INSTRUCTOR’S RESOURCE GUIDE
EDWARD B. SAFFVanderbilt University
A. DAVID SNIDERUniversity of South Florida
FUNDAMENTALS
OF DIFFERENTIAL EQUATIONSSIXTH EDITION
FUNDAMENTALS OF
DIFFERENTIAL EQUATIONS
AND BOUNDARY VALUE PROBLEMS
FOURTH EDITION
R. Kent NagleEdward B. SaffA. 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 priorwritten 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 1
Software Supplements 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 8
Answers to Even-Numbered Problems 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 lecture-based, 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 [email protected]
A. D. Snider [email protected]
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
b. From calculus recall that dy
dx= tanα . Use this, the fact that tan ,
yx
β= and the double angle formula to
show that
( )2
2.
1
dydx
dydx
y
x=
−
c. Now show that the curve satisfies the differential equation.
(1) dy
dx=
−x + x2 + y2
y
d. Solve equation (1).
e. Describe the solutions and identify the type of collector obtained.
SUPPLEMENTAL GROUP PROJECTS 9
_____________ †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.
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) ( ) ( )03
6 ,500
x t x t t′ = − −
( ) 00 for ,0x t t t = ∈ −
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 ( )0t t− 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
(2) ( ) ( ) ,x t ax t b′ = −
where a and b are constants, has a solution of
the form ( ) ,stx t Ce= for any constant C,
provided s satisfies the transcendental equation
.bss ae−=
b. A solution to (2) for t > 0 can also be found using the method of steps. Assume that
( ) ( ) for 0.x t f t b t= − ≤ ≤ For 0 ,t b≤ ≤
equation (2) becomes
( ) ( ) ( ) ,x t ax t b a f t b′ = − = −
and so
( ) ( ) ( )1
00 .x t a f v b dv x= − +∫
Now that we know x(t) on [0, b], we can repeat this procedure to obtain
( ) ( ) ( )1
bx t ax v b dv x b= − +∫
for b ≤ t ≤ 2b. This process can be continued indefinitely.
Use the method of steps to show that the solution to the initial value problem
( ) ( ) ( ) [ ]1 , 1 on 1,0x t x t x t′ = − − = −
is given by
( ) ( ) ( )0
11 , for 1 ,
!
knk
k
t kx t n t n
k=
− − = − − ≤ ≤∑
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 0 3.t =
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 0h → (cf. Section 3.6). From theoretical considerations assume we know, more precisely, that (1) y(x; h) = φ(x) + hpap(x) + O(hp +1),
where y(x; h) is the approximation to φ(x) using step size h and ap (x) is some function that is independent
of h (typically we do not know a formula for ap (x),
only that it exists). Our goal is to obtain approximations that converge at the faster rate O(h p+1).
We start by replacing h by h
2 in (1) to get
y x;h
2
= φ(x) +
hp
2p ap(x) + O(h p+1).
If we multiply both sides by 2p
and subtract equation (1), we find
2py x;
h
2
− y(x; h) = (2
p − 1)φ(x) + O(hp+1
).
Solving for φ(x) yields
( )2 1
2 ; ( ; )( ) O( ).
2 1
p hp
p
y x y x hx hφ +−
= +−
Hence
( )2
2 ; ( ; )* ; :
2 2 1
p h
p
y x y x hhy x
− = −
has a rate of convergence of O(hp +1).
a. Assuming
y * x;h
2
= φ(x) + hp +1ap +1(x) + O(h p+2 ),
show that
( ) ( )14 21
2 * ; * ;** ; :
4 2 1
p h h
p
y x y xhy x
+
+
− = − has a
rate of convergence of O(hp+2
).
b. Assuming
y ** x;h
4
= φ(x) + h p+2ap +2(x) + O(hp+3),
show that
( ) ( )28 42
2 ** ; ** ;*** ; :
8 2 1
p h h
p
y x y xhy x
+
+
− = −
has a rate of convergence of O(hp +3).
c. The results of using Euler’s method
with h = 1,1
2,
1
4,
1
8
to approximate the
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)
by computing y *** 1;1
8
. [Hint: Compute
y * 1;1
2
, y * 1;
1
4
, and y * 1;
1
8
; then
compute y ** 1;1
4
, and y ** 1;
1
8
.
d. Table 1.2 also contains Euler’s approximation
for y(1) when h =1
16. Use this additional
information to compute the next step in the extrapolation procedure; that is, compute
y **** 1;1
16
.
SUPPLEMENTAL GROUP PROJECTS 11
_________________________ † 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
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 predator-prey model given by the system
(1) ,dx
Ax Bxydt
= −
(2) ,dy
Cy Dxydt
= − +
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
( ) ( )0 0
1 1: , :
T Tx x t dt y y t dt
T T= =∫ ∫
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
( )( ) ( )( )0 0
0.]T T x t
A By t dt dtx 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 x A x Bxydt
= − − ε = − ε −
(4) ( ) ,dy
Cy Dxy y C y Dxydt
δ δ= − + − = − + +
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?
12 SUPPLEMENTAL GROUP PROJECTS
_________________ †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.
Limit Cycles
In the study of triode vacuum tubes, one encounters the van der Pol equation:†
( )2 0,y y y yµ″ − 1− ′ + =
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 0.y y y yµ ″ − 1− ′ ′ + =
SUPPLEMENTAL GROUP PROJECTS 13
_________________________ † 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
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
.. .200
.. 20
.. 200
4 ( ) ( – )
3 ( )
( ) ( ) ,
x z y y z x
y x z
z y x y z x
I I I I I I
I I I
I I I I I I
φ ω φ ω ψ
θ ω θ
ψ ω ψ ω φ
= − − − −
= − −
= − − + − −
(where ω0 =GM
r 3 is the angular frequency of the
orbit and the positive constants Ix, Iy , Iz 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
1 2
3 4
0 0 1 00 0 0 1
0 00 0
dc cdt
c c
φ φψ ψφ φ
ψθ
=
and
5
0 10
dcdt
θ θθ θ
=
b. Show that the origin is asymptotically stable for the first system in (a) if
22 4 3 1 1 3( ) 4 0,c c c c c c+ + − > c1c3 > 0, and
2 4 3 1 0c c c c+ + > and hence deduce that Iy > Ix > Iz 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 Ix, Iy , Iz and 0 .ω Phobos maintains
Iy > Ix > Iz in its orientation with respect to
Mars, and has angular frequency of orbit
ω0 = .82 rad/hr. If ( )
.23,x z
y
I I
I
−= show that
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.
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. 4 ,dx
kxdt
= where k is the proportionality constant
16. 2 ,dA
kAdt
= where k is the proportionality constant
Exercises 1.2 (page 14)
4. Yes
6. No
8. Yes
10. Yes
12. Yes
20. a. –1, –5
b. 0, –1, –2
22. a. 1 25 1
,3 3
c c= =
b. 1 2
1 22
,3 3
e ec c
−= =
24. Yes
26. No
28. Yes
16 CHAPTER 1 INTRODUCTION
Exercises 1.3 (page 22)
2. a. y = –2 – 2x
Figure 1 Solution to Problem 2(a)
Figure 2 Solution to Problem 2(b)
c. 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. 2
b. For x > 1 we have x + sin y > 0, so 0y ′ > and hence y is increasing.
c. 2
2( sin ) 1 cos
1 ( sin ) cos1
1 cos sin(2 )2
d y dx y y y
dxdxx y y
x y y
′= + = +
= + +
= + +
d. For x = y = 0 we have 0y ′ = and, from the formula in part (c), it follows that 2
21 0
d y
dx= > when
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 17
8. a. 7 b. 2
3 3 2 22
2 2 3 3
2 2 3 5
( ) 3 3
3 3 ( )3 3 3
d x d dxt x t x
dt dtdtt x t xt x t x
= − = −
= − −= − +
c. No, because if 1 < x < 2 and t > 2.5, then 3 3 0,dx
t xdt
= − > and so the particle moves to the right,
away from x = 1.
10.
12.
Figure 5
18 CHAPTER 1 INTRODUCTION
14.
Figure 6
16.
18. Every solution approaches zero.
EXERCISES 1.4 19
Exercises 1.4 (page 28)
2. nx 0.1 0.2 0.3 0.4 0.5
ny 4.000 3.998 3.992 3.985 3.975
(rounded to three decimal places)
4. nx 0.1 0.2 0.3 0.4 0.5
ny 1.00 1.220 1.362 1.528 1.721
6. nx 1.1 1.2 1.3 1.4 1.5
ny 0.1 0.209 0.32463 0.44409 0.56437
8. N 1 2 4 8
( )φ π π 2π 1.20739 1.14763
10. nx ny 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
20 CHAPTER 1 INTRODUCTION
14. h = 0.5
nx 0.5 1.0 1.5 2.0
ny 1.000 1.500 3.750 24.844
h = 0.1
nx 0.5 1.0 1.5 2.0
ny 1.231 3.819 1,023,041.61 overflow
h = 0.05
nx 0.5 1.0 1.5 2.0
ny 1.278 5.936 overflow overflow
h = 0.01
nx 0.5 1.0 1.5 2.0
ny 1.321 18.299 overflow overflow
16. T(1) ≈ 82.694°; T(2) ≈ 76.446°
21
CHAPTER 2
Exercises 2.2 (page 46)
2. No
4. Yes
6. Yes
8. 4 4 lny x C= +
10. 3tx Ce=
12. 2 8 / 34 1v Cx−= +
14. 3tan( )y x C= +
16. 22ln(1 ) xy e C−+ = +
18. tan ln(cos )3
y xπ = −
20. 3 211 1 4 8 8
2y x x x
= − − + + +
22. 3
43
xy = −
24. 4ln 4 3y x= −
26. ( )21 ln 1y x= − +
28. 223 ;t ty e −= maximum value is y(1) = 3e.
32. 79.95 mCi
34. a. ktT Ce M= +
b. 70.77
36. 20°
38. / 20( ) 196 186 tv t e−= − m/sec; limiting velocity
is 196 m/sec.
Exercises 2.3 (page 54)
2. Neither
4. Linear
6. Linear
8. 22 lny x x x Cx= + +
10. 3 2y x Cx− −= − +
12. 3 4
4
3
xxx e
y Ce−
−= +
14. 3 2
32
6 5
x xy x Cx−= − + +
16.
5 4 25 2
2 2( 1)
x x x x Cy
x
+ + − +=
+
18. 4
3
xxe
y e−
−= +
20. 2 33 9
5 2 10
x x xy
−= − +
22. y = 1 – x cot x + csc x
24. a. 10 20( ) 5 35 ;t ty t e e− −= + the term 105 te−
eventually dominates.
b. 10 10( ) 50 40t ty t te e− −= +
26. 0.860
30. b. 1/ 3
61
2 12xx
y Ce− = − +
22 CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS
32. 2
6 2
1 0 3
1 (2 1) 3
x
x
e xy
e e x
−
−
− ≤ ≤= − + − <
Figure 8
36. b. 3( )hy x x−=
d. 6
( )6
xv x =
e. 3
3( )6
xy x Cx−= +
Exercises 2.4 (page 65)
2. Linear with y as dep. var.
4. Separable
6. Separable and linear with y as dep. var.
8. Exact
10. 2 2x xy y C+ − =
12. 3 1/ 3sinxe y x y C− + =
14. ( 1)
1
t
t
C e ty
e
+ −=
+
16. xy xe C
y− =
18. 2 2 sin( ) yx xy x y e C+ − + − =
20. 2 / 33
2 arcsin sin( )2
yx xy C+ − =
22. 1xy xe e
y− = −
24. 1t
e tx
e
−=−
26. x tan y + ln y – 2x = 0
28. a. x cos(xy) + g(y)
b. 4 ( )xyxe x g y− +
where g is a function of y only.
30. b. 2x y
c. 5 2 6 3 4 3x y x y x y C+ + =
CHAPTER 2 REVIEW 23
34. 2 2 2 2;x y x y Cyµ −= + =
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. 4 2 3 1,y x y y Cµ − − −= − = and 0.y ≡
10. 2xµ −= (also linear); 4
ln3
xy x x Cx= − +
12. Exact; 2 3 lnx y x y C+ − =
14. 3 2 ,x yµ = 4 2 5 33 ,x y x y C+ = but not 0.y ≡
16. y = Cx
Exercises 2.6 (page 0)
2. Homogeneous
4. Bernoulli
6. Homogeneous
8. ( )y G ax by′ = +
10. ln
xy
C x=
− and 0.y ≡
12. 3 23x xy C+ =
14. sin lny
Cθθ
− =
16. Cxy xe=
18. y = –x – 2 + tan(x + C)
20. tan(x – y) + sec(x – y) – x = C
22. 2
2 2
2
xx e
y Ce− −= − and 0.y ≡
24. 2 1/ 2 2[( 2) ( 2) ]y x C x −= − + − and y ≡ 0
26.
1/ 333
4
xxe
y Ce− = +
28. 2 21
2xy x Ce− = − − + and y ≡ 0.
30. 2 2 3ln[( 3) ( 2) ] 2 arctan
2
yy x C
x
− − + + − = +
32. 2 26 6 8 8
5 5 5 5x x y y C
+ + + + − + =
34. 2 2t x Ct+ =
36. 4
3y
Cx x=
− and y ≡ 0.
38. ( )2ln
4
Cy
θ θ +=
40. cos( ) sin( ) x yx y x y Ce −+ + + =
42. 2 2
sec tany y
Cxx x
+ =
46. b. 5
5xy x
C x= +
−
Chapter 2 Review (page 81)
2. 2 48 4 1 xy x x Ce= − − − +
4. 3 2
34 3
6 5 4
x x xCx−− + +
6. 2 22 ln 1y x C− = − + and y ≡ 0.
24 CHAPTER 2 FIRST-ORDER DIFFERENTIAL EQUATIONS
8. 2 3 1( 2 )y Cx x −= − and y ≡ 0.
10. 1/ 22 arctan( )x y y x y C+ + + + =
12. 2 32 x xye y e C+ =
14. 2 ( 1)
( 1) 3( 1) ln 1 ( 1)2
t tx t t t t C t
−= + − + − − + −
16. (cos ) ln cos cosy x x C x= +
18. ( )1 2 2 tan 2y x x C= − + +
20.
1/ 323 12
5y C
θθ − = −
22. 4(3 2 9)( 2)y x y x C− + + − =
24. 2 sin cosxy x y C+ − =
26. 2 / 2xy Ce−=
28. 2 2( 3) 2( 3)( 2) ( 2)y y x x C+ + + + − + =
30. 4 1
4xy Ce x= − −
32. 2 2 2 2ln( ) 16y x x x= +
34. 2
22
2sin
xy x x= +
π
36. 3 2
sin(2 ) sin 23 3
yxx y e+ − + = +
38. 2
12 arctan
4 2
xy
= −
40. 4
8
1 3 4xe x−− −
25
CHAPTER 3
Exercises 3.2 (page 98)
2. 3 / 252.5 2 ,te−− 5.78 min
4. 8 3( 100) 10 ( 100)
,5 5
t t −+ +− 18.92 min
6. 43.8 min
8. 3 /125 3(0.2)(1 ) g/cm ;te−− 28.88 sec
14. 187,500
16. Since 0
( ) ( )lim ( )h
p t h p tp t
h→
+ − ′= and
0
( ) ( )lim ( ),h
p t h p tp t
h→
− − ′=−
adding these two
equations and dividing by 2 yields the given formula.
