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280 Copyright © 2015 Pearson Education, Inc.

CHAPTER 5

LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS

Along with Chapter 4, this chapter is designed to offer considerable flexibility in the treatment of linear systems, depending on the background in linear algebra that students are assumed to have. Sections 4.1 and 4.2 of the previous chapter can stand alone as a brief introduction to linear sys-tems without the use of linear algebra and matrices. But this chapter employs the notation and terminology of elementary linear algebra. For ready reference and review, Section 5.1 includes a complete and self-contained account of the needed background of determinants, matrices, and vectors. The additional linear algebra that is needed in subsequent sections is introduced along the way.

SECTION 5.1

MATRICES AND LINEAR SYSTEMS

The first half-dozen pages of this section are devoted to a review of matrix notation and termi-nology. With students who've had some prior exposure to matrices and determinants, this review material can be skimmed rapidly. In this event serious study of the section can begin with the subsections on matrix-valued functions and first-order linear systems. About all that’s actually needed for this purpose is some acquaintance with determinants, with matrix multiplication and inverse matrices, and with the fact that a square matrix is invertible if and only if its determinant is nonzero.

1. (a) 4 6 9 12 13 18

2 38 14 15 3 23 17

A B

(b) 6 9 6 8 0 1

3 212 21 10 2 2 19

A B

(c) 9 11

47 9

AB (d)

10 37

14 8

BA

2. 9 11 0 2 33 7 2 3 12 10

47 9 3 1 27 103 4 7 3 9

AB C A BC

2 3 3 2 18 4 9 11 9 7

4 7 8 0 68 8 47 9 21 1

A B C AB AC

Section 5.1 281

Copyright © 2015 Pearson Education, Inc.

3. 1 8

46 1

AB ;

11 12 14

14 0 7

0 8 13

BA

4. 2

2

2 cos

3 4sin 5cos

t t

t t t

Ay ;

2 3

14

6 2

t

t

t e

t

t e

Bx

The products Ax and By are not defined, because in neither case is the number of col-umns of the first factor equal to the number of rows of the second factor.

5. (a)

21 14 7 0 12 8 21 2 1

7 4 0 28 21 4 16 12 4 44 9

35 14 49 8 20 4 27 34 45

A B

(b)

9 6 3 0 15 10 9 21 13

3 5 0 12 9 5 20 15 5 8 24

15 6 21 10 25 5 25 19 26

A B

(c)

0 6 1

10 31 15

16 58 23

AB (d)

10 8 5

18 12 10

11 22 6

BA

(e)

3 2 1 0 0 3 2 1

0 4 3 0 0 0 4 3

5 2 7 0 0 5 2 7

t t

t t t

t t

A I

6. (a) 1 2

5 10

4 8

A B A B (b)

1 2 2 4 0 0

2 4 1 2 0 0

AB 0

7. det 0 7 0 det det AB A B

8. det det 144 AB BA (with AB and BA as in Problem 5)

9. 2 3 2 3 4 2 2 3

3 4 2 3 4 2 3 2 3

4 6 4 8 1 8 18 1 2 12 32

3 4 3 3 4 8 3 4

t t t t t t t t t t t t

t t t t t t t t t t t

AB

282 MATRICES AND LINEAR SYSTEMS


2 3 22 32

2 3 2 2 3

2 3 2 3 3 3

2 2 3

2 3 2 3

1 2 2 11 1 1 1

1 13 4 6 123

1 6 1 8 7 12 12 24

3 3 3 4 3 3 6 12

1 8 18 1 2 12 32

3 3 4 8 3 4

t tt t

t t t tt tt t

t t t t t t t t t

t t t t t t t t

t t t t t

t t t t t

A B AB

10. 3 2

4 3

3 3 2 9 3 2 2

3 3

3 24 2 12 24 2

t t t t t

t t

t e te t e e te

t

t t e t e

AB

2

2 3

2 2

3 3

2

3

1 2 3 0

1 0 0 2 0 2 2

8 0 3 3 8 1 3

6 3 2 3 2

3 6

24 9 3 2

9 3 2 2

3

12 24 2

t t

t t

t t t

t

t t t

t

e t e t t

e t e

t t t t

t e e t te

t t e

t e e te

t e

A B AB

11. x

y

x ; 0 3

3 0t

P ; 0

0t

f

12. x

y

x ; 3 2

2 1t

P ; 0

0t

f

13. x

y

x ; 2 4

5 1t

P ; 2

3 tet

t

f

14. x

y

x ; 2

t

t

t et

e t

P ; cos

sin

tt

t

f

Section 5.1 283


15.

x

y

z

x ; 0 1 1

1 0 1

1 1 0

t

P ; 0

0

0

t

f

16.

x

y

z

x ; 2 3 0

1 1 2

0 5 7

t

P ; 0

0

0

t

f

17.

x

y

z

x ; 3 4 1

1 0 3

0 6 7

t

P ; 2

3

t

t t

t

f

18.

x

y

z

x ; 2

3

1

2 1

3

t

t

t e

t t

e t t

P ; 0

0

0

t

f

19.

1

2

3

4

x

x

x

x

x ;

0 1 0 0

0 0 2 0

0 0 0 3

4 0 0 0

t

P ;

0

0

0

0

t

f

20.

1

2

3

4

x

x

x

x

x ;

0 1 1 0

0 0 1 1

1 0 0 1

1 1 0 0

t

P ; 2

3

0

tt

t

t

f

21. 2

3

2

20

3

t tt

t t

e eW t e

e e

;

1 1

2 2 2

2 22 2 2

221 2

1 1 2 2 1 2 221 2

4 22 2 2

3 13 3 3

4 22

3 12

22

33

t t t

t t t

t t t

t t t

t tt t

t tt t

e e e

e e e

e e e

e e e

c e c ee et c c c c

c e c ee e

x Ax

x Ax

x x x



In most of Problems 22–30, we omit the verifications of the given solutions. In each case, this is simply a matter of calculating both the derivative ix of the given solution vector and the product

iAx (where A is the coefficient matrix in the given differential equation) to verify that i i x Ax

(just as in the verification of the solutions 1x and 2x in Problem 21 above).

22. 3 2

3 2

25 0

3

t tt

t t

e eW t e

e e

3 23 2

1 21 1 2 2 1 2 3 23 2

1 2

22

33

t tt t

t tt t

c e c ee et c c c c

c e c ee e

x x x

23. 2 2

2 24 0

5

t t

t t

e eW t

e e

2 2

2 2 1 21 1 2 2 1 2 2 2

1 2

1 1

1 5 5

t tt t

t t

c e c et c c c e c e

c e c e

x x x

24. 3 2

5

3 20

2

t tt

t t

e eW t e

e e

3 2

3 2 1 21 1 2 2 1 2 3 2

1 2

1 1

1 2 2

t tt t

t t

c e c et c c c e c e

c e c e

x x x

25. 2 5

3

2 5

37 0

2 3

t tt

t t

e eW t e

e e

2 52 5

1 21 1 2 2 1 2 2 52 5

1 2

33

2 32 3

t tt t

t tt t

c e c ee et c c c c

c e c ee e

x x x

26. 3 5

5 9

3 5

2 2 2

2 0 2 16 0

t t t

t t t

t t t

e e e

W t e e e

e e e

3 5

1 2 33 5 5

1 1 2 2 3 3 1 2 3 1 33 5

1 2 3

2 2 2 2 2 2

2 0 2 2 2

1 1 1

t t t

t t t t t

t t t

c e c e c e

t c c c c e c e c e c e c e

c e c e c e

x x x x

Section 5.1 285


27. 2

2

2

0

0 3 0

t t

t t

t t t

e e

W t e e

e e e

2

1 22 2

1 1 2 2 3 3 1 2 3 1 32

1 2 3

1 1 0

1 0 1

1 1 1

t t

t t t t t

t t t

c e c e

t c c c c e c e c e c e c e

c e c e c e

x x x x

1 1 2 2

3 3

2 0 1 1 1 1 0 1 1 1

2 1 0 1 1 ; 0 1 0 1 0 ;

2 1 1 0 1 1 1 1 0 1

0 0 1 1 0

1 1 0 1 1

1 1 1 0 1

t t t t

t t

e e e e

e e

x Ax x Ax

x Ax

28. 3 4

3 4

3 4

1 2

6 3 2 84 0

13 2

t t

t t t

t t

e e

W t e e e

e e

3 4

1 2 33 4 3 4

1 1 2 2 3 3 1 2 3 1 2 33 4

1 2 3

1 2 1 2

6 3 2 6 3 2

13 2 1 13 2

t t

t t t t

t t

c c e c e

t c c c c c e c e c c e c e

c c e c e

x x x x

29. 2 3

2 3 2

2

3

2 0

2 0

t t t

t t t t

t t

e e e

W t e e e e

e e

2 3

1 2 32 3 2 3

1 1 2 2 3 3 1 2 3 1 2 32

1 2

3 1 1 3

2 1 1 2

2 1 0 2

t t t

t t t t t t

t t

c e c e c e

t c c c c e c e c e c e c e c e

c e c e

x x x x

30. 1

0 00

0 0 0 00 0 1 0

0 0 3 22 0

0 2 0

t tt t

t tt t

t t t tt t

t t

e ee e

e eW t e e

e e e ee e

e e



1 4

31 2 3 4

2 4

1 3

1 0 0 1

0 0 1 0

0 1 0 3 3

1 0 2 0 2

t t

tt t t t

t t

t t

c e c e

c et c e c e c e c e

c e c e

c e c e

x

In Problems 31–34 (and similarly in Problems 35–40) we give first the scalar components 1x t

and 2x t of a general solution, then the equations in the coefficients 1c and 2c that are obtained

when the given initial conditions are imposed, and finally the resulting particular solution of the given system.

