Differential Equations and Linear Algebra 4th Edition edwards Solutions Manual by sandrayd74

INEAR

EDITION

Henry Edwards David E. Penney The University of Georgia David T. Calvis Baldwin Wallace University Differential Equations and Linear Algebra 4th Edition edwards Solutions Manual Full Download: http://testbanktip.com/download/differential-equations-and-linear-algebra-4th-edition-edwards-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

INSTRUCTOR’S SOLUTIONS MANUAL D IFFERENTIAL E QUATIONS & L

A LGEBRA FOURTH

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.

Reproduced by Pearson from electronic files supplied by the author.

Publishing as Pearson, 330 Hudson Street, NY NY 10013

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.

ISBN-13: 978-0-13-449825-6

ISBN-10: 0-13-449825-9

1 FIRST-ORDER DIFFERENTIAL EQUATIONS 1.1 Differential Equations and Mathematical Models 1 1.2 Integrals as General and Particular Solutions 8 1.3 Slope Fields and Solution Curves 16 1.4 Separable Equations and Applications 28 1.5 Linear First-Order Equations 44 1.6 Substitution Methods and Exact Equations 62 Chapter 1 Review Problems 86 2 MATHEMATICAL MODELS AND NUMERICAL METHODS 2.1 Population Models 101 2.2 Equilibrium Solutions and Stability 117 2.3 Acceleration-Velocity Models 128 2.4 Numerical Approximation: Euler's Method 138 2.5 A Closer Look at the Euler Method 146 2.6 The Runge-Kutta Method 158 3 LINEAR SYSTEMS AND MATRICES 3.1 Introduction to Linear Systems 173 3.2 Matrices and Gaussian Elimination 177 3.3 Reduced Row-Echelon Matrices 183 3.4 Matrix Operations 192 3.5 Inverses of Matrices 199 3.6 Determinants 208 3.7 Linear Equations and Curve Fitting 219

CONTENTS

iv Copyright © 2018 Pearson Education, Inc. 4 VECTOR SPACES 4.1 The Vector Space 3 R 229 4.2 The Vector Space R n and Subspaces 235 4.3 Linear Combinations and Independence of Vectors 241 4.4 Bases and Dimension for Vector Spaces 249 4.5 Row and Column Spaces 256 4.6 Orthogonal Vectors in R n 262 4.7 General Vector Spaces 268 5 HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS 5.1 Introduction: Second-Order Linear Equations 275 5.2 General Solutions of Linear Equations 282 5.3 Homogeneous Equations with Constant Coefficients 290 5.4 Mechanical Vibrations 298 5.5 Nonhomogeneous Equations and Undetermined Coefficients 309 5.6 Forced Oscillations and Resonance 322 6 EIGENVALUES AND EIGENVECTORS 6.1 Introduction to Eigenvalues 335 6.2 Diagonalization of Matrices 349 6.3 Applications Involving Powers of Matrices 361 7 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 7.1 First-Order Systems and Applications 379 7.2 Matrices and Linear Systems 388 7.3 The Eigenvalue Method for Linear Systems 395 7.4 A Gallery of Solution Curves of Linear Systems 427 7.5 Second-Order Systems and Mechanical Applications 433

v Copyright © 2018 Pearson Education, Inc. 7.6 Multiple Eigenvalue Solutions 445 7.7 Numerical Methods for Systems 464 8 MATRIX EXPONENTIAL METHODS 8.1 Matrix Exponentials and Linear Systems 473 8.2 Nonhomogeneous Linear Systems 483 8.3 Spectral Decomposition Methods 491 9 NONLINEAR SYSTEMS AND PHENOMENA 9.1 Stability and the Phase Plane 511 9.2 Linear and Almost Linear Systems 520 9.3 Ecological Applications: Predators and Competitors 538 9.4 Nonlinear Mechanical Systems 553 10 LAPLACE TRANSFORM METHODS 10.1 Laplace Transforms and Inverse Transforms 565 10.2 Transformation of Initial Value Problems 570 10.3 Translation and Partial Fractions 579 10.4 Derivatives, Integrals, and Products of Transforms 588 10.5 Periodic and Piecewise Continuous Input Functions 595 11 POWER SERIES METHODS 11.1 Introduction and Review of Power Series 609 11.2 Power Series Solutions 615 11.3 Frobenius Series Solutions 628 11.4 Bessel Functions 642 APPENDIX A Existence and Uniqueness of Solutions 649

