t −
cos
t
12
24
24
5 + 10 12
5 − 10
5 + 10
5 − 10
Exercises 5.6 (page 291)
1
2. q (t ) = e −4t cos(6t ) + 3 cos(2t ) + sin(2t ) coulombs
2
10
10
4. I (t ) = cos 5t − cos 50t amps
33
33
8. L = 0.01 henrys, R = 0.2 ohms, C =
25
2
farads, and E (t ) = cos 8t volts
32
5
5
1
1
9
1
3
10. I1 = − e−2t − e−2t / 3 + ; I 2 = e−2t − e−2t / 3 + ;
2
2
4
4
4
4
1
3
I 3 = − e −2 t − e −2 t / 3 + 2
2
2
5 5
4 4
12. I1 = 1 − e−900t ; I 2 = − e−900t ; I3 = − e−900t
9 9
9 9
CHAPTER 5 REVIEW
Exercises 5.7 (page 301)
2. ( x0 , v0 ) = (−1.5, 0.5774)
( x1 , v1 ) = (−1.9671, − 0.5105)
( x2 , v2 ) = (−0.6740, 0.3254)
#
( x20 , v20 ) = (−1.7911, − 0.5524)
The limit set is the ellipse ( x + 1.5)2 + 3v 2 = 1 .
4. ( x0 , v0 ) = (0, 10.9987)
( x1 , v1 ) = (−0.00574, 10.7298)
( x2 , v2 ) = (−0.00838, 10.5332)
#
( x20 , v20 ) = (−0.00029, 10.0019)
The attractor is the point (0, 10).
12. For F = 0.2, attractor is the point (–0.319, –0.335). For F = 0.28, attractor is the point (–0.783, 0.026).
14.Attractor consists of two points: (–1.51, 0.06) and (–0.22, –0.99).
Figure 30
53
54
CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
Chapter 5 Review Problems (page 304)
2. x = −(c1 + c2 )e −t cos 2t + (c1 − c2 )e −t sin 2t
y = 2c1e −t cos 2t + 2c2 e −t sin 2t
c
c
4. x = c1t + c2 + e−t , y = 1 t 3 + 2 t 2 + c3 t + c4
6
2
6. x = e 2t + e −t , y = e2t + e−t , z = e2t − 2e −t
8. With x1 = y , x2 = y ′, we obtain x1′ = x2 , x2′ =
1
(sin t − 8 x1 + tx2 ).
2
10. With x1 = x, x2 = x ′, x3 = y, x4 = y ′, we get x1′ = x2 , x2′ = x1 − x3 , x3′ = x4 , x4′ = − x2 + x3 .
dy 2 − x
=
are given implicitly by ( x − 2)2 + ( y − 2) 2 = const. Critical point
dx y − 2
is at (2, 2), which is a stable center.
12. Solutions to phase plane equation
Figure 31
(2 j + 1) π (2k + 1) π
,
14. Critical points are ( mπ , nπ ) , m, n integers and
, j, k integers. Equation for integral
2
2
dy cos x sin y tan y
=
=
, with solutions sin y = C sin x.
curves is
dx sin x cos y tan x
CHAPTER 5 REVIEW
16. Origin is saddle (unstable).
Figure 32
18. Natural angular frequencies are
2, 2 3.
( )
( )
( 2t ) + c4 sin ( 2t ) ,
c
1
y (t ) = − c1 cos ( 2 3t ) − 2 sin ( 2 3t ) + 3c3 cos ( 2t ) + 3c4 sin ( 2t ) .
3
3
General solution is x(t ) = c1 cos 2 3t + c2 sin 2 3t + c3 cos
55
CHAPTER 6
Exercises 6.1 (page 324)
2.
(0, ∞ )
4. (–1, 0)
6. (0, 1)
8. Lin. dep.; 0
10. Lin. ind.; −2 tan 3 x − sin x cos x − sin 2 x tan x − 2 tan x
12. Lin. dep.; 0
14. Lin. indep.; ( x + 2)e x
16. c1e x + c2 cos 2 x + c3 sin 2 x
18. c1e x + c2 e− x + c3 cos x + c4 sin x
20. a.
b.
22. a.
b.
c1 + c2 x + c3 x3 + x 2
2 − x3 + x 2
c1e x cos x + c2 e x sin x + c3 e − x cos x + c4 e − x sin x
e x cos x + cos x
24. a.
7 cos 2x + 1
b.
−6 cos 2 x −
26. y ( x ) =
11
3
n +1
∑ γ j −1 y j ( x)
j =1
34. The Wronskian W [ f1 , f 2 , f3 ]( x)
Exercises 6.2 (page 331)
2. c1e x + c2 e − x + c3 e3 x
4. c1e− x + c2 e −5 x + c3 e4 x
6. c1e− x + c2 e x cos x + c3 e x sin x
57
58
CHAPTER 6 THEORY OF HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS
8. c1e x + c2 xe x + c3 e−7 x
10. c1e− x + c2 e
( −1+ 7 )x + c e( −1− 7 )x
3
12. c1e x + c2 e−3 x + c3 xe −3 x
14. c1 sin x + c2 cos x + c3 e− x + c4 xe− x
16. (c1 + c2 x)e− x + (c3 + c4 x + c5 x 2 )e6 x + c6 e −5 x + c7 cos x + c8 sin x + c9 sin 2 x + c10 cos 2 x
3x
3x
−x / 2
2 −3 x
18. (c1 + c2 x + c3 x 2 )e x + c4 e 2 x + c5 e− x / 2 cos
sin
cos x
+ c6 e
+ (c7 + c8 x + c9 x )e
2
2
+ (c10 + c11 x + c12 x 2 )e−3 x sin x
20. e− x − e−2 x + e−4 x
22. x(t ) = c1e 3t + c2 e − 3t + c3 cos 2t + c4 sin 2t
2
2
y (t ) = − c1e 3t − c2 e− 3t + c3 cos 2t + c4 sin 2t
5
5
28. c1e1.879 x + c2 e −1.532 x + c3 e−0.347 x
5 − 10
5 + 10
34. x(t ) =
t
t
10 cos 4 + 10 + 10 cos 4 − 10 ,
5 − 2 10
5 + 2 10
y (t ) =
t +
+
cos
4
10
cos 4 − 10 t
10
10
Exercises 6.3 (page 337)
2. c1e− x + c2 cos x + c3 sin x
4. c1 + c2 xe x + c3 x 2 e x
1
3
1
6. c1e x + c2 xe x + c3 e−3 x + e− x + cos x + sin x
8
20
20
1 4
1
8. c1e x + c2 e− x cos x + c3 e− x sin x − − xe x + x 2 e x
2 25
10
EXERCISES 6.3
59
5 −3 x
1
1
1
10. c1e2 x + c2 e −3 x cos 2 x + c3 e −3 x sin 2 x +
cos 2 x − xe −3 x sin 2 x − x −
xe
116
58
26
676
12. D3
14. D – 5
16. D3 ( D − 1)
18. [( D − 3)2 + 25]2
20. D 4 ( D − 1)3 ( D 2 + 16)2
22. c3 e3 x + c4 cos x + c5 sin x
24. c3 xe x + c4 x 2 e x
26. c3 x 2 + c4 x + c5
28. c3 e3 x + c4 + c5 x
30. c4 xe x + c5
32. 5 + e x sin 2 x − e3 x
7 −t / 2
7t 7
7t
1 1
3
3 −t / 2
38. x(t ) = c1 + + t et + − c2 −
cos
sin
c3 e
+
c2 − c3 e
+ t + 1
2 4
2
2
2
2 2
2
7t
7t 1
1
−t / 2
y (t ) = c1 + t et + c2 e −t / 2 cos
sin
+ c3 e
+
4
2
2 2
2187 t 3 t
t
40. I1 (t ) =
sin − sin − 18 sin
40 8 40 72
24
t
t
243 27
I 2 (t ) =
sin − sin
40 8 40 72
243 t 3 t
t
I3 (t ) =
sin + sin − 18 sin
5
8
5
72
24
60
CHAPTER 6 THEORY OF HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS
Exercises 6.4 (page 341)
1
2. x 2 + 2 x
2
1
4. x3 e x
6
6. sec θ – sin θ tan θ + θ sin θ + (cos θ)ln(cos θ)
8. c1 x + c2 x ln x + c3 x3 − x 2
x −1
x4
x
10.
∫ g ( x) dx + ∫ x −5 g ( x)dx − ∫ x −2 g ( x)dx
10
15
6
Chapter 6 Review Problems (page 344)
2. a.
Lin. indep.
b.
Lin. indep.
c.
Lin. dep.
4. a.
c1e−3 x + c2 e− x + c3 e x + c4 xe x
( − 2 + 5 ) x + c e( − 2 − 5 ) x
b.
c1e x + c2 e
c.
c1e x + c2 cos x + c3 sin x + c4 x cos x + c5 x sin x
d.
x
x 1
c1e x + c2 e − x + c3 e2 x − e x + +
2
2 4
3
x
x
x
x
x/ 2
x/ 2
−x / 2
2
6. c1e− x / 2 cos
sin
cos
sin
+ c2 e
+ c3 e
+ c4 e
+ sin( x )
2
2
2
2
8. a.
c3 xe− x + c4 + c5 x + c6 x 2
b.
c4 xe− x + c5 x 2 e − x
c.
c5 + c6 x + c7 x 2 + c8 cos 3 x + c9 sin 3 x
d.
c4 cos x + c5 sin x + c6 x cos x + c7 x sin x
10. a.
b.
c1 x1/ 2 + c2 x −1/ 2 + c3 x
c1 x −1 + c2 x cos
(
)
3 ln x + c3 x sin
(
3 ln x
)
CHAPTER 7
Exercises 7.2 (page 359)
2.
2
s
4.
6.
8.
3
Exercises 7.3 (page 0)
, s>0
2.
1
, s>3
4.
, s>0
6.
( s − 3)2
s
s 2 + b2
s
2
2
( s + 1) + 4
8.
, s > –1
10.
12.
1 − e 6 − 3 s e −3 s
, s>2
+
s−2
s
12.
14.
5
1
12
−
+ , s>2
s s − 2 s3
14.
18.
20.