31. 3 2 3 21 1 2 2 1 22 , 3t t t tx t c e c e x t c e c e

1 2 1 2

3 2 3 21 2

2 0, 3 5

2 2 , 6t t t t

c c c c

x t e e x t e e

32. 2 2 2 21 1 2 2 1 2, 5t t t tx t c e c e x t c e c e

1 2 1 2

2 2 2 21 2

5, 5 3

7 2 , 7 10t t t t

c c c c

x t e e x t e e

33. 3 2 3 21 1 2 2 1 2, 2t t t tx t c e c e x t c e c e

1 2 1 2

3 2 3 21 2

11, 2 7

15 4 , 15 8t t t t

c c c c

x t e e x t e e

34. 2 5 2 51 1 2 2 1 23 , 2 3t t t tx t c e c e x t c e c e

1 2 1 2

2 5 2 51 2

3 8, 2 3 0

8 489 2 ,

7 7t t t t

c c c c

x t e e x t e e

35. 3 5 5 3 51 1 2 3 2 1 3 3 1 2 32 2 2 , 2 2 ,t t t t t t t tx t c e c e c e x t c e c e x t c e c e c e

1 2 3 1 3 1 2 3

3 5 5 3 51 2 3

2 2 2 0, 2 2 0, 4

2 4 2 , 2 2 , 2t t t t t t t t

c c c c c c c c

x t e e e x t e e x t e e e

36. 2 2 21 1 2 2 1 3 3 1 2 3, ,t t t t t t tx t c e c e x t c e c e x t c e c e c e

1 2 1 3 1 2 3

2 2 21 2 3

10, 12, 1

7 3 , 7 5 , 7 8t t t t t t

c c c c c c c

x t e e x t e e x t e e

Section 5.1 287


37. 2 3 2 3 21 1 2 3 2 1 2 3 3 1 23 , 2 , 2t t t t t t t tx t c e c e c e x t c e c e c e x t c e c e

1 2 3 1 2 3 1 2

2 3 2 3 21 2 3

3 1, 2 2, 2 3

9 3 5 , 6 3 5 , 6 3t t t t t t t t

c c c c c c c c

x t e e e x t e e e x t e e

38. 2 3 2 3 21 1 2 3 2 1 2 3 3 1 23 , 2 , 2t t t t t t t tx t c e c e c e x t c e c e c e x t c e c e

1 2 3 1 2 3 1 2

2 3 2 3 21 2 3

3 5, 2 7, 2 11

6 15 4 , 4 15 4 , 4 15t t t t t t t t

c c c c c c c c

x t e e e x t e e e x t e e

39. 1 1 4 2 3 3 2 4 4 1 3, , 3 , 2t t t t t t tx t c e c e x t c e x t c e c e x t c e c e

1 4 3 2 4 1 3

1 2 3 4

1, 1, 3 1, 2 1

3 2 , , 7 6 , 3 2t t t t t t t

c c c c c c c

x t e e x t e x t e e x t e e

40. 1 1 4 2 3 3 2 4 4 1 3, , 3 , 2t t t t t t tx t c e c e x t c e x t c e c e x t c e c e

1 4 3 2 4 1 3

1 2 3 4

1, 3, 3 4, 2 7

13 12 , 3 , 40 36 , 13 6t t t t t t t

c c c c c c c

x t e e x t e x t e e x t e e

41. (a) 2 1tx x , so neither is a constant multiple of the other.

(b) 1 2, 0W x x , whereas Theorem 2 would imply that 0W if 1x and 2x were inde-

pendent solutions of a system of the indicated form.

42. If 12 11x t cx t and 22 21x t cx t , then

11 22 12 21 11 21 11 21 0W t x t x t x t x t cx t x t cx t x t .

43. Suppose 11 22 12 21 0W a x a x a x a x a . Then the coefficient determinant of

the homogeneous linear system

1 11 2 12 1 21 2 220, 0c x a c x a c x a c x a

vanishes. The system therefore has a non-trivial solution 1 2,c c such that a x 0 . It

therefore follows (by uniqueness of solutions) that t x 0 , that is, 1 1 2 2c t c t x x 0

with 1c and 2c not both zero. Thus the solution vectors 1x and 2x are linearly depend-

ent.

44. The argument is precisely the same, except with n solution vectors each having n compo-nent functions (rather than 2 solution vectors each having 2 component functions).



45. Suppose that 1 1 2 2 n nc t c t c t x x x 0 . Then the ith scalar component of this

vector equation is 1 1 2 2 0i i n inc x t c x t c x t . Hence the fact that the scalar

functions 1ix t , 2ix t , inx t are linearly independent implies that

1 2 0nc c c . Consequently the vector functions 1 2, , , nt t tx x x are linear-

ly independent.

SECTION 5.2

THE EIGENVALUE METHOD FOR HOMOGENEOUS LINEAR SYSTEMS

In each of Problems 1–16 we give the characteristic equation, the eigenvalues 1 and 2 of the

coefficient matrix of the given system, the corresponding equations determining the associated

eigenvectors T

1 1 1a bv and T2 2 2a bv , these eigenvectors, and the resulting scalar com-

ponents 1x t and 2x t of a general solution 1 21 1 2 2

t tt c e c e x v v of the system. Finally,

the figure for each Problem shows a direction field and some typical solution curves for the sys-tem.

1. Characteristic equation 2 2 3 0 ;

Eigenvalues 1 1 and 2 3 ;

Eigenvector equations 1

1

2 2 0

2 2 0

a

b

and 2

2

2 2 0

2 2 0

a

b

;

Eigenvectors T

1 1 1 v and T2 1 1v ;

3 31 1 2 2 1 2,t t t tx t c e c e x t c e c e

Section 5.2 289


−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 1

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 2




1

3 3 0

2 2 0

a

b

and 2

2

2 3 0

2 3 0

a

b

;

Eigenvectors T

1 1 1 v and T

2 3 2v ;

4 41 1 2 2 1 23 , 2t t t tx t c e c e x t c e c e




1

4 4 0

3 3 0

a

b

and 2

2

3 4 0

3 4 0

a

b

;

Eigenvectors T

1 1 1 v and T

1 4 3v ;

6 61 1 2 2 1 24 , 3t t t tx t c e c e x t c e c e ;

The equations

1 1 2 2 1 20 4 1, 0 3 1x c c x c c

yield 1

1

7c and 2

2

7c , so the desired particular solution is given by

6 61 2

1 18 , 6

7 7t t t tx t e e x t e e .

290 THE EIGENVALUE METHOD FOR HOMOGENEOUS LINEAR SYSTEMS


−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 3

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 4




1

6 1 0

6 1 0

a

b

and 2

2

1 1 0

6 6 0

a

b

;

Eigenvectors T

1 1 6 v and T

2 1 1v ;

2 5 2 51 1 2 2 1 2, 6t t t tx t c e c e x t c e c e




1

7 7 0

1 1 0

a

b

and 2

2

1 7 0

1 7 0

a

b

;

Eigenvectors T

1 1 1v and T

2 7 1v ;

5 51 1 2 2 1 27 ,t t t tx t c e c e x t c e c e

Section 5.2 291


−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 6




1

6 5 0

6 5 0

a

b

and 2

2

5 5 0

6 6 0

a

b

;

Eigenvectors T1 5 6 v and T

2 1 1 v ;

3 4 3 41 1 2 2 1 25 , 6t t t tx t c e c e x t c e c e

The initial conditions yield 1 1c and 2 6c , so

3 4 3 41 25 6 , 6 6t t t tx t e e x t e e




1

4 4 0

6 6 0

a

b

and 2

2

6 4 0

6 4 0

a

b

;

Eigenvectors T

1 1 1v and T

2 2 3 v ;

9 91 1 2 2 1 22 , 3t t t tx t c e c e x t c e c e



−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 7

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 8

8. Characteristic equation 2 4 0 ;

Eigenvalue 2i ;

Eigenvector equation 1 2 5 0

1 1 2 0

i a

i b

;

Eigenvector T5 1 2i v ;

2

5cos2 5 sin 2

cos2 2sin 2 (sin 2 2cos2 )it

t i tt e

t t i t t

x v ;

1 1 2

2 1 2 1 2 1 2

5 cos2 5 sin 2 ,

cos2 2sin 2 sin 2 2cos2 2 cos2 2 sin 2

x t c t c t

x t c t t c t t c c t c c t


Eigenvalue 4i ;


4 2 4 0

i a

i b

;


The real and imaginary parts of

4

5cos4 5 sin 4

2cos4 4sin 4 2sin 4 4cos4it

t i tt e

t t i t t

x v

yield the general solution

Section 5.2 293


1 1 2

2 1 2

5 cos4 5 sin 4 ,

2cos4 4sin 4 2sin 4 4cos4

x t c t c t

x t c t t c t t

The initial conditions 1 0 2x and 2 0 3x give 1

2

5c and 2

11

20c , so the desired

particular solution is

1 2

11 12cos4 sin 4 , 3cos4 sin 4

4 2x t t t x t t t

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 9

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 10


Eigenvalue 3i ;


9 3 3 0

i a

i b

;


3

2cos3 2 sin 3

3cos3 3sin 3 3sin 3 3cos3it

t i tt e

t t i t t

x v ;

1 1 2

2 1 2

1 2 2 1

2 cos3 2 sin 3 ,

3cos3 3sin 3 3cos3 3sin 3

3 3 cos3 3 3 sin 3

x t c t c t

x t c t t c t t

c c t c c t


Eigenvalue 1 2i ;



Eigenvector equation 2 2 0

2 2 0

i a

i b

;

Eigenvector T1 iv ;

The real and imaginary parts of

T T T1 cos2 sin 2 cos2 sin 2 sin 2 cos2t t tt i e t i t e t t ie t t x

yield the general solution

1 1 2 2 1 2cos2 sin 2 , sin 2 cos2t tx t e c t c t x t e c t c t

The particular solution with 1 0 0x and 2 0 4x is obtained with 1 0c and 2 4c ,

so

1 24 sin 2 , 4 cos2t tx t e t x t e t .