PREFACE

This is a solutions manual to accompany the textbook DIFFERENTIAL EQUATIONS & LINEAR ALGEBRA (4th edition, 2018) by C. Henry Edwards, David E. Penney, and David T. Calvis. We include solutions to most of the problems in the text. The corresponding Student’s Solutions Manual contains solutions to most of the odd-numbered solutions in the text.

Our goal is to support teaching of the subject of differential equations with linear algebra in every way that we can. We therefore invite comments and suggested improvements for future printings of this manual, as well as advice regarding features that might be added to increase its usefulness in subsequent editions. Additional supplementary material can be found at the Expanded Applications website listed below.

Henry Edwards David Calvis

h.edwards@mindspring.com dcalvis@bw.edu

http://goo.gl/UYnW2g

CHAPTER 1

FIRST-ORDER DIFFERENTIAL EQUATIONS

SECTION 1.1

DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS

The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of differential equations, and to show the student what is meant by a solution of a differential equation. Also, the use of differential equations in the mathematical modeling of real-world phenomena is outlined.

Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the given differential equations. We include here just some typical examples of such verifications.

8. If 1 coscos2 y xx and 2 sincos2 y xx , then 1 sin2sin2, y xx 

cos4cos2, y xx

3. If 1 cos2 y x  and 2 sin2 y x  , then 1 2sin2 y x   2 2cos2 y x   , so 11 4cos24 y xy   and 22 4sin24 y xy   . Thus 1140yy   and 2240yy   4. If 3 1 x ye  and 3 2 x ye  , then 3 1 3 x ye  and 3 2 3 x ye  , so 3 11 99 x yey   and 3 22 99 x yey   . 5. If x x y ee , then x x y ee   , so  2.xxxxx yy eeeee   Thus 2. x y ye   6. If 2 1 x ye  and 2 2 x yxe  , then 2 1 2 x ye   , 2 1 4 x ye   , 22 2 2 x x yexe   , and 22 2 44. x x yexe   Hence  222 1114444240 xxx yyyeee  and 

22244444240.







22222

xxxxx yyyexeexexe



 2 cos2sin2 y

  , and 2 sin4cos2. y xx   Hence

11 cos4cos2coscos23cos2 yyxxxxx   and

22 sin4cos2sincos23cos2. yyxxxxx  

14.

15. Substitution of rx y e  into 20

, which simplifies to

16. Substitution of rx y e  into 3340 yyy

gives the equation 2 3340, rxrxrx reree which simplifies to 2 3340

. The quadratic formula then gives the solutions

The verifications of the suggested solutions in Problems 17-26 are similar to those in Problems 1-12. We illustrate the determination of the value of C only in some typical cases. However, we illustrate typical solution curves for each of these problems.

2 Chapter 1: First-Order Differential Equations Copyright © 2018 Pearson Education, Inc. 11. If 2 1 yyx  , then 3 2 y x   and 4 6, y x   so  224325465240. xyxyyxxxxx  If 2 2 ln yyxx  , then 332ln y xxx   and 44 56ln y xxx   , so   2244332 22222 5456ln52ln4ln 556104ln0. x yxyyxxxxxxxxxx xxxxxx   13. Substitution of rx y e  into 32yy   gives the equation 32rxrx ree 

simplifies



r  .

, which

to 32. r

Thus 2/3



 

rxrx ree  , which simplifies to 2 41. r  Thus 1/2 r 

Substitution of rx y e

into 4 yy

gives the equation 2 4

yyy gives the equation 2 20rxrxrx



2 2(2)(1)0.rrrr Thus 2 r  or 1 r  .

reree







r 

3576

17.