18. a. ( ) exp bt aP t Ce
b− = +
b. 0( ) exp ln( ) bta aP t P e
b b− = − +
c. b > 0: /( ) ;a bP t e→
If b < 0 and 0ln( ) ,a
Pb
> then P(t) →
If b < 0 and 0ln( ) ,a
Pb
< then P(t) → 0;
If b < 0 and 0ln( ) ,a
Pb
= then /( ) a bP t e≡ .
20. 6.9315 light years
22. 90 min., vanishes at t =
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
12. The impatient friend
14. 127.5°F
Exercises 3.4 (page 115)
2. (0.8)40 50 50tt e−+ − ft; 13.75 sec
4. 1.67 sec
6. 212.45 4.91 12.45te t− + − m, 22.9 sec
8. 206.8 sec, 86.8 sec
26 CHAPTER 3 MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS
10. 5
50( 15.5)
1.96 0.39( 1) 0 15.5( )
30 0.098( 15.5) 0.035[1 ] 15.5
t
t
t e tx t
t e t
−
− −
+ − ≤ ≤= + − + − ≤
0.098 m/sec
12. 16.48 sec; 1535 m
14. 1/ nmg
k
16. ( )2// / 2 ,
T kk t Ie k T Ceω ω −− = where ( )2
0//
0T kkC e k Tω ω= −
18. 21.61t m, 5.65 m/sec
20. 0 arctanα µ=
22. 2
13/ 5
6 / 5
5 5 16 ln 5 0.805
2 22( )5 37 1
10 2 ln 5 ln 56 26
t
t
t tex t
t te
+ − ≤ ≈ = + − − >
10 m/sec
24. 0
0( ) ln ,
mv t gt
m tβ
α
= − − 2 0
00
1( ) ( ) ln
2
mx t t gt m t
m t
ββ αα α
= − − − −
Exercises 3.5 (page 122)
2. capacitor voltage = 1,000,00010, 000 100, 000, 000 10, 000cos100 sin100
100, 000, 001 100, 000, 001 100, 000, 001tt t e V−− + +
resistor voltage = 1,000,00010, 000 1 10, 000cos100 sin100
100, 000, 001 100, 000, 001 100, 000, 001tt t e V−+ −
current = 1,000,00010, 000 1 10, 000cos100 sin100
100, 000, 001 100, 000, 001 100, 000, 001tt t e A−+ −
4. From (2), ( )1.I E t dt
L= ∫ From the derivative of (4), .
dEI C
dt=
6. Multiply (2) by I to derive 2 21
2
dLI RI EI
dt + =
(power generated by the voltage source equals the power
inserted into the inductor plus the power dissipated by the resistor). Multiply the equation above (4) by I,
replace I by dq
dt and then replace q by CEC in the capacitor term, and derive 2 21
2 Cd
RI CE EIdt
+ = (power
generated by the voltage source equals the power inserted into the capacitor plus the power dissipated by the resistor).
8. In cold weather, 96.27 hours. In (extremely) humid weather, 0.0485 seconds.
EXERCISES 3.6 27
Exercises 3.6 (page 132)
6. For k = 0, 1, 2, …, n, let kx kh= and ( )k kz f x= where 1
.hn
=
a. 0 1 2 1( )nh z z z z −+ + + +"
b. 0 1 2 1( 2 2 2 )2 n nh
z z z z z− + + + + +
"
c. Same as (b)
8. nx 1.2 1.4 1.6 1.8
ny 1.48 2.24780 3.65173 6.88712
10. nx ny
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. 4( ) ( ; 2 ) 1.09589yφ −π ≈ π π =
14. 2.36 at x = 0.78
16. x = 1.26
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)
2. 2
2 21 ( ) ( ( 2 )( ))
2n n n n n n n n n n nh
y y h x y y y x y x y y+
= + − + + − −
4. 2 3 4
2 2 2 21 ( ) (2 ) (2 2 ) (2 2 )
2! 3! 4!n n n n n n n n n n n n nh h h
y y h x y x x y x x y x x y+
= + + + + + + + + + + + + +
6. Order 2: φ (1) §
order 4: φ (1) §
EXERCISES 3.7 29
8. 0.63211
10. 0.70139 with h = 0.25
12. –0.928 at x = 1.2
14. 1.00000 with 16
hπ=
16. nx ny
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. 4(10) 2.23597 10x −≈ ×
31
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. 30 25
cos 3 sin 361 61
t t− −
10. a. 2
2 2 2 2 2 2 2 2cos sin cos sin
( ) ( )
k m bA t B t t t
k m b k m b
− Ω ΩΩ + Ω = Ω + Ω− Ω + Ω − Ω + Ω
b.
c.
Figure 10
32 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
Exercises 4.2 (page 167)
2. 21 2
t tc e c e− +
4. 3 31 2
t tc e c te− −+
6. 2 31 2
t tc e c e+
8. / 2 2 / 31 2
t tc e c e−+
10. / 2 / 21 2
t tc e c te+
12. ( ) ( )11 205 / 6 11 205 / 6
1 2t t
c e c e− + − −
+
14. 3 te−−
16. 34
3 3
t te e−
18. 3
3 72
3
tt te
e +
20. 2 2(2 ) tt e −−
22. 7
3
tce
24. 11 / 3tce−
26. a. 2 cos t
b. 22 cos sin ,t c t+ where 2c is arbitrary.
28. linearly independent
30. linearly independent
32. linearly dependent
38. If 1 0,c ≠ then 21 2
1.
cy y
c
= −
42. 61 2 3
t t tc e c e c e− + +
44. 3 51 2 3
t t tc e c e c e− + +
46. ( ) ( )1 3 1 3
1 2 3t ttc e c e c e
− + − −+ +
48. 2t te e+
Exercises 4.3 (page 177)
2. 1 2cos sinc t c t+
4. 5 51 2cos sint tc e t c e t+
6. 2 21 2cos 3 sin 3t tc e t c e t+
8. / 2 / 21 2
5 5cos sin
2 2t tt t
c e c e− − +
10. 2 21 2cos 2 sin 2t tc e t c e t− −+
12. 1 2cos 7 sin 7c t c t+
14. 1 2cos 5 sin 5t tc e t c e t+
16. ( ) ( )3 53 / 2 3 53 / 2
1 2t t
c e c e+ −
+
18. 7 / 21 2
t tc e c e− +
20. 1 2 3cos sint t tc e c e t c e t− + +
22. cos 4te t−
24. 1
sin 3 cos 33
t t +
26. 3t te te−
28. b = 5: 44
3
t te ey
− −−=
b = 4: 2(1 2 ) ty t e−= +
b = 2: 1/ 2cos 3 3 sin 3t ty e t e t− − −= +
32. a. y(t) = 0.3 cos 5t – 0.02 sin 5t m
b. 5
2π
34. 2
( ) sin 255
tI t e t− =
36. a. ( 1 ) ( 1 )3 3( ) 1 1
2 2i t i ty t i e i e− + − − = − + +
40. 71 2c x c x−+
EXERCISES 4.5 33
42. 2 21 1sin 2 ln cos 2 lnc x x c x x +
44. [ ] [ ]1 11 1sin 2 ln cos 2 lnc x x c x x− −+
Exercises 4.4 (page 186)
2. Yes
4. Yes
6. Yes
8. Yes
10. 10py = −
12. 23 12 24px t t= − +
14. ( )122 ln 2 1x
py− = +
16. sin cos
2
t t tpθ += −
18. 2 cos 2py t t= −
20. 22 cos 2 sin 2py t t t t= − +
22. 42 tpx t e=
24. 2 sin cospy x x x x= +
26. 2 sin cost tpy t e t te t− −= +
28. ( )4 3 2 tAt Bt Ct Dt E e+ + + +
30. sin cost tAe t Be t+
32. ( )6 5 4 3 2 3tt At Bt Ct Dt Et Ft e−+ + + + +
34. 4
te−−
36. sin
4
t−
Exercises 4.5 (page 192)
2. a. 1 1
sin 24 8 4
tt
− +
b. 1 3
sin 22 4 4
tt
− −
c. 11 11
3sin 24 8
tt− −
4. 1 2tc c e t−+ +
6. 2 3 21 2
x x xc e c e e x− −+ + +
8. 2 21 2 tanx xc e c e x−+ +
10. Yes
12. No
14. Yes
16. Yes
18. 3 21 2
4 1
3 9t t t
c e c e t−+ − + +
20. 1 2sin cos
cos 2 sin 23
c cθ θθ θ −+ +
22. 3 3 4 21 2cos sin 2x xc e x c e x x x− −+ + − +
24. 3 3t t− +
26. cos 3t + 2 sin 3t + 3
28. 4 2 31 7
60 12 10 6 6
t t t te e e e−+ − − +
30. 2 274 7
4 4 2
t t te e tet t
− −− + − − −
32. 2 2( ) tAt Bt C e+ +
34. A cos t + B sin t + C cos 2t + D sin 2t
36. 2 4 3 2( )te At Bt Ct+ +
34 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
38. 2 sin cost t− +
40. 23 2t t− +
42. a. ( ) sin cos ,hy t A t B t yβ β= + + where 2
2 2 2 2,
( )
k mA
k m b
ββ β−
=− +
2 2 2 2( )
bB
k m b
ββ β−=
− + and
2 2/ 2
1 24 4
cos sin2 2
bt mh
mk b mk by e c t c t
m m−
− − = +
b. In each case, 0hy → as .t →∞ So as ,t →∞ y(t) approaches the function ( ) sin cos .py t A t B tβ β= +
44. sin 3 cos 3 sin 2tt t e t−− +
46. If 2 1λ = the general solution is 1 2cos
sin cos ;2
t tc t c t− + + if 2 4,λ = 9, ... the general solution is
1 22
sinsin cos .
1
tc t c tλ λ
λ+ +
− In neither case is there a solution satisfying the boundary conditions.
Exercises 4.6 (page 197)
2. 1 2cos sin sin (cos ) ln cosc t c t t t t t+ + +
4. 1 2 2 4t tc e c e t−+ − −
6. 2
1 2 2
tt t t e
c e c te−
− −+ +
8. 1 2(sin 3 ) ln sec 3 tan 3 1
cos 3 sin 39
t t tc t c t
+ −+ +
10. 2 22 2
2 21 2
3ln
2 4
ttt t t et ec e c te t
−−− − + + −
12. 3
1 2cos sin 1 (cos ) ln sec tan10
tec t c t t t t+ + − − +
14. 1 2sec
cos sin sin tan2
c cθθ θ θ θ+ + −
16. 2 3 21 2
193 5
6t tc e c e t t− −+ + − +
18. 3
3 31 2 2
tt t e
c e c tet
+ +
22. 2 3 31 2
1ln
6c t c t t t+ + +
24. 2
1 2 ln4
tt t t e
c e c e t+ +
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
2
10
( )y
c t= >
− run away).
4. Suppose 31 22 2 2
0(3 ) (2 ) (1 )
aa a
t t t+ + ≡
− − − for –1 < t < 1. By taking limits as 1,t → we conclude 3 0.a =
Therefore 1 22 2
0,(3 ) (2 )
a a
t t+ ≡
− − so 2 2 2
1 2 1 2 1 2 1 2(2 ) (3 ) ( ) ( 4 6 ) 4 9 0.a t a t a a t a a t a a− + − = + + − − + + ≡
Thus 1 2 0a a+ = and 1 24 6 0a a− − = and 1 24 9 0,a a+ = which implies 1 2 0.a a= = Hence 1 2 3 0a a a= = =
and they are linearly independent.
6. 2 ,2
k d ky y y
m dy m ′′ = − = −
so 2 21( ) constant,
2 2
ky y
m′ + = or 2 2( ) 2 (constant).m y ky m′ + = ⋅
8. By eq. (21), sin cos .g gd
dθ θ θ
θ ′′ = − = A A
Therefore 2( )
cos constant.2
gθ θ′
− =A
10. At t = 0, (0) 0θ ′ = and ( )0 ,θ α= so 2 2( ) (0)
cos cos2 2
cos .
g g
g
θ θθ α
α
′ ′− = −
= −A A
A
Therefore, 2( )
cos ( ) cos cos ,2
tt
g
θθ α α′
= + ≥A and since cosθ is a decreasing function for
( )0 , .tθ π θ α≤ ≤ ≤
16. 2 4
3 ,2 4
y ydy y y
dy
′′ = − − = − −
so
2 2 4( ).
2 2 4
y y yK
′+ + =
Therefore, 2 2 4( )
,2 2 4
y y yK K
′≤ − − ≤ and hence 2 .y K≤
Exercises 4.8 (page 219)
2. 1 1
cos 5 sin 54 541
sin(5 ),20
y t t
t φ
= − −
= +
where 5
arctan 2.2464
φ = − π = − .
41 2amp.= , period ,
20 5
π= 5
freq. .2
=π
Passes through equilibrium at time ( )5
4arctan
0.449 sec.5
π−≈
36 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
4. b = 0: y(t) = cos 8t
Figure 11 (b = 0)
b = 10:
( )
5 5
5
5( ) cos 39 sin 39
398
sin 39 ,39
t t
t
y t e t e t
e t φ
− −
−
= +
= +
where 39
arctan 0.896.5
φ
= ≈
Figure 12 (b = 10)
EXERCISES 4.8 37
b = 16: 8( ) (1 8 ) ty t t e−= +
Figure 13 (b = 16)
b = 20: 4 164 1( )
3 3t ty t e e− − = −
Figure 14 (b = 20)
38 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
6. k = 2: ( ) ( )2 2 2 21 2 1 2
( )2 2
t ty t e e
− + − − + −= +
Figure 15 (k = 2)
k = 4: 2( ) (1 2 ) ty t t e−= +
Figure 16 (k = 4)
EXERCISES 4.9 39
k = 6:
( )2 2
2( ) cos 2 2 sin 2
3 sin 2 ,
t t
ty t e t e t
e t φ
− −−
= += +
where 2
arctan 0.6152
φ
= ≈ .