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 11

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 12


Eigenvalue 2 2i ;


1 1 2 0

i a

i b

;


(2 2 ) 2

5cos2 5 sin 2

cos2 2sin 2 sin 2 2cos2i t t

t i tt e e

t t i t t

x v ;

Section 5.2 295


21 1 2

22 1 2

21 2 1 2

5 cos2 5 sin 2 ,

cos2 2sin 2 2cos2 sin 2

2 cos2 2 sin 2

t

t

t

x t e c t c t

x t e c t t c t t

e c c t c c t


Eigenvalue 2 3i ;


2 3 3 0

i a

i b

;

Eigenvector T3 1 i v ;

(2 3 ) 2

3cos3 3 sin 3

cos3 sin 3 cos3 sin 3i t t

t i tt e e

t t i t t

x v

2 21 1 2 2 1 2 1 23 cos3 sin 3 , cos3 sin 3t tx t e c t c t x t e c c t c c t

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 13

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 14


Eigenvalue 3 4i ;

Eigenvector equation 4 4 0

4 4 0

i a

i b

;

Eigenvector T1 i v ;

(3 4 ) 3 cos4 sin 4

sin 4 cos4i t t t i t

t e et i t

x v ;



3 31 1 2 2 1 2cos4 sin 4 , sin 4 cos4t tx t e c t c t x t e c t c t


Eigenvalue 5 4i ;


4 2 4 0

i a

i b

;


5 4 5

5cos4 5 sin 4

2cos4 4sin 4 4cos4 2sin 4i t t

t i tt e e

t t i t t

x v ;

5 51 1 2 2 1 2 1 25 cos4 sin 4 , 2 4 cos4 4 2 sin 4t tx t e c t c t x t e c c t c c t

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 15

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 16




1

40 20 0

100 50 0

a

b

and 2

2

50 20 0

100 40 0

a

b

;

Eigenvectors T

1 1 2v and T

2 2 5 v ;

10 100 10 1001 1 2 2 1 22 , 2 5t t t tx t c e c e x t c e c e


Eigenvalues 1 9 , 2 6 , 3 0 ;

Eigenvector equations

Section 5.2 297


1 2 3

1 2 3

1 2 3

5 1 4 0 2 1 4 0 4 1 4 0

1 2 1 0 , 1 1 1 0 , 1 7 1 0

4 1 5 0 4 1 2 0 4 1 4 0

a a a

b b b

c c c

;

Eigenvectors T

1 1 1 1v , T

2 1 2 1 v , T

3 1 0 1 v ;

9 6 9 6 9 61 1 2 3 2 1 2 3 1 2 3, 2 ,t t t t t tx t c e c e c x t c e c e x t c e c e c


Eigenvalues 1 9 , 2 6 , 3 0 ;


1 2 3

1 2 3

1 2 3

8 2 2 0 5 2 2 0 1 2 2 0

2 2 1 0 , 2 1 1 0 , 2 7 1 0

2 1 2 0 2 1 1 0 2 1 7 0

a a a

b b b

c c c

;

Eigenvectors T

1 1 2 2v , T

2 0 1 1 v , T

3 4 1 1 v ;

9 9 6 9 61 1 3 2 1 2 3 3 1 2 34 , 2 , 2t t t t tx t c e c x t c e c e c x t c e c e c

19. Characteristic equation 3 212 45 54 0 ;

Eigenvalues 1 6 , 2 3 , 3 3 ;


1 2 3

1 2 3

1 2 3

2 1 1 0 1 1 1 0 1 1 1 0

1 2 1 0 , 1 1 1 0 , 1 1 1 0

1 1 2 0 1 1 1 0 1 1 1 0

a a a

b b b

c c c

;

Eigenvectors T

1 1 1 1v , T

2 1 2 1 v , T

3 1 0 1 v ;

6 3 3 6 3 6 3 31 1 2 3 2 1 2 3 1 2 3, 2 ,t t t t t t t tx t c e c e c e x t c e c e x t c e c e c e


Eigenvalues 1 9 , 2 6 , 3 2 ;


1 2 3

1 2 3

1 2 3

4 1 3 0 1 1 3 0 3 1 3 0

1 2 1 0 , 1 1 1 0 , 1 5 1 0

3 1 4 0 3 1 1 0 3 1 3 0

a a a

b b b

c c c

;

Eigenvectors T

1 1 1 1v , T

2 1 2 1 v , T

3 1 0 1 v ;



9 6 2 9 6 9 6 21 1 2 3 2 1 2 3 1 2 3, 2 ,t t t t t t t tx t c e c e c e x t c e c e x t c e c e c e

21. Characteristic equation 3 0 ;

Eigenvalues 1 0 , 2 1 , 3 1 ;


1 2 3

1 2 3

1 2 3

5 0 6 0 4 0 6 0 6 0 6 0

2 1 2 0 , 2 2 2 0 , 2 0 2 0

4 2 4 0 4 2 5 0 4 2 3 0

a a a

b b b

c c c

;

Eigenvectors T

1 6 2 5v , T

2 3 1 2v , T

3 2 1 2v ;

1 1 2 3 2 1 2 3 3 1 2 36 3 2 , 2 , 5 2 2t t t t t tx t c c e c e x t c c e c e x t c c e c e


Distinct eigenvalues 1 2 , 2 1 , 3 3 ;


1 2 3

1 2 3

1 2 3

5 2 2 0 2 2 2 0 0 2 2 0

5 2 2 0 , 5 5 2 0 , 5 7 2 0

5 5 5 0 5 5 2 0 5 5 0 0

a a a

b b b

c c c

;

Eigenvectors T

1 0 1 1 v , T

2 1 1 0 v , T

3 1 1 1 v ;

3 2 3 2 31 2 3 2 1 2 3 3 1 3, ,t t t t t t tx t c e c e x t c e c e c e x t c e c e


Eigenvalues 1 2 , 2 2 , 3 3 ;


1 2 3

1 2 3

1 2 3

1 1 1 0 5 1 1 0 0 1 1 0

5 5 1 0 , 5 1 1 0 , 5 6 1 0

5 5 1 0 5 5 5 0 5 5 0 0

a a a

b b b

c c c

;

Eigenvectors T

1 1 1 0 v , T

2 0 1 1 v , T

3 1 1 1 v ;

2 3 2 2 3 2 31 1 3 2 1 2 3 3 2 3, ,t t t t t t tx t c e c e x t c e c e c e x t c e c e


Eigenvalues 1 and 2i ;

Section 5.2 299


With 1 the eigenvector equation 1

1

1

1 1 1 0

4 4 1 0

4 4 1 0

a

b

c

gives the eigenvector

T

1 1 1 0 v . To find an eigenvector Ta b cv associated with 2i we

must find a nontrivial solution of the equations

2 2 0,

4 3 2 0,

4 4 2 2 0.

i a b c

a i b c

a b i c

Subtraction of the first two equations yields

6 2 4 2 0i a i b ,

so we take 2a i and 3b i . Then the first equation gives 3c i . Thus

T2 3 3i i i v . Finally

2

2

2 2cos2 sin 2 cos2 2sin 2 ,

3 3cos2 sin 2 3sin 2 cos2 ,

it

it

i e t t i t t

i e t t i t t

so the solution is

1 1 2 3

2 1 2 3

3 2 3

2cos2 sin 2 cos2 2sin 2 ,

3cos2 sin 2 cos2 3sin 2 ,

3cos2 sin 2 3sin 2 cos2 .

t

t

x t c e c t t c t t

x t c e c t t c t t

x t c t t c t t


Eigenvalues 0 and 2 3i ;


1

1

5 5 2 0

6 6 5 0

6 6 5 0

a

b

c


T

1 1 1 0 v . With 2 3i we solve the eigenvector equation

3 3 5 2 0

6 8 3 5 0

6 6 3 3 0

i a

i b

i c

to find the complex-valued eigenvector T

1 1 2 2i v . The corresponding com-

plex-valued solution is



2 3 2

cos3 sin 3 cos3 sin 3

2cos3 2 sin 3

2cos3 2 sin 3

i t t

t t i t t

t e e t i t

t i t

x v .

The scalar components of the resulting general solution are

21 1 2 3 2 3

22 1 2 3

23 2 3

cos3 sin 3 ,

2 cos3 sin 3 ,

2 cos3 sin 3 .

t

t

t

x t c e c c t c c t

x t c e c t c t

x t e c t c t


Eigenvalues 3 and 1 i ;


1

1

0 0 1 0

9 4 2 0

9 4 4 0

a

b

c


T

1 4 9 0v . With 1 i we solve the eigenvector equation

4 0 1 0

9 2 0

9 4 0

i a

i b

i c

to find the complex-valued eigenvector T

1 1 2 4i i v . The corresponding

complex-valued solution is

1

cos sin

2cos sin cos 2sin

4cos sin cos 4sin

i t t

t i t

t e e t t i t t

t t i t t

x v

with real and imaginary parts 2 tx and 3 tx . Assembling the general solution

1 1 2 2 3 3c c c x x x x , we get the scalar equations

31 1 2 3

32 1 2 3 2 3

3 2 3 2 3

4 cos sin ,

9 2 cos 2 sin ,

4 cos 4 sin .

t t

t t

t

x t c e e c t c t

x t c e e c c t c c t

x t e c c t c c t

Finally, the given initial conditions yield the values 1 1c , 2 4c , and 3 1c , so the

desired particular solution is

31

32

3

4 4cos sin ,

9 9cos 2sin ,

17 cos .

t t

t t

t

x t e e t t

x t e e t t

x t e t

Section 5.2 301


27. The coefficient matrix 0.2 0

0.2 0.4

A has characteristic equation

2 0.6 0.08 0 with eigenvalues 1 0.2 and 2 0.4 . We find easily that the

associated eigenvectors are T

1 1 1v and T

2 0 1v , so we get the general solution

0.2 0.2 0.41 1 2 1 2,t t tx t c e x t c e c e .

The initial conditions 1 0 15x , 2 0 0x give 1 15c and 2 15c , so we get

0.2 0.2 0.41 215 , 15 15t t tx t e x t e e .

To find the maximum value of 2x t , we solve the equation 2x t for 5ln 2t , which

gives the maximum value 2 5ln 2 3.75 lbx . The figure shows the graphs of 1x t and

2x t .

0 5 10 15 200

5

10

15

x 1

x 2

t

x

Problem 27

0 5 10 15 20

0

5

10

15

x1

x2

t

x

Problem 28

28. The coefficient matrix 0.4 0

0.4 0.25

A has characteristic equation

2 0.65 0.10 0 with eigenvalues 1 0.4 and 2 0.25 . We find easily that

the associated eigenvectors are T

1 3 8 v and T

2 0 1v , so we get the general so-

lution

0.4 0.4 0.251 1 2 1 23 , 8t t tx t c e x t c e c e .

The initial conditions 1 0 15x , 2 0 0x give 1 5c and 2 40c , so we get

0.2 0.2 0.41 2 1 215 , 8t t tx t e x t c e c e 0.4 0.4 0.25

1 215 , 40 40t t tx t e x t e e .



To find the maximum value of 2x t , we solve the equation 2 0x t for 20 8

ln3 5mt ,

which gives the maximum value 2 6.85 lbmx t . The figure shows the graphs of 1x t

and 2x t .

29. The coefficient matrix 0.2 0.4

0.2 0.4

A has eigenvalues 1 0 and 2 0.6 , with

eigenvectors T

1 2 1v and T

2 1 1 v that yield the general solution

0.6 0.61 1 2 2 1 22 ,t tx t c c e x t c c e .