 18.

−4 0 4 −4 0 4 x y (0, 2) Problem 17 −5 0 5 −5 0 5 x y (0, 3) Problem 18

2 C

3 C 

Section 1.1: Differential Equations and Mathematical Models 3



C 





010



110 C 

11 C  . 21. 7 C 

 ()ln yxxC 





ln0 C 

1 C  .

, then 05 y  gives 15 C  , so 6

20. If

1 x yxCex

, then

gives

, or

22. If

, then

00 y

gives

, so





 . 24.

 . −5 0 5 −10 −5 0 5 10 x y (0, 5) Problem 19 −10 −5 0 5 10 −20 0 20 x y (0, 10) Problem 20 −2 −1 0 1 2 −10 −5 0 5 10 x y (0, 7) Problem 21 −20 −10 0 10 20 −5 0 5 x y (0, 0) Problem 22

23. If 52 1 4 () yxxCx

, then 21 y  gives 11 48321 C

, or 56 C

17 C

4 Chapter 1: First-Order Differential Equations

25. If  3 tan y xC , then 01 y  gives the equation tan1 C  . Hence one value of C is /4 C   , as is this value plus any integral multiple of  .

26. Substitution of x   and 0 y  into  cos yxCx  yields  01

27. y x y

28. The slope of the line through  ,x y and 

, so

, so the differential equation is 2 xyy

x is

C  

C 

 

2,0

0 2 /2 y yyx xx  

 

0 1 2 3 −30 −20 −10 0 10 20 30 x y (2, 1) Problem

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −30 −20 −10 0 10 20 30 x y (1, 17) Problem

−2 −1 0 1 2 −4 −2 0 2 4 x y (0, 1) Problem

0 5 10 −10 −5 0 5 10 x y (, 0) Problem 26

29. If m y   is the slope of the tangent line and m  is the slope of the normal line at (,), xy then the relation 1 mm   yields

for y

then gives the differential equation  1

30. Here m y   and 2 ()2 x mDxkx   , so the orthogonality relation 1 mm   gives the differential equation 21. xy  

31. The slope of the line through  ,x y and (,) y x is

, so the differential equation is (). xyyy x  

In Problems 32-36 we get the desired differential equation when we replace the “time rate of change” of the dependent variable with its derivative with respect to time t, the word “is” with the = sign, the phrase “proportional to” with k, and finally translate the remainder of the given sentence into symbols.

37. The second derivative of any linear function is zero, so we spot the two solutions

1 yx

and () y xx

of the differential equation

38. A function whose derivative equals itself, and is hence a solution of the differential equation yy   , is () x y xe  .

39. We reason that if 2 y kx  , then each term in the differential equation is a multiple of 2 x The choice 1 k  balances the equation and provides the solution 2 () y xx  .

40. If y is a constant, then 0 y   , so the differential equation reduces to 2 1 y  . This gives the two constant-valued solutions ()1yx  and ()1yx 

41. We reason that if x y ke  , then each term in the differential equation is a multiple of x e . The choice 1 2 k  balances the equation and provides the solution 1 2 () x yxe 

42. Two functions, each equaling the negative of its own second derivative, are the two solutions  cos yxx  and ()sin y xx  of the differential equation yy   .

Section 1.1: Differential

Equations and Mathematical Models

 110myyx   . Solving

yyx   .



 yxyyx  

32. dPdtkP  33. 2 dvdtkv  34.  250 dvdtkv  35. dNdtkPN  36. dNdtkNPN 





y   .



43. (a) We need only substitute  ()1 x tCkt  in both sides of the differential equation

2 x kx

 for a routine verification.