2
s3
24
s5
−
−
3
s2
6
s3
−
−
6
( s + 1)2 + 9
1
s2
s+2
2
( s + 2) + 3
−
+
2
s2 + 2
2
( s + 2)
3
, s>0
16.
, s>0
18.
, s > –2
20.
3
72
s
1 e− s − 1
+
10.
, s>0
s
s2
16.
6
5
−
1
s−2
−
4
s
3
+
2
2
( s + 2) + 4
30.
lim F ( s ) = 0
s →∞
34.
( s − 3)3
[( s − 2) 2 + 25]2
3
2
s + 36
2
( s − 7)[( s − 7)2 + 4]
3s 2 + 4
s 2 ( s 2 + 4)2
s
2
2
2[ s + (n + m) ]
+
s
2
2[ s + (n − m)2 ]
20( s 2 + 9)(s 2 + 49) − 40s 2 (2s 2 + 58)
( s 2 + 9)2 ( s 2 + 49)2
24. Piecewise continuous
28. Continuous
2
( s − 2)2 − 25
30. H ( s ) =
32.
+
1
2
1
+
+
s s +1 s + 2
22. Piecewise continuous
26. Neither
1
s
1
2
s + 5s + 6
e−2s
s
e−πs
s2 + 1
61
62
CHAPTER 7 LAPLACE TRANSFORMS
Exercises 7.4 (page 374)
32. F1 ( s ) = F2 ( s ) = F3 ( s ) =
$−1
1 t
= e = f3 (t )
s − 1
2. sin 2t
4
4. sin 3t
3
36.
1
8. t 4
24
sin t
t
Exercises 7.5 (page 383)
47t 5 −t / 4
47t
1
10. e−t / 4 cos
e
−
sin
4 2 47
4
2
12.
2
3
−
s +1 s − 2
14.
1
2
11
+
−
s +1 s −1 s − 2
2. e2t − 3e−t
4. et − e −5t − e−t
6. 2e3t + e2t sin t
8. 2 − t + t 2 − 2 cos 2t + 2 sin 2t
s
2
3
16. −
+2
−3
2
2
s+2
s +9
s +9
20.
e3t − e 4t
t
34.
3
6. e −5t / 2 t 2
16
18.
10. −t − e−2t + 2te−2t + e 2t
12. 10 + 6t + et +1 + 2tet +1
1 2
3
1
+
+ −
2
3
s s
s +1
s
14. t + π cos t + sin t
1 1
1
1
1 1
+
−
4 s − 1 2 ( s − 1) 2 4 s + 1
16.
22. 3et − 2e−3t
18.
−s3 − s 2 + 2
s3 ( s 2 + 6)
s 3 − 2s 2 − s + 3
( s − 1)( s − 2)( s 2 − 2 s − 1)
24. 5et + 2e 2t cos 3t − 5e2t sin 3t
20.
3
26. 1 − t 2 + 6e 2t
2
28. −
30. −
2 5e−t 4e 2t 13e−3t
+
+
−
3
6
15
30
−t
1
;
1− s
−t
7e
3te
−
4
2
+
7e
4
t
22.
24.
6
5
s ( s + 3)
2 s 3 − 17 s 2 + 28s − 12
( s − 1)3 ( s − 5)
s 3 + 2 s 2 + s + 2se −3s + e −3s
s 2 ( s − 1)(s + 1)
26. 2 + et − 3e−2t + e−3t
EXERCISES 7.6
28. −2e−t + et + e−t cos 2t
30. −
6 t 5a b 1 − t a b
1 −5t
+ + + + e − + +
e
25 5 4 4 4
4 4 100
32. (3a − b + 6)e2t + 6te2t + 3t 2 e2t + (b − 2a − 6)e3t
36. 2 – t
38. 3t
40. − ae
µt / 2 I
4 Ik − µ 2 t
µ
+
cos
2
I
2
4 Ik − µ
4 Ik − µ 2 t
sin
I
2
Exercises 7.6 (page 395)
2.
4.
e − s − e −4 s
s
e− s ( s + 1)
s2
6. (t + 1)u(t – 2);
e−2 s (3s + 1)
s2
– πs / 2
s
π e
8. (sin t )u t − ;
2
2 s +1
10. (t − 1)2 u (t − 1);
e− s 2
s3
12. (t – 3)u(t – 3)
sin 3(t − 3)
14.
u (t − 3)
3
1
16. [sin 2(t − 1)]u (t − 1)
2
18. [cos(t − 1) + 2et −1 ]u (t − 1)
1
20. sin t + sin 2t + cos 2t + sin 2t − sin t u (t − 2π)
2
63
64
22.
CHAPTER 7 LAPLACE TRANSFORMS
1 − e−( s −1)
( s − 1)(1 − e − s )
Figure 33
24.
se−2 s + e −2 s − 2e− s − se − s + 1
s 2 (1 − e−2 s )
Figure 34
26.
28.
1 − e− as − ase− as
as 2 (1 − e− as )
1
( s + 1)(1 − e−πs )
2
EXERCISES 7.6
30. cos t + [1 – cos(t – 2)]u(t – 2) + [1 – cos(t – 4)]u(t – 4)
Figure 35
32. cos t – sin 2t – 2(sin t) u (t − 2π ) + (sin 2t) u (t − 2π )
Figure 36
1
1
34. y (t ) = [1 − e −2(t −π) − 2(t − π)e −2(t −π) ]u (t − π) − [1 − e−(t −2 π) − 2(t − 2π)e−2(t − 2 π) ]u (t − 2π)
4
4
7 1
3
5
36. e−2t − e−3t + + (t − 2) − e−2(t − 2) + e−3(t − 2) u (t − 2)
16
6
4
9
2
1
38. 1 − 2e−t cos 3t − e−t sin 3t + 1 − e−(t −10) cos[3(t − 10)] − e−(t −10) sin[3(t − 10)] u (t − 10)
3
3
2 −(t − 20)
− (t − 20)
− 2 − 2e
cos[3(t − 20)] − e
sin[3(t − 20)] u (t − 20)
3
65
66
CHAPTER 7 LAPLACE TRANSFORMS
40. 2e−t + te−t + [(1 − 2e−3 )e−(t −3) + e−3 (t − 3)e−(t −3) + (2e−3 − 1)e−2(t −3) ]u (t − 3)
42. c.
Figure 37
46. h(t) – h(t – 1)u(t – 1), where h(t ) =
∞
48.
1
s2
∑ (−1)n+1
n =1
∞
50.
∑
k =0
1 + n e−t (e2( n +1) − 1) e−2t (e4( n +1) − 1)
−
+
, for 2n < t < 2(n + 1).
2
e2 − 1
2(e4 − 1)
n
(−1)k (2k )!
k ! s 2 k +1
et −3 e−(t −3)
et − 7
+
− 1 u (t − 3) −
+ e−(t −7) − 1 u (t − 7) + e −t
60.
2
2
2
0.6 − 0.4e−3t /125
62. The concentration of salt: 0.4 − 0.4e−3t /125 + 0.2e( −3 /125)(t −10)
0.6 − 0.4e−3t /125 + 0.2e( −3 /125)(t −10) − 2e( −3 /125)(t −20)
0 ≤ t ≤ 10
10 ≤ t ≤ 20
t ≥ 20
Exercises 7.7 (page 405)
1 t
2. cos 3t + ∫ sin[3(t − v)]g (v) dv
3 0
4.
t
∫0 g (v) sin(t − v)dv + sin t
6. e−t − e−2t
10. cos t − 1 +
t2
2
1
12. (t sin t + sin t − t cos t )
2
14. ( s 2 + 1)−1 ( s − 1)−1
16. cos t + sin t – 1
1
t
8. sin 2t − cos 2t
16
8
18. 2 – 2 cos t
EXERCISES 7.8
3t 1 t / 2
3t
1
1
20. e−t − et / 2 cos
e sin
+
3
3
3
2
2
22.
t
3
3
+ 2e−t + e 2t −
2
4
4
24. H ( s ) =
26. H ( s ) =
28. H ( s ) =
1
s2 − 9
; h(t ) =
e3t − e−3t
1 t
; yk (t ) = e3t + e−3t ; y (t ) = ∫ [e3(t −v) − e −3(t −v) ]g (v)dv + e3t + e −3t
6
6 0
e3t − e−5t
1
1 t
; h(t ) =
; yk (t ) = e3t − e −5t ; y (t ) = ∫ [e3(t −v) − e −5(t −v) ]g (v )dv + e3t − e −5t
( s − 3)( s + 5)
8
8 0
1
( s − 2) 2 + 1
30.
1 t −4(t −v)
e
(sin 5(t − v))e(v)dv + 2e −4t cos 5t
50 ∫0
32.
t5
60
38. d.
t
; h (t ) = e2t sin t ; yk (t ) = e2t sin t ; y (t ) = ∫ e 2(t −v ) (sin(t − v )) g (v )dv + e 2t sin t
0
87.403 ≈ 100 3 − 5 ≤ b ≤ 100 3 + 5 ≈ 228.825
Exercises 7.8 (page 412)
2. 1
4. e2
6. 1
8. 3e− s
10. 27e−3s
12. e3−3s
14. e−t cos t + 2 sin t − e−(t −π) (sin t )u (t − π)
1
1
16. e3t + e−t + [e3(t −1) − e −(t −1) ]u (t − 1) + [e−(t −3) − e3(t −3) ]u (t − 3)
2
4
4
1
5
18. e2t − et − e−t + (e2t − 2 − e1−t )u (t − 1)
3
2
6
20. e−2t + e−3t + e−2(t −1) − e−3(t − 4 / 3) u (t − 2)
67
68
CHAPTER 7 LAPLACE TRANSFORMS
π
22. sin t − (cos t )u t −
2
Figure 38
24. sin t − (sin t ) u (t − π ) − (sin t ) u (t − 2π )
Figure 39
1
26. e3t sin 2t
2
28.
( et − e − t )
2
30. y (t ) =
∞
∑ (sin t )u (t − 2k π)
k =1
Notice that the magnitude of the oscillations of y(t) becomes unbounded as t → ∞.