The initial conditions 1 0 15x , 2 0 0x give 1 2 5c c , so we get

0.6 0.61 210 5 , 5 5t tx t e x t e .

The figure shows the graphs of 1x t and 2x t .

0 5 10 150

5

10

15

x1

x2

t

x

Problem 29

0 5 10 15

0

5

10

15

x1

x2

t

x

Problem 30

30. The coefficient matrix 0.4 0.25

0.4 0.25

A has eigenvalues 1 0 and 2 0.65 , with

eigenvectors T

1 5 8v and T

2 1 1 v that yield the general solution

0.65 0.651 1 2 2 1 25 , 8t tx t c c e x t c c e .

The initial conditions 1 0 15x , 2 0 0x give 1

15

13c and 2

120

13c , so we get

0.65 0.651 2

1 175 120 , 120 120

13 13t tx t e x t e .

Section 5.2 303


The figure shows the graphs of 1x t and 2x t .

31. The coefficient matrix

1 0 0

1 2 0

0 2 3

A

has as eigenvalues its diagonal elements 1 1 , 2 2 , and 3 3 . We find readily

that the associated eigenvectors are T

1 1 1 1v , T

2 0 1 2v , and

T

3 0 0 1v . The resulting general solution is given by

2 2 31 1 2 1 2 3 1 2 3, , 2t t t t t tx t c e x t c e c e x t c e c e c e .

The initial conditions 1 0 27x and 2 20 0 0x x give 1 3 27c c and 2 27c ,

so we get

2 2 31 2 327 , 27 27 , 27 54 27t t t t t tx t e x t e e x t e e e .

The equation 3 0x t simplifies to the equation

23 4 1 3 1 1 0t t t te e e e ,

with positive solution ln 3mt . Thus the maximum amount of salt ever in tank 3 is

3 ln 3 lb4x . The figure shows the graphs of 1x t , 2x t , and 3x t .

0 50

5

10

15

20

25

x1

x2

x3

t

x

Problem 31

0 5

0

5

10

15

20

25

30

35

40

45

x1

x2 x

3

t

x

Problem 32




3 0 0

3 2 0

0 2 1

A



1 1 3 3 v , T

2 0 1 2 v , and

T


3 3 2 3 21 1 2 1 2 3 1 2 3, 3 , 3 2t t t t t tx t c e x t c e c e x t c e c e c e .

The initial conditions 1 0 45x and 2 20 0 0x x give 1 45c , 2 135c , and

3 135c , so we get

3 3 2 3 21 2 345 , 135 135 , 135 270 135t t t t t tx t e x t e e x t e e e .


23 4 1 3 1 1 0t t t te e e e ,

with positive solution ln 3mt . Thus the maximum amount of salt ever in tank 3 is

3 ln 3 20 lbx . The figure shows the graphs of 1x t , 2x t , and 3x t .


4 0 0

4 6 0

0 6 2

A



1 1 2 6 v , T

2 0 2 3 v , and

T


4 4 6 4 6 21 1 2 1 2 3 1 2 3, 2 2 , 6 3t t t t t tx t c e x t c e c e x t c e c e c e .


3 135c , so we get

4 4 6 4 6 21 2 345 , 90 90 , 270 135 135t t t t t tx t e x t e e x t e e e .


4 2 2 23 4 1 3 1 1 0t t t te e e e ,

Section 5.2 305


with positive solution 1

ln 32mt . Thus the maximum amount of salt ever in tank 3 is

3

1ln 3 20 lb

2x

. The figure shows the graphs of 1x t , 2x t , and 3x t .

0 20

5

10

15

20

25

30

35

40

45

x1

x2

x3

t

x

Problem 33

0 4

0

5

10

15

20

25

30

35

40

x1

x2 (dashed)

x3

t

x

Problem 34


3 0 0

3 5 0

0 5 1

A



1 4 6 15 v , T

2 0 4 5 v , and

T


3 3 5 3 51 1 2 1 2 3 1 2 34 , 6 4 , 15 5t t t t t tx t c e x t c e c e x t c e c e c e .


3 75c , so we get

3 3 5 3 51 2 340 , 60 60 , 150 75 75t t t t t tx t e x t e e x t e e e .


4 2 2 25 6 1 5 1 1 0t t t te e e e ,



with positive solution 1

ln52mt . Thus the maximum amount of salt ever in tank 3 is

3

1 lbln5 21.4663

2x

. The figure shows the graphs of 1x t , 2x t , and 3x t .


6 0 3

6 20 0

0 20 3

A

has characteristic equation

3 229 198 18 11 0 ,

with eigenvalues 0 0 , 1 18 , and 2 11 . We find that associated eigenvectors

are T

0 10 3 20v , T

1 1 3 4 v , and T

2 3 2 5 v . The resulting gen-

eral solution is given by

18 111 0 1 2

18 112 0 1 2

18 113 0 1 2

10 3

3 3 2

20 4 5 .

t t

t t

t t

x t c c e c e

x t c c e c e

x t c c e c e

The initial conditions 1 0 33x and 2 20 0 0x x give 1 1c , 2

55

7c , and

3

72

7c , so we get

18 111

18 112

18 113

110 55 216

71

3 165 1447

120 220 360 .

7

t t

t t

t t

x t e e

x t e e

x t e e

Thus the limiting amounts of salt in tanks 1, 2, and 3 are 10 lb, 3 lb, and 20 lb. The fig-ure shows the graphs of 1x t , 2x t , and 3x t .

Section 5.2 307


0 10

5

10

15

20

25

30

x1

x2

x3

t

x

Problem 35

0 5 10

0

5

10

15

x1

x2

x3

t

x

Problem 36


1 10

2 21 1

02 5

1 10

5 2

A

has characteristic equation 3 26 90

5 20 , with eigenvalues 0 0 ,

1

32

10i , and 2

32

10i . The eigenvector equation

1 2 0 1 2 0

1 2 1 5 0 0

0 1 5 1 2 0

a

b

c

associated with the eigenvalue 0 0 yields the associated eigenvector T

0 1 5 2 1v

and consequently the constant solution 0 0t x v . Then the eigenvector equation

.

1 11 3 0

10 2 01 1

(4 3 ) 0 02 10

01 10 (1 3 )

5 10

ia

i b

ci

.



associated with 1

32

10i yields the complex-valued eigenvector

T

1

1 11 3 1 3 1

2 2i i

v .

The corresponding complex-valued solution is

6 3 10 3 /51 1

3 3 3 3cos 3sin 3cos sin

10 10 10 10

1 3 3 3 3cos 3sin 3cos sin

2 10 10 10 10

3 32cos 2 sin

10 10

i t t

t t t ti

t t t tt e e i

t ti

x v .

The scalar components of the resulting general solution 0 0 1 1 2 1Re Imc c c x x x x

are given by

3 511 0 1 2 1 22

3 512 0 1 2 1 22

3 53 0 1 2

3 33 cos 3 sin

10 10

5 3 33 cos 3 sin

2 10 10

3 3cos sin .

10 10

t

t

t

t tx t c e c c c c

t tx t c e c c c c

t tx t c e c c

When we impose the initial conditions 1 0 18x and 2 20 0 0x x we find that

0 4c , 1 4c , and 2 8c . This finally gives the particular solution

3 51

3 52

3 53

3 34 14cos 2sin

10 10

3 310 10cos 10sin

10 10

3 34 4cos 8sin .

10 10

t

t

t

t tx t e

t tx t e

t tx t e

Thus the limiting amounts of salt in tanks 1, 2, and 3 are 4 lb, 10 lb, and 4 lb. The figure below shows the graphs of 1x t , 2x t , and 3x t .


1 0 2

1 3 0

0 3 2

A

Section 5.2 309


has characteristic equation 3 26 11 0 , with eigenvalues 0 0 , 1 3 2i ,

and 1 3 2i . The eigenvector equation

1 0 2 0

1 3 0 0

0 3 2 0

a

b

c

associated with the eigenvalue 0 0 yields the associated eigenvector T

0 6 2 3v ,

and consequently the constant solution 0 0t x v . Then the eigenvector equation

2 2 0 2 0

1 2 0 0

00 3 1 2

i a

i b

ci

associated with 1 3 2i yields the complex-valued eigenvector

T

1

1 12 2 1 2 1

3 3i i

v .

The corresponding complex-valued solution is

3 2 31 1

2cos 2 2 sin 2 2 cos 2 2sin 2

1cos 2 2 sin 2 2 cos 2 sin 2

3

3cos 2 3 sin 2

i t t

t t i t t

t e e t t i t t

t i t

x v .

The scalar components of resulting general solution 0 0 1 1 2 1Re Imc c c x x x x are

given by

31 0 1 2 1 2

32 0 1 2 1 2

33 0 1 2

16 2 2 cos 2 2 2 sin 2

31

2 2 cos 2 2 sin 23

3 cos 2 sin 2 .

t

t

t

x t c e c c t c c t

x t c e c c t c c t

x t c e c t c t

When we impose the initial conditions 1 0 55x and 2 20 0 0x x we find that

0 5c , 1 15c , and 2

45

2c . This finally gives the particular solution



31

32

33

30 25cos 2 10 2 sin 2

2510 10cos 2 2 sin 2

2

4515 15cos 2 2 sin 2 .

2

t

t

t

x t e t t

x t e t t

x t e t t

Thus the limiting amounts of salt in tanks 1, 2, and 3 are 30 lb, 10 lb, and 15 lb. The fig-ure shows the graphs of 1x t , 2x t , and 3x t .

0 20

5

10

15

20

25

30

35

40

45

50

55

x1

x2

x3

t

x

Problem 37

In Problems 38-41 the Maple command with(linalg):eigenvects(A), the Mathematica command Eigensystem[A], or the MATLAB command [V,D] = eig(A) can be used to find the eigenvalues and associated eigenvectors of the given coefficient matrix A.