Figure 17 (k = 6)
8. Never 10. / 127 0.755 me−π ≈ 12. 0.080 sec 16. 18
kg7
Exercises 4.9 (page 227)
2. 2 2 2
1( )
(3 2 ) 9M γ
γ γ=
− +
Figure 18 Response curve for Problem 2
4. 5
( ) 1 sin2
y t t t = +
40 CHAPTER 4 LINEAR SECOND-ORDER EQUATIONS
8. 33 9( ) 3;
2 2t ty t e e− − = − +
lim ( ) 3,t
y t→∞
= which is the displacement of a simple spring with spring constant k
= 6 when a force 0 18F = is applied.
Figure 19
12. 5 / 4 5 / 4759 759 3( ) (0.047) cos (0.058) sin sin( ) m,
4 4 10 9241t tt t
y t e e t θ− − = + + +
where 96
arctan 1.519;5
θ = ≈
3671.078
4 2≈
π
Chapter 4 Review (page 230)
2. / 7 / 71 2
t tc e c te− −+
4. 5 / 3 5 / 31 2
t tc e c te+
6. ( ) ( )4 30 4 30
1 2t t
c e c e− + − −
+
8. 2 / 5 2 / 51 2
t tc e c te− −+
10. ( ) ( )1 2cos 11 sin 11c t c t+
12. 5 / 2 2 21 2 3
t t tc e c e c te− + +
14. ( ) ( )2 21 2cos 3 sin 3t tc e t c e t+
CHAPTER 4 REVIEW 41
16. ( ) ( )1 2 3cos 2 sin 2te c c t c t− + +
18. 3 2 51 2 3 4c t c t c t c t+ + + +
20. / 2 / 21 2
4 1cos sin
9 3t tc e c e t t t− + − −
22. 11 31 2
1092 4641cos sin
305 305t tc e c e t t− + + −
24. / 2 3 / 5 / 21 2
1 1 1
3 9 11t t tc e c e t te− + − − −
26. ( ) ( )3 3 21 2
1cos 6 sin 6 5
31t t tc e t c e t e− − + + +
28. 3 31 2cos(2 ln ) sin(2 ln )c x x c x x+
30. 3 2 sine eθ θθ θ− −+ +
32. / 2 / 2cos 6 sint te t e t−
34. 7 24t te e− +
36. 2 / 3 2 / 33 t te te− −− +
38. 1 1
( ) cos 5 sin 5 ,4 5
y t t t= − + amp. 0.320 m,≈ 2
period ,5
π= 5
freq. ,2
=π
equil1 5
arctan 0.179 sec.5 4
t = ≈
43
CHAPTER 5
Exercises 5.2 (page 250)
2. 2 3 2 31 2 1 2
3;
2t t t tx c e c e y c e c e− − = − = +
4. 31 2
1 1;
2 2t tx c e c e−
= − + 3
1 2t ty c e c e−= +
6. 1 2 2 1 7 1cos 2 sin 2 cos sin ;
2 2 10 10t tc c c c
x e t e t t t+ − = + + −
1 211 7
cos 2 sin 2 cos sin10 10
t ty c e t c e t t t = + + +
8. 1115 4 26;
4 11 121tc
x e t = − − −
111
1 45
11 121ty c e t
= + +
10. 1 2cos sin ;x c t c t= + 2 1 1 23 3 1 1cos sin
2 2 2 2t tc c c c
y t t e e−− + = − + −
12. 2 21 2 1;t tu c e c e−= + + 2 2
1 2 32 2 2t tv c e c e t c−= − + + +
14. 1 2sin cos 2 1;x c t c t t= − + + − 21 2cos sin 2y c t c t t= + + −
16. 2 41 2
2;
9t t tx c e c e e− = + +
2 41 2 3
1 12
2 36t t ty c e c e c e− = − − + −
18. 23 4 1 2
1( ) 4 3 ;
2t t tx t t t c c e c te c e−
= − − − + + − − 2
1 2 3( ) 2 t ty t t t c e c e c−= − − − + +
20. 33 1( ) ,
2 2t tx t e e= − 3 23 1
( )2 2
t t ty t e e e= − − +
22. 39 5( ) 1 ,
4 4t tx t e e−= + − 33 5
( ) 12 2
t ty t e e−= + −
24. No solutions
26. 21 2 1 2 3
1( ) cos ( ) sin ;
2t tx e c c t c c t c e
= − + + + 1 2( cos sin );ty e c t c t= +
21 2 1 2 3
3( ) cos ( ) sin
2t tz e c c t c c t c e
= − + + +
28. 4 31 2 3( ) 2 ,tx t c c e c e t−= + + 4 3
1 2 3( ) 6 2 3 ,t ty t c c e c e−= − + 4 31 2 3( ) 13 2t tz t c c e c e−= − − −
30. λ
44 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
32. ( ) ( )7 19 /100 7 19 /10020 10 19 20 10 19
20;19 19
t tx e e
− + − −− − −= + +
( ) ( )7 19 /100 7 19 /10050 5020
19 19
t ty e e
− + − −−= + +
34. b. 1 2 1 2 13 1 1 3
cos 3 sin 3 5 L;2 2 2 2
t tV c c e t c c e t− − = − − + + 2 1 2cos 3 sin 3 18 Lt tV c e t c e t− −= + +
c. As t → + 1 5 LV → and 2 18 L.V →
36. 400
36.4 F11
≈ °
38. A runaway arms race.
40. a. 2 33x x= b. 2 36 3 2x x x+ − c. 2 33 2x x+ d. 2 36 3 2x x x+ − e. 2 2D D+ − f. 2 36 3 2x x x+ −
Exercises 5.3 (page 261)
2. 21 2 2 1 1, cos( );x x x x t x′ ′= = + − 1(0) 1,x = 2 (0) 0x =
4. 1 2 2 3 3 4 4 5 5 6, , , , ,x x x x x x x x x x′ ′ ′ ′ ′= = = = = 2 26 2 1( ) sin ;tx x x e′ = − + 1 6(0) (0) 0x x= = ="
6. Setting 1 2 3, , ,x x x x x y′= = = 4 ,x y ′= we obtain 1 2 ,x x′ = 2 1 35 2
,3 3
x x x′ = − + 3 4 ,x x′ = 4 1 33 1
2 2x x x′ = − .
8. 1 2(0) , (0) (0)x a x p b= =
10. i it ( )iy t
1 0.250 0.96924
2 0.500 0.88251
3 0.750 0.75486
4 1.000 0.60656
12. i it ( )iy t
1 1.250 0.80761
2 1.500 0.71351
3 1.750 0.69724
4 2.000 0.74357
14. (8) 24.01531y ≈
16. x(1) ≈ 127.773; y(1) ≈ –423.476
18. Conventional troops
EXERCISES 5.3 45
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
28. a. 1 2x x′ = 1(0) 1x =
12 2 2 3/ 2
1 3( )
xx
x x
−′ =+
2 (0) 0x =
3 4x x′ = 3 (0) 0x =
34 2 2 3/ 2
1 3( )
xx
x x
−′ =+
4 (0) 1x =
b. i it 1( ) ( )i ix t x t≈ 3 ( ) ( )i ix t y t≈
10 0.628 0.80902 0.58778
20 1.257 0.30902 0.95106
30 1.885 –0.30902 0.95106
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
30. a. it il iθ
1.0 5.27015 0.0
2.0 4.79193 0.0
3.0 4.50500 0.0
4.0 4.67318 0.0
5.0 5.14183 0.0
6.0 5.48008 0.0
7.0 5.37695 0.0
8.0 4.92725 0.0
9.0 4.54444 0.0
10.0 4.58046 0.0
b. it il iθ
1.0 5.13916 0.45454
2.0 4.10579 0.28930
3.0 2.89358 –0.10506
4.0 2.11863 –0.83585
5.0 2.13296 –1.51111
6.0 3.18065 –1.64163
7.0 5.10863 –1.49843
8.0 6.94525 –1.29488
9.0 7.76451 –1.04062
10.0 7.68681 –0.69607
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. 3 2 2x x y y c−− − =
10. Critical points are (1, 0) and (–1, 0). Integral curves:
for y > 0, 21, 1;x y c x> = −
for y > 0, 21, 1 ;x y c x< = −
for y < 0, 21, 1;x y c x> = − −
for y < 0, 21, 1 ;x y c x< = − −
all with 0.c ≥
If c = 1, 21y x= ± − are semicircles ending at
(1, 0) and (–1, 0).
EXERCISES 5.4 47
12. 2 29 4x y c+ =
Figure 21
14. 2 / 3y cx=
Figure 22
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 49
20. y vv y′ =′ = − ; (0, 0) is a center.
Figure 25
22. 3 ;y vv y′ =
′ = − (0, 0) is a center.
Figure 26
50 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
24. 3 ;y vv y y′ =
′ = − +
(0, 0) is a center. (–1, 0) is a saddle point. (1, 0) is a saddle point.
Figure 27
26. 22 4
;2 4 2
yx xc+ + = all solutions are bounded.
Figure 28
28. (0, 0) is a center. (1, 0) is a saddle.
EXERCISES 5.4 51
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′ ≡ →
b. ( ) ( ) ( )( *, *)
( ) ( )2
( *, *)2
t
Tx t x d x T
f x yt T x T
tf x y C
τ τ′= +
> − +
≡ +
∫
c. If f(x*, y*) > 0, f(x*, y*)t→ LPSO\LQJ x(t)→
d, e. similar
32. 2 4( )
2 4
y yc
′+ = by Problem 30.
Thus, 4 2( )
,4 2
y yc c
′= − ≤ so 4 4 .y c≤
34. a. Peak will occur when 0,dI
dt= that is when aSI – bI = 0 or
bS
a= provided (0)
bS
a >
so that can in fact attain the value .b
Sa
b. ( ) 0I t′ > if I > 0 and ,b
Sa
> while ( ) 0I t′ < if I > 0 and .b
Sa
<
c. It steadily decreases to zero.
36.
Figure 29
52 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
38. a. 2 2 2[ ] 0;d
x y zdt
+ + = the magnitude of the angular velocity is constant.
b. All points on the axes are critical points: (x, 0, 0), (0, y, 0), (0, 0, z).
c. From (a), 2 2 2x y z K+ + = (sphere). Also 2
,dy x
dx y
−= so 2
2
2
yx c+ = (cylinder).
d. The solutions are periodic.
e. The critical point on the y-axis is unstable. The other two are stable.
Exercises 5.5 (page 284)
2. ( ) ( )1 210
( ) 1 10 cos 1 10 cos ,20
x t r t r t = − − + 1 23 10
( ) [cos cos ],20
y t r t r t= − where 1 4 10 ,r = +
2 4 10.r = −
4. 1 1 2 ( ),m x k x k y x′′ = − + − 2 2 ( )m y k y x by′′ ′= − − −
6. b. 1 2 3 437
( ) cos sin cos 2 sin 2 cos 340
x t c t c t c t c t t = + + + +
c. 1 2 3 4111
( ) 2 cos 2 sin cos 2 sin 2 cos 320
y t c t c t c t c t t = + − − −
d. 23 9 37
( ) cos cos 2 cos 3 ,8 5 40
x t t t t = − +
23 9 111
( ) cos cos 2 cos 34 5 20
y t t t t = + −
8. 19.8 9.8
( ) cos cos ;12 125 10 5 10
t t tθ π π= + + −
210 9.8 10 9.8
( ) cos cos24 245 10 5 10
t t tθ π π= − + −
Exercises 5.6 (page 291)
2. 41( ) cos(6 ) 3 cos(2 ) sin(2 ) coulombs
2tq t e t t t− = + +
4. 10 10
( ) cos 5 cos 50 amps33 33
I t t t = −
8. L = 0.01 henrys, R = 0.2 ohms, 25
farads,32
C = and 2
( ) cos 8 volts5
E t t =
10. 2 2 / 31
1 9 5;
4 4 2t tI e e− − = − − +
2 2 / 32
1 3 1;
4 4 2t tI e e− − = − +
2 2 / 33
1 32
2 2t tI e e− − = − − +
12. 9001 1 ;tI e−= − 900
25 5
;9 9
tI e− = −
9003
4 4
9 9tI e−
= −
CHAPTER 5 REVIEW 53
Exercises 5.7 (page 301)
2. 0 0( , ) ( 1.5, 0.5774)x v = −
1 1( , ) ( 1.9671, 0.5105)x v = − −
2 2( , ) ( 0.6740, 0.3254)x v = −
#
20 20( , ) ( 1.7911, 0.5524)x v = − −
The limit set is the ellipse 2 2( 1.5) 3 1x v+ + = .
4. 0 0( , ) (0, 10.9987)x v =
1 1( , ) ( 0.00574, 10.7298)x v = −
2 2( , ) ( 0.00838, 10.5332)x v = −
#
20 20( , ) ( 0.00029, 10.0019)x v = −
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
54 CHAPTER 5 INTRODUCTION TO SYSTEMS AND PHASE PLANE ANALYSIS
Chapter 5 Review Problems (page 304)
2. 1 2 1 2( ) cos 2 ( ) sin 2t tx c c e t c c e t− −= − + + −
1 22 cos 2 2 sin 2t ty c e t c e t− −= +
4. 1 2 ,tx c t c e−= + + 3 21 23 46 2
c cy t t c t c= + + +
6. 2 2, ,t t t tx e e y e e− −= + = + 2 2t tz e e−= −
8. With 1 2, ,x y x y ′= = we obtain 1 2 2 1 21
, (sin 8 ).2
x x x t x tx′ ′= = − +
10. With 1 2 3 4, , , ,x x x x x y x y′ ′= = = = we get 1 2 2 1 3 3 4, , ,x x x x x x x′ ′ ′= = − = 4 2 3.x x x′ = − +
12. Solutions to phase plane equation 2
2
dy x
dx y
−=−
are given implicitly by 2 2( 2) ( 2) const.x y− + − = Critical point
is at (2, 2), which is a stable center.
Figure 31
14. Critical points are ( ), ,m nπ π m, n integers and (2 1) (2 1)
, ,2 2
j k+ π + π
j, k integers. Equation for integral
curves is cos sin tan
,sin cos tan
dy x y y
dx x y x= = with solutions sin y = C sin x.
CHAPTER 5 REVIEW 55
16. Origin is saddle (unstable).
Figure 32
18. Natural angular frequencies are 2, 2 3.
General solution is ( ) ( ) ( ) ( )1 2 3 4( ) cos 2 3 sin 2 3 cos 2 sin 2x t c t c t c t c t= + + + ,
( ) ( ) ( ) ( )21 3 4
1( ) cos 2 3 sin 2 3 3 cos 2 3 sin 2 .