(b) The zero-valued function ()0xt  obviously satisfies the initial value problem

44. (a) The figure shows typical graphs of solutions of the differential equation

(b) The figure shows typical graphs of solutions of the differential equation

We see that—whereas the graphs with

appear to “diverge to infinity”—each solution with 1

appears to approach 0

from the resulting solution

Indeed, we see from the Problem

Substitution of 1 P

 and 10 P  into the differential equation

Problem 43(a) yields a solution of the form

so we get the solution

PkP

gives

We now find readily that 100 P  when 49 t  and that 1000 P  when 49.9 t  . It appears that P grows without bound (and thus “explodes”) as t approaches 50.

6 Chapter 1: First-Order Differential Equations



 

2 x kx

, (0)0 x 

2 1 2 x x   .

2 1 2 x x  

1 2





as . t 

43(a) solution  1 2 ()1 x tCt  that ()xt  as 2 tC  . However, with 1 2 k  it is

 1 2 ()1 x tCt  that () x t remains bounded on any

()0xt 



45.



 

1 100 , k  so

 1 100 ()1 PtCt  .

(0)2 P 

1 2 , C 

1100 () 1 50 2100 Pt t t 

clear

bounded interval, but

as t

The initial condition

now yields

0 1 2 3 4 0 1 2 3 4 5 t x Problem 44a 0 1 2 3 4 0 1 2 3 4 5 6 t x Problem 44b

46. Substitution of 1 v   and 5 v  into the differential equation 2 vkv   gives

Problem 43(a) yields a solution of the form  ()125 vtCt  . The initial condition (0)10 v  now yields

, C  so we get the solution

We now find readily that 1 v  when 22.5 t  and that 0.1 v  when 247.5 t  . It appears that v approaches 0 as t increases without bound. Thus the boat gradually slows, but never comes to a “full stop” in a finite period of time.

47. (a) (10)10 y  yields 10110 C  , so 10110 C  .

(b) There is no such value of C, but the constant function ()0yx  satisfies the conditions 2 y y   and (0)0 y  .

(c) It is obvious visually (in Fig. 1.1.8 of the text) that one and only one solution curve passes through each point (,) ab of the xy-plane, so it follows that there exists a unique solution to the initial value problem

48. (b) Obviously the functions

() vxx  both satisfy the differential equation 4.xyy   But their derivatives

() uxx  and

and

match at

, where both are zero. Hence the given piecewise-defined function 

is differentiable, and therefore satisfies the differential equation because 

and vx do so (for 0 x  and 0 x  , respectively).

satisfies the given differential equation for every real number value of C

Section 1.2: Integrals as General and Particular Solutions 7

1 25 , k

150 () 1 52 1025 vt t t    .

 so

1 10

2 y y   , () y ab  .

3 ()4 uxx  

3 ()4 vxx  



0 x

 4 4 if 0 if 0 Cxx yx Cxx       

SECTION 1.2

INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS

This section introduces general solutions and particular solutions in the very simplest situation — a differential equation of the form

— where only direct integration and evaluation of the constant of integration are involved. Students should review carefully the elementary concepts of velocity and acceleration, as well as the fps and mks unit systems.

8 Chapter 1: First-Order Differential



 

1. Integration of 21 y x   yields  2 ()21 y xxdxxxC   . Then substitution of 0 x  , 3 y  gives 300 CC , so  2 3 yxxx 2. Integration of  2 2 yx   yields   23 1 3 22 yxxdxxC   . Then substitution of 2 x  , 1 y  gives 10 CC , so   3 1 3 21 yxx 3. Integration of yx   yields  3/2 2 3 y xxdxxC   . Then substitution of 4 x  , 0 y  gives 16 3 0 C  , so   3/2 2 3 8 yxx . 4. Integration of 2 y x   yields  2 1 y xxdxxC   . Then substitution of 1 x  , 5 y  gives 51 C  , so  16yxx . 5. Integration of  12 2 yx   yields  12 222 yxxdxxC   . Then substitution of 2 x  , 1 y  gives 122 C  , so  225yxx 6. Integration of 12 2 9 yxx   yields    221232 1 3 99 y xxxdxxC   . Then substitution of 4 x  , 0 y  gives 3 1 3 0(5) C  , so  3/2 2 1 3 9125 yxx  7. Integration of 2 10 1 y x    yields  1 2 10 10tan 1 yxdxxC x    . Then substitution of 0 x  , 0 y  gives 0100 C  , so  1 10tan yxx  . 8. Integration of cos2 y x   yields  1 2 cos2sin2 y xxdxxC   . Then substitution of 0 x  , 1 y  gives 10 C  , so  1 2 sin21yxx .

