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.


Created by MedConnoisseur ©2025/2026

, 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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Created by MedConnoisseur ©2025/2026

, 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 2 x and y2  sin 2 x , then y1   2sin 2 x y2  2 cos 2 x , so
y1  4 cos 2 x  4 y1 and y2  4sin 2 x  4 y2 . Thus y1  4 y1  0 and y2  4 y2  0 .

4. If y1  e 3 x and y 2  e 3 x , then y1  3 e3 x and y2   3 e 3 x , so y1  9e 3 x  9 y1 and
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y2  9e 3 x  9 y2 .


5. If y  e x  e x , then y  e x  e  x , so y   y   e x  e  x    e x  e  x   2 e  x . Thus
y  y  2 e x .
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6. If y1  e 2 x and y2  x e 2 x , then y1   2 e 2 x , y1  4 e 2 x , y 2  e 2 x  2 x e 2 x , and
y 2   4 e 2 x  4 x e 2 x . Hence
y1  4 y1  4 y1   4 e 2 x   4  2 e 2 x   4  e 2 x   0
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and
y2  4 y2  4 y2    4e 2 x
 4 x e 2 x   4  e 2 x  2 x e 2 x   4  x e 2 x   0.

8. If y1  cos x  cos 2 x and y2  sin x  cos 2 x , then y1   sin x  2sin 2 x,
y1   cos x  4 cos 2 x, y2  cos x  2sin 2 x , and y2   sin x  4 cos 2 x. Hence
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y1  y1    cos x  4 cos 2 x    cos x  cos 2 x   3cos 2 x
and
y2  y2    sin x  4cos 2 x    sin x  cos 2 x   3cos 2 x.
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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
x 2 y   5 x y   4 y  x 2  6 x 4   5 x  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
x 2 y   5 x y  4 y  x 2  5 x 4  6 x 4 ln x   5 x  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 3 y   2 y gives the equation 3r erx  2 erx , which simplifies
to 3 r  2. Thus r  .

14. Substitution of y  erx into 4 y  y gives the equation 4r 2 e rx  e rx , 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 r 2 e rx  r e rx  2 e rx  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
3r 2e rx  3r e rx  4 e rx  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
illustrate typical solution curves for each of these problems.
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17. C2 18. C 3
Problem 17 Problem 18
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(0, 3)
(0, 2)



y y
0 0
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−4 −5
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−4 0 4 −5 0 5
x x


Copyright © 2018 Pearson Education, Inc.