3 3
cy t c t t c t c t= − − + +
57
CHAPTER 6
Exercises 6.1 (page 324)
2. ( )0, ∞
4. (–1, 0)
6. (0, 1)
8. Lin. dep.; 0
10. Lin. ind.; 3 22 tan sin cos sin tan 2 tanx x x x x x− − − −
12. Lin. dep.; 0
14. Lin. indep.; ( 2) xx e+
16. 1 2 3cos 2 sin 2xc e c x c x+ +
18. 1 2 3 4cos sinx xc e c e c x c x−+ + +
20. a. 3 21 2 3c c x c x x+ + +
b. 3 22 x x− +
22. a. 1 2 3 4cos sin cos sinx x x xc e x c e x c e x c e x− −+ + +
b. cos cosxe x x+
24. a. 7 cos 2x + 1
b. 11
6 cos 23
x− −
26. 1
11
( ) ( )n
j jj
y x y xγ+
−=
= ∑
34. The Wronskian 1 2 3[ , , ]( )W f f f x
Exercises 6.2 (page 331)
2. 31 2 3
x x xc e c e c e−+ +
4. 5 41 2 3
x x xc e c e c e− −+ +
6. 1 2 3cos sinx x xc e c e x c e x− + +
58 CHAPTER 6 THEORY OF HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS
8. 71 2 3
x x xc e c xe c e−+ +
10. ( ) ( )1 7 1 7
1 2 3x xxc e c e c e
− + − −− + +
12. 3 31 2 3
x x xc e c e c xe− −+ +
14. 1 2 3 4sin cos x xc x c x c e c xe− −+ + +
16. 2 6 51 2 3 4 5 6 7 8 9 10( ) ( ) cos sin sin 2 cos 2x x xc c x e c c x c x e c e c x c x c x c x− −+ + + + + + + + +
18. 2 2 / 2 / 2 2 31 2 3 4 5 6 7 8 9
2 310 11 12
3 3( ) cos sin ( ) cos
2 2
( ) sin
x x x x x
x
x xc c x c x e c e c e c e c c x c x e x
c c x c x e x
− − −
−
+ + + + + + + +
+ + +
20. 2 4x x xe e e− − −− +
22. 3 31 2 3 4( ) cos 2 sin 2t tx t c e c e c t c t−= + + +
3 31 2 3 4
2 2( ) cos 2 sin 2
5 5t ty t c e c e c t c t− = − − + +
28. 1.879 1.532 0.3471 2 3
x x xc e c e c e− −+ +
34. 5 10 5 10
( ) cos 4 10 cos 4 10 ,10 10
x t t t − += + + −
5 2 10 5 2 10( ) cos 4 10 cos 4 10
10 10y t t t
− += + + −
Exercises 6.3 (page 337)
2. 1 2 3cos sinxc e c x c x− + +
4. 21 2 3
x xc c xe c x e+ +
6. 31 2 3
1 3 1cos sin
8 20 20x x x xc e c xe c e e x x− − + + + + +
8. 21 2 3
1 4 1cos sin
2 25 10x x x x xc e c e x c e x xe x e− − + + − − +
EXERCISES 6.3 59
10. 2 3 3 3 31 2 3
5 1 1 1cos 2 sin 2 cos 2 sin 2
116 58 26 676x x x x xc e c e x c e x xe x xe x x− − − − + + + − − −
12. 3D
14. D – 5
16. 3 ( 1)D D −
18. 2 2[( 3) 25]D − +
20. 4 3 2 2( 1) ( 16)D D D− +
22. 33 4 5cos sinxc e c x c x+ +
24. 23 4
x xc xe c x e+
26. 23 4 5c x c x c+ +
28. 33 4 5
xc e c c x+ +
30. 4 5xc xe c+
32. 35 sin 2x xe x e+ −
38. / 2 / 21 2 3 2 3
1 1 3 7 7 7 3 7( ) cos sin 1
2 4 2 2 2 2 2 2t t tt t
x t c t e c c e c c e t− − = + + + − − + − + +
/ 2 / 21 2 3
1 7 7 1( ) cos sin
4 2 2 2t t tt t
y t c t e c e c e− − = + + + +
40. 12187 3
( ) sin sin 18sin40 8 40 72 24
t t tI t
= − −
2243 27
( ) sin sin40 8 40 72
t tI t
= −
3243 3
( ) sin sin 18sin5 8 5 72 24
t t tI t
= + −
60 CHAPTER 6 THEORY OF HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS
Exercises 6.4 (page 341)
2. 212
2x x
+
4. 31
6xx e
6. sec θ – sin θ tan θ + θ sin θ + (cos θ)ln(cos θ)
8. 3 21 2 3lnc x c x x c x x+ + −
10. 1 4
5 2( ) ( ) ( )10 15 6
x x xg x dx x g x dx x g x dx
−− − + −
∫ ∫ ∫
Chapter 6 Review Problems (page 344)
2. a. Lin. indep.
b. Lin. indep.
c. Lin. dep.
4. a. 31 2 3 4
x x x xc e c e c e c xe− −+ + +
b. ( ) ( )2 5 2 5
1 2 3x xxc e c e c e
− + − −+ +
c. 1 2 3 4 5cos sin cos sinxc e c x c x c x x c x x+ + + +
d. 21 2 3
1
2 2 4x x x xx x
c e c e c e e− + + − + +
6. / 2 / 2 / 2 / 2 21 2 3 4cos sin cos sin sin( )
2 2 2 2x x x xx x x x
c e c e c e c e x− − + + + +
8. a. 23 4 5 6
xc xe c c x c x− + + +
b. 24 5
x xc xe c x e− −+
c. 25 6 7 8 9cos 3 sin 3c c x c x c x c x+ + + +
d. 4 5 6 7cos sin cos sinc x c x c x x c x x+ + +
10. a. 1/ 2 1/ 21 2 3c x c x c x−+ +
b. ( ) ( )11 2 3cos 3 ln sin 3 lnc x c x x c x x− + +
61
CHAPTER 7
Exercises 7.2 (page 359)
2. 3
2, 0s
s>
4. 2
1,
( 3)s − s > 3
6. 2 2
,s
s b+ s > 0
8. 2
2,
( 1) 4s + + s > –1
10. 2
1 1,
se
s s
− −+ s > 0
12. 6 3 31
,2
s se e
s s
− −− +−
s > 2
14. 3
5 1 12,
2s s s− +
− s > 2
16. 3 2 2
2 3 6,
( 1) 9s s s− −
+ + s > 0
18. 5 3 2 2
24 6 1 2,
2s s s s− − +
+ s > 0
20. 2 3
2 2,
( 2) 3 ( 2)
s
s s
+ −+ + +
s > –2
22. Piecewise continuous
24. Piecewise continuous
26. Neither
28. Continuous
30. lim ( ) 0s
F s→∞
=
Exercises 7.3 (page 0)
2. 3
6 1
2ss−
−
4. 5 3
72 4 1
ss s− +
6. 2 3
2 2
( 2) 4 ( 3)s s+
+ + −
8. 1 2 1
1 2s s s+ +
+ +
10. 2
2 2
( 2) 25
[( 2) 25]
s
s
− −
− +
12. 2
3
36s +
14. 2
2
( 7)[( 7) 4]s s− − +
16. 2
2 2 2
3 4
( 4)
s
s s
+
+
18. 2 2 2 22[ ( ) ] 2[ ( ) ]
s s
s n m s n m+
+ + + −
20. 2 2 2 2
2 2 2 2
20( 9)( 49) 40 (2 58)
( 9) ( 49)
s s s s
s s
+ + − +
+ +
30. 2
1( )
5 6H s
s s=
+ +
32. 2se
s
−
34. 2 1
se
s
−π
+
62 CHAPTER 7 LAPLACE TRANSFORMS
Exercises 7.4 (page 374)
2. sin 2t
4. 4
sin 33
t
6. 5 / 2 23
16te t−
8. 41
24t
10. / 4 / 41 47 5 47cos sin
2 4 42 47
t tt te e− − −
12. 2 3
1 2s s−
+ −
14. 1 2 11
1 1 2s s s+ −
+ − −
16. 2 2
2 32 3
2 9 9
s
s s s− + −
+ + +
18. 2 3
1 2 3 1
1s ss s+ + −
+
20. 2
1 1 1 1 1 1
4 1 2 4 1( 1)s ss+ −
− +−
22. 33 2t te e−−
24. 2 25 2 cos 3 5 sin 3t t te e t e t+ −
26. 2 231 6
2tt e
− +
28. 2 32 5 4 13
3 6 15 30
t t te e e− −− + + −
30. 7 3 7
4 2 4
t t te te e− −− − +
32. 1 2 31
( ) ( ) ( ) ;1
F s F s F ss
= = =−
13
1( )
1te f t
s− = = − $
34. 3 4t te e
t
−
36. sin t
t
Exercises 7.5 (page 383)
2. 2 3t te e−−
4. 5t t te e e− −− −
6. 3 22 sint te e t+
8. 22 2 cos 2 2 sin 2t t t t− + − +
10. 2 2 22t t tt e te e− −− − + +
12. 1 110 6 2t tt e te+ ++ + +
14. cos sint t tπ+ +
16. 3 2
3 2
2
( 6)
s s
s s
− − +
+
18. 3 2
2
2 3
( 1)( 2)( 2 1)
s s s
s s s s
− − +
− − − −
20. 5
6
( 3)s s +
22. 3 2
3
2 17 28 12
( 1) ( 5)
s s s
s s
− + −
− −
24. 3 2 3 3
2
2 2
( 1)( 1)
s ss s s se e
s s s
− −+ + + +
− +
26. 2 32 3t t te e e− −+ − +
EXERCISES 7.6 63
28. 2 cos 2t t te e e t− −− + +
30. 56 5 1 1
25 5 4 4 4 4 4 100t tt a b a b
e e− − − + + + + − + +
32. 2 2 2 2 3(3 6) 6 3 ( 2 6)t t t ta b e te t e b a e− + + + + − −
36. 2 – t
38. 3t
40. 2 2
/ 2
2
4 4cos sin
2 24
t I Ik t Ik tae
I IIk
µ µ µµ
µ
− − − + −
Exercises 7.6 (page 395)
2. 4s se e
s
− −−
4. 2
( 1)se s
s
− +
6. (t + 1)u(t – 2); 2
2
(3 1)se s
s
− +
8. (sin )2
t u tπ −
; – / 2
2 1
se s
s
π
+
10. 23
2( 1) ( 1);
set u t
s
−− −
12. (t – 3)u(t – 3)
14. sin 3( 3)
( 3)3
tu t
− −
16. 1
[sin 2( 1)] ( 1)2
t u t − −
18. 1[cos( 1) 2 ] ( 1)tt e u t−− + −
20. 1
sin sin 2 cos 2 sin 2 sin ( 2 )2
t t t t t u t + + + − − π
64 CHAPTER 7 LAPLACE TRANSFORMS
22. ( 1)1
( 1)(1 )
s
s
e
s e
− −
−−
− −
Figure 33
24. 2 2
2 2
2 1
(1 )
s s s s
s
se e e se
s e
− − − −
−+ − − +
−
Figure 34
26. 2
1
(1 )
as as
as
e ase
as e
− −
−− −
−
28. 2
1
( 1)(1 )ss e−π+ −
EXERCISES 7.6 65
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) ( )2u t π− + (sin 2t) ( )2u t π−
Figure 36
34. 2( ) 2( ) ( 2 ) 2( 2 )1 1( ) [1 2( ) ] ( ) [1 2( 2 ) ] ( 2 )
4 4t t t ty t e t e u t e t e u t− −π − −π − − π − − π = − − − π − π − − − − π − π
36. 2 3 2( 2) 3( 2)7 1 3 5( 2) ( 2)
16 6 4 9t t t te e t e e u t− − − − − − − + + − − + −
38. ( 10) ( 10)
( 20) ( 20)
2 11 2 cos 3 sin 3 1 cos[3( 10)] sin[3( 10)] ( 10)
3 32
2 2 cos[3( 20)] sin[3( 20)] ( 20)3
t t t t
t t
e t e t e t e t u t
e t e t u t
− − − − − −
− − − −
− − + − − − − − − − − − − −
66 CHAPTER 7 LAPLACE TRANSFORMS
40. 3 ( 3) 3 ( 3) 3 2( 3)2 [(1 2 ) ( 3) (2 1) ] ( 3)t t t t te te e e e t e e e u t− − − − − − − − − − −+ + − + − + − −
42. c.
Figure 37
46. h(t) – h(t – 1)u(t – 1), where 2( 1) 2 4( 1)
2 4
( 1) ( 1)1( ) ,
2 1 2( 1)
t n t ne e e enh t
e e
− + − +− −+= − +− −
for 2n < t < 2(n + 1).
48. 12
1
1( 1)
nn
n s
∞+
=
−
∑
50. 2 1
0
( 1) (2 )!
!