EXERCISES 7.9
Exercises 7.9 (page 416)
2. x = e3t − 2e 2t ; y = −2e3t + 2e 2t
7
2
7
1
4. x = − et cos 2t + et sin 2t + cos t − sin t ;
10
5
10
10
11
3
11
7
y = − et cos 2t − et sin 2t + cos t + sin t
10
10
10
10
1
1 3
6. x = −e 2t + t + 1 ; y = −e2t − t −
2
2 2
3
3
1
8. x = e −2t − e 2t + ; y = 3e −2t + e2t
4
2
2
10. x = et + cos t − 1; y = −et + cos t + 1
1 1
1 1
2
1
1
12. x = − − e2t + e−t + + (t − 3) + e 2t −6 − e−t +3 u (t − 3) ;
2 6
3
12
3
4 2
1 1
1 1
4
1
2
y = − + e 2t + e −t + − + (t − 3) − e 2t −6 − e−t +3 u (t − 3)
2
6
3
4
2
12
3
14. x = cos t + cosh t – 1 – [cosh(t – 1) – 1]u(t – 1); y = –cos t + cosh t – [cosh(t – 1) + 1]u(t – 1)
16. x(t ) = cos t + sin t +
y (t ) = cos t +
13 + 17
e
3 + 17
(t −π)
2
2 17
−12 + 2 17
e
3 + 17
(t −π)
2
17
+
−
13 − 17
e
3 − 17
(t −π)
2
;
2 17
12 + 2 17
e
3 − 17
(t −π)
2
17
18. x(t ) = t 2 ; y (t ) = t ; z (t ) = t 2 − 2
20. x = e −2t + e2t + 1;
y = 2e −2t − 2e2t + 2t + 2e2 − 2e −2 + 2
22. See answer to Exercises 5.5, Problem 1 on pages B-14 of text.
15 20
5
10
5
24. I1 = − e−1000t − e −4000t ; I 2 = 5 − e −1000t − e−4000t ;
2 3
6
3
3
5 10 −1000t 5 −4000t
+ e
I3 = − e
2 3
6
69
70
CHAPTER 7 LAPLACE TRANSFORMS
Chapter 7 Review Problems (page 418)
2.
4.
6.
1 − e−5( s +1) e−5s
−
s +1
s
4
( s − 3)2 + 16
7( s − 2)
2
( s − 2) + 9
−
70
( s − 7)2 + 25
8. 2 s −3 + 6 s −2 − ( s − 2)−1 − 6( s − 1) −1
10.
s
e−πs / 2
+
s 2 + 1 (1 + s 2 )(1 − e −πs )
12. 2e2t cos
( 2t ) +
3 2t
e sin
2
( 2t )
14. e−t − 3e−3t + 3e2t
16.
sin 3t − 3t cos 3t
54
18.
f (t ) =
∞
∞
(−1)n t 2 n
(−1)n (2 n)! 1
=
F
s
;
(
)
∑ n!
∑ n!
s 2 n +1
n =0
n=0
20. (t − 3)e−3t
10
23
15
22. e2t − cos 3t + sin 3t
13
13
13
24. 6et + 4tet + t 2 et + 2te2t − 6e2t
26. (1 + Ct 3 )e−t , where C is an arbitrary constant
7t 1 −t / 2
7t 1
3 −t / 2
28.
sin
cos
− e
−
e
2 7
2 2
2 2
1
π π
30. sin 2t + sin 2 t − u t −
2
2 2
1 1
1 1
4
1
2
32. x = − + e 2t + e −t + − + (t − 3) − e 2t −6 − e −t +3 u (t − 3)
2
6
3
4
2
12
3
1 1
1 1
2
1
1
y = − − e2t + e −t + + (t − 3) + e2t −6 − e −t +3 u (t − 3)
2
6
3
4
2
12
3
CHAPTER 8
Exercises 8.1...(page 430)
2. 2 + 4 x + 8 x 2 +
"
1
1
1
4. x 2 + x3 − x5 +
2
6
20
1
1 5
6. x − x3 +
x +
6
120
8. 1 −
10. a.
b.
"
"
(sin 1) x 2 (cos1)(sin 1) x 4
+
+
2
24
p3 ( x) =
ε3 =
"
1 x x 2 x3
+ +
+
2 4 8 16
1
24
4! (2 − ξ )5
4
1 1
1
≤
2
3 5 24
(2)
=
2
35
≈ 0.00823
c.
2
1
1
− p3 =
≈ 0.00260
3
2 384
d.
Figure 40
12. The differential equation implies y(x), y ′( x) , and y ′′( x) exist and are continuous. Furthermore y ′′′( x) can be
obtained by differentiating the other terms: y ′′′ = − py ′′ − p ′y ′ − qy ′ − q ′y + g ′. Since p, q, and g have derivatives
of all orders, subsequent differentiations display the fact that, in turn, y′′′, y (iv) , y (v) , etc. all exist.
71
72
CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
1
1
t + t 2 − t3 +
2
6
14. a.
b.
c.
16. 1 −
"
"
1
For r = 1, y (t ) = 1 − t 2 − 4t 4 + .
2
1
49
For r = –1, y (t ) = 1 + t 2 − t 4 +
2
12
For small t in part (b), the hard spring recoils but the soft spring extends.
t2 t4
+ +
2 4!
"
Exercises 8.2 (page 438)
2.
( −∞, ∞ )
4. [2, 4]
6. (–3, –1)
8. a.
1 1
− ,
2 2
b.
1 1
− ,
2 2
c.
( −∞, ∞ )
d.
( −∞, ∞ )
e.
( −∞, ∞ )
f.
1
1
,
−
2
2
∞
10.
".
2n +3
∑ (n + 3)! +
n = 0
(n + 2)2
n
( x − 1)
2n +1
2
2
12. x − x3 + x5 +
3
15
"
14. 1
1
1
1
16. 1 − x + x 2 − x3 +
2
4
24
"
EXERCISES 8.3
∞
18.
(−1)k 2k
x = cos x
k = 0 (2k )!
20.
∑ n(n − 1)an x n−2
∑
∞
n=2
(−1)k x 2k +1
∑
k = 0 (2k + 1)!(2k + 1)
∞
22.
∞
24.
∑ (k − 2)(k − 3)ak −2 xk
k =4
∞
26.
ak −3 k
x
k =4 k
30.
∑ (−1)n ( x − 1)n
∑
∞
n=0
(−1)n +1 n
x
n
n =1
∞
32.
∑
1
1
1
5
4
34. 1 + ( x − 1) − ( x − 1) 2 + ( x − 1)3 −
( x − 1) +
2
8
16
128
36. a.
Always true
b.
Sometimes false
c.
Always true
d.
Always true
∞
38.
(−1)n 2n + 2
1
1
= x 2 − x 4 + x6 –
x
+
n
1
2
3
n=0
∑
Exercises 8.3 (page 449)
2. 0
4. –1, 0
6. –1
8. No singular points
10. x ≤ 1 and x = 2
"
"
73
74
CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
1
1
12. y = a0 1 + x + x 2 + x3 +
2
3!
∞ n
x
= a0 ∑
n =0 n !
= a0 e x
x2
+
14. a0 1 −
2
" + a1 x − x6 + "
3
x 2 x3
−
+
16. a0 1 −
2
3
1
18. a0 1 + x 2 +
6
∞
20. a0
∑ (−1)k
k =0
"
" + a1 x + x2 + x2 + "
3
= a0 e x + (a1 − a0 ) xe x
" + a1 x + 271 x3 + "
∞
(−1)k x 2 k +1
x 2k
+ a1 ∑
= a0 cos x + a1 sin x
(2k )!
k = 0 (2k + 1)!
22. a3k + 2 = 0, k = 0, 1, …
∞
1⋅ 4
a0 1 + ∑
k =1
" (3k − 5)(3k − 2) x3k + a1 x + ∑∞ 2 ⋅ 5" (3k − 4)(3k − 1) x3k +1
(3k )!
k =1
∞
1
[(−1)k (2k − 3)2 (2k − 5)2
1
24. a0 1 − x 2 + x 4 + ∑
2
(2k )!
24
k =3
1
1
1
26. x + x 2 + x3 + x 4 +
2
2
3
"
1
1
1
28. −1 − x 2 − x3 − x 4 +
2
3
8
"
5
1
30. −1 + x + x 2 − x3
2
6
32. a.
d.
If a1 = 0, then y(x) is an even function.
a0 = 0, a1 > 0
9
36. 3 − t 2 + t 3 + t 4 +
2
"
(3k + 1)!
" 32 ] x2k + a1 x
EXERCISES 8.4
Exercises 8.4 (page 456)
2. infinite
4. 2
6. 1
10
8. a0 1 − 2( x + 1) + 3( x + 1)2 − ( x + 1)3 +
3
1
1
10. a0 1 − ( x − 2) 2 − ( x − 2)3 +
4
24
1
2
12. a0 1 + ( x + 1) 2 + ( x + 1)3 +
3
2
5
14. 1 + x + x 2 + x3 +
6
"
" + a1 ( x − 2) + 14 ( x − 2)2 − 121 ( x − 2)3 + "
" + a1 ( x + 1) + 2( x + 1)2 + 73 ( x + 1)3 + "
"
1
1
1
16. −t + t 3 + t 4 + t 5 +
3
12
24
2
"
4
π 1
π 1
π
18. 1 + x − + x − − x − +
2 2
2 24
2
"
1
22. a0 1 − x 2 +
2
" + x + 12 x2 − 16 x3 + "
3
24. a0 1 − x 2 +
2
" + a1 x − 16 x3 + " + 23 − x2
1
26. a0 [1 − x 2 ] + a1 x − x3 +
6
1
28. a0 1 + x3 +
6
" + 12 x2 + "
" + a1[ x + "] + 12 x2 + "
1
1
1
30. 1 − t 2 + t 3 − t 4 +
2
2
4
"
75
76
CHAPTER 8
SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Exercises 8.5 (page 460)
2. c1 x −5 / 2 + c2 x −3
4. c1 x
(
)
− 1+ 13 / 2
6. c1 x cos
(
+ c2 x
(
)
− 1− 13 / 2
)
3 ln x + c2 x sin
(
3 ln x
)
5
5
−1/ 2
8. c1 + c2 x −1/ 2 cos
sin
ln x
2 ln x + c3 x
2
10. c1 x −2 + c2 x −2 ln x + c3 x −2 (ln x)2
12. c1 ( x + 2)1/ 2 cos[ln( x + 2)] + c2 ( x + 2)1/ 2 sin[ln( x + 2)]
14. c1 x + c2 x −1/ 2 + x ln x − 2 x −2 ln x
16. 3 x −2 + 13 x −2 ln x
Exercises 8.6 (page 472)
2. 0 is regular.