yfx

Section 1.2: Integrals as General and Particular Solutions 9 Copyright © 2018 Pearson Education, Inc. 9. Integration of 2 1 1 y x   yields 1 2 1 ()sin 1 yxdxxC x   . Then substitution of 0 x  , 0 y  gives 00 C  , so  1 sin yxx  . 10. Integration of x y xe   yields  11 xuux y xxedxueduuexeC   , using the substitution ux  together with Formula #46 inside the back cover of the textbook. Then substituting 0 x  , 1 y  gives 11, C  so ()(1)2. x yxxe

If  50 at  , then  0 50505010vtdttvt   . Hence  22 0 50102510251020xttdtttxtt   12. If  20 at  , then  0 20202015vtdttvt   . Hence  22 0 2015101510155 x ttdtttxtt   13. If  3 att  , then  3322 0 22 35vttdttvt  . Hence  233 3 11 0 222555 x ttdtttxtt   . 14. If  21att , then  22 0 217vttdtttvtt   . Hence  2331111 0 3232 7774 x tttdttttxttt   . 15. If  2 43att , then    233 44 33 433337vttdttCt  (taking 37 C  so that 01 v  ). Hence     411344 33333733733726 xttdtttCtt   . 16. If  1 4 at t   , then  1 24245 4 vtdttCt t    (taking 5 C  so that 01 v  ). Hence     3/23/2 29 44 2454545333 xttdtttCtt   (taking 293 C  so that 01 x  ).

11.

17. If

, then

). Hence

Students should understand that Problems 19-22, though different at first glance, are solved in the same way as the preceding ones, that is, by means of the fundamental theorem of calculus in the form

cited in the text. Actually

19. The graph of vt shows that

, leading to the graph of

shown.

Alternate solution for Problem 19 (and similarly for 20-22): The graph of

The graph of

is shown.

10 Chapter

Education, Inc.

 3 1



   322111 111222vttdttCt  (taking 1 2 C  so that 00 v  ). Hence     11111211 222221111 xttdtttCtt  (taking 1 2 C  so that 00 x  ).

 50sin5 att 

 50sin510cos510cos5 vttdttCt   (taking 0 C  so that 010



 10cos52sin52sin510xttdttCt   (taking 10 C  so

08 x 

1: First-Order Differential Equations

2018 Pearson

att

18. If

, then

that

 0 0 t t x txtvsds  

 0 , t x tvsds   since 0t and  0 x t are each given to be zero.

in these problems

 5if 05 10if 510 t vt tt       , so that  1 2 1 2 2 5if 05 10if 510 tCt xt ttCt       . Now 1 0 C  because 00 x  , and continuity of  x t requires that  5 x tt  and  2 1 2 2 10 x tttC 

5 t 

that 25 2 2 C 

 x t

agree when

. This implies

 5if 05 10if 510 t vt tt       .

t  ,  0 t x tvsds   is given by 0 55 t dst   , whereas for 510 t  we

 5 005 222 5 510 7525 2510251010. 22222 tt st s xtvsdsdssds stt stt         



shows that

Thus for 05

have

 x

20. The graph of vt shows that  if 05 5if 510

21. The graph of vt shows that

Section 1.2: Integrals as General and Particular Solutions 11

t      

that  2 1 1 2 2 if 05 5if 510 tCt xt tCt       . Now 1 0 C  because 00 x  , and continuity of  x t requires that  2 1 2 x tt  and  2 5 x ttC  agree when 5 t  . This implies that 25 2 2 C  , leading to the graph of  x t shown.