k
kk
k
k s
∞
+=
−∑
60. 3 ( 3) 7
( 7)1 ( 3) 1 ( 7)2 2 2
t t tt te e e
u t e u t e− − − −
− − − + − − − + − − +
62. The concentration of salt:
3 /125
3 /125 ( 3 /125)( 10)
3 /125 ( 3 /125)( 10) ( 3 /125)( 20)
0.6 0.4 0 10
0.4 0.4 0.2 10 20
0.6 0.4 0.2 2 20
t
t t
t t t
e t
e e t
e e e t
−
− − −
− − − − −
− ≤ ≤ − + ≤ ≤ − + − ≥
Exercises 7.7 (page 405)
2. 0
1cos 3 sin[3( )] ( )
3
tt t v g v dv
+ − ∫
4. 0
( ) sin( ) sintg v t v dv t− +∫
6. 2t te e− −−
8. 1
sin 2 cos 216 8
tt t
−
10. 2
cos 12
tt − +
12. 1
( sin sin cos )2
t t t t t + −
14. 2 1 1( 1) ( 1)s s− −+ −
16. cos t + sin t – 1
18. 2 – 2 cos t
EXERCISES 7.8 67
20. / 2 / 21 1 3 1 3cos sin
3 3 2 23t t tt t
e e e− − +
22. 23 32
2 4 4t tt
e e− + + −
24. 3 3
2
1( ) ; ( ) ;
69
t te eH s h t
s
−−= =−
3 3( ) ;t tky t e e−= + 3( ) 3( ) 3 3
0
1( ) [ ] ( )
6
t t v t v t ty t e e g v dv e e− − − −= − + +∫
26. 3 51
( ) ; ( ) ;( 3)( 5) 8
t te eH s h t
s s
−−= =− +
3 5( ) ;t tky t e e−= − 3( ) 5( ) 3 5
0
1( ) [ ] ( )
8
t t v t v t ty t e e g v dv e e− − − −= − + −∫
28. 22
1( ) ; ( ) sin ;
( 2) 1
tH s h t e ts
= =− +
2( ) sin ;tky t e t= 2( ) 2
0( ) (sin( )) ( ) sin
t t v ty t e t v g v dv e t−= − +∫
30. 4( ) 40
1(sin 5( )) ( ) 2 cos 5
50
t t v te t v e v dv e t− − −− +∫
32. 5
60
t
38. d. 87.403 100 3 5 100 3 5 228.825b≈ − ≤ ≤ + ≈
Exercises 7.8 (page 412)
2. 1
4. 2e
6. 1
8. 3 se−
10. 327 se−
12. 3 3se −
14. ( )cos 2 sin (sin ) ( )t te t t e t u t− − −π+ − − π
16. 3 3( 1) ( 3) 3( 3)( 1)1 1[ ] ( 1) [ ] ( 3)
2 4t t t t tte e e e u t e e u t− − − − −− − + + − − + − −
18. 2 2 2 14 1 5( ) ( 1)
3 2 6t t t t te e e e e u t− − − − − + − −
20. 2 3 2( 1) 3( 4 / 3) ( 2)t t t te e e e u t− − − − − − + + − −
68 CHAPTER 7 LAPLACE TRANSFORMS
22. sin (cos )2
t t u tπ − −
Figure 38
24. ( ) ( ) ( ) ( )sin sin sin 2t t u t t u tπ π− − − −
Figure 39
26. 31sin 2
2te t
28. ( )
2
t te e−−
30. 1
( ) (sin ) ( 2 )k
y t t u t k∞
== − π∑
Notice that the magnitude of the oscillations of y(t) becomes unbounded as .t →∞
EXERCISES 7.9 69
Exercises 7.9 (page 416)
2. 3 2 3 22 ; 2 2t t t tx e e y e e= − = − +
4. 7 2 7 1
cos 2 sin 2 cos sin ;10 5 10 10
t tx e t e t t t = − + + −
11 3 11 7cos 2 sin 2 cos sin
10 10 10 10t ty e t e t t t
= − − + +
6. 2 11
2tx e t
= − + + ; 2 1 3
2 2ty e t
= − − −
8. 2 2 2 23 1 3; 3
2 2 4t t t tx e e y e e− − = − + = +
10. cos 1; cos 1t tx e t y e t= + − = − + +
12. 2 2 6 31 1 2 1 1 1 1( 3) ( 3)
2 6 3 4 2 12 3t t t tx e e t e e u t− − − + = − − + + + − + − −
;
2 2 6 31 1 4 1 1 1 2( 3) ( 3)
2 6 3 4 2 12 3t t t ty e e t e e u t− − − + = − + + + − + − − − −
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. 3 17 3 17
2 2( ) ( )13 17 13 17
( ) cos sin ;2 17 2 17
t tx t t t e e
+ −−π −π+ −= + + −
3 17 3 172 2
( ) ( )12 2 17 12 2 17( ) cos
17 17
t ty t t e e
+ −−π −π− + += + +
18. 2 2( ) ; ( ) ; ( ) 2x t t y t t z t t= = = −
20. 2 2 1;t tx e e−= + + 2 2 2 22 2 2 2 2 2t ty e e t e e− −= − + + − +
22. See answer to Exercises 5.5, Problem 1 on pages B-14 of text.
24. 1000 40001
15 20 5
2 3 6t tI e e− − = − −
; 1000 40002
10 55
3 3t tI e e− − = − −
;
1000 40003
5 10 5
2 3 6t tI e e− − = − +
70 CHAPTER 7 LAPLACE TRANSFORMS
Chapter 7 Review Problems (page 418)
2. 5( 1) 51
1
s se e
s s
− + −− −+
4. 2
4
( 3) 16s − +
6. 2 2
7( 2) 70
( 2) 9 ( 7) 25
s
s s
− −− + − +
8. 3 2 1 12 6 ( 2) 6( 1)s s s s− − − −+ − − − −
10. / 2
2 21 (1 )(1 )
s
s
s e
s s e
−π
−π+
+ + −
12. ( ) ( )2 232 cos 2 sin 2
2t te t e t
+
14. 3 23 3t t te e e− −− +
16. sin 3 3 cos 3
54
t t t−
18. 2
0
( 1)( ) ;
!
n n
n
tf t
n
∞
=
−= ∑ 2 1
0
( 1) (2 )! 1( )
!
n
nn
nF s
n s
∞
+=
−= ∑
20. 3( 3) tt e−−
22. 210 23 15cos 3 sin 3
13 13 13te t t
− +
24. 2 2 26 4 2 6t t t t te te t e te e+ + + −
26. 3(1 ) ,tCt e−+ where C is an arbitrary constant
28. / 2 / 23 7 1 7 1sin cos
2 2 2 22 7t tt t
e e− − − −
30. 1
sin 2 sin 22 2 2
t t u t π π + − −
32. 2 2 6 31 1 4 1 1 1 2( 3) ( 3)
2 6 3 4 2 12 3t t t tx e e t e e u t− − − + = − + + + − + − − − −
2 2 6 31 1 2 1 1 1 1( 3) ( 3)
2 6 3 4 2 12 3t t t ty e e t e e u t− − − + = − − + + + − + − −
71
CHAPTER 8
Exercises 8.1...(page 430)
2. 22 4 8x x+ + + "
4. 2 3 51 1 1
2 6 20x x x
+ − + "
6. 3 51 1
6 120x x x
− + + "
8. 2 4(sin1) (cos1)(sin1)
12 24
x x− + +"
10. a. 2 3
31
( )2 4 8 16
x x xp x = + + +
b.
( )4
3 5 5 432
5
1 24 1 1 1
4! 2(2 ) 2
2
30.00823
εξ
= ≤ −
=
≈
c. 32 1 1
0.002603 2 384
p − = ≈
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 ′′′ (iv) (v), ,y y etc. all exist.
72 CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
14. a. 2 31 1
2 6t t t
+ − + "
b. For r = 1, 2 41( ) 1 4
2y t t t
= − − + " .
For r = –1, 2 41 49( ) 1
2 12y t t t
= + − + " .
c. For small t in part (b), the hard spring recoils but the soft spring extends.
16. 2 4
12 4!
t t− + +"
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. 23
10
( 2)2( 1)
( 3)! 2
nn
nn
nx
n
+∞
+=
++ −
+ ∑
12. 3 52 2
3 15x x x
− + + "
14. 1
16. 2 31 1 11
2 4 24x x x
− + − + "
EXERCISES 8.3 73
18. 2
0
( 1)cos
(2 )!
kk
k
x xk
∞
=
− =∑
20. 2
2
( 1) nn
n
n n a x∞
−
=−∑
22. 2 1
0
( 1)
(2 1)!(2 1)
k k
k
x
k k
+∞
=
−+ +∑
24. 24
( 2)( 3) kk
k
k k a x∞
−=
− −∑
26. 3
4
k k
k
ax
k
∞−
=∑
30. 0
( 1) ( 1)n n
n
x∞
=− −∑
32. 1
1
( 1)nn
n
xn
+∞
=
−∑
34. 2 3 41 1 1 51 ( 1) ( 1) ( 1) ( 1)
2 8 16 128x x x x
+ − − − + − − − + "
36. a. Always true
b. Sometimes false
c. Always true
d. Always true
38. 2 2 2 4 6
0
( 1) 1 1–
1 2 3
nn
n
x x x xn
∞+
=
− = − ++∑ "
Exercises 8.3 (page 449)
2. 0
4. –1, 0
6. –1
8. No singular points
10. 1x ≤ and x = 2
74 CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
12. 2 30
00
0
1 11
2 3!
!
n
nx
y a x x x
xa
n
a e
∞
=
= + + + +
=
=
∑
"
14. 2 3
0 112 6
x xa a x
− + + − +
" "
16. 2 3 3
20 11
2 3 2
x x xa a x x
− − + + + + +
" " 0 1 0( )x xa e a a xe= + −
18. 2 30 1
1 11
6 27a x a x x
+ + + + + " "
20. 2 12
0 10 0
( 1)( 1)
(2 )! (2 1)!
k kkk
k k
xxa a
k k
+∞ ∞
= =
−− ++∑ ∑ 0 1cos sina x a x= +
22. 3 2 0,ka + = k = 0, 1, …
3 3 10 1
1 1
1 4 (3 5)(3 2) 2 5 (3 4)(3 1)1
(3 )! (3 1)!k k
k k
k k k ka x a x x
k k
∞ ∞+
= =
⋅ − − ⋅ − −+ + + + ∑ ∑" "
24. 2 2 2
2 4 20 1
3
[( 1) (2 3) (2 5) 3 ]1 11
2 24 (2 )!
kk
k
k ka x x x a x
k
∞
=
− − − − + + + ∑ "
26. 2 3 41 1 1
2 2 3x x x x
+ + + + "
28. 2 3 41 1 11
2 3 8x x x
− − − − + "
30. 2 35 11
2 6x x x
− + + −
32. a. If 1 0,a = then y(x) is an even function.
d. 0 10, 0a a= >
36. 2 3 493
2t t t
− + + + "
EXERCISES 8.4 75
Exercises 8.4 (page 456)
2. infinite
4. 2
6. 1
8. 2 30
101 2( 1) 3( 1) ( 1)
3a x x x
− + + + − + + "
10. 2 3 2 30 1
1 1 1 11 ( 2) ( 2) ( 2) ( 2) ( 2)
4 24 4 12a x x a x x x
− − − − + + − + − − − + " "
12. 2 3 2 30 1
1 2 71 ( 1) ( 1) ( 1) 2( 1) ( 1)
2 3 3a x x a x x x
+ + + + + + + + + + + + " "
14. 2 351
6x x x
+ + + + "
16. 3 4 51 1 1
3 12 24t t t t
− + + + + "
18. 2 41 1
12 2 2 24 2
x x xπ π π + − + − − − +
"
22. 2 2 30
1 1 11
2 2 6a x x x x
− + + + − + " "
24. 2 3 20 1
3 1 21
2 6 3a x a x x x
− + + − + + − " "
26. 2 3 20 1
1 1[1 ]
6 2a x a x x x
− + − + + + " "
28. 3 20 1
1 11 [ ]
6 2a x a x x
+ + + + + + " " "
30. 2 3 41 1 11
2 2 4t t t
− + − + "
76 CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Exercises 8.5 (page 460)
2. 5 / 2 31 2c x c x− −+
4. ( ) ( )1 13 / 2 1 13 / 2
1 2c x c x− + − −
+
6. ( ) ( )1 2cos 3 ln sin 3 lnc x x c x x+
8. 1/ 2 1/ 21 2 3
5 5cos ln sin ln
2 2c c x x c x x− −
+ +
10. 2 2 2 21 2 3ln (ln )c x c x x c x x− − −+ +
12. 1/ 2 1/ 21 2( 2) cos[ln( 2)] ( 2) sin[ln( 2)]c x x c x x+ + + + +
14. 1/ 2 21 2 ln 2 lnc x c x x x x x− −+ + −
16. 2 23 13 lnx x 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. 21 23 2 0; 1, 2r r r r+ + = = − = −
14. 21 20; 1, 0r r r r− = = =
16. 2 5 30;
4 4
rr − − = 1 2
5 73 5 73,
8 8r r
+ −= =
18. 21 2
3 3 10; ,
4 2 2r r r r
− − = = = −
EXERCISES 8.6 77
20. 2 30
1 1 11
3 15 35a x x x
− − − + "
22. 2 30
161 4 4
9a x x x
+ + + + "
24. 1/ 3 4 / 3 7 / 3 10 / 30
1 1 1
3 18 162a x x x x
+ + + + "
26. 1
0 00
( 1)
!
n nx
n
xa a xe
n
+∞−
=
− =∑
28. 1/ 3
1/ 30
1
( 1)
![10 13 (3 7)]
n n
n
xa x
n n
+∞
=
−+
⋅ + ∑
"
30. 20
4 11
5 5a x x
+ +
32. 1
0 00
;!
nx
n
xa a xe
n
+∞
==∑ yes, 0 0a <
34. 01
1 ;2
a x +
yes, 0 0a <
36. 2 3 40
1 1 1
20 1960 529, 200a x x x x
+ + + + "
38. 5 / 6 11/ 6 17 / 6 23/ 60
31 2821 629083
726 2517768 (9522)(2517768)a x x x x
+ + + + "
40. 20[1 2 2 ]a x x+ +
42. The transformed equation is 2
2 22
18 (4 1) (6 1) 9(4 1)(96 40 3) 32 0d y dy
z z z z z z ydzdz
− − + − − + + = .
Also, 296 40 3
( )2(4 1)(6 1)
z zzp z
z z
− +=− −
and 22
32( )
18(4 1) (6 1)
zz q z
z z=
− − are analytic at z = 0; hence z = 0 is a regular
singular point.
1 2 31 0
32 1600 241664( ) 1
27 243 6561y x a x x x− − − = + + + +
"
78 CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Exercises 8.7 (page 482)
2. 1 1 2 2( ) ( ),c y x c y x+ where 21
1 1( ) 1
3 15y x x x
= − − + " and 1/ 2 1/ 2
2 ( ) .y x x x−= −
4. 1 1 2 2( ) ( ),c y x c y x+ where 21( ) 1 4 4y x x x= + + +" and 2 3
2 1176
( ) ( ) ln 8 1227
y x y x x x x x = − − − +
" .
6. 1/ 3 4 / 3 7 / 3 21 2
1 1 1 11
3 18 2 10c x x x c x x
+ + + + + + + " "
8. 1 1 2 2( ) ( ),c y x c y x+ where 2 31
1( )
2y x x x x
= − + + " and 2 3 4
2 13 11
( ) ( ) ln .4 36
y x y x x x x x = + − + +
"
10. 1/ 3 4 / 3 7 / 3 2 11 2
1 1 1 1
10 260 4 8c x x x c x x− − − + + + + + +
" "
12. 1 1 2 2( ) ( ),c y x c y x+ where 21
4 1( ) 1
5 5y x x x
= + + and 4 3 2
2 ( ) 4 5y x x x x− − −= + + .
14. 1 1 2 2( ) ( ),c y x c y x+ where 2 31
1( )
2y x x x x
= + + + " and 2 3 4
2 13 11
( ) ( ) ln4 36
y x y x x x x x = − − − +
" .
16. 11 2
1 1 91 ln 2 ln ;
2 2 4c x c x x x x x− + + − − − − + +
" has a bounded solution near the origin, but not all
solutions are bounded near the origin.
18. 1 1 2 2 3 3( ) ( ) ( ),c y x c y x c y x+ + where 2 31
1 1( ) ,
20 1960y x x x x
= + + + "
2 / 3 5 / 3 8 / 32
3 9( ) ,
26 4940y x x x x
= + + + " 1/ 2 1/ 2 3/ 2
32
( ) 25
y x x x x− = + + + " .
20. 1 1 2 2 3 3( ) ( ) ( ),c y x c y x c y x+ + where 5 / 6 11/ 6 17 / 61
31 2821( ) ,
726 2517768y x x x x
= + + + "
22
1( ) 1 ,
28y x x x
= + + + " and 2 3
3 23 437
( ) ( ) ln 998 383292
y x y x x x x x = − − + +
" .