4. 0 is regular.
6. ±2 are regular.
8. 0 and 1 are regular.
10. 0 and 1 are regular.
12. r 2 + 3r + 2 = 0; r1 = −1, r2 = −2
14. r 2 − r = 0; r1 = 1, r2 = 0
16. r 2 −
5r 3
5 + 73
5 − 73
− = 0; r1 =
, r2 =
4 4
8
8
3
1
3
18. r 2 − r − = 0; r1 = , r2 = −
4
2
2
EXERCISES 8.6
1
1
1
20. a0 1 − x − x 2 − x3 +
15
35
3
16
22. a0 1 + 4 x + 4 x 2 + x3 +
9
"
"
1
1
1 10 / 3
+
24. a0 x1/ 3 + x 4 / 3 + x 7 / 3 +
x
3
18
162
"
(−1)n x n +1
= a0 xe− x
n
!
n=0
∞
26. a0 ∑
∞
(−1)n x n +1/ 3
28. a0 x1/ 3 + ∑
n =1 n ![10 ⋅ 13 (3n + 7)]
"
4
1
30. a0 1 + x + x 2
5
5
x n +1
= a0 xe x ; yes, a0 < 0
n
!
n=0
∞
32. a0 ∑
1
34. a0 1 + x ; yes, a0 < 0
2
1
1
4
1 3
36. a0 x + x 2 +
x +
x +
20
1960
529,
200
"
23/ 6
629083
31 11/ 6 2821 17 / 6
38. a0 x5 / 6 +
+
+
+
x
x
x
726
2517768
(9522)(2517768)
"
40. a0 [1 + 2 x + 2 x 2 ]
42. The transformed equation is 18 z (4 z − 1)2 (6 z − 1)
Also, zp ( z ) =
d2 y
dz
2
+ 9(4 z − 1)(96 z 2 − 40 z + 3)
2
dy
+ 32 y = 0 .
dz
96 z − 40 z + 3
32 z
are analytic at z = 0; hence z = 0 is a regular
and z 2 q ( z ) =
2(4 z − 1)(6 z − 1)
18(4 z − 1)2 (6 z − 1)
singular point.
32
1600 −2 241664 −3
y1 ( x ) = a0 1 + x −1 +
x +
x +
243
6561
27
"
77
78
CHAPTER 8
SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Exercises 8.7 (page 482)
1
1
2. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = 1 − x − x 2 +
3
15
4. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = 1 + 4 x + 4 x 2 +
1
1
6. c1 x1/ 3 + x 4 / 3 + x 7 / 3 +
3
18
" and y2 ( x) = x−1/ 2 − x1/ 2 .
3
" and y2 ( x) = y1 ( x) ln x − 8x − 12 x2 − 176
x +" .
27
" + c2 1 + 12 x + 101 x2 + "
1
8. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = x − x 2 + x3 +
2
1
1 7/3
10. c1 x1/ 3 − x 4 / 3 +
+
x
10
260
" and y2 ( x) = y1 ( x) ln x + x2 − 34 x3 + 1136 x4 + " .
" + c2 x−2 + 14 x−1 + 81 + "
4
1
12. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = 1 + x + x 2 and y2 ( x) = x −4 + 4 x −3 + 5 x −2 .
5
5
1
14. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = x + x 2 + x3 +
2
4
" and y2 ( x) = y1 ( x) ln x − x2 − 34 x3 − 11
x +" .
36
1
1 9
16. c1 1 + x + c2 − x −1 − x ln x − 2 ln x − + x +
2 4
2
solutions are bounded near the origin.
" ; has a bounded solution near the origin, but not all
"
1
1 3
18. c1 y1 ( x) + c2 y2 ( x) + c3 y3 ( x), where y1 ( x ) = x + x 2 +
x + ,
20
1960
3
9 8/ 3
2
+ , y3 ( x) = x −1/ 2 + 2 x1/ 2 + x3/ 2 +
y2 ( x ) = x 2 / 3 + x 5 / 3 +
x
26
4940
5
"
".
31 11/ 6 2821 17 / 6
20. c1 y1 ( x) + c2 y2 ( x) + c3 y3 ( x), where y1 ( x ) = x5 / 6 +
+
+
x
x
726
2517768
1
3
437 3
y2 ( x) = 1 + x + x 2 + , and y3 ( x) = y2 ( x ) ln x − 9 x − x 2 +
x +
28
98
383292
"
α2 2 α4 3 α6 4
x +
x −
x +
2
12
144
5α 4 2 5α 6 3
y2 ( x) = −α 2 y1 ( x ) ln x + 1 + α 2 x −
x +
x + .
4
18
22. c1 y1 ( x) + c2 y2 ( x), where y1 ( x ) = x −
"
26. d.
a1 = −
2+i
2+i
, a2 =
5
20
" , and
",
".
EXERCISES 8.8
Exercises 8.8 (page 493)
1
2. c1 F 3, 5; ;
3
11 17 5
x + c2 x 2 / 3 F ,
; ;
3 3 3
x
3
1 5 1
4. c1 F 1, 3; ; x + c2 x −1/ 2 F , ; ; x
2
2 2 2
6. F (α , β ; β ; x) =
∞
∑ (α )n
n=0
xn
= (1 − x)−α
n!
∞
(−1)n 2 n
3
1
8. F , 1; ; − x 2 = ∑
x
2
2
n = 0 2n + 1
= x −1 arctan x
1 1
10. y1 ( x ) = F , ; 2; x
2 2
1
3
25 3
= 1 + x + x2 +
x +
8
64
1024
and
5
1
y2 ( x) = y1 ( x) ln x + 4 x −1 + x + x 2 +
16
8
"
".
14. c1 J 4 / 3 ( x) + c2 J −4 / 3 ( x)
16. c1 J 0 ( x) + c2 Y0 ( x)
18. c1 J 4 ( x) + c2 Y4 ( x)
1 3
17 4
20. J 2 ( x ) ln x − 2 x −2 − − x 2 +
x +
2 16
1152
"
26. J −3 / 2 ( x) = − x −1 J −1/ 2 ( x ) − J1/ 2 ( x)
2 −3 / 2
2 −1/ 2
x
x
cos x −
sin x
π
π
=−
J 5 / 2 ( x) =
3 − x2
=3
x
2
J1/ 2 ( x) − 3 x −1 J −1/ 2 ( x)
2 −5 / 2
2
2 −1/ 2
x
x
sin x − 3 x −3 / 2 cos x −
sin x
π
π
π
∞
1
x 2k .
−
k
2
1
k =1
36. y1 ( x ) = x and y2 ( x) = 1 − ∑
79
80
CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
38. 2 x 2 − 1, 4 x3 − 3 x, 8 x 4 − 8 x 2 + 1
2 c n /2+1
40. c. c1 J ( n + 2) −1
x
+ c2 Y( n + 2) −1
n+2
2 c n /2+1
x
n+2
Chapter 8 Review (page 497)
2. a.
b.
4. a.
b.
6. a. c1 x
b.
8. a.
b.
c.
10. a.
b.
±2 are irregular singular points.
nπ , where n is an integer, are regular singular points.
"
∞
[( −3)( −1)(1) (2k − 5)] 2 k
2
a0 1 + ∑
x + a1 x − x3
k
3
2 k!
k =1
"
"
∞
∞
[3 ⋅ 5 ⋅15 (4k 2 − 10k + 9)] 2k
[3 ⋅ 9 ⋅ 23 (4k 2 − 6k + 5)] 2k +1
a0 1 + ∑
x + a1 x + ∑
x
k =1
k =1
2k (2k )!
2k (2k + 1)!
( −3+
)
105 / 4
+ c2 x
( −3−
)
105 / 4
c1 x −1 + c2 x −1 ln x + c3 x 2
y1 ( x ) =
y1 ( x ) =
y1 ( x ) =
∞
∑ an x n+ 2 and y2 ( x) = y1 ( x) ln x +
n=0
∞
∑ an x n
n=0
∞
∑ an x n
n=0
and y2 ( x) =
∞
∑ bn x n−2 .
n=0
∞
∑ bn xn−(3 / 2) .
n =0
∞
and y2 ( x) = y1 ( x) ln x + ∑ bn x n .
1
7 5 3
c1 F 3, 2; ; x + c2 x1/ 2 F , ; ; x
2
2 2 2
c1 J1/ 3 (θ ) + c2 J −1/ 3 (θ )
n =1
CHAPTER 9
Exercises 9.1 (page 507)
12. a.
The equations produce the format
1
x1 − x2 = 0
2
0 = 1.
b.
The equations produce the format
1
+ x3 = 0
x1
2
11
x2 + x3 = 0
2
0 = 1.
x ′ 0 1 x
2. =
y −1 0 y
x1 ′ 1 −1 1 −1 x1
x 1 0 0 1 x
2
2
4. =
x3 π 0 −1 0 x3
x4 0 0 0 0 x4
0
0 x1
x1 ′ cos 2t
6. x2 = 0
sin 2t 0 x2
x3 1
−1 0 x3
x ′ 0
8. 1 = −2
x2 1−t 2
1 x
1
2t
2 x2
1−t
0
x ′
10. 1 =
n2
x2 −1 + t 2
1 x
1
− 1t x2
1 0 0 x1
x1 ′ 0
x 0 –3 –2
1 x2
12. 2 =
x3 0 0 0
1 x3
x
0
–1
–1
–3
x4
4
Exercises 9.2 (page 512)
14. For r = –1, the unique solution is
x1 = x2 = x3 = 0. For r = 2 the solutions are
s
s
x1 = − , x2 = , x3 = s ( −∞ < s < ∞ ) .
2
4
Exercises 9.3 (page 521)
2. a.
b.
4. a.
b.
3 −1 7
A +B =
2 4 −1
10 4 27
7 A − 4B =
14 −5 15
2 5 −1
AB = 0 12 4
−1 8 4
3 2
BA =
−1 15
1
2
2. x1 = 0, x2 = , x3 = , x4 = 0
3
3
6. a.
