Differential Equations and Boundary Value Problems – Instructor’s Solutions Manual (6th Edition) | ISBN 9780137540365 | Edwards, Penney & Calvis | Complete Worked Solutions

INSTRUCTOR’S SOLUTIONS MANUAL

DIFFERENTIAL EQUATIONS AND BOUNDARY VALUE PROBLEMS
6TH EDITION

CHAPTER NO. 01: 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 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.

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.
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  e3 x and y2  e 3 x , then y1  3 e3 x and y2   3 e 3 x , so y1  9e3 x  9 y1 and
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 .

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

,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.

If y  y2  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.


13. Substitution of y  erx into 3 y   2 y gives the equation 3r e rx  2 e rx , 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
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 ,
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 2 e rx  3r e rx  4 e rx  0
, which simplifies to 3r 2  3r  4  0 . The quadratic formula then 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.

17. C2 18. C 3

Problem 17 Problem 18
4 5


(0, 3)
(0, 2)



y y
0 0




−4 −5
−4 0 4 −5 0 5
x x

,19. If y  x   Ce x  1 , then y  0   5 gives C  1  5 , so C  6 .

20. If y  x   C e x  x  1 , then y  0   10 gives C  1  10 , or C  11 .

Problem 19 Problem 20
10 20



5 (0, 5) (0, 10)



y y
0 0



−5



−10 −20
−5 0 5 −10 −5 0 5 10
x x

21. C  7.

22. If y ( x)  ln  x  C  , then y  0   0 gives ln C  0 , so C  1 .

Problem 21 Problem 22
10 5
(0, 7)

5


y y
0 0
(0, 0)


−5



−10 −5
−2 −1 0 1 2 −20 −10 0 10 20
x x

23. If y ( x )  14 x 5  C x 2 , then y  2   1 gives 14 32  C  81  1 , or C  56 .

24. C  17 .

, Problem 21 Problem 22
10 5
(0, 7)

5


y y
0 0
(0, 0)


−5



−10 −5
−2 −1 0 1 2 −20 −10 0 10 20
x x

23. If y ( x )  14 x 5  C x 2 , then y  2   1 gives 14  32  C  81  1 , or C  56 .

24. C  17 .

Problem 23 Problem 24
30 30

20 20 (1, 17)

10 10

y (2, 1) y
0 0

−10 −10

−20 −20

−30 −30
0 1 2 3 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
x x


25. If y  tan  x 3  C  , then y  0   1 gives the equation tan C  1 . Hence one value of C is
C   / 4 , as is this value plus any integral multiple of  .