22. 1 1 2 2( ) ( ),c y x c y x+ where 2 4 6
2 3 41( ) ,
2 12 144y x x x x x
α α α= − + − +" and
4 62 2 2 3
2 15 5
( ) ( ) ln 1 .4 18
y x y x x x x xα αα α= − + + − + +"
26. d. 1 22 2
,5 20
i ia a
+ += − =
EXERCISES 8.8 79
Exercises 8.8 (page 493)
2. 2 / 31 2
1 11 17 53, 5; ; , ; ;
3 3 3 3c F x c x F x
+
4. 1/ 21 2
3 1 5 11, 3; ; , ; ;
2 2 2 2c F x c x F x− +
6. 0
( , ; ; ) ( ) (1 )!
n
nn
xF x x
nαα β β α
∞−
== = −∑
8. 2 2
01
( 1)1 3, 1; ;
2 2 2 1
arctan
nn
n
F x xn
x x
∞
=−
− − = +
=
∑
10. 1
2 3
1 1( ) , ; 2;
2 21 3 25
18 64 1024
y x F x
x x x
= = + + + +
"
and
1 22 1
5 1( ) ( ) ln 4 .
16 8y x y x x x x x− = + + + +
"
14. 1 4 / 3 2 4 / 3( ) ( )c J x c J x−+
16. 1 0 2 0( ) ( )c J x c Y x+
18. 1 4 2 4( ) ( )c J x c Y x+
20. 2 2 42
1 3 17( ) ln 2
2 16 1152J x x x x x− − − − + +
"
26. 13 / 2 1/ 2 1/ 2
3/ 2 1/ 2
( ) ( ) ( )
2 2cos sin
J x x J x J x
x x x x
−− −
− −
= − −
= − −π π
21
5 / 2 1/ 2 1/ 22
5 / 2 3/ 2 1/ 2
3( ) ( ) 3 ( )
2 2 23 sin 3 cos sin
xJ x J x x J x
x
x x x x x x
−−
− − −
−= −
= − −π π π
36. 1 ( )y x x= and 22
1
1( ) 1 .
2 1k
k
y x xk
∞
== −
−∑
80 CHAPTER 8 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
38. 2 3 4 22 1, 4 3 , 8 8 1x x x x x− − − +
40. c. 1 1/2 1 /2 1
1 2( 2) ( 2)2 2
2 2n n
n nc c
c J x c Y xn n
− −+ +
+ +
+ + +
Chapter 8 Review (page 497)
2. a. ±2 are irregular singular points.
b. ,nπ where n is an integer, are regular singular points.
4. a. 2 30 1
1
[( 3)( 1)(1) (2 5)] 21
32 !
kk
k
ka x a x x
k
∞
=
− − − + + − ∑ "
b. 2 2
2 2 10 1
1 1
[3 5 15 (4 10 9)] [3 9 23 (4 6 5)]1
2 (2 )! 2 (2 1)!
k kk k
k k
k k k ka x a x x
k k
∞ ∞+
= =
⋅ ⋅ − + ⋅ ⋅ − ++ + +
+ ∑ ∑" "
6. a. ( ) ( )3 105 / 4 3 105 / 4
1 2c x c x− + − −
+
b. 1 1 21 2 3lnc x c x x c x− −+ +
8. a. 21
0
( ) nn
n
y x a x∞
+
== ∑ and 2
2 10
( ) ( ) ln nn
n
y x y x x b x∞
−
== + ∑ .
b. 10
( ) nn
n
y x a x∞
== ∑ and (3 / 2)
20
( ) .nn
n
y x b x∞
−
== ∑
c. 10
( ) nn
n
y x a x∞
== ∑ and 2 1
1
( ) ( ) ln .nn
n
y x y x x b x∞
== + ∑
10. a. 1/ 21 2
1 7 5 33, 2; ; , ; ;
2 2 2 2c F x c x F x
+
b. 1 1/ 3 2 1/ 3( ) ( )c J c Jθ θ−+
81
CHAPTER 9
Exercises 9.1 (page 507)
2. 0 1
1 0
x x
y y
′ = −
4.
1 1
2 2
3 3
4 4
1 1 1 1
1 0 0 1
0 1 0
0 0 0 0
x x
x x
x x
x x
′ − − = π −
6. 1 1
2 2
3 3
cos 2 0 0
0 sin 2 0
1 1 0
x t x
x t x
x x
′ = −
8. 2 2
1 122
2 21 1
0 1t
t t
x x
x x−− −
′ =
10. 2
2
1 11
2 2
0 1
1 ntt
x x
x x
′ = − + −
12.
1 1
2 2
3 3
4 4
0 1 0 0
0 –3 –2 1
0 0 0 1
0 –1 –1 –3
x x
x x
x x
x x
′ =
Exercises 9.2 (page 512)
2. 1 2 3 41 2
0, , , 03 3
x x x x= = = =
4. 1 2 3 41 2
, , 0, 03 3
x x x x= = = =
6. 1 2 3, ,4 4
s sx x x s= − = = ( )s−∞ < < ∞
8. 1 2 3, ,x s t x s x t= − + = = ( ),s t−∞ < < ∞
10. 1 2 32 2 4
, 0,5 5
i ix x x
− + += = =
12. a. The equations produce the format
1 21
02
0 1.
x x− =
=
b. The equations produce the format
1 3
2 3
10
211
02
0 1.
x x
x x
+ =
+ =
=
14. For r = –1, the unique solution is
1 2 3 0.x x x= = = For r = 2 the solutions are
1 2 3, ,2 4
s sx x x s= − = = ( ).s−∞ < < ∞
Exercises 9.3 (page 521)
2. a. 3 1 7
+2 4 1
− = −
A B
b. 10 4 27
7 414 5 15
− = −
A B
4. a.
2 5 1
0 12 4
1 8 4
− = −
AB
b. 3 2
1 15
= −
BA
6. a. 2 7
1 5
=
AB
b. 9 1
( )6 1
− =
AB C
c. 6 1
( )5 5
+ = −
A B C
10. 9 131 31
5 431 31
− −
82 CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
12.
1 0 1
1 1 2
1 1 1
− − − −
14.
1 12 2
1 1 13 2 61 13 3
1
0
− − −
16. c.
2 1
1 1 ,
0 1
x c
− = + −
with c arbitrary.
18. ( )( )12
12
sin 2 cos 2
cos 2 sin 2
t t
t t
−
20.
( ) 319
13 9
13
0 0
1
0 1
t
t
e
t
− − − −
22. 0
24. 11
26. 54
28. 1, 6
30. b. 0
c.
1
1
1
c
= −
x
d. 1 2 3 1c c c= = = −
32. 3 cos 3 sin 3
3 cos 3 sin 3
t t
t t
e t e t
e t e t
− −
− −
− − +
34.
2
2
2
2 cos 2 2 sin 2 2
2 cos 2 4 sin 2 6
6 cos 2 2 sin 2 2
t
t
t
t t e
t t e
t t e
−
−
−
− − − − − − −
40. a. ( )( )
212 11 12
21 21 222
3
t
t
t e c c
c ct e
−
−
− + −
b. 1 1
1 1
1 1
1 3 3
e e
e e
− −
− −
− + − + − − +
c. 3 3
3 3
3 3 9
3 3 3 9
t t t t
t t t t
e e e e
e e e e
− − − −
− − − −
− + − − − + − −
42. In general, ( ) .T T T=AB B A Thus
( ) ( ) ,T T T T T T= =A A A A A A so TA A is
symmetric. Similarly, one shows that TAA is symmetric.
Exercises 9.4 (page 530)
2. ( ) 2 0 ( ) sin
( ) 1 1 ( ) 1
r t r t t
t tθ θ′
= + ′ −
4.
1 1 1
2 1 3
1 0 5
dxdt
dydt
dzdt
x
y
z
= −
6. 1 12
2 2
0( ) ( )0 1
( ) ( )1 0
x t x t
x t x t t
′ = + ′ −
8. 1 1
2 2
3 3
( ) 0 1 0 ( ) 0
( ) 0 0 1 ( ) 0
( ) 1 1 0 ( ) cos
x t x t
x t x t
x t x t t
′ ′ = + ′ −
10. 1 1 2( ) 2 ( ) ( ) ,tx t x t x t te′ = + +
2 1 2( ) ( ) 3 ( ) tx t x t x t e′ = − + +
12. 1 2( ) ( ) 3x t x t t′ = + +
2 3( ) ( ) 1x t x t t′ = − +
3 1 2 3( ) ( ) ( ) 2 ( ) 2x t x t x t x t t′ = − + + +
14. Linearly independent
16. Linearly dependent
18. Linearly independent
EXERCISES 9.5 83
20. Yes; 4
4
3;
2
t t
t t
e e
e e
−
−
−
4
1 2 4
3
2
t t
t t
e ec c
e e
−
−
+ −
22. Linearly independent; fundamental matrix is
sin cos
cos sin .
sin cos
t
t
t
e t t
e t t
e t t
− −
The general solution is
1 2 3
sin cos
cos sin .
sin cos
t
t
t
e t t
c e c t c t
t te
− + + −
24.
3 33
3 31 2 3
3 3
5 1
0 2
0 4 2
t tt
t t
t t
e ee t
c c e c e t
te e
−
−
−
− − + + + − + +
28. ( ) ( )
( ) ( )1 12 21
5 51 12 2
( ) ;
t t
t t
e et
e e
−− −
− =
X
5
5
2( )
2
t t
t t
e et
e e
−
−
+ = − +
x
32. Choosing 0 col(1, 0, 0, , 0),= …x then
0 col(0, 1, 0, , 0),= …x and so on, the
corresponding solutions 1 2, , , n…x x x will
have a nonvanishing Wronskian at the initial point 0 .t Hence 1 , , n…x x is a fundamental
solution set.
Exercises 9.5 (page 541)
2. Eigenvalues are 1 3r = and 2 4r = with
associated eigenvectors 11
1s
=
u and
23
.2
s
=
u
4. Eigenvalues are 1 4r = − and 2 2r = with
associated eigenvectors 111
s = − u and
25 .1
s = u
6. Eigenvalues are 1 2 1r r= = − and 3 2r = with
associated eigenvectors 1
11 ,0
s−
=
u
2
10 ,1
v−
=
u and 3
11 .1
s =
u
8. Eigenvalues are 1 1r = − and 2 2r = − with
associated eigenvectors 1
124
s =
u and
2
11 .1
s =
u
10. Eigenvalues are 1 21, 1 ,r r i= = + and 3 1 ,r i= −
with associated eigenvectors 10 ,0
s =
1u
1 21 ,
is
i
− − =
2u and 1 2
1 ,i
si
− + = −
3u where
s is any complex constant.
12. 7 51 2
1 1
2 2t tc e c e−
+ −
14. 2 31 2 3
1 1 1
0 1 4
1 3 3
t t tc e c e c e− −−
+ − +
16. 10 5 51 2 3
2 0 1
0 1 0
1 0 2
t t tc e c e c e− + + −
84 CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
18. a. Eigenvalues are 1 1,r = − and 2 3r = − with associated eigenvectors 1
,1
s
=
1u and 1
.1
s
= − 2u
b. 1
2
t
t
x e
x e
−
−
=
= c.
31
32
t
t
x e
x e
−
−
= −
= d.
31
32
t t
t t
x e e
x e e
− −
− −
= −
= +
Figure 41 Figure 42 Figure 43
20. 4
4
4t t
t t
e e
e e
− −
22.
2 2
2
2
0
0 3
t t t
t t
t t
e e e
e e
e e
−
24.
41 0 0
4 0 0 0
0 0 3
0 0
t
t t
t t
e
e e
e e
−
−
26. 51 2( ) 2t tx t c e c e−= +
51 2( ) 2 t ty t c e c e−= +
9.5 EXERCISES 85
28.
0.4679 3.8794 1.6527
0.4679 3.8794 1.6527
0.4679 3.8794 1.6527
0.2931 0.4491 0.6527
0.5509 0.1560
0.7733
t t t
t t t
t t t
e e e
e e e
e e e
− − − − −
30.
0.6180 1.6180
0.6180 1.6180
0.5858 3.4142
0.5858 3.4142
0.6180 0 0
0.6180 0 0
0 0 0.2929
0 0 0.5858
t t
t t
t t
t t
e e
e e
e e
e e
−
−
−
32. 3 4
3 4
2 12
2 8
t t
t t
e e
e e
− −
34.
2
2
2
2
3
t t
t t
t t
e e
e e
e e
−
−
−
− + −
36. 4
2 2 2 31 2
1 1
1 1 1
t t tc e c te e + +
38.
142
1 2 312
1 1 0 1 0
1 1 0 1 0 02
0 0 1 0 1
tt t t t tt e
c e c te e c te e
− + + + + +
40. 1 2 3
1 0 1 0
0 2 2 1
0 3 3 1
t t t tc e c e c te e
+ + +
44. 51 2
1 2
2 1c c t− −
+
46. 3
1( ) kg,10
tex t
−= 3
23
( ) ( 1) kg10
t tx t e eααα
−−= −
The mass of salt in tank A is independent of .α The maximum mass of salt in tank B is 3 /3
0.13
αα−
kg.
50. b. 33 1( ) 1
2 2t tx t e e− −= + +
3( ) 1 ty t e−= −
33 1( ) 1
2 2t tz t e e− −= − +
86 CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
Exercises 9.6 (page 549)
2. 1 25 cos 5sin
2 cos sin 2 sin cos
t tc c
t t t t
− − + − +
4.
2 2
2 2 2 2 21 2 3
2 2 2
5 cos 5 sin 0
2 cos sin 2 sin cos
5 cos 5 sin
t t
t t t t t
t t t
e t e t
c e t e t c e t e t c e
e t e t e
− + + − − + −
6. cos 2 sin 2
sin 2 cos 2 sin 2 cos 2
t t
t t t t
− − −
8. 2 2
2 2
0 0
0 0
0 0 cos 3 sin 3
0 0 (2 cos 3 3sin 3 ) (2 sin 3 3cos 3 )
t t
t t
t t
t t
e e
e e
e t e t
e t t e t t
−
−
−
− +
10.
2 2 2 2
2 2 2 2
2 2 2 2
2
0.0209 cos 3 0.0041 sin 3 0.0209 sin 3 0.0041 cos 3
0.0296 cos 3 0.0710 sin 3 0.0296 sin 3 0.0710 cos 3
0.1538 cos 3 0.2308 sin 3 0.1538 sin 3 0.2308 cos 3
co
t t t t t t
t t t t t t
t t t t t t
t
e t e t e t e t e e
e t e t e t e t e e
e t e t e t e t e e
e
−
−
−
− + − − −
− + − −
+ − −2s 3 sin 3t t tt e t e e−
12.