 if 05 10if 510 tt vt tt       , so that  2 1 1 2 2 1 2 2 if 05 10if 510 tCt xt ttCt       . Now 1 0 C  because 00 x  , and continuity of  x t requires that  2 1 2 x tt  and  2 1 2 2 10 x tttC  agree when 5 t  . This implies that 2 25 C  , leading to the graph of  x t shown. 22. For 03 t  , 5 3 () vtt  , so  2 5 1 6 x ttC  . Now 1 0 C  because 00 x  , so  2 5 6 x tt  on this first interval, and its right-endpoint value is  15 2 3 x  . For 37 t  ,  5 vt  , so  2 5 x ttC  Now 15 2 (3) x  implies that 15 2 2 C  , so  15 2 5 xtt on this second interval, and its right-endpoint value is  55 2 7 x  For 710 t  ,  5 3 57vt , so  550 33vtt . Hence  2 550 3 63 x tttC  , and 55 2 (7) x  implies that 290 3 6 C  . Finally,  2 1 6 (5100290)xttt on this third interval, leading to the graph of  x t shown. 0 2 4 6 8 10 0 10 20 30 40 t x (5, 25) Problem 19 0 2 4 6 8 10 0 10 20 30 40 t x (5, 12.5) Problem 20

, so

23.  9.849vtt , so the ball reaches its maximum height ( 0 v  ) after 5 t  seconds. Its maximum height then is

2 54.95495122.5 meters y  .

24. 32 vt  and 2 16400yt , so the ball hits the ground ( 0 y  ) when 5 sec t  , and then  325160 ft/sec v 

25. 2 10 m/s a  and 0 100 km/h27.78 m/s v  , so 1027.78vt , and hence  2 527.78 x ttt  . The car stops when 0 v  , that is 2.78 s t  , and thus the distance traveled before stopping is 2.7838.59 meters x 

26. 9.8100vt and 2 4.910020ytt .

(a) 0 v  when 1009.8 s t  , so the projectile's maximum height is

2 1009.84.91009.81001009.820530 y  meters.

(b) It passes the top of the building when  2 4.91002020yttt , and hence after 1004.920.41 t  seconds.

are 0.20,20.61 t  . Hence the projectile is in the air 20.61 seconds.

. The ball hits the ground when 0 y  and 9.81060 m/s vt

ytty

. Hence the height of the building is

12 Chapter 1: First-Order

Differential Equations



  

27.

a 





 2 0 4.95.10105.10178.57



0 2 4 6 8 10 0 10 20 30 40 t x (5, 12.5) Problem

0 2 4 6 8 10 0 10 20 30 40 t x (3, 7.5) (7, 27.5) Problem 22



2 9.8 m/s

, so 9.810vt

and 2

4.910





so 5.10 s t

m y

28. 3240vt and 2 1640555ytt . The ball hits the ground ( 0 y  ) when 4.77 s t  , with velocity 4.77192.64 ft/s vv , an impact speed of about 131 mph.

29. Integration of 2 0.120.6 dvdttt  with 00 v  gives  0.040.332 vttt  . Hence

1070 ft/s v  . Then integration of 0.040.332 dxdttt  with 00 x  gives

 0.010.143 x ttt  , so 10200 ft x  . Thus after 10 seconds the car has gone 200 ft and is traveling at 70 ft/s.

30. Taking 0 0 x  and 0 60 mph88 ft/s v  , we get 88 vat , and 0 v  yields 88 ta  . Substituting this value of t, as well as 176 ft x  , into 2 288 x att leads to 2 22 ft/s a  . Hence the car skids for 88224s t  .