38. Characteristic equation: 1 2 3 4 0

Eigenvalues and associated eigenvectors:

T T T T

1 2 3 4

1 2 3 4 0 1 3 6 0 0 1 4 0 0 0 1

v

Scalar solution equations:

2

1 1

22 1

33 1 2 3

2 3 44 1 2

2

3 4

3 3

4 6 4

2t t t

t

t t

t

t

t t

x t c e

x t c e c

x t c e c e c e

x t c e c e c e

e

c e

Section 5.2 311


39. Characteristic equation: 2 21 4 0


T T T T

1 1 2 2

3 2 4 1 0 0 1 0 0 1 0 0 1 1 0 0

v


21 1 4

2 22 1 3 4

3 1 2

4 1

3

2

4

t t

t t t

t t

t

x t c e c e

x t c e c e c e

x t c e c e

x t c e

40. Characteristic equation: 2 24 25 0


T T T T

2 2 5 5

1 3 0 0 0 3 0 1 0 0 1 3 0 1 0 0

v


21 1

2 2 52 1 2 4

53 3

2 54 2 3

3 3

3

t

t t t

t

t t

x t c e

x t c e c e c e

x t c e

x t c e c e

41. The eigenvalues and respective eigenvalues are given by the following table:

T T T T

3 6 10 15

1 0 0 1 0 1 1 0 2 1 1 2 1 2 2 1

v

Hence the general solution has scalar component functions

3 10 151 1 3 4

6 10 152 2 3 4

6 10 153 2 3 4

103 151 44 3

2

2

2

2

t t t

t t t

t t t

tt t

x t c e c e c e

x t c e c e c e

x t c e c e

c e c e

c e

x t c e

The given initial conditions are satisfied by choosing 1 2 0c c , 3 1c , and 4 1c , so

the desired particular solution is given by

10 151 4

10 152 3

2

2

t t

t t

x t e e x t

x t e e x t



In Problems 42–50 we give a general solution in the form 1 21 1 2 2

t tt c e c e x v v that ex-

hibits explicitly the eigenvalues 1 2, , and corresponding eigenvectors 1 2, ,v v of the given

coefficient matrix A.

42. 2 51 2 3

3 1 2

1 1 3

2 1 1

t tt c c e c e

x

43. 2 4 81 2 3

3 1 1

1 1 1

5 1 3

t t tt c e c e c e

x

44. 3 6 121 2 3

3 7 5

2 1 3

2 5 3

t t tt c e c e c e

x

45. 3 3 61 2 3 4

1 1 2 1

1 2 1 1

1 1 1 2

1 1 1 1

t t tt c e c c e c e

x

46. 4 2 4 81 2 3 4

3 1 1 3

2 2 1 2

1 2 1 3

1 1 1 3

t t t tt c e c e c e c e

x

47. 3 3 6 91 2 3 4

2 1 2 1

2 2 1 1

1 1 1 2

1 1 1 1


x

48. 16 32 48 641 2 3 4

1 2 3 1

2 5 1 1

1 1 1 2

2 1 2 3


x

Section 5.2 313


49. 3 3 6 91 2 3 4 5

1 0 1 0 2

0 3 7 1 0

3 0 1 0 5

1 1 1 1 2

1 1 1 1 1

t t t tt c e c c e c e c e

x

50. 7 4 3 5 9 111 2 3 4 5 6

0 1 0 0 1 0

1 0 1 0 1 0

1 0 0 1 0 1

1 0 1 0 0 1

0 1 0 1 0 1

1 1 1 0 1 0

t t t t t tt c e c e c e c e c e c e

x

SECTION 5.3

SOLUTION CURVES OF LINEAR SYSTEMS

This section emphasizes the connection between the algebraic properties of the matrix A—specifically, its eigenvalues and eigenvectors—and the characteristic pattern of the phase dia-gram of the system x Ax .

In Problems 1-16 the eigenvalues, eigenvectors and phase portraits appear in the solutions to Section 5.2. Thus here we simply categorize each phase portrait according to the gallery in Fig. 5.3.16.

1. Saddle point (real eigenvalues of opposite sign)





6. Improper nodal source (distinct positive real eigenvalues)


8. Center (pure imaginary eigenvalues)


314 SOLUTION CURVES OF LINEAR SYSTEMS



11. Spiral source (complex conjugate eigenvalues with positive real part)





16. Improper nodal sink (distinct real eigenvalues)

In Problems 17-28 we “pigeonhole” each phase portrait according to the gallery in Fig. 5.3.16 and give the nature of the eigenvalues of the matrix A. Where appropriate we further give ap-proximate values of the corresponding eigenvectors.

17. Center; pure imaginary eigenvalues

18. Improper nodal source; distinct positive real eigenvalues; T

1 0 1v , T

2 1 1 v

19. Saddle point; real eigenvalues of opposite sign; T

1 0 1v corresponds to the negative

eigenvalue and T

2 1 1 v to the positive one.

20. Spiral source; complex conjugate eigenvalues with positive real part

21. Proper nodal source; repeated positive real eigenvalue with linearly independent eigen-vectors

22. Parallel lines; one zero and one negative real eigenvalue

23. Spiral sink; complex conjugate eigenvalues with negative real part

24. Improper nodal sink; distinct negative real eigenvalues; T

1 1 1v , T

2 1 4 v .

25. Saddle point; real eigenvalues of opposite sign; T

1 1 1v corresponds to the positive

eigenvalue and T

2 4 1 v to the negative one.

26. Center; pure imaginary eigenvalues

Section 5.3 315


27. Improper nodal source; distinct positive real eigenvalues; T

1 2 3v , T

2 2 1 v

28. Spiral sink; complex conjugate eigenvalues with negative real part

29. a) If 0 0v , then 0v t for all t, so that the point 20, ,0tu v u e (in oblique coor-

dinates) lies on the u-axis. The reverse argument applies if 0 0u .

b) If both 0u and 0v are nonzero, then solving 20

tu u e for t gives 0

1ln

2

ut

u , where-

as solving 50

tv v e for t gives 0

1ln

5

vt

v . We can thus eliminate t to conclude that

0 0

1 1ln ln

2 5

u v

u v . Then solving for v gives

0 0

5ln ln

2

v u

v u , or

5 2

0 0 0

5exp ln

2

v u u

v u u

, or finally 5 2

5 2 5 2 5 20 0 0

0

uv v v u u Cu

u

, where

5 20 0C v u .

30. The chain rule for vector-valued functions, together with the definition of tx as tx ,

gives

1d d d

t t t t t t tdt dt dt

x x x x x x

for all t. Because tx is a solution of the system x Ax , t x can be replaced with

tAx , which (by definition of x ) is tAx . Then the above displayed equation says

that t t x Ax for all t, which means that tx is a solution of the system

x Ax .

31. If is an eigenvalue of A with associated eigenvector v, then A I v 0 . Taking

the negative of both sides of this equation then gives A I v 0 0 . However,

A I can be written as A I , and so A I v 0 as well. It fol-

lows that is an eigenvalue of the matrix A with associated eigenvector v. This means that if A has positive eigenvalues 2 10 with associated eigenvectors 1v and

2v , then A has negative eigenvalues 1 2 0 associated to these same eigenvec-

tors 1v and 2v .

32. If we suppose that the system x Ax has a nonzero constant solution x, then

0d

dt x x for all t, which means that 0 Ax . Hence x is a nonzero constant vector



with Ax 0 . Conversely, if we assume that there exists a constant vector x 0 with Ax 0 , then x 0 Ax , so that the system x Ax has constant solutions other than

the zero solution.

33. a) Let 1v and 2v denote the two linearly independent eigenvectors of A associated with

the eigenvalue , so that 1 1Av v and 2 2Av v . If v is any two-dimensional vector,

then the fact that 1v and 2v are linearly independent implies that v can be written as a

linear combination of 1v and 2v , so that 1 1 2 2c c v v v for some scalars 1c and 2c . But

then the linearity of matrix multiplication gives

1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2c c c c c c c c Av A v v Av Av v v v v v ,

proving that v is an eigenvector of A.

b) Let a b

c d

A . Part a) implies that Av v for all vectors v, so in particular, if we

take T1 0v , then T T

0a c Av v , proving that a and 0c . Sim-

ilarly, if we take T0 1v , then T T

0b d Av v , proving that 0b and

d . Thus A is given by Eq. (22).

34. Substituting the expressions for 1x t and 2x t into the expressions for u and v gives

24cos10 sin10

5u t t 1

2cos10 2sin105

t t 10cos10 2 5 cos10

5t t

and

14cos10

5v t 2

sin10 2cos105

t t 52sin10 sin10 5 sin10

5t t t .

35. Write the given equation as 1 2 1 1 2 2, , 0M x x dx N x x dx , where

1 2 2 1, 6 8M x x x x and 1 2 1 2, 6 17N x x x x . The equation is exact because

2 1

6M N

x x

. Its general solution is therefore given by 1 2,F x x k , where 1 2,F x x

(as discussed in Section 1.6) satisfies the conditions 1

FM

x

and 2

FN

x

and k is a

constant. The first condition implies that

21 2 1 1 2 1 2, 6 4F x x M dx x x x h x ,

which specifies F up to the unknown function 2h x . Then the second condition gives

21 2 1 2 1 2 1 2

2

6 4 6 6 17x x x h x x h x x xx

,

Section 5.3 317


or simply 2 217h x x , which means that up to a constant, 22 2

17

2h x x . Altogeth-

er then, the general solution of the given equation is

2 21 2 1 2 1 2

17, 6 4

2F x x x x x x k .

36. Taking 4A , 6B , and 17

2C gives 2 2 17

4 6 4 4 100 02

B AC . Thus

the nontrivial solution curves are indeed elliptical.

37. With the same values of A, B, and C we find that 6 6 4

17 9 2 342

B

A C

. Thus it

suffices to confirm that 2

arctan4

satisfies 4

tan 23

. However, this is readily veri-

fied using the double-angle formula for the tangent function:

22

222 tan 44tan 2

1 tan 321

4

.

38. a) Let z r si , so that

3 53 5 3 5 5 3

44 4 4

Tr si ii r s i r s

z r sir si r si

v v

has real and imaginary parts 3 5 4T

r s r a and 5 3 4T

r s s b , respectively.

These are perpendicular if and only if

3 5 5 3 16 0r s r s rs a b ,

that is,

215 25r sr 9rs 215 16s rs 0 ,

or simply 2 2r s . This is the same as to say that 1z r ir r i .

b) If indeed s r , then 3 5 4T

r s r a is either 2 4T

r r or 8 4T

r r , each of

which is parallel to an axis of the elliptical trajectory shown in Fig. 5.3.12. The same is

true of 5 3 4T

r s s b , which becomes either 8 4T

r r or 2 4T

r r if s r .