2 2 2 2
2 2
0 0 0 0
0 0 0
0 0 0
0.0690 cos 5 0.1724 sin 5 0.0690 sin 5 0.1724 cos 5 0 0 0
cos 5 sin 5 0 0 0
t
t t
t t
t t t t
t t
e
e e
e e
e t e t e t e t
e t e t
−
−
− − − −
− −
− − + − −
14. a. 2
sin 2 cos
2
cos 2 sin
t t
t
t t
e t e t
e
e t e t
− − −
b. 2( )
sin
cos
t
t
t
e t
e
e t
+π
+π
+π
−
18. 1 11 2
cos(3 ln ) sin(3 ln )
3sin(3 ln ) 3cos(3 ln )
t tc t c t
t t− −
+ −
EXERCISES 9.7 87
20. 1( ) cos cos 3x t t t= −
2 ( ) cos cos 3x t t t= +
22. 2 81
16 1( ) 2
5 5t tI t e e− −= − +
2 82 ( ) 4 2t tI t e e− −= − +
2 83
4 4( )
5 5t tI t e e− −= − +
Exercises 9.7 (page 555)
2. 3
1 23 22 2
t t
t t
e e tc c
e e
−
−
+ + −
4. 4
1 2 4
1 2 sin
1 2 sin cos
t
t
e tc c
t te
− + + − +
6. 3tp t e= + +x a b c
8. 2p t t= + +x a b c
10. t tp e te− −= +x a b
12. 41 2
2 1 1
3 1 1t tc e c e−
+ + − −
14. 1 2 2
2 1cos sin
sin cos 2
tt tc c
t t t
− − + + −
16. 1 2cos sin 4 sin
sin cos 4 cos 4 sin
t t t tc c
t t t t t
+ + − −
18. 1 2 3
1 0
0 1 0
0 1 1
t t
t t t
t
t te e
c e c e c e t
t e
− − + + + + −
20.
( ) ( ) ( )( ) ( )
( ) ( )( ) ( )
2 8 17 11 2 3 4 15 225 8
2 8 881 2 3 4 15 225
2 8 321 2 3 4 15 225
2 8 1521 2 3 4 15 225
sin 2 cos 2 8
2 cos 2 2 sin 2 2 8
4 sin 2 4 cos 2 4 8
8 cos 2 8 sin 2 8 8 1
t t t t
t t t t
t t t t
t t t t
c t c t c e c e te e
c t c t c e c e te e
c t c t c e c e te e
c t c t c e c e te e
−
−
−
−
− + − + + + − − − + + + −
− − + + + + + + + + −
88 CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
22. a. 4 2
4 2
3
6 2
t t
t t
e e
e e t
−
−
+ − + −
b. ( ) ( )
( ) ( )4( 2) 2( 2)74
3 3
4( 2) 2( 2)8 73 3
2
t t
t t
e e
e e t
− − −
− − −
− + + −
24. 4 2( ) 3 t tx t e e−= + 4 2( ) 6 2t ty t e e t−= − + −
26. a. 2
t
t
e
e
−
b. t
t
te
te
28. 1
1
t
t
− + − −
30. 132
1 2 23
1 3
2 4c t c t−
+ +
34. a. Neither wins.
b. The 1x force wins.
c. The 2x force wins.
Exercises 9.8 (page 566)
2. a. r = 2; k = 2 b. 2 1
1t t t
et t
− − +
4. a. r = 2; k = 3 b.
2
22
1 3
0 1
0 0 1
t
t
t t
e t
− −
6. a. r = –1; k = 3 b.
2 2
2 2
2 2
22 2
22 2
22 2
1
1
3 1 2
t t
t t t
t t
t t t
e t t t
t t t t
−
+ + + − + − −
− + − + − +
8. ( ) ( ) ( ) ( )
( ) ( )3 31 1 1 1
2 2 4 4
3 31 12 2
t t t t
t t t t
e e e e
e e e e
− −
− −
+ − − +
EXERCISES 9.8 89
10.
4 2 4 2 4 2
4 2 4 2 4 2
4 2 4 2 4 2
21
23
2
t t t t t t
t t t t t t
t t t t t t
e e e e e e
e e e e e e
e e e e e e
− − −
− − −
− − −
+ − − − + − − − +
12.
2 2 2
2 2 2 2 2 2
2 2 2 2 2 2
3 6 3 3 3 31
4 4 6 4 5 3 4 4 39
2 2 6 2 2 3 2 7 3
t t t t t t
t t t t t t t t t
t t t t t t t t t
e e e e e e
e e te e e te e e te
e e te e e te e e te
− − −
− − −
− − −
+ − + − + − + + + + − + − + − − − + −
14.
0 0 0 0
0 0 0
0 0 0
0 0 0 cos sin
0 0 0 sin cos
t
t t t
t t t
e
e te te
te e te
t t
t t
− −
− − −
+ − − −
16.
2 2 2
2 2 2
0 0 0 0
0 0 0
0 0 0
0 0 0 2
0 0 0 4 2
t
t t t
t t t
t t t
t t t
e
e te te
te e te
e te te
te e te
−
− − −
− − −
− − −
− − −
+ − −
+
− −
18. 1 2 3
1 0 1
0 1 2
0 0 1
t tc c e c e t
+ +
20.
2
21 2 3
1 2 44 3 4
1
0 2 4
t t t
t tt
c e c e t c e t t
t
− +− − + + − − −
22.
( ) ( )( ) ( ) ( )( ) ( ) ( )
24 13 3
2 216 16 19 9 3
2 28 19 19 9 3
t t
t t t
t t t
e e
e e te
e e te
−
−
−
− + − +
+ −
24.
cos sin 1
sin cos 1
cos sin
t
t
t
e t t t
e t t
e t t
+ − − − − − − − +
90 CHAPTER 9 MATRIX METHODS FOR LINEAR SYSTEMS
Chapter 9 Review (page 570)
2. 2 21 2
2 cos 3 2 sin 3
cos 3 3sin 3 sin 3 3cos 3t tt t
c e c et t t t
− − + + −
4. 21 2 3
0 1
0 0 1
1 0 0
t tt
c e c e c
+ +
6.
5
5 5
5 5
0 0
3 0
0 3
t
t t
t t
e
e e
e e
−
−
−
8. 115364
1 2 13518
1 2
2
tt
t
ec c e
e
−
−
+ + −
10.
2 5 / 2 5 / 21 2
5 / 2 5 / 23
3 71 117 7
0 cos 2 sin 2 72 2
0 4 0
3 7117 7
sin 2 cos 2 72 2
4 0
t t t
t t
t tc e c e e
t tc e e
− + − − −
− + − + −
( )1 113 16
1458
te− − + + − −
12. ( ) ( )2 2 23 1
2 2
2 2 2
sin 2 cos 2
2 cos 2 3 sin 2
t t t
t t t
e t e t e
e t e t te
+ + − −
14. 3 21 2 3
1 1 1
1 1 1
2 0 1
c t c t c t−−
+ + −
16.
21 4
0 1 2
0 0 1
t t t
t
+
91
CHAPTER 10
Exercises 10.2 (page 587)
2. 10 2 5
10 2
( 1) (1 )x xe e e ey
e e
− + −=
−
4. y = 2 sin 3x
6. No solution
8. 1 1x xy e xe− −= +
10. 2(2 1)
4nnλ −
= and (2 1)
cos ,2n n
n xy c
− = where n = 1, 2, 3, … and the nc ’s are arbitrary.
12. 24n nλ = and cos(2 ),n ny c nx= where n = 0, 1, 2, … and the nc ’s are arbitrary.
14. 2 1n nλ = + and sin( ),xn ny c e nx= where n = 1, 2, 3, … and the nc ’s are arbitrary.
16. 27 147 507( , ) sin 3 5 sin 7 2 sin13t t tu x t e x e x e x− − −= + −
18. 48 108 300( , ) sin 4 3 sin 6 sin10t t tu x t e x e x e x− − −= + −
20. 2 3 1
( , ) sin 9 sin 3 sin 21 sin 7 sin 30 sin109 7 30
u x t t x t x t x = − + −
22. 2 7
( , ) cos 3 sin cos 6 sin 2 cos 9 sin 3 sin 9 sin 3 sin15 sin 53 15
u x t t x t x t x t x t x = − + + −
24. 1
2 21
( 1)1( , ) cos 4 sin 4 sin
4
n
n
u x t nt nt nxn n
+∞
=
− = +
∑
Exercises 10.3 (page 603)
2. Even
4. Neither
6. Odd
10. 2
0
4 1( ) cos(2 1)
2 (2 1)k
f x k xk
∞
=
π − +π +
∑
12. 2 22
2 31
2( 1) 2( 1) ( 1) 2( ) ~ cos sin
6
n n n
n
nf x nx nx
n n
∞
=
− − − π − −π + + π
∑
92 CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
14. 1
1( ) ~ sin 2
2 n
f x nxn
∞
=
π + ∑
16. 1
2( ) ~ 1 cos sin
2n
nf x nx
n
∞
=
π − π ∑
18. The 2π periodic function g(x), where ( ) ,g x x= –π x π
20. The 2π periodic function g(x), where 2
2
0 0
( ) 0
2
x
g x x x
x
−π < ≤= ≤ < ππ = ±π
22. The 2π periodic function g(x), where0
( ) 0
0,2
x xg x x x
x
+ π −π < <= < < π
π = ± π
24. The 2π periodic function g(x), where
0 –2
1
2 2
1 02
( ) 0 0
1 02
1
2 2
02
x
x
x
g x x
x
x
x
−π π ≤ < π− = −
−π− < <= =π < <
π =
π < ≤ π
30. 0 1 21 5
, 0,2 8
a a a= = =
EXERCISES 10.4 93
Exercises 10.4 (page 611)
2. a. The π periodic function ( ) sin 2f x x= for x kπ where k is an integer.
b. The 2π periodic function 0 ( ) sin 2f x x= for x kπ where k is an integer.
c. The 2π periodic function ( ),ef x where sin 2 0( )sin 2 0e
x xf xx x
< < π= − −π < <
4. a. The π periodic function ( ),f x where ( ) , 0f x x x= π − < < π
b. The 2π periodic function 0 ( ),f x where 00( )
0x xf x
x xπ − < < π= −π − −π < <
c. The 2π periodic function ( ),ef x where 0( )– 0e
x xf xx x
π − < < π= π + π < <
6. 2
1
8( ) ~ sin(2 )
(4 1)k
kf x kx
k
∞
= π −∑
8. 1
2( ) ~ sin
n
f x nxn
∞
=∑
10. 2 2
1
2 [1 ( 1) ]( ) ~ sin( )
1
n
n
n ef x n x
n
∞
=
π − − π+ π
∑
12. 2
0
4 1( ) ~ 1 cos(2 1)
2 (2 1)k
f x k xk
∞
=
π + − + π +∑
14. 1 12 2
1
2( ) ~ 1 [1 ( 1) ]cos( )
1
n
n
f x e e n xn
∞− −
=− + − − π
+ π∑
16. 2 2
1
1 1( ) ~ cos(2 )
6 k
f x k xk
∞
=− π
π∑
18. 25(2 1)
30
8( , ) sin(2 1)
(2 1)
k t
k
u x t e k xk
∞− +
== +
+ π∑
94 CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
Exercises 10.5 (page 624)
2. 21
31
2 4( , ) ( 1) [( 1) 1] sinn n n t
n
u x t e nxn n
∞+ −
=
π= − + − − π
∑
4. 2 28
2 21
1 1( , ) cos 2
6k t
k
u x t e kxk
∞− π
== − π
π∑
6. 27 28
21
2 4( , ) 1 2 cos cos 2
(4 1)
t k t
k
u x t e x e kxk
∞− −
== − + +
π π −∑
8. 2
1
( 1)( , ) 3 6 sin
nn t
n
u x t x e nxn
∞−
=
−= + ∑
10. 22
3 3 33
2
2( 1)1 1( , ) sin sin
18 18 3 3
nt n t
n
u x t x x e x e nxn
∞− −
=
−π = − + + ∑
12. 2
1
( , ) sin ,ntn n
n
u x t a e xµ µ∞
−
== ∑ where 1 n nµ ∞
= is the increasing sequence of positive real numbers that are
solutions to tan ,n nµ π µ= − and 0sin(2 )
2 4
1( ) sin
nn na f x x dx
µµ
π
ππ=
− ∫ .
14. 22 3(2 1)
30
5 5 20 1( , ) 1 sin(2 1)
6 6 3 (2 1)
k t
k
u x t x x e k xk
∞− +
=
π = + − − + π +∑
16. 2 5 25( , , ) cos sin 4 cos 2 sin 3 cos 3 sin 4t t tu x y t e x y e x y e x y− − −= + −
18. 2[(2 1) 1]
20
4 1( , , ) sin cos[(2 1) ]sin
2 (2 1)
t k t
k
u x y t e y e k x yk
∞− − + +
=
π = − + π +∑
EXERCISES 10.6 95
Exercises 10.6 (page 636)
2. 1
( , ) [ cos 4 sin 4 ]sin ,n nn
u x t a nt b nt nx∞
== +∑ where
2
0 even2 1
odd4
n
nna
nnn
= − π −
and
2
2
1 even
4 ( 1)1
oddn
nnb
nn
− π −= π
4. 3
0
( 1)4( , ) cos12 sin 4 7 cos15 sin 5 sin 3(2 1) sin(2 1)
3 (2 1)
k
k
u x t t x t x k t k xk
∞
=
−= + + + +
π +∑
6. 3
03 3
1
2 1( , ) sin sin sin
( ) n
v L n tn a n xu x t
L L La L a n
α
α
∞
=
ππ π=π −
∑
8. 2 2
2
2( 1)( , ) (sin cos ) sin [sin sin ]sin
( 1)
n
n
u x t t t t x nt n t nxn n
∞
=
−= − + −
−∑
10. 1 2 11
( , ) ( ) cos sin sin ,n nn
n tx n t n xu x t U U U a b
L L L L
α α∞
=
π π π = + − + + ∑ where na ’s and nb ’s are chosen that
1 2 11
1
( ) ( ) sin
( ) sin
nn
nn
x n xf x U U U a
L L
n n xg x b
L L
α
∞
=∞
=
π − − − =
π π =
∑
∑
14. 2 2 2( , )u x t x tα= +
16. u(x, t) = sin 3x cos 3 t tα +
18. u(x, t) = cos 2x cos 2 t t xtα + −
96 CHAPTER 10 PARTIAL DIFFERENTIAL EQUATIONS
Exercises 10.7 (page 649)
2. cos sinh( ) 2 cos 4 sinh(4 4 )
( , )sinh( ) sinh( 4 )
x y x yu x y
− π − π= −
−π − π
4. sin sinh( ) sin 4 sinh(4 4 )
( , )sinh( ) sinh( 4 )
x y x yu x y
− π − π= +
−π − π
8. 21
( , ) cos 22 8
ru r θ θ= +
12. 5
3 3 5 56 10
27 3( , ) [ ]cos 3 [ ]cos 5
3 1 3 1u r r r r rθ θ θ− −= − + −
− −
14. 1
( , ) [ cos sin ],nn n
n
u r C r a n b nθ θ θ∞
−
== + +∑ where C is arbitrary and for n = 1, 2, …
1( ) cosna f n d
nθ θ θ
π−π
−=π ∫
1( ) sinnb f n d
nθ θ θ
π−π
−=π ∫
16. 1
( ) (ln ln )( , ) sinh sin ,
ln 2 ln 2nn
n n ru r a
θθ∞
=
π − π π − π = ∑ where
( )2
2
ln 2
(ln ln )2 1sin sin
ln 2ln 2 sinhn
n
n ra r dr
r
πππ
π − π− = ∫ .