1
2
4. x1 = , x2 = , x3 = 0, x4 = 0
3
3
b.
9 −1
( AB)C =
6 1
s
s
6. x1 = − , x2 = , x3 = s ( −∞ < s < ∞ )
4
4
c.
1
6
( A + B )C =
5 −5
8. x1 = − s + t , x2 = s, x3 = t ( −∞ < s, t < ∞ )
10. x1 =
−2 + i
2 + 4i
, x2 = 0, x3 =
5
5
2 7
AB =
1 5
9 − 1
31
31
10.
4
− 5
31
31
81
82
CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
1 0 −1
12. 1 −1 2
−1 1 −1
1 − 1
1
2
2
1
1
1
−2
14.
6
3
1
− 1
0
3
3
2
−1
16. c. x = 1 + c −1 , with c arbitrary.
0
1
sin 2t
18.
cos 2t
( 12 ) cos 2t
− ( 12 ) sin 2t
0 0
20. 1 −t
0 1
( 19 ) e
−3t
t − 1
3 9
− 13
22. 0
t
3t
()
()
− 12 e −2t c
+ 11
−2t c21
1
− 2 e
b.
−e −1 + 1 −e −1 + 1
−1
−1
e − 1 −3e + 3
c.
−e−t + 3e−3t
−t
−3t
−3e + 3e
c12
c22
−3e−t − 9e−3t
−3e−t − 9e−3t
42. In general, ( AB )T = BT AT . Thus
( AT A )T = AT ( AT )T = AT A, so AT A is
symmetric. Similarly, one shows that AAT is
symmetric.
Exercises 9.4 (page 530)
r ′(t ) 2 0 r (t ) sin t
2.
=
+
θ ′(t ) 1 −1 θ (t ) 1
dx
dt 1 1 1 x
dy
4. dt = 2 −1 3 y
dz 1 0 5 z
dt
x ′ (t ) 0 1 x1 (t ) 0
6. 1 =
+ 2
x2′ (t ) −1 0 x2 (t ) t
24. 11
26. 54
28. 1, 6
30. b.
40. a.
0
c.
1
x = c 1
−1
d.
c1 = c2 = c3 = −1
3e −t cos 3t − e −t sin 3t
32.
−3e−t cos 3t + e−t sin 3t
2 cos 2t −2 sin 2t −2e−2t
34. −2 cos 2t −4 sin 2t −6e−2t
6 cos 2t −2 sin 2t −2e−2t
x1′ (t ) 0 1 0 x1 (t ) 0
8. x2′ (t ) = 0 0 1 x2 (t ) + 0
x3′ (t ) −1 1 0 x3 (t ) cos t
10. x1′ (t ) = 2 x1 (t ) + x2 (t ) + tet ,
x2′ (t ) = − x1 (t ) + 3 x2 (t ) + et
12. x1′ (t ) = x2 (t ) + t + 3
x2′ (t ) = x3 (t ) − t + 1
x3′ (t ) = − x1 (t ) + x2 (t ) + 2 x3 (t ) + 2t
14. Linearly independent
16. Linearly dependent
18. Linearly independent
EXERCISES 9.5
3e−t
20. Yes;
2e−t
3e−t
e 4t
; c1
+ c2
2e −t
−e 4t
e 4t
−e 4t
22. Linearly independent; fundamental matrix is
et
et
et
sin t
cos t
− sin t
− cos t
sin t .
cos t
The general solution is
et
c1 et + c2
et
e3t
24. c1 0 + c2
3t
e
28. X (t ) =
−1
sin t
cos t + c
3
− sin t
−e3t
e3t + c3
0
− cos t
sin t .
cos t
−e −3t
5t + 1
3
−
t
−e + 2t
e−3t 4t + 2
( 12 ) et − ( 12 ) et ;
( 12 ) e−5t ( 12 ) e−5t
2e−t + e5t
x (t ) =
−2e−t + e5t
32. Choosing x 0 = col(1, 0, 0, … , 0), then
4. Eigenvalues are r1 = −4 and r2 = 2 with
1
associated eigenvectors u1 = s and
−1
5
u2 = s .
1
6. Eigenvalues are r1 = r2 = −1 and r3 = 2 with
−1
associated eigenvectors u1 = s 1 ,
0
−1
1
u 2 = v 0 , and u3 = s 1 .
1
1
8. Eigenvalues are r1 = −1 and r2 = −2 with
1
associated eigenvectors u1 = s 2 and
4
1
u 2 = s 1 .
1
10. Eigenvalues are r1 = 1, r2 = 1 + i, and r3 = 1 − i,
1
with associated eigenvectors u1 = s 0 ,
0
−1 + 2i
−1 − 2i
u 2 = s 1 , and u 3 = s 1 , where
−i
i
s is any complex constant.
x 0 = col(0, 1, 0, … , 0), and so on, the
corresponding solutions x1 , x 2 , … , x n will
have a nonvanishing Wronskian at the initial
point t0 . Hence {x1 , … , x n } is a fundamental
solution set.
Exercises 9.5 (page 541)
2. Eigenvalues are r1 = 3 and r2 = 4 with
1
associated eigenvectors u1 = s and
1
3
u2 = s .
2
83
1
1
12. c1e7t + c2 e−5t
2
−2
−1
14. c1e−t 0 + c2 e −2t
1
16. c1e
−10t
2
0 + c e5t
2
−1
1
−1 + c e3t
3
3
1
4
3
0
1 + c e5t
3
0
1
0
2
84
CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
18. a.
1
1
Eigenvalues are r1 = −1, and r2 = −3 with associated eigenvectors u1 = s , and u 2 = s .
−1
1
b.
x1 = e−t
−t
x2 = e
x1 = −e−3t
c.
−3t
x2 = e
Figure 41
et
20.
−et
4e 4t
−e 4t
e 2t − e 2t
et
22. 0
e 2t
et
e 2t
0 3et
1 e 4t
4 0
24.
0 0
0 0
0
0
3et
et
0
0
e −t
e −t
26. x(t ) = c1e −5t + 2c2 et
y (t ) = 2c1e−5t + c2 et
Figure 42
x1 = e−t − e−3t
d.
−t
−3t
x2 = e + e
Figure 43
9.5 EXERCISES
−0.2931e0.4679t
28. −0.5509e0.4679t
e0.4679t
e0.6180t
0.6180e0.6180t
30.
0
0
0.4491e3.8794t
−0.1560e3.8794t
e3.8794t
−0.6527e1.6527t
e1.6527t
−0.7733e1.6527t
−0.6180e−1.6180t
0
e−1.6180t
0
0
0.5858t
e
0.5858e0.5858t
0
0
0.2929e3.4142t
e3.4142t
0
2e3t − 12e4t
32.
2e3t − 8e4t
e 2t − 2e − t
34. e2t + 3e−t
e 2t − e −t
1
36. c1e2t + c2
1
2t 1
2t
te + e
1
1
t
38. c1e 1 + c2
0
t
te
1
40. c1e 0 + c2 et
0
t
1
44. c1 + c2 t −5
2
46. x1 (t ) =
1
1 + et
0
4
3
1
0
2 t
0 + c t e
3
2
1
0
2 + c tet
3
3
1
2 + et
3
1
1 + tet
0
0
0 + et
1
− 1
4
0
1
2
0
1
1
−2
1
e−3t
3 − α αt
kg, x2 (t ) =
(e − 1)e −3t kg
10α
10
3 −α
The mass of salt in tank A is independent of α . The maximum mass of salt in tank B is 0.1
3
50. b.
3
1
x(t ) = 1 + e −t + e−3t
2
2
y (t ) = 1 − e−3t
3
1
z (t ) = 1 − e −t + e−3t
2
2
3 /α
kg.
85
86
CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
Exercises 9.6 (page 549)
−5 cos t
2. c1
+ c2
2 cos t − sin t
−5 sin t
2 sin t + cos t
5e2t cos t
4. c1 −2e2t cos t + e2t sin t + c2
5e2t cos t
cos 2t
6.
sin
2
t − cos 2t
et
et
8.
0
0
e −t
− e −t
0
0
5e2t sin t
−2e 2t sin t − e2t cos t + c3
5e2t sin t
0
2t
e
2t
−e
sin 2t
− sin 2t − cos 2t
0
0
e2t cos 3t
e2t sin 3t
e2t (2 cos 3t − 3 sin 3t ) e2t (2 sin 3t + 3 cos 3t )
0
−0.0209e 2t cos 3t + 0.0041e 2t sin 3t
−0.0296e2t cos 3t + 0.0710e2t sin 3t
10.
0.1538e2t cos 3t + 0.2308e2t sin 3t
e2t cos 3t
0
−0.0209e 2t sin 3t − 0.0041e 2t cos 3t
−e − t
−0.0296e2t sin 3t − 0.0710e2t cos 3t
e −t
0.1538e2t sin 3t − 0.2308e2t cos 3t
− e− t
e2t sin 3t
e −t
0
0
0
0
12.