39. a) The characteristic equation of A is given by det 0 A I , that is



2 0a b

a d bc a d ad bcc d

.

b) The quadratic formula shows that if the solutions to the characteristic equation are pure imaginary, then the coefficient a d of must vanish. Hence the trace

0T a d A , which means that d a . For the same reason, the constant term

ad bc must be positive. Substituting d a then gives 2 0ad bc a bc , which is impossible if 0c .

40. From the result of Problem 39, the characteristic equation of the matrix 5 17

8 7

A is

2 2 101 0 , whose roots (the eigenvalues of A) are 2 4 404

1 102

i .

An eigenvector Ta bv of A corresponding to 1 10i satisfies A I v 0 , or

5 1 10 17 0

8 7 1 10 0

i a

i b

,

or

6 10 17 0

8 6 10 0

i a

i b

,

which is equivalent to the system of equations

6 10 17 0,

8 6 10 0.

i a b

a i b

Both of these equations are satisfied if 17a and 6 10b i , and so we can take

17 6 10T

i v , with real and imaginary parts 17 6Ta and 0 10

Tb . Thus

by Eq. (5) of this Section, the general solution of the system x Ax is given by

1 2

17 0 0 17cos10 sin10 cos10 sin10

6 10 10 6t tt c e t t c e t t

x ,

or in scalar form,

1 1 2

2 1 2 2 1

17 cos10 17 sin10 ,

6 10 cos10 6 10 sin10 .

t

t

x t e c t c t

x t e c c t c c t

The initial condition 0 4 2Tx implies that 117 4c and 1 26 10 2c c , leading to

1

4

17c and 2

1

17c . All told, the solution of the initial value problem in Eq. (59) is

Section 5.3 319


1

2

4cos10 sin10 ,

2cos10 2sin10 ,

t

t

x t e t t

x t e t t

in agreement with Eq. (61).

SECTION 5.4

SECOND-ORDER SYSTEMS AND MECHANICAL APPLICATIONS

This section uses the eigenvalue method to exhibit realistic applications of linear systems. If a computer system like Maple, Mathematica, MATLAB, or even a TI-85/86/89/92/Nspire calcula-tor is available, then a system of more than three railway cars, or a multistory building with four or more floors (as in the project), can be investigated. However, the problems in the text are in-tended for manual solution.

Problems 1–7 involve the system

1 1 1 2 1 2 2

2 2 2 1 2 3 2

m x k k x k x

m x k x k k x

with various values of 1 2,m m and 1 2 3, ,k k k . In each problem we divide the first equation by 1m

and the second one by 2m to obtain a second-order linear system x Ax in the standard form

of Theorem 1 in this section. If the eigenvalues 1 and 2 are both negative, then the natural

(circular) frequencies of the system are 1 1 and 2 2 , and—according to Eq. (11)

in Theorem 1 of this section—the eigenvectors 1v and 2v associated with 1 and 2 determine

the natural modes of oscillations at these frequencies.

1. The matrix 2 2

2 2

A has eigenvalues 0 0 and 1 4 with associated eigenvec-

tors T

0 1 1v and T1 1 1 v . Thus we have the special case described in Eq. (12)

of Theorem 1, and a general solution is given by

1 1 2 1 2

2 1 2 1 2

cos2 sin 2 ,

cos2 sin 2 .

x t a a t b t b t

x t a a t b t b t

The natural frequencies are 1 0 and 2 2 . In the degenerate natural mode with

“frequency” 1 0 the two masses move by translation without oscillating. At frequen-

cy 2 2 they oscillate in opposite directions with equal amplitudes.

320 SECOND-ORDER SYSTEMS AND MECHANICAL APPLICATIONS


2. The matrix 5 4

5 5

A has eigenvalues 1 1 and 2 9 with associated eigen-

vectors T

1 1 1v and T2 1 1 v . Hence a general solution is given by

1 1 2 1 2

2 1 2 1 2

cos sin cos3 sin 3 ,

cos sin cos3 sin 3 .

x t a t a t b t b t

x t a t a t b t b t

3. The matrix 3 2

1 2

A has eigenvalues 1 1 and 2 4 , with associated eigen-

vectors T


1 1 2 1 2

2 1 2 1 2

cos sin 2 cos2 2 sin 2 ,

cos sin cos2 sin 2 .

x t a t a t b t b t

x t a t a t b t b t

The natural frequencies are 1 1 and 2 2 . In the natural mode with frequency 1 ,

the two masses 1m and 2m move in the same direction with equal amplitudes of oscilla-

tion. In the natural mode with frequency 2 they move in opposite directions with the

amplitude of oscillation of 1m twice that of 2m .

4. The matrix 3 2

2 3


vectors T


1 1 2 1 2

2 1 2 1 2

cos sin cos 5 sin 5 ,

cos sin cos 5 sin 5 .

x t a t a t b t b t

x t a t a t b t b t



tion. At frequency 2 they move in opposite directions with equal amplitudes.

5. The matrix 3 1

1 3


vectors T


1 1 2 1 2

2 1 2 1 2

cos 2 sin 2 cos2 sin 2 ,

cos 2 sin 2 cos2 sin 2 .

x t a t a t b t b t

x t a t a t b t b t

Section 5.4 321





6. The matrix 6 4

2 4


vectors T


1 1 2 1 2

2 1 2 1 2

cos 2 sin 2 2 cos 8 2 sin 8 ,

cos 2 sin 2 cos 8 sin 8 .

x t a t a t b t b t

x t a t a t b t b t

The natural frequencies are 1 2 and 2 8 . In the natural mode with frequency

1 , the two masses 1m and 2m move in the same direction with equal amplitudes of os-

cillation. In the natural mode with frequency 2 they move in opposite directions with

the amplitude of oscillation of 1m twice that of 2m .

7. The matrix 10 6

6 10

A has eigenvalues 1 4 and 2 16 with associated ei-

genvectors T


1 1 2 1 2

2 1 2 1 2

cos2 sin 2 cos4 sin 4 ,

cos2 sin 2 cos4 sin 4 .

x t a t a t b t b t

x t a t a t b t b t




8. Substitution of the trial solution 1 1 cos5x c t , 2 2 cos5x c t in the system

1 1 2 2 1 25 4 96cos5 , 4 5x x x t x x x

yields 1 5c and 2 1c , so a general solution is given by

1 1 2 1 2

2 1 2 1 2

cos sin cos3 sin 3 5cos5 ,

cos sin cos3 sin 3 cos5 .

x t a t a t b t b t t


Imposition of the initial conditions 1 2 1 20 0 0 0 0x x x x now yields 1 2a ,

2 0a , 1 3b , and 2 0b . The resulting particular solution is



1

2

2cos 3cos3 5cos5 ,

2cos 3cos3 cos5 .

x t t t t

x t t t t

We have a superposition of three oscillations, in which the two masses move

in the same direction with frequency 1 1 and equal amplitudes;

in opposite directions with frequency 2 3 and equal amplitudes;

in opposite directions with frequency 3 5 and with the amplitude of motion of

1m being 5 times that of 2m .

9. Substitution of the trial solution 1 1 cos3x c t , 2 2 cos3x c t in the system

1 1 2 2 1 23 2 2 2 4 120cos3x x x x x x t


1 1 2 1 2

2 1 2 1 2

cos sin 2 cos2 2 sin 2 3cos3 ,

cos sin cos2 sin 2 9cos3 .



Imposition of the initial conditions 1 2 1 20 0 0 0 0x x x x now yields 1 5a ,


1

2

5cos 8cos2 3cos3 ,

5cos 4cos2 9cos3 .

x t t t t

x t t t t




1m being twice that of 2m ;


2m being 3 times that of 1m .

10. Substitution of the trial solution 1 1 cosx c t , 2 2 cosx c t in the system

1 1 2 2 1 210 6 30cos , 6 10 60cosx x x t x x x t


1 1 2 1 2

2 1 2 1 2

cos2 sin 2 cos4 sin 4 14cos ,

cos2 sin 2 cos4 sin 4 16cos .



Imposition of the initial conditions 1 2 1 20 0 0 0 0x x x x now yields 1 1a


Section 5.4 323


1

2

cos2 15cos4 14cos ,

cos2 15cos4 16cos .

x t t t t

x t t t t


in the same direction with frequency 1 1 and with the amplitude of motion of

2m being 8

7 times that of 2m ;


in opposite directions with frequency 3 4 and equal amplitudes.

11. (a) The matrix 40 8

12 60

A has eigenvalues 1 36 and 2 64 with associated

eigenvalues T

1 2 1v and T

2 1 3 v . Hence a general solution is given by

1 2 1 2

1 2 1 2

2 cos6 2 sin 6 cos8 sin8 ,

cos6 sin 6 3 cos8 3 sin8 .

x t a t a t b t b t

y t a t a t b t b t

The natural frequencies are 1 6 and 2 8 . In mode 1 the two masses oscillate in the

same direction with frequency 1 6 and with the amplitude of motion of 1m being

twice that of 2m . In mode 2 the two masses oscillate in opposite directions with frequen-

cy 2 8 and with the amplitude of motion of 2m being 3 times that of 1m .

(b) Substitution of the trial solution 1 cos7x c t , 2 cos7y c t in the system

40 8 195cos7 , 12 60 195cos7x x y t y x y t


1 2 1 2

1 2 1 2

2 cos6 2 sin 6 cos8 sin8 19cos7 ,

cos6 sin 6 3 cos8 3 sin8 3cos7 .


y t a t a t b t b t t

Imposition of the initial conditions 0 19x , 0 12x , 0 3y , and 0 6y now

yields 1 0a , 2 1a , 1 0b , and 2 0b . The resulting particular solution is

2sin 6 19cos7 ,

sin 6 3cos7 .

x t t t

y t t t

Thus the expected oscillation with frequency 2 8 is missing, and we have a superposi-

tion of (only two) oscillations, in which the two masses move


1m being twice that of 2m ;




1m being 19

3 times that of 2m .


2 1 0

1 2 1

0 1 2

A has characteristic polynomial

3 2 26 10 4 2 4 2 .

Its eigenvalues 1 2 , 2 2 2 , and 3 2 2 have associated eigenvectors

T

1 1 0 1 v , T

2 1 2 1 v , and T

3 1 2 1 v . Hence the system’s three

natural modes of oscillation have

Natural frequency 1 2 with amplitude ratios 1: 0 : 1 ;

Natural frequency 2 2 2 with amplitude ratios 1: 2 :1 .