24. 1
( , ) sin ,nyn
n
u x y A e nx∞
−
== ∑ where
0
2( ) sin .nA f x nx dx
π=
π ∫
97
CHAPTER 11
Exercises 11.2 (page 671)
2. sin x – cos x + x + 1
4. 2 (sin 2 cos 2 )xce x x− +
6. 2 x xe e x−+ −
8. 1 2sin 2 cos 2 1c x c x+ +
10. No solution
12. No solution
14. 2 2
;9nnλ π= ( ) sin cos ,
3 3n n nnx nx
y x b cπ π = +
n = 0, 1, 2, …
16. 2(2 1)
2 ;4n
nλ += +
(2 1)( ) sin ,
2n nn x
y x c+ =
n = 0, 1, 2, …
18. 20 0 ,λ µ= − where 0 0tanh( ) 2 ;µ µπ = 0 0 0( ) sinh( ),y x c xµ=
2 ,n nλ µ= where tan( ) 2 ;n nµ µπ = ( ) sin( ),n n ny x c xµ= n = 1, 2, 3, …
20. 2 2 ;n nλ = π ( ) cos( ln ),n ny x c n x= π n = 0, 1, 2, …
22. 1 2 34.116, 24.139, 63.659λ λ λ= = =
24. No nontrivial solutions
26. 2 ,n nλ µ= where cot n nµ µπ = for 0;nµ > ( ) sin ,n n ny x c xµ= n = 1, 2, 3, …
28. 4 ;n nλ µ= − 0nµ > and cos( ) cosh( ) 1;n nL Lµ µ = −
sin sinhsin sinh (cos cosh ) ,
cos coshn n
n n n n n nn n
L Ly c x x x x
L L
µ µµ µ µ µ
µ µ +
= − − − + n = 1, 2, …
34. b. 0 01, ( )X x cλ = − = and 2 21 , ( ) cos( )n nn X x c nxλ = − + π = π
98 CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
Exercises 11.3 (page 682)
2. 1 0y x yλ −′′ + =
4. 1( ) 0xy xy x yλ −′ ′ + − =
6. 2((1 ) ) 0x y yλ′ ′− + =
8. Yes
10. No
18. a. 1 1 1
, sin , cos ,3 36 3 3
n x n xπ π
n = 1, 2, 3, …
b. 1
1
( 1) 6sin
3
n
n
n x
n
+∞
=
− π π
∑
20. a. 2 1
sin ,2
n x + π
n = 0, 1, 2, …
b. 2
0
( 1) 8 1sin
2(2 1)
n
n
n xn
∞
=
− + π +∑
22. a. 00
0 02 sinh( )
sinh(2 ) 2x
µµ
µ µπ − π where 0 0tanh( ) 2 ;µ µπ =
2 sin( )2 sin(2 )
nn
n nx
µµ
µ µπ− π where tan( ) 2 ,n nµ µπ = n = 1, 2, 3, …
b. 00
0 0 1
4( 2) cosh( ) 4(2 ) cos( )sinh( ) sin( )
sinh(2 ) 2 2 sin(2 )n
nn nn
x xµ µ
µ µµ µ µ µ
∞
=
π − π − π π+
π − π π− π∑
24. a. 01
( ) ;1
y xe
=−
( ) 2 cos( ln ),ny x n x= π n = 1, 2, 3, …
b. 2 2
1
2[( 1) 1]1 cos( ln )
1
n
n
en x
n
∞
=
− −+ π+ π
∑
Exercises 11.4 (page 692)
2. 2[ ] (4 sin ) (2 2 cos )L y x y x x y x x y+ ′′ ′= + − + + −
4. 2[ ] 6 7L y x y xy y+ ′′ ′= + +
6. [ ] (sin ) (2 cos ) ( sin 1)x xL y x y x e y x e y+ ′′ ′= + + + − + +
EXERCISES 11.6 99
8. [ ] 4 5 ;L y y y y+ ′′ ′= + + 2( ) [0, 2 ] : (0) (2 ) 0D L y C y y+ = ∈ π = π =
10. 2 5[ ] 2
4L y x y xy y+ ′′ ′= + +
2( ) [1, ] : (1) ( ) 0D L y C e y y e+ π π= ∈ = =
12. 0;y y y′′ ′− + = y(0) = y (0) ( )y y′ ′= π
14. 0; (0) ; (0)2 2
y y y y yπ π ′′ ′ ′= = =
16. 2 2 0;x y xy′′ ′+ = y(1) = 4y(2), (1) 4 (2)y y′ ′=
18. 2 20
( ) sin 0xh x e x dxπ − =∫
20. 1/ 21
( ) sin(ln ) 0e
h x x x dxπ
− =∫
22. Unique solution for each h
24. / 2
0( ) 0h x dx
π=∫
26. 2
1
3( ) 1 0h x
x − = ∫
Exercises 11.5 (page 698)
2. 1 1
sin11 sin 4122 17
x x−
4. 1 5
cos 7 cos1049 100
x x+π− π−
6. 1 1
cos 5 cos 419 10
x x−
8. 3 2
0
8sin[(2 1) ]
(2 1) (4 4 1)n
n xn n n
∞
=
− +π + + −
∑
10.
( )2102
1sin ,
27
n
n
n xn
γ∞
=
+ − +∑ where
0
2 1( ) sin
2n f x n x dxγπ = + π ∫ .
12. Let 1
2( ) sin( ln ) .
en f x n x dxγ
π
=π ∫ If 1 0,γ ≠
there is no solution. If 1 0,γ = then
22
sin(ln ) sin( ln )1
n
n
c x n xn
γ∞
=+
−∑ is a solution.
14. 2 2
0
cos( ln ),1
n
n
n xn
γ∞
=π
− − π∑ where
1
1 21
( ) cos( ln ).
cos ( ln )
e
n e
f x n x dx
x n x dxγ
−
π=
π
∫∫
Exercises 11.6 (page 706)
2.
sinh sinh( 1)0sinh1
( , )sinh sinh( 1)
1sinh1
s xs x
G x sx s
x s
−− ≤ ≤= −− ≤ ≤
4. cos sin 0( , )cos sin
s x s xG x sx s x s
− ≤ ≤= − ≤ ≤ π
100 CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
6. (sin cos ) sin 0( , )(sin cos ) sin
s s x s xG x sx x s x s− ≤ ≤= − ≤ ≤ π
8.
2 2 2 2
2 2 2 2
( )( 16 )168
( , )
( )( 16 )2
68
s s x xs x
G x s
x x s sx s
− −
− −
+ −− ≤ ≤= + −− ≤ ≤
10.
5 6 5
6
5 6 5
6
( )(5 )030 6
( , )
( )(5 )1
30 6
s s x x
x x s s
e e e es xe
G x s
e e e ex s
e
− −
− −
− + ≤ ≤+= − + ≤ ≤ +
12.
( )0
( , )( )
s xs x
G x sx s
x s
− − π ≤ ≤π= − − π
≤ ≤ ππ
4 3
12
x xy
− π=
14. 0( , ) s s xG x sx x s
≤ ≤= ≤ ≤ π
3 44
12
x xy
π −=
16.
sin sin( 2)0sin 2
( , )sin sin( 2)
2sin 2
s xs x
G x sx s
x s
−− ≤ ≤= −− ≤ ≤
12[sin sin( 2) sin 2]
sin 2y x x= − − −
18.
cosh cosh( 1)0sinh1
( , )cosh cosh( 1)
1sinh1
s xs x
G x sx s
x s
− ≤ ≤= −
≤ ≤
y = –24
20.
1 21 1 1
( , )1 2
1 1 2
s xs x
G x s
x sx s
− − − ≤ ≤ = − − − ≤ ≤
2 ln 2ln
4
xy
x = +
24. a. (1 ) 0( , )(1 ) 1
x s
x se s x s xK x se x s x s
−
− − ≤ ≤=
− ≤ ≤
b. 1( 1) 1x xy x e xe −= − − +
26. a.
2 2 3
3 2 2
( )(3 16 )176
( , )( )(3 16 )
276
s s x xs x
K x sx x s s
x s
− −
− −
− + ≤ ≤= − +
≤ ≤
b. 34(1 ln 2)1ln ( )
4 19y x x x x−+= + −
EXERCISES 11.8 101
28.
2 2
2 2
( )( 2 )06
( , )( – )( 2 )
6
x s s s xs x
H x ss x x x s
x s
π − − π + ≤ ≤π= π − π +
≤ ≤ π π
30.
2 3 2 3
3
2 3 2 3
3
[( 3 )( 3 ) 2 ( 3 )]012
( , )[( 3 )( 3 ) 2 ( 3 )]
12
x x s s x ss x
H x ss s x x s x
x s
− π − π + π − ≤ ≤π
= − π − π + π −
≤ ≤ ππ
Exercises 11.7 (page 715)
2. 3 31
( )n nn
b J xα∞
=∑ where 3 nα is the increasing sequence of real zeros of 3J and
13 30
12 23 3 30
( ) ( )
( ) ( )
nn
n n
f x J x dxb
J x x dx
α
µ α α=
−
∫∫
4. 0
( )n nn
b P x∞
=∑ where
1
11 21
( ) ( )
[ ( 1)] ( )
nn
n
f x P x dxb
n n P x dxµ
−
−
=− +
∫∫
6. 20
( )n nn
b P x∞
=∑ where
120
1 220
( ) ( )
[ 2 (2 1)] ( )
nn
n
f x P x dxb
n n P x dxµ=
− +
∫∫
16. c. 0
( )n nn
b L x∞
=∑ where 0
20
( ) ( )
( ) ( )
nn
xn
f x L x dxb
n L x e dxµ
∞
∞ −=
−
∫∫
Exercises 11.8 (page 725)
2. No; sin x has a finite number of zeros on any closed bounded interval.
4. No; it has an infinite number of zeros on the interval 1 1
, .2 2
−
8. 3
π
10. Between 25
1
e
λ
−π
+ and
26
26 sin 5λπ
+
102 CHAPTER 11 EIGENVALUE PROBLEMS AND STURM-LIOUVILLE EQUATIONS
Chapter 11 Review (page 729)
2. a. 7 7( ) 0x xe y e yλ′ ′ + =
b. ( )2 23 / 2 3 / 2 0x xe y e yλ− −′′ + =
c. ( ) 0x xxe y e yλ− −′ ′ + =
4. a. 0; (0) 0, (1) 0y xy y y′′ ′ ′ ′− = = =
b. 2 2 3 0; (1) 0,x y xy y y′′ ′+ − = = ( ) 0y e′ =
6. 2 2
0
2 14 cos 2 cos[(2 1) ]
6 (2 2 1)(2 1)k
x k xk k k
∞
=
π + + +π + − +∑
8. a. 7 71
( )n nn
b J xα∞
=∑ where 7 nα is the increasing sequence of real zeros of 7J and
17 70
12 27 7 70
( ) ( )
( ) ( )
nn
n n
f x J x dxb
J x x dx
α
µ α α=
−
∫∫
b. 0
( )n nn
b P x∞
=∑ where
1
11 21
( ) ( )
[ ( 1)] ( )
nn
n
f x P x dxb
n n P x dxµ
−
−
=− +
∫∫
10. Between 6
π and
3.
5π
103
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
104 CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
16. Asymptotically stable spiral point.
Figure 45
18. Unstable improper node
Figure 46
EXERCISES 12.3 105
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.
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 12.4 107
Exercises 12.4 (page 774)
2. G(x) = sin x + C; 21( , ) sin
2E x v v x= +
4. 2 4 61 1 1( ) ;
2 24 720G x x x x C= − + + 2 2 4 61 1 1 1
( , )2 2 24 720
E x v v x x x= + − +
6. ( ) ;xG x e x C= − + 21( , ) 1
2xE x v v e x= + − −
8.
Figure 50
108 CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
10.
Figure 51
12.
Figure 52
EXERCISES 12.4 109
14. 2( , ) ,vh x v v= so energy is decreasing along a trajectory.
Figure 53
16. 2( , )vh x v v= , so energy is decreasing along a trajectory.
Figure 54
110 CHAPTER 12 STABILITY OF AUTONMOUS SYSTEMS
18.
Figure 55
Exercises 12.5 (page 782)
2. Asymptotically stable
4. Stable
6. Unstable
8. Asymptotically stable
10. Stable
12. Stable
14. Stable
EXERCISES 12.6 111
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
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
4. Unstable
6. Asymptotically stable
8. a. 12
( ) ctc
= x
b. 1 21( )0 1
tt c c = + x
10. Stable
12. Asymptotically stable
14. Asymptotically stable
16. The equilibrium solution corresponding to the critical point (–3, 0, 1) is unstable.
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 115
8.
Figure 63
10. Unstable
12. Asymptotically stable
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
117
CHAPTER 13
Exercises 13.1 (page 812)
2. ( ) sin( ( ))x
y x t y t dtπ
= +∫
4. ( )0
( ) 1x y ty x e dt= + ∫
6. 0.3775396
8. 2.2360680
10. 1.9345632
12. 1 ( ) 1 ,y x x= + 2 32
1( ) 1
3y x x x x
= + + +
14. 1 2( ) ( ) siny x y x x= =
16. 21
3 1( ) ,
2 2y x x x
= − +
2 32
5 3 1( )
3 2 6y x x x x
= − + −
Exercises 13.2 (page 820)
2. No
4. Yes
6. Yes
14. No; let
2
2
10
1 2( ) 2
20 1
n
n x x n
y x n n x xn n
xn
≤ ≤= − ≤ ≤ ≤ ≤
.
Then, lim ( ) 0,nn
y x→∞
= but
1
0lim ( ) 1 0.n
ny x dx
→∞= ≠∫
Exercises 13.3 (page 826)
2. [–2, 1)
4. (0, 3]
6. ( ),−∞ ∞
Exercises 13.4 (page 832)
2. 210 e−
4. 1 / 22 210 ee
−−
6. 210 e−
8. sin11
24e
10. 6
e
Chapter 13 Review (page 835)
2. 0.7390851
4. 2 3 20
9 [ ( ) ( )]x
t y t y t dt+ −∫
6. 21 2( ) 1 2 , ( ) 1 2 2y x x y x x x= − + = − + −
8. No
10. ,2 2
π π −
12. 6
e