0
0
−2t
−2t
−2t
−2t
−0.0690e cos 5t + 0.1724e sin 5t −0.0690e sin 5t − 0.1724e cos 5t
e−2t cos 5t
e−2t sin 5t
14. a.
b.
et sin t − 2et cos t
2e 2t
t
t
−e cos t − 2e sin t
et +π sin t
e2(t +π)
−et +π cos t
cos(3 ln t )
sin(3 ln t )
+ c2 t −1
18. c1t −1
3 sin(3 ln t )
−3 cos(3 ln t )
et
et
et
et
0
0
e −t
et
− e −t
et
0
0
0
0
et
0
0
0
0
EXERCISES 9.7
20. x1 (t ) = cos t − cos 3t
x2 (t ) = cos t + cos 3t
22. I1 (t ) =
16 −2t 1 −8t
e − e +2
5
5
I 2 (t ) = 4e−2t − e−8t + 2
4
4
I3 (t ) = − e−2t + e−8t
5
5
Exercises 9.7 (page 555)
e3t
+ c2
2. c1
2e3t
1
4. c1 + c2
−1
e −t t
+
−2e−t 2
e4t −2 sin t
+
e4t 2 sin t + cos t
6. x p = ta + b + e3t c
8. x p = t 2 a + tb + c
10. x p = e −t a + te −t b
2
12. c1e4t + c2 e −t
3
cos t
14. c1
+ c2
sin t
1 1
−1 + −1
− sin t 2t − 1
cos t + 2
t − 2
cos t
16. c1
+ c2
− sin t
4t sin t
sin t
cos t + 4t cos t − 4 sin t
1
t
18. c1e 0 + c2 et
0
0
1 + c et
3
1
t
t
t −te − e
t +
0
t
1 + t −e
( )
( ) ()
( ) ( )
( ) ( )
( ) ( )
−c1 sin 2t + c2 cos 2t − c3 e−2t + 8c4 et + 8 tet + 17 et − 1
15
225
8
−2c cos 2t − 2c sin 2t + 2c e −2t + 8c et + 8 tet − 88 et
1
2
3
4
15
225
20.
t
t
t
−2t
8
32
4c1 sin 2t − 4c2 cos 2t − 4c3 e + 8c4 e + 15 te + 225 e
8 tet + 152 et − 1
8c1 cos 2t + 8c2 sin 2t + 8c3 e−2t + 8c4 et + 15
225
87
88
CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
3e−4t + e2t
−6e−4t + e2t − 2t
22. a.
b.
( )
()
( 73 ) e2(t −2)
( ) e2(t −2) − 2t
− 4 e −4(t −2) +
3
8 e −4(t −2) + 7
3
3
24. x(t ) = 3e−4t + e 2t
y (t ) = −6e−4t + e2t − 2t
et
−2et
26. a.
tet
b.
tet
−t + 1
28.
−t − 1
1
30. c1t −2 + c2
2
34. a.
1
3 3
+
4 t 2
3
Neither wins.
b.
The x1 force wins.
c.
The x2 force wins.
Exercises 9.8 (page 566)
2. a.
4. a.
6. a.
r = 2; k = 2
1 − t −t
b. e2t
1 + t
t
r = 2; k = 3
1 t 3t − t 2
2
2t
−t
b. e 0 1
1
0 0
r = –1; k = 3
()
()
1 e3t + 1 e−t
2
2
8.
3t
−t
e −e
1 + t + t 2
2
2
−t
t
b. e − 2
2
−t + t2
( 14 ) e3t − ( 14 ) e−t
( 12 ) e3t + ( 12 ) e−t
2
2
t
1+ t − t
t− 2
2
−3t + t 2 1 − 2t + t2
t + t2
t2
2
EXERCISES 9.8
e4t + 2e −2t
1 4t
e − e−2t
10.
3
e 4t − e−2t
e4t − e −2t
e 4t + 2e −2t
e 4t − e −2t
3e−t + 6e2t
1
−4e −t + 4e 2t + 6te 2t
12.
9
−2e−t + 2e 2t − 6te 2t
et
0
14. 0
0
0
e −t
0
16. 0
0
0
4e −t + 5e 2t + 3te 2t
2e−t − 2e 2t − 3te 2t
0
et + te−t
te−t
−te −t
e −t − te −t
0
0
0
0
0
e−t + te−t
te−t
−te
e
−t
4e −t − 4e 2t + 3te 2t
2e−t + 7e 2t − 3te2t
0
0
0
0
0
cos t sin t
− sin t cos t
− te
0
0
−t
0
0
0
e−2t + 2te−2t
0
0
−4te −2t
0
1 + c et
3
0
−4
t
20. c1e 1 + c2 et
0
−3e−t + 3e2t
0
0
1
18. c1 0 + c2 et
0
22.
−3e−t + 3e2t
0
−t
e4t − e −2t
e 4t − e−2t
e 4t + 2e−2t
1
2t
1
3 − 4t
t + c et
3
2
() ()
16 e −t − 16 e 2t + 1 te 2t
( 9 ) ( 9 ) (3)
( 89 ) e−t + (199 ) e2t − ( 13 ) te2t
− 43 e−t + 13 e2t
et + cos t − sin t − 1 − t
24. et − sin t − cos t − 1
et − cos t + sin t
1 − 2t + 4t 2
−t − t 2
−4t
0
0
te−2t
−2t
−2t
e − 2te
0
89
90
CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
Chapter 9 Review (page 570)
−2 cos 3t
2t
2. c1e2t
+ c2 e
+
cos
3
3
sin
3
t
t
0
4. c1e2t 0 + c2 et
1
0
6. 3e−5t
−e −5t
10.
t
1
0
0
e5t
3e5t
e5t
0
0
1
8. c1 + c2
2
1
0 + c
3
0
−2 sin 3t
sin 3t − 3 cos 3t
2e−5t
+ e 4t
−e−5t
1
c1e2t 0 + c2
0
11
36
13
18
−3 7
11
7t 5t / 2
7t
5t / 2
cos
sin
−2 − e
−2 7
e
2 4
2
0
−3
11
7t 5t / 2
7t
5t / 2
+ c3 e
− +
−
sin
cos
2 e
2
2 4
2 0
()
()
e2t sin 2t + 3 e2t cos 2t + 1 e2t
2
2
12.
2e2t cos 2t − 3e2t sin 2t − te2t
−1
14. c1t 1 + c2 t 3
2
1
1 + c t −2
3
0
1 t 4t + t 2
2t
16. 0 1
0 0
1
1
−1
1
()
1 −t 11
7 − 3 e + 16
− 14
7 +
5
−
8
CHAPTER 10
Exercises 10.2 (page 587)
2. y =
(e10 − 1)e x + (1 − e 2 )e5 x
e10 − e2
4. y = 2 sin 3x
6. No solution
8. y = e x −1 + xe x −1
10. λn =
(2n − 1) 2
(2n − 1) x
and yn = cn cos
, where n = 1, 2, 3, … and the cn ’s are arbitrary.
4
2
12. λn = 4n 2 and yn = cn cos(2nx), where n = 0, 1, 2, … and the cn ’s are arbitrary.
14. λn = n 2 + 1 and yn = cn e x sin( nx), where n = 1, 2, 3, … and the cn ’s are arbitrary.
16. u ( x, t ) = e−27t sin 3 x + 5e−147t sin 7 x − 2e−507t sin 13 x
18. u ( x, t ) = e−48t sin 4 x + 3e−108t sin 6 x − e−300t sin 10 x
2
3
1
20. u ( x, t ) = − sin 9t sin 3 x + sin 21t sin 7 x − sin 30t sin 10 x
9
7
30
2
7
22. u ( x, t ) = cos 3t sin x − cos 6t sin 2 x + cos 9t sin 3 x + sin 9t sin 3 x − sin 15t sin 5 x
3
15
∞
1
24. u ( x, t ) = ∑
2
n =1
n
(−1)n +1
sin 4nt sin nx
cos 4 nt +
2
4n
Exercises 10.3 (page 603)
2. Even
4. Neither
6. Odd
10.
f ( x) π 4 ∞
1
− ∑
cos(2k + 1) x
2 π k =0 (2k + 1) 2
12.
f ( x) ~
2(−1)n − n 2 π2 (−1)n − 2
π2 ∞ 2(−1)n
+∑
cos nx +
sin nx
6 n =1 n 2
n3 π
91
92
CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
14.
f ( x) ~
16.
f ( x) ~
π ∞ 1
+ ∑ sin 2nx
2 n =1 n
∞
2
nπ
sin nx
2
∑ nπ 1 − cos
n =1
18. The 2π periodic function g(x), where g ( x) = x , –π x π
0
20. The 2π periodic function g(x), where g ( x) = x 2
π2
2
−π < x ≤ 0
0≤x<π
x = ±π
x + π −π < x < 0
22. The 2π periodic function g(x), where g ( x) = x
0<x<π
π
x = 0, ± π
2
0
1
−
2
−1
24. The 2π periodic function g(x), where g ( x) = 0
1
1
2
0
30. a0 =
1
5
, a1 = 0, a2 =
2
8
–π ≤ x <
x=−
π
2
−π
2
−π
<x<0
2
x=0
π
0< x<
2
π
x=
2
π
<x≤π
2
EXERCISES 10.4
Exercises 10.4 (page 611)
2. a.
The π periodic function f ( x) = sin 2 x for x kπ where k is an integer.
b.
The 2π periodic function f 0 ( x) = sin 2 x for x kπ where k is an integer.
c.
The 2π periodic function f e ( x), where f e ( x) =
4. a.
{
b.
π− x
0< x<π
The 2π periodic function f 0 ( x ), where f 0 ( x) =
−π − x −π < x < 0
c.
The 2π periodic function f e ( x), where f e ( x) =
∞
f ( x) ~
8.
f ( x) ~
8k
∑
k =1 π(4k
∞
2
− 1)
sin(2kx)
2
∑ n sin nx
n =1
f ( x) ~
∞
2πn[1 − e(−1) n ]
n =1
1 + π2 n 2
∑
f ( x) ~ 1 +
14.
f ( x ) ~ 1 − e −1 + ∑
∞
2
n =1 1 + π
f ( x) ~
sin(n πx)
π 4 ∞
1
− ∑
cos(2k + 1) x
2 π k =0 (2k + 1)2
12.
16.
0< x<π
−π < x < 0
The π periodic function f ( x), where f ( x) = π − x, 0 < x < π
6.
10.
{sin− sin2x2x
2 2
n
[1 − (−1)n e−1 ] cos(n πx)
1 ∞ 1
cos(2k πx)
−∑
6 k =1 k 2 π2
18. u ( x, t ) =
∞
∑
k = 0 (2k
8
3
+ 1) π
2
e−5(2 k +1) t sin(2 k + 1) x
{ππ −+ xx
0< x<π
–π < x < 0
93
94
CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
Exercises 10.5 (page 624)
∞
2
2π
4
[( −1)n − 1] e−n t sin nx
2. u ( x, t ) = ∑ ( −1)n +1 +
3
n π
n =1 n
4. u ( x, t ) =
2 2
1 ∞ 1
e−8k π t cos 2πkx
−∑
2
2
6 k =1 k π
6. u ( x, t ) = 1 −
∞
2
2
4
+ 2e−7t cos x + ∑
e−28k t cos 2 kx
2
π
k =1 π(4k − 1)
∞
(−1)n − n 2t
e
sin nx
n =1 n
8. u ( x, t ) = 3 x + 6 ∑
∞
π2
2(−1) n −3n2t
1
1
e
sin nx
10. u ( x, t ) =
x − x3 + e −3t sin x + ∑
3
18
3
n = 2 3n
18
∞
12. u ( x, t ) = ∑ an e− µn t sin µn x, where {µn }∞
n =1 is the increasing sequence of positive real numbers that are
2
n =1
solutions to tan µ nπ = − µ n , and an =
1
∫
π
π − sin(2 πµn ) 0
4
2
f ( x) sin µn x dx .
