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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.


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


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

.


_________________________ † 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− ′ ′ + =


_________________________ † 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


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


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


14.

Figure 6

16.

18. Every solution approaches zero.

EXERCISES 1.4 19


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


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


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.


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−= +


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


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


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


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) →


Pb

< then P(t) → 0;


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


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


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

= − − − −


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


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


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


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


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+


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− −+


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−


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+ +


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.


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


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≤


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−≈


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)


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


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 = +


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


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


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+ −


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



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


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


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


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.


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= − + −


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


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


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


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


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

− += + + −


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


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

−− − + −

∫ ∫ ∫


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


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→∞

=


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


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


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

− − − + −


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


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

− − − − − −

− − − −

− − + − − − − − − − − − − −


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

−

− − −

− − − − −

− ≤ ≤ − + ≤ ≤ − + − ≥


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≈ − ≤ ≤ + ≈


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− − − − − − + + − −


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


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− − = − +



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− + +"


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

∞+

=

− = − ++∑ "


2. 0

4. –1, 0

6. –1

8. No singular points

10. 1x ≤ and x = 2


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


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

− + − + "



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− −+


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− − − = + + + +

"



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


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

∞

== −

−∑


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

− −+ +

+ +

+ + +


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


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

′ =


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−∞ < < ∞


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.


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.


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


s = − u and

25 .1

s = u

6. Eigenvalues are 1 2 1r r= = − and 3 2r = with


11 ,0

s−

=

u

2

10 ,1

v−

=

u and 3

11 .1

s =

u

8. Eigenvalues are 1 1r = − and 2 2r = − with


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− + + −


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− −= − +



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− −= − +


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




−

−

−

−

− + − + + + − − − + + + −

− − + + + + + + + + −


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.


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

+ − − − − − − − +



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


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

+∞

=

− = +

∑


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


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∑



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


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 + −



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


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


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

∑


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 − = ∫


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

∫∫


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


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


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

∞

∞ −=

−

∫∫


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

+



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


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


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.


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


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


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


18.

Figure 55


2. Asymptotically stable

4. Stable

6. Unstable


10. Stable

12. Stable

14. Stable

EXERCISES 12.6 111


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


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



4. Unstable


8. a. 12

( ) ctc

= x

b. 1 21( )0 1

tt c c = + x

10. 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.


2. (0, 0) is an unstable saddle point.

Figure 60


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



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


117

CHAPTER 13


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

= − + −


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

→∞= ≠∫


2. [–2, 1)

4. (0, 3]

6. ( ),−∞ ∞


2. 210 e−

4. 1 / 22 210 ee

−−

6. 210 e−

8. sin11

24e

10. 6

e


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

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