Natural frequency 3 2 2 with amplitude ratios 1: 2 :1 .


4 2 0

2 4 2

0 2 4

A has characteristic polynomial

3 2 212 40 32 4 8 8 .

Its eigenvalues 1 4 , 2 4 2 2 , and 3 4 2 2 have associated eigenvectors

T

1 1 0 1 v , T

2 1 2 1 v , and T

3 1 2 1 v . Hence the system’s three

natural modes of oscillation have

Natural frequency 1 2 with amplitude ratios 1: 0 : 1 ;

Natural frequency 2 4 2 2 with amplitude ratios 1: 2 :1 .

Natural frequency 3 4 2 2 with amplitude ratios 1: 2 :1.

14. The equations of motion of the given system are

1 1 2 1

2 2 2 1

50 10 5cos10 ,

10 .

x x x x t

m x x x

When we substitute 1 cos10x A t , 2 cos10x B t and cancel cos10t throughout we get

the equations

Section 5.4 325


2

40 10 5,

10 10 100 0.

A B

A m B

If 2 0.1m (slug), then it follows that 0A , so the mass 1m remains at rest.

15. First we need the general solution of the homogeneous system x Ax with

50 25 2

50 50

A .

The eigenvalues of A are 1 25 and 2 75 , so the natural frequencies of the system

are 1 5 and 2 5 3 . The associated eigenvectors are T

1 1 2v and

T

2 1 2 v , so the complementary solution c tx is given by

1 1 2 1 2

2 1 2 1 2

cos5 sin5 cos 5 3 sin 5 3 ,

2 cos5 2 sin5 2 cos 5 3 2 sin 5 3 .

x t a t a t b t b t

x t a t a t b t b t

When we substitute the trial solution T

1 2 cos10p t c c tx in the nonhomogeneous

system, we find that 1

4

3c and 2

16

3c , so a particular solution p tx is described by

1 2

4 16cos10 , cos10

3 3x t t x t t .

Finally, when we impose the zero initial conditions on the solution c pt t t x x x

we find that 1

2

3a , 2 0a , 1 2b , and 2 0b . Thus the solution we seek is described

by

1

2

2 4cos5 2cos 5 3 cos10 ,

3 34 16

cos5 4cos 5 3 cos103 3

x t t t t

x t t t t

We have a superposition of two oscillations with the natural frequencies 1 5 and

2 5 3 and a forced oscillation with frequency 10 . In each of the two natural os-

cillations the amplitude of motion of 2m is twice that of 1m , while in the forced oscilla-

tion the amplitude of motion of 2m is four times that of 1m .

16. The characteristic equation of A is

21 2 1 2 1 2 0c c c c c c ,

whence the given eigenvalues and eigenvectors follow readily.



17. With 1 2 2c c , it follows from Problem 16 that the natural frequencies and associated

eigenvectors are 1 0 , T

1 1 1v and 2 2 , T

2 1 1 v . Hence Theorem 1 gives

the general solution

1 1 1 2 2

2 1 1 2 2

cos2 sin 2 ,

cos2 sin 2 .

x t a b t a t b t

x t a b t a t b t

The initial conditions 1 00x v , 1 2 20 0 0 0x x x yield 1 2 0a a , 01 2

vb ,

and 02 4

vb , so

0 01 22 sin 2 , 2 sin 2

4 4

v vx t t t x t t t ,

while 02 1 2sin 2 0

4

vx x t ; that is, until

2t

. Finally, 1 02

x

and

2 02x v

.

18. With 1 6c and 2 3c , it follows from Problem 16 that the natural frequencies and asso-

ciated eigenvectors are 1 0 , T

1 1 1v and 2 3 , T

2 2 1 v . Hence Theorem

1 gives the general solution

1 1 1 2 2

2 1 1 2 2

2 cos3 2 sin 3 ,

cos3 sin 3 .

x t a b t a t b t

x t a b t a t b t

The initial conditions 1 00x v , 1 2 20 0 0 0x x x yield 1 2 0a a , 01 3

vb ,

and 02 9

vb , so

0 01 23 2sin 3 , 3 sin 3

9 9


while 02 1 3sin 3 0

9


3t

. Finally, 01 3 3

vx

and

02

2

3 3

vx

.

19. With 1 1c and 2 3c , it follows from Problem 16 that the natural frequencies and asso-

ciated eigenvectors are 1 0 , T

1 1 1v and 2 2 , T

2 1 3 v . Hence Theorem

1 gives the general solution

Section 5.4 327


1 1 1 2 2

2 1 1 2 2

cos2 sin 2 ,

3 cos2 3 sin 2 .

x t a b t a t b t

x t a b t a t b t

The initial conditions 1 00x v , 1 2 20 0 0 0x x x yield 1 2 0a a , 01

3

4

vb ,

and 02 8

vb , so

0 01 26 sin 2 , 6 3sin 2

8 8


while 02 1 4sin 2 0

8


2t

. Finally, 01 2 2

vx

and

02

3

2 2

vx

.

The method of solution in each of Problems 20–23 is the same as that in Example 2 in this sec-tion. Thus, looking at the equations in (26), we need to solve the equations

1 2 3 1

1 3 2

1 2 3 3

2 4 0

12 0

2 4 0

b b b x

b b x

b b b x

for the coefficients 1b , 2b , 3b after inserting given initial values 1 0x , 2 0x , 3 0x of the

three railway cars.

20. With 1 00x v , 2 0 0x , and 3 00x v , substitution of the resulting coefficient

values 1 2 3, ,b b b in (25) gives the railway car displacement functions

1 0 2 3 0

1 1sin 2 , 0, sin 2

2 2x t v t x t x t v t

and their velocities

1 0 2 3 0cos2 , 0, cos2x t v t x t x t v t .

We then see that

2 1 3 2 0

1sin 2

2x t x t x t x t v t

remains negative until 2

t , at which time the cars separate with velocities

1 0 2 3 0, 0,2 2 2

x v x x v

.

Thus the car in the center remains fixed thereafter and the first and third cars rebound from the collision with the same speeds with which they approached it.



21. With 1 00 2x v , 2 0 0x , and 3 00x v , substitution of the resulting coefficient


1 0

2 0

3 0

112 24sin 2 sin 4 ,

321

12 3sin 4 ,321

12 24sin 2 sin 4 .32

x t v t t t

x t v t t

x t v t t t

We then see (substituting sin 4 2sin 2 cos2t t t ) that

2 1 0 0

1 1sin 4 6sin 2 sin 2 cos2 3

8 4x t x t v t t v t t


t (as does 3 2x t x t , similarly) at which time the cars

separate with velocities

1 0 2 3 0, 0, 22 2 2

x v x x v

.

Thus the car in the center remains fixed thereafter, whereas the first and third cars re-bound in opposite directions, having exchanged their original velocities.

22. With 1 00x v , 2 00x v , and 3 00 2x v , substitution of the resulting coefficient


1 0

2 0

3 0

14 24sin 2 3sin 4 ,

321

4 9sin 4 ,321

4 24sin 2 3sin 4 .32

x t v t t t

x t v t t

x t v t t t


2 1 0 0

3 32sin 2 sin 4 sin 2 1 cos2





1 0 2 0 3 02 , ,2 2 2

x v x v x v

.

Section 5.4 329


Thus the car in the center proceeds thereafter with the same velocity it had originally, whereas the first and third cars rebound in opposite directions, having exchanged their original velocities.

23. With 1 00 3x v , 2 00 2x v , and 3 00 2x v , substitution of the resulting coefficient


1 0

2 0

3 0

176 8sin 2 sin 4 ,

321

76 3sin 4 ,321

76 8sin 2 sin 4 .32

x t v t t t

x t v t t

x t v t t t


2 1 0 0

1 12sin 2 sin 4 sin 2 1 cos2





1 0 2 0 3 02 , 2 , 32 2 2

x v x v x v

.

Thus the car in the center proceeds thereafter with the same velocity it had originally, whereas the first and third cars rebound in opposite directions, having exchanged their original velocities.

24. With 1 3 4c c and 2 16c the characteristic equation of the matrix

4 4 0

16 32 16

0 4 4

A

is

3 240 144 4 0 .

The resulting eigenvalues, natural frequencies, and associated eigenvectors are

T

1 1 1

T

2 2 2

T

3 3 3

0, 0, 1 1 1 ,

4, 2, 1 0 1 ,

36, 6, 1 8 1 .

v

v

v

Theorem 1 then gives the general solution



1 1 1 2 2 3 3

2 1 1 3 3

3 1 1 2 2 3 3

cos2 sin 2 cos6 sin 6 ,

8 cos6 8 sin 6 ,

cos2 sin 2 cos6 sin 6 .

x t a b t a t b t a t b t

x t a b t a t b t

x t a b t a t b t a t b t

The initial conditions yield 1 2 3 0a a a , 01

4

9

vb , 0

2 4

vb , and 0

3 108

vb , so

01

02

03

48 27sin 2 sin 6 ,108

48 8sin 6 ,108

48 27sin 2 sin 6 ,108

vx t t t t

vx t t t

vx t t t t

while

3 32 1 3 218sin 2 3 2sin 2 0, 9 4sin 2 0x x t t x x t ,

that is, until 2

t . Finally

0 0 01 2 3

8 8, ,

2 9 2 9 2 9

v v vx x x

.

25. (a) The matrix

160 3 320 3

8 116

A

has eigenvalues 1 41.8285 and 2 127.5049 , so the natural frequencies are

1 26.4675rad sec 1.0293Hz, 11.2918rad sec 1.7971Hz .

(b) Resonance occurs at the two critical speeds

1 21 2

20 2041ft sec 28mi h , 72ft sec 49 mi hv v

.

26. With 1 2k k k and 1 2 2

LL L the equations in (42) reduce to

2mx kx and 2

2

kLI .

The first equation yields 1

2k

m and the second one yields

2

2 2

kL

I .

In Problems 27–29 we substitute the given physical parameters into the equations in (42):

Section 5.4 331


2 2

1

1 2 1 1 2 2

1 12 21 22

,

.

mx k k x k L k L

I k L k L x k L k L

As in Problem 25, a critical frequency of rad sec yields a critical velocity of 20

ft secv

.

27. 100 4000x x , 800 100000 .

Obviously the matrix 40 0

0 125

A has eigenvalues 1 40 and 2 125 .

Up-and-down: 1 40 , 1 40.26ft sec 27 mi hv ;

Angular: 2 125 , 2 71.18ft sec 49 mi hv .