2
20 ∞
1
5π
5
e −3(2k +1) t sin(2k + 1) x
14. u ( x, t ) = 1 + x − x 2 −
∑
3
3π k =0 (2k + 1)
6
6
16. u ( x, y, t ) = e−2t cos x sin y + 4e−5t cos 2 x sin y − 3e−25t cos 3 x sin 4 y
2
4 ∞
1
π
e−[(2 k +1) +1]t cos[(2 k + 1) x]sin y
18. u ( x, y, t ) = e−t sin y − ∑
2
π k =0 (2k + 1)
2
EXERCISES 10.6
95
Exercises 10.6 (page 636)
2. u ( x, t ) =
∞
∑ [an cos 4nt + bn sin 4nt ] sin nx,
where
n =1
n even
0
1
an = 2 n
−
n odd
π n 2 − 4 n
and
−1
2
bn = 4π(n − 1)
1
πn 2
n even
n odd
4. u ( x, t ) = cos12t sin 4 x + 7 cos15t sin 5 x +
6. u ( x, t ) =
2v0 L3
3
∞
∑
1
π α a ( L − a ) n =1 n
3
sin
8. u ( x, t ) = (sin t − t cos t ) sin x +
4 ∞ (−1)k
sin 3(2 k + 1)t sin(2 k + 1) x
∑
3π k =0 (2k + 1)3
n πα t
n πa
n πx
sin
sin
L
L
L
∞
2(−1) n
n=2
n 2 (n 2 − 1)
∑
[sin nt − n sin t ]sin nx
∞
n πα t
n πα t
n πx
x
+ bn sin
sin
, where an ’s and bn ’s are chosen that
10. u ( x, t ) = U1 + (U 2 − U1 ) + ∑ an cos
L
L
L
L n =1
∞
n πx
x
f ( x) − U1 − (U 2 − U1 ) = ∑ an sin
L
L
n =1
∞
n πx
n πα
g ( x) = ∑ bn
sin
L
L
n =1
14. u ( x, t ) = x 2 + α 2 t 2
16. u(x, t) = sin 3x cos 3α t + t
18. u(x, t) = cos 2x cos 2α t + t − xt
96
CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
Exercises 10.7 (page 649)
2. u ( x, y ) =
cos x sinh( y − π) 2 cos 4 x sinh(4 y − 4π)
−
sinh(−π)
sinh(−4π)
4. u ( x, y ) =
sin x sinh( y − π) sin 4 x sinh(4 y − 4π)
+
sinh(−π)
sinh(−4π)
8. u ( r , θ ) =
1 r2
+
cos 2θ
2 8
12. u ( r , θ ) =
27
36 − 1
[ r 3 − r −3 ] cos 3θ +
35
310 − 1
[ r 5 − r −5 ] cos 5θ
∞
14. u ( r , θ ) = C + ∑ r −n [an cos nθ + bn sin nθ ], where C is arbitrary and for n = 1, 2, …
n =1
−1 π
an =
f (θ ) cos nθ d θ
n π ∫−π
−1 π
bn =
f (θ ) sin nθ dθ
n π ∫−π
16. u ( r , θ ) =
∞
n π(θ − π)
n π(ln r − ln π)
sin
, where
ln 2
ln 2
∑ an sinh
n =1
an =
−2
( )
ln 2 sinh nlnπ2
2
∞
2π
∫π
n π(ln r − ln π) 1
sin r sin
r dr .
ln 2
24. u ( x, y ) = ∑ An e −ny sin nx, where An =
n =1
2 π
f ( x) sin nx dx.
π ∫0
CHAPTER 11
Exercises 11.2 (page 671)
2. sin x – cos x + x + 1
4. ce −2 x (sin 2 x + cos 2 x)
6. 2e x + e− x − x
8. c1 sin 2 x + c2 cos 2 x + 1
10. No solution
12. No solution
14. λn =
π2 n 2
πnx
πnx
; yn ( x) = bn sin
+ cn cos
, n = 0, 1, 2, …
9
3
3
16. λn = 2 +
(2n + 1) 2
(2n + 1) x
; yn ( x) = cn sin
, n = 0, 1, 2, …
4
2
18. λ0 = − µ02 , where tanh( µ0 π) = 2 µ0 ; y0 ( x) = c0 sinh( µ0 x),
λn = µ n2 , where tan(µ n π) = 2µ n ; yn ( x) = cn sin( µn x ), n = 1, 2, 3, …
20. λn = π2 n 2 ; yn ( x) = cn cos( πn ln x), n = 0, 1, 2, …
22. λ1 = 4.116, λ2 = 24.139, λ3 = 63.659
24. No nontrivial solutions
26. λn = µ n2 , where cot πµ n = µ n for µn > 0; yn ( x) = cn sin µn x, n = 1, 2, 3, …
28. λn = − µ n4 ; µn > 0 and cos( µ n L) cosh( µ n L) = −1;
sin µn L + sinh µn L
yn = cn sin µn x − sinh µ n x −
(cos µn x − cosh µ n x ) , n = 1, 2, …
cos µ n L + cosh µ n L
34. b.
λ0 = −1, X 0 ( x) = c and λn = −1 + n 2 π2 , X n ( x ) = c cos(πnx )
97
98
CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
Exercises 11.3 (page 682)
2. y ′′ + λ x −1 y = 0
4. ( xy ′) ′ + xy − λ x −1 y = 0
6. ((1 − x 2 ) y ′) ′ + λ y = 0
8. Yes
10. No
18. a.
1
n πx 1
n πx
sin
cos
,
, n = 1, 2, 3, …
3
3
3
3
1
,
6
(−1)n +1 6
n πx
sin
πn
3
n =1
∞
b.
∑
2
1
sin n + x , n = 0, 1, 2, …
π
2
20. a.
∞
b.
22. a.
b.
24. a.
b.
(−1)n 8
1
sin n + x
2
n = 0 π(2n + 1)
∑
2
2
µ0
sinh( µ0 x) where tanh( µ0 π) = 2 µ0 ;
sinh(2 µ0 π) − 2µ0 π
2
µn
sin( µ n x) where tan(µ n π) = 2µ n , n = 1, 2, 3, …
2 µn π − sin(2 µn π)
∞ 4(2 − π) cos( µ π)
4(π − 2) cosh( µ0 π)
n
sinh( µ0 x) + ∑
sin( µ n x)
π
−
µ
µ
sinh(2 µ0 π) − 2 µ0 π
2
sin(2
n
n π)
n =1
y0 ( x) =
1
e −1
; yn ( x) = 2 cos(πn ln x ), n = 1, 2, 3, …
∞
2[(−1)n e − 1]
n =1
1 + π2 n 2
1+ ∑
cos(πn ln x)
Exercises 11.4 (page 692)
2. L+ [ y ] = x 2 y ′′ + (4 x − sin x) y ′ + (2 x + 2 − cos x) y
4. L+ [ y ] = x 2 y ′′ + 6 xy ′ + 7 y
6. L+ [ y ] = (sin x) y ′′ + (2 cos x + e x ) y ′ + (− sin x + e x + 1) y
EXERCISES 11.6
8. L+ [ y ] = y ′′ + 4 y ′ + 5 y;
D( L+ ) = { y ∈ C 2 [0, 2 π] : y (0) = y (2 π) = 0}
1
1
cos 5 x − cos 4 x
19
10
8.
∑
∞
5
10. L [ y ] = x y ′′ + 2 xy ′ + y
4
+
6.
2
n=0
−8
3
π(2n + 1) (4n 2 + 4n − 1)
99
sin[(2n + 1) x]
D( L+ ) = { y ∈ C 2 [1, eπ ] : y (1) = y (eπ ) = 0}
∞
12. y ′′ − y ′ + y = 0; y(0) = y
y ′(0) = y ′(π)
10.
n=0 7 −
γn =
π
π
14. y ′′ = 0; y (0) = y ; y ′(0) = y ′
2
2
16. x 2 y ′′ + 2 xy ′ = 0; y(1) = 4y(2), y ′(1) = 4 y ′(2)
18.
20.
2π
∫0
eπ
∫1
h ( x )e
−2 x
h( x ) x
−1/ 2
(n + )
1 2
2
2 π
1
f ( x) sin n + x dx .
π ∫0
2
2 eπ
f ( x) sin( n ln x) dx. If γ 1 ≠ 0,
π ∫1
there is no solution. If γ 1 = 0, then
c sin(ln x) +
14.
24.
26.
π/2
∫0
2
∞
∑
γn
n =2 1 − n
∞
γn
n=0
−1 − π2 n 2
γn
∫
= 1
∑
e
22. Unique solution for each h
1
sin n + x , where
2
12. Let γ n =
sin x dx = 0
sin(ln x)dx = 0
γn
∑
2
sin(n ln x) is a solution.
cos(πn ln x), where
f ( x) cos(πn ln x)dx
e −1
x cos 2 ( πn ln
1
∫
.
x) dx
h( x)dx = 0
Exercises 11.6 (page 706)
3
∫1 h( x) 1 − x = 0
Exercises 11.5 (page 698)
2.
1
1
sin 11x − sin 4 x
122
17
4.
