Instructor’s Solutions Manual — Differential Equations & Linear Algebra, 4th Edition — C. Henry Edwards, David E. Penney & David T. Calvis

Differential Equations and
Linear Algebra – 4th Edition
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INSTRUCTOR’S
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SOLUTIONS
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MANUAL
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C. Henry Edwards
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David E. Penney

David T. Calvis
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Comprehensive Solutions Manual for
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Instructors and Students
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© C. Henry Edwards, David E. Penney & David T. Calvis
All rights reserved. Reproduction or distribution without permission is prohibited.


©STUDYSTREAM

, TABLE OF CONTENTS
Differential Equations and Linear Algebra – 4th
Edition
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C. Henry Edwards, David E. Penney & David T. Calvis
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Chapter 1. First- Order Differential Equations
Chapter 2. Mathematical Models and Numerical Methods
Chapter 3. Linear Systems and Matrices
Chapter 4. Vector Spaces
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Chapter 5. Higher- Order Linear Differential Equations
Chapter 6. Eigenvalues and Eigenvectors
Chapter 7. Linear Systems of Differential Equations
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Chapter 8. Matrix Exponential Methods
Chapter 9. Nonlinear Systems and Phenomena
Chapter 10. Laplace Transform Methods
Chapter 11. Power Series Methods
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, CHAPTER 1
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FIRST-ORDER DIFFERENTIAL EQUATIONS
SECTION 1.1
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DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS

The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of dif-
ferential 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.
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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.
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3. If y1  cos 2x and y2  sin 2x , then y1   2sin 2x y2  2 cos 2x , so
y1  4cos 2x  4y1 and y2  4sin 2x  4 y2 . Thus y1 4 y1  0 and y2  4 y2  0 .

4. If y1  e3x and y 2  e3x , then y1  3 e3x and y 2   3 e3x , so y1  9e3x  9 y 1 and
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y2  9e3x  9 y 2 .


5. If y  ex  ex , then y  ex  ex , so y  y  ex  ex   ex  ex   2 e x . Thus
y  y  2 ex .
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6. If y1  e2x and y 2  x e2 x , then y1   2 e2 x , y1  4 e2 x , y2  e2 x  2x e2 x , and
y2   4 e2 x  4x e2 x . Hence

y1 4 y1  4 y1   4 e2x   4 2 e2x   4 e2x   0
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and
y2  4 y2  4 y2    4 e2x  4x e2x   4 e2x  2x e2x   4  x e2x   0.
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8. If y1  cos x  cos 2x and y2  sin x  cos 2x , then y1   sin x  2sin 2x,
y1   cos x  4 cos 2x, y2  cos x  2sin 2x , and y2   sin x  4 cos 2x. Hence
y1 y1    cos x  4 cos 2x  cos x  cos 2x  3cos 2x
and
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y2 y2  sin x  4cos 2x   sin x  cos 2x  3cos 2x.



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Copyright © 2018 Pearson Education, Inc.

, 2 Chapter 1: First-Order Differential Equations

11. If y  y1  x 2 , then y  2 x 3 and y  6 x4 , so

  
x2 y  5x y  4 y  x2 6 x4  5x 2 x3  4 x2     0.
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If y  y 2  x2 ln x , then y  x3  2 x3 ln x and y   5 x4  6 x4 ln x , so

    
x2 y  5x y  4 y  x2 5 x4  6 x4 ln x  5x x 3  2 x3 ln x  4 x 2 ln x 
 5 x2  5 x2   6 x2 10 x2  4 x2 ln x  0.
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13. Substitution of y  erx into 3y  2 y gives the equation 3r erx  2erx , which simplifies
to 3 r  2. Thus r  2/ 3.

14. Substitution of y  erx into 4 y  y gives the equation 4r 2 erx  erx , which simplifies to
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4 r 2  1. Thus r   .

15. Substitution of y  erx into y  y  2 y  0 gives the equation r2erx  r erx  2 erx  0 ,
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which simplifies to r 2  r  2  (r  2)(r 1)  0. Thus r  2 or r  1.

16. Substitution of y  erx into 3 y  3 y  4 y  0 gives the equation
3r2erx  3r erx  4erx  0, which simplifies to 3r 2  3r  4  0 . The quadratic formula then
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gives the solutions r  3  57   6.

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
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illustrate typical solution curves for each of these problems.

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