31. If 2 20 m/s a  and 0 0 x  , then the car's velocity and position at time t are given by 0 20 vtv  and 2 0 10 x tvt . It stops when 0 v  (so 0 20 vt  ), and hence when

22 75102010 x tttt

Thus

32. Starting with 0 0 x  and 4 0 50km/h510m/h v

, we find by the method of Problem 30 that the car's deceleration is

. Then, starting with 0 0 x  and

. Thus doubling the initial velocity quadruples the distance the car skids.

when 0 v

33. If 0 0 v  and 0 20 y  , then vat

yields

34. On Earth: 0 32 vtv  , so 0 32 tv  at maximum height (when 0 v  ). Substituting this value of t and 144 y  in 2 0 16 ytvt  , we solve for 0 96ft/s v  as the initial speed with which the person can throw a ball straight upward.

On Planet Gzyx: From Problem 33, the surface gravitational acceleration on planet Gzyx is 2 10ft/s a  , so 1096vt

. Therefore 0 v  yields 9.6s t  and so  max 9.6460.8ft yy is the height a ball will reach if its initial velocity is 96ft/s .

and 2 596 y tt

Section 1.2: Integrals as General and Particular Solutions 13







7.5s t  , so 0 207.554.77m/s197km/hr v

25310m/h72



5 0 100km/h10m/h v 

0 tva 

2 1 0 2 x atvt and

x 





, we substitute

into

find that 60m



2 1 2 20 yat . Substitution of 2 t  , 0 y 

2 10ft/s a  .

0 0 v  and 0 200 y  , then 10 vt  and 2 5200yt . Hence 0 y 



and

when 40210s t  and 201063.25ft/s v



Differential Equations and Linear Algebra 4th Edition edwards Solutions Manual Full Download: http://testbanktip.com/download/differential-equations-and-linear-algebra-4th-edition-edwards-solutions-manual/

35. If 0 0 v  and 0y h  , then the stone’s velocity and height are given by vgt  and 2 0.5 y gth , respectively. Hence 0 y  when 2 thg  , so 22 vghggh

36. The method of solution is precisely the same as that in Problem 30. We find first that, on Earth, the woman must jump straight upward with initial velocity 0 12ft/s v  to reach a maximum height of 2.25 ft. Then we find that, on the Moon, this initial velocity yields a maximum height of about 13.58 ft.

37. We use units of miles and hours. If

, then the car’s velocity and position after t hours are given by vat

and

, respectively. Since 60 v

when 56 t

, the velocity equation yields . Hence the distance traveled by 12:50 pm is

38. Again we have vat  and

at  . But now 60 v  when 35 x  . Substitution of 60 at  (from the velocity equation) into the position equation yields

39. Integration of

, and the initial condition 120 y  gives 3 CvS  . Hence the swimmers trajectory is

 3 3341 S yxvxx Substitution of 121 y  now gives 6mph vS 

40. Integration of  4 3116 y x   yields  5 3485 yxxC  , and the initial condition

120 y  gives 65 C  . Hence the swimmers trajectory is

  5 1515486yxxx , and so his downstream drift is 122.4miles y 

41. The bomb equations are 32 a  , 32 vt  , and 2 16800 B sst

with 0 t  at the instant the bomb is dropped. The projectile is fired at time 2, t  so its corresponding equations are 32 a  ,

(the arbitrary constant vanishing because 20 sP  ). Now the condition

16800400 B stt gives 5 t  , and then the further requirement that 5400 sP  yields 0 544/3181.33ft/s v  for the projectile’s needed initial velocity.

14 Chapter 1: First-Order

Differential Equations



xv



2 1 2 x at 



 2 1 2

x 



725625miles

2 1 2

 2 1 2

ttt



356030

, whence 76h t

, that is, 1:10 pm.

  2

 

  3 334 S y vxxC

 

914 S y vx

yields







vtv 

 2 0 1622 P sstvt  for 2 t 

0 322

, and

 2

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Differential Equations and Linear Algebra 4th Edition edwards Solutions Manual

PREFACE

CHAPTER 1

FIRST-ORDER DIFFERENTIAL EQUATIONS

SECTION 1.1

DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS

4 Chapter 1: First-Order Differential Equations

SECTION 1.2

INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS

Articles inside

4 Chapter 1: First-Order Differential Equations

FIRST-ORDER DIFFERENTIAL EQUATIONS

PREFACE