1
5
cos 7 x +
cos10 x
π − 49
π − 100
sinh s sinh( x − 1)
−
sinh 1
2. G ( x, s ) =
− sinh x sinh( s − 1)
sinh 1
4. G ( x, s ) =
s sin x
{−− cos
cos x sin s
0≤s≤ x
x ≤ s ≤1
0≤s≤ x
x≤s≤π
100
CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
6. G ( x, s ) =
s − cos s ) sin x
{(sin
(sin x − cos x ) sin s
0≤ s≤ x
x≤s≤π
( s 2 + s −2 )( x 2 − 16 x −2 )
−
1≤ s ≤ x
68
8. G ( x, s ) =
( x 2 + x −2 )( s 2 − 16 s −2 )
−
x≤s≤2
68
(e5 s − e− s )(5e6− x + e5 x )
30e6 + 6
10. G ( x, s ) =
(e5 x − e − x )(5e6 − s + e5 s )
30e6 + 6
− s ( x − π)
π
12. G ( x, s ) =
− x ( s − π)
π
y=
4
0≤s≤ x
0≤s≤ x
x ≤ s ≤1
1 2
− 1 − s 1 − x 1 ≤ s ≤ x
20. G ( x, s ) =
− 1 − 1 1 − 2 x ≤ s ≤ 2
x s
y=
2 ln 2
x
+ ln
x
4
e x − s s (1 − x) 0 ≤ s ≤ x
24. a. K ( x, s ) = x − s
e x(1 − s) x ≤ s ≤ 1
b. y = ( x − 1)e x − xe x −1 + 1
x≤s≤π
3
{sx
0≤s≤ x
x ≤ s ≤1
x −π x
12
14. G ( x, s ) =
cosh s cosh( x − 1)
sinh 1
18. G ( x, s ) =
cosh x cosh( s − 1)
sinh 1
y = –24
( s 2 − s −2 )(3x + 16 x −3 )
1≤ s ≤ x
76
26. a. K ( x, s ) =
( x − x −3 )(3s 2 + 16s −2 )
x≤s≤2
76
0≤s≤ x
x≤s≤π
4π3 x − x 4
y=
12
sin s sin( x − 2)
−
0≤s≤ x
sin 2
16. G ( x, s ) =
sin x sin( s − 2)
x≤s≤2
−
sin 2
12
y=
[sin x − sin( x − 2) − sin 2]
sin 2
b. y =
4(1 + ln 2) −3
1
x ln x +
( x − x)
4
19
EXERCISES 11.8
x(π − s )( s 2 − 2πs + x 2 )
6π
28. H ( x, s ) =
s (π – x)( x 2 − 2πx + s 2 )
6π
101
0≤s≤x
x≤s≤π
x 2 [( x − 3π)( s3 − 3πs 2 ) + 2π3 ( x − 3s )]
0≤s≤ x
12π3
30. H ( x, s ) =
s 2 [( s − 3π)( x3 − 3πx 2 ) + 2π3 ( s − 3 x)]
x≤s≤π
12π3
Exercises 11.7 (page 715)
1
∞
2.
∑ bn J3 (α3n x)
n =1
1
∞
4.
∑ bn Pn ( x)
where bn =
n=0
1
[ µ − n(n + 1)]∫ Pn2 ( x) dx
1
∑ bn P2n ( x)
n=0
where bn =
∫0 f ( x) P2n ( x) dx
1
[ µ − 2n(2n + 1)]∫ P22n ( x) dx
0
∞
16. c.
∫−1 f ( x) Pn ( x) dx
−1
∞
6.
where {α 3n } is the increasing sequence of real zeros of J 3 and bn =
∞
∑ bn Ln ( x)
n=0
where bn =
∫0
f ( x) Ln ( x) dx
∞
( µ − n) ∫ L2n ( x)e− x dx
0
Exercises 11.8 (page 725)
2. No; sin x has a finite number of zeros on any closed bounded interval.
1 1
4. No; it has an infinite number of zeros on the interval − , .
2 2
8.
π
3
10. Between π
26
e−25
and π
λ +1
λ + 26 sin 5
∫0 f ( x) J3 (α3n x) dx
1
( µ − α32n ) ∫ J 32 (α 3n x ) x dx
0
102
CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
Chapter 11 Review (page 729)
2. a.
b.
(e
c.
( xe − x y ′) ′ + λ e− x y = 0
4. a.
b.
6.
(e7 x y ′) ′ + λ e7 x y = 0
−3 x 2 / 2
)
2
y ′ ′ + λ e −3 x / 2 y = 0
y ′′ − xy ′ = 0; y ′(0) = 0, y ′(1) = 0
x 2 y ′′ + 2 xy ′ − 3 y = 0; y (1) = 0, y ′(e) = 0
π
2 ∞
1
+ 4 cos 2 x + ∑
cos[(2k + 1) x]
π k =0 (2k 2 + 2k − 1)(2k + 1)2
6
∞
8. a.
∑ bn J 7 (α 7n x)
n =1
where {α 7 n } is the increasing sequence of real zeros of J 7 and
1
bn =
∫0 f ( x) J 7 (α7n x)dx
1
( µ − α 72n ) ∫ J 72 (α 7 n x ) x dx
0
1
∞
b.
∑ bn Pn ( x)
n=0
10. Between
where bn =
∫−1 f ( x) Pn ( x) dx
1
[ µ − n(n + 1)]∫ Pn2 ( x) dx
−1
π
6
and π
3
.
5
CHAPTER 12
Exercises 12.2 (page 753)
2. Unstable proper node
4. Unstable improper node
6. Stable center
8. (–1, –1) is an asymptotically stable spiral point.
10. (2, –2) is an unstable spiral point.
12. (5, 1) is an asymptotically stable improper node.
14. Unstable proper node
Figure 44
103
104
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
16. Asymptotically stable spiral point.
Figure 45
18. Unstable improper node
Figure 46
EXERCISES 12.3
20. Stable center
Figure 47
Exercises 12.3 (page 764)
2. Asymptotically stable improper node
4. Asymptotically stable spiral point
6. Unstable saddle point
8. Asymptotically stable improper node
10. (0, 0) is indeterminant; (–1, 1) is an unstable saddle point.
12. (2, 2) is an asymptotically stable spiral point; (–2, –2) is an unstable saddle point.
105
106
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
14. (3, 3) is an unstable saddle point; (–2, –2) is an asymptotically stable spiral point.
Figure 48
16. (0, 0) is an unstable saddle point; (–4, –2) is an asymptotically stable spiral point.
Figure 49
EXERCISES 12.4
Exercises 12.4 (page 774)
2. G(x) = sin x + C; E ( x, v ) =
4. G ( x) =
1 2
v + sin x
2
1 2 1 4
1 6
1
1
1 4
1 6
x −
x +
x + C ; E ( x, v ) = v 2 + x 2 −
x +
x
2
24
720
2
2
24
720
6. G ( x) = e x − x + C ; E ( x, v ) =
8.
Figure 50
1 2
v + ex − x −1
2
107
108
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
10.
Figure 51
12.
Figure 52
EXERCISES 12.4
14. vh ( x, v ) = v 2 , so energy is decreasing along a trajectory.
Figure 53
16. vh( x, v) = v 2 , so energy is decreasing along a trajectory.
Figure 54
109
110
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
18.
Figure 55
Exercises 12.5 (page 782)
2. Asymptotically stable
10. Stable
4. Stable
12. Stable
6. Unstable
14. Stable
8. Asymptotically stable
EXERCISES 12.6
Exercises 12.6 (page 791)
4. b.
Clockwise
6. r = 0 is an unstable spiral point.
r = 2 is a stable limit cycle.
r = 5 is an unstable limit cycle.
Figure 56
8. r = 0 is an unstable spiral point.
Figure 57
111
112
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
10. r = 0 is an unstable spiral point.
r = 2 is a stable limit cycle.
r = 3 is an unstable limit cycle.
Figure 58
12. r = 0 is an unstable spiral point.
r = n n = 1, 2, 3, …, is a limit cycle that is stable for n odd and unstable for n even.
Figure 59
CHAPTER 12 REVIEW
113
Exercises 12.7 (page 798)
2. Asymptotically stable
10. Stable
4. Unstable
12. Asymptotically stable
6. Asymptotically stable
14. Asymptotically stable
8. a.
b.
c
x(t ) = 1
c2
16. The equilibrium solution corresponding to
the critical point (–3, 0, 1) is unstable.
t
1
x(t ) = c1 + c2
0
1
18. The equilibrium solutions corresponding
to the critical points (0, 0, 0) and (0, 0, 1)
are unstable.
Chapter 12 Review (page 801)
2. (0, 0) is an unstable saddle point.
Figure 60
114
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
4. (0, 0) is a stable center.
Figure 61
6. (0, 0) is an unstable improper node.
Figure 62
CHAPTER 12 REVIEW
8.
Figure 63
10. Unstable
12. Asymptotically stable
115
116
CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
14. r = 0 is an asymptotically stable spiral point.
r = 2 is an unstable limit cycle.
r = 3 is a stable limit cycle.
r = 4 is an unstable limit cycle.
Figure 64
16. No
18. Asymptotically stable
CHAPTER 13
Exercises 13.1 (page 812)
x
2. y ( x ) = ∫ sin(t + y (t ))dt
Exercises 13.3 (page 826)
2. [–2, 1)
π
4. (0, 3]
x y (t )
e dt
0
4. y ( x ) = 1 + ∫
6.
( −∞, ∞ )
6. 0.3775396
Exercises 13.4 (page 832)
8. 2.2360680
2. 10−2 e
10. 1.9345632
1
12. y1 ( x ) = 1 + x, y2 ( x) = 1 + x + x 2 + x3
3
14. y1 ( x) = y2 ( x) = sin x
3
1
16. y1 ( x ) = − x + x 2 ,
2
2
5 3
1
y2 ( x ) = − x + x 2 − x 3
3 2
6
Exercises 13.2 (page 820)
2. No
−1 / 2
4. 10−2 e 2e
6. 10−2 e
1
8. esin1
24
10.
e
6
Chapter 13 Review (page 835)
2. 0.7390851
4. Yes
x
4. 9 + ∫ [t 2 y 3 (t ) − y 2 (t )]dt
0
6. Yes
1
2
0≤ x≤ n
n x
1
2
14. No; let yn ( x) = 2n − n 2 x n ≤ x ≤ n .
2
0
n ≤ x ≤1
Then, lim yn ( x) = 0, but
lim
n→∞
n→∞
1
y ( x) dx
0 n
∫
6. y1 ( x ) = −1 + 2 x, y2 ( x ) = −1 + 2 x − 2 x 2
8. No
π π
10. − ,
2 2
12.
e
6
= 1 ≠ 0.
117