28. 100 4000 4000x x , 1000 4000 104000x .

The matrix 40 40

4 104

A has eigenvalues 1 2, 4 18 74 .

1 6.1311 , 1 39.03ft sec 27 mi hv ;

2 10.3155 , 1 65.67ft sec 45mi hv .

29. 100 3000 5000x x , 800 5000 75000x .

The matrix 30 50

25 4 375 4

A has eigenvalues 1 2

5, 99 3401

8 .

1 5.0424 , 1 32.10ft sec 22 mi hv ;

2 9.9158 , 2 63.13ft sec 43mi hv .

SECTION 5.5

MULTIPLE EIGENVALUE SOLUTIONS

In each of Problems 1–6 we give first the characteristic equation with repeated (multiplicity 2)

eigenvalue . In each case we find that 2 A I 0 . Then T1 0w is a generalized ei-

genvector and v A I w 0 is an ordinary eigenvector associated with . We give finally

the scalar component functions 1x t , 2x t of the general solution

1 2t tt c e c t e x v v w

of the given system x Ax .

332 MULTIPLE EIGENVALUE SOLUTIONS


1. Characteristic equation 2 6 9 0 ; repeated eigenvalue 3 ; generalized eigen-

vector T1 0w ;

1 1 1 1

1 1 0 1

v A I w ;

3 31 1 2 2 2 1 2 , t tx t c c c t e x t c c t e .

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 1

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 2


vector T1 0w ;

1 1 1 1

1 1 0 1

v A I w ;

2 21 1 2 2 2 1 2 , t tx t c c c t e x t c c t e .


vector T1 0w ;

2 2 1 2

2 2 0 2

v A I w ;

3 31 1 2 2 2 1 2 2 2 2 2,t tx t c c c t e x t c c t e .

Section 5.5 333


−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 3

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 4


vector T1 0w ;

1 1 1 1

1 1 0 1

v A I w ;

4 41 1 2 2 2 1 2 ,t tx t c c c t e x t c c t e .


vector T1 0w ;

2 1 1 2

4 2 0 4

v A I w ;




−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 5

−5 −4 −3 −2 −1 0 1 2 3 4 5

−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Problem 6


vector T1 0w ;

4 4 1 4

4 4 0 4

v A I w ;


In each of Problems 7–10 the characteristic polynomial is easily calculated by expansion along the row or column of A that contains two zeros. The matrix A has only two distinct eigenvalues, so we write 1 2 3, , with either 1 2 or 2 3 . Nevertheless, we find that it has 3 linearly

independent eigenvectors 1v , 2v , and 3v . We list also the scalar components 1 2 3, ,x t x t x t

of the general solution 31 21 1 2 2 3 3

tt tt c e c e c e x v v v of the system.

7. Characteristic equation 23 213 40 36 2 9 ;

Eigenvalues 2,2,9 ;

Eigenvectors T1 1 0 , T

1 0 1 , T0 1 0 ;

2 21 1 2

2 92 1 3

23 2

t t

t t

t

x t c e c e

x t c e c e

x t c e


Eigenvalues 7,13,13 ;

Section 5.5 335



0 0 1 , T1 1 0 ;

7 131 1 3

7 132 1 3

7 133 1 2

2

3

t t

t t

t t

x t c e c e

x t c e c e

x t c e c e


Eigenvalues 5,5,9 ;


7 0 2 , T3 0 1 ;

5 5 91 1 2 3

52 1

5 93 2 3

7 3

2

2

t t t

t

t t

x t c e c e c e

x t c e

x t c e c e


Eigenvalues 3,3,7 ;


3 0 1 , T2 1 0 ;

5 3 71 1 2 3

3 72 1 3

33 2

5 3 2

2

t t t

t t

t

x t c e c e c e

x t c e c e

x t c e

In each of Problems 11-14, the characteristic equation is 33 23 3 1 1 . Hence

1 is a triple eigenvalue of defect 2, and we find that 3 A I 0 . In each problem we

start with T3 1 0 0v and then calculate 2 3 v A I v and 1 2 0 v A I v . It fol-

lows that 2 3

1 2 3 A I v A I v A I v 0 , so the vector 1v (if nonzero) is an ordi-

nary eigenvector associated with the triple eigenvalue . Hence 1 2 3, ,v v v is a length 3 chain

of generalized eigenvectors, and the corresponding general solution is described by

2

1 1 2 1 2 3 1 2 32t t

t e c c t c t

x v v v v v v .

We give the scalar components 1x t , 2x t , 3x t of tx .

11. T

1 0 1 0v , T

2 2 1 1 v , T

3 1 0 0v ;



1 2 3 3 2 1 2 2 3 3 2

2

3 3, ,2

2 2 t t tx t e c ct

c t x t e c c c t c t c x t e c c t

.

12. T

1 1 1 0v , T

2 0 0 1v , T

3 1 0 0v ;

1 1 3 2 3 2 1 2 3

2

3 2 3

2

, ,2 2

t t tx t e c c c t c x t et t

c c t c x t e c c t

.

13. Here we are stymied initially, because if T

3 1 0 0v , then 3 A I v 0 does not

qualify as a (nonzero) generalized eigenvector. We therefore make a fresh start with

T

3 0 1 0v , and now we get the desired nonzero generalized eigenvectors upon suc-

cessive multiplication by A I .

T

1 1 0 0v , T

2 0 2 1v , T

3 0 1 0v ;

1 1 2 3 2 2 3 2 3

2

3 3, 2 ,2

2 t t tx e c ct

t t c x e c c c x t e c c tt t

.

14. T

1 5 25 5 v , T

2 1 5 4 v , T

3 1 0 0v ;

2

1 1 2 3 2 3 3

2 1 2 2 3 3

3 1 2 2 3 3

2

2

2 5 5 5

25 5 25 5 25

2

2 5 4 5 4 5

t

t

t

x t e c c c c t c t c

x t e c c c t c t c

x t e c c c t c t c

t

t

t

In each of Problems 15-18, the characteristic equation is 33 23 3 1 1 . Hence

1 is a triple eigenvalue of defect 1, and we find that 2 A I 0 . First we find the two

linearly independent (ordinary) eigenvectors 1u and 2u associated with . Then we start with

T

2 1 0 0v and calculate 1 2 v A I v 0 . It follows that

2

1 2 A I v A I v 0 , so 1v is an ordinary eigenvector associated with . However,

1v is a linear combination of 1u and 2u , so 1tev is a linear combination of the independent solu-

tions 1teu and 2

teu . But 1 2,v v is a length 2 chain of generalized eigenvectors associated with

, so 1 2tt ev v is the desired third independent solution. The corresponding general solution

is described by

Section 5.5 337


1 1 2 2 3 1 2 tt e c c c t x u u v v .

We give the scalar components 1x t , 2x t , 3x t of tx .

15. T

1 3 1 0 u , T

2 0 0 1u ;

T

1 3 1 1 v , T

2 1 0 0v ;

1 1 3 3 2 1 3 3 2 3 3 3 , ,t t tx t e c c c t x t e c c t x t e c c t .

16. T

1 3 2 0 u , T

2 3 0 2 u ;

T

1 0 2 2 v , T

2 1 0 0v ;

1 1 2 3 2 1 3 3 2 3, 3 3 2 2 – , 2 2t t tx t e c c c x t e c c t x t e c c t .

17. T

1 2 0 9 u , T

2 1 3 0 u ;

T

1 0 6 9 v ; T

2 0 1 0v ;

(Either T

2 1 0 0v or T

2 0 0 1v can be used also, but they yield different

forms of the solution than given in the book’s answer section.)

1 1 2 2 2 3 3 3 1 3 2 –3 6 9, , 9t t tx t e c c x t e c c c t x t e c c t .

18. T

1 1 0 1 u , T

2 2 1 0 u ;

T

1 0 1 2 v , T

2 1 0 0v ;

1 1 2 3 2 2 3 3 1 3 2 2, , t t tx t e c c c x t e c c t x t e c c t .


Double eigenvalue 1 with eigenvectors T

1 1 0 0 1v and

T

2 0 0 1 0v ;

Double eigenvalue 1 with eigenvectors T

3 0 1 0 2 v and

T

4 1 0 3 0v ;

General solution

1 1 2 2 3 3 4 4 t tt e c c e c c x v v v v ;

Scalar components

1 1 4 2 3 3 2 4 4 1 3 , 3, 2, t t t t t t tx t c e c e x t c e x t c e c e x t c e c e .



20. Characteristic equation 44 3 28 24 32 16 2 0 ;

Eigenvalue 2 with multiplicity 4 and defect 3.

We find that 3 A I 0 but 4 A I 0 . We therefore start with

T

4 0 0 0 1v and define 3 4 v A I v , 2 3 v A I v , 2 3 v A I v , and

1 2 v A I v 0 . This gives the length 4 chain 1 2 3 4, , ,v v v v with

T T T T

1 2 3 41 0 0 1 , 0 1 0 0 , 1 0 1 0 , 0 0 0 1 v v v v .

The corresponding general solution is given by

2 3 2

1 1 2 1 2 3 1 2 3 4 1 2 3 4 2 6 2

t t t tt e c c t c t c t

x v v v v v v v v v v ,

with scalar components

21 1 3 2 4 3 4

22 2 3 4

23 3 4

24

2

4

3

2

2

2

6t

t

t

t

x t e c c c t c t c c

x t e

t

c c t c

x t e c c t

x

t

t

t

e c

21. Characteristic equation 44 3 24 6 4 1 1 0 ;

Eigenvalue 1 with multiplicity 4 and defect 2.

We find that 2 A I 0 but 3 A I 0 . We therefore start with

T

3 1 0 0 0v and define 2 3 v A I v and 1 2 v A I v 0 , thereby ob-

taining the length 3 chain 1 2 3, ,v v v with

T T T

1 2 30 0 0 1 , 2 1 1 0 , 1 0 0 0 v v v .

Then we find the second ordinary eigenvector T

4 0 0 1 0v . The corresponding

general solution

2

1 1 2 1 2 3 1 2 3 4 42 tt e c

tc t c t c

x v v v v v v v

has scalar components

Section 5.5 339


1 2 3 3

2 2 3

3 2 4 3

4 2 3

2

1

2 2

2

t

t

t

t

x t e c c c t

x t e c c t

x t

t

e c c c t

x t e c c t c

22. Same eigenvalue and chain structure as in Problem 21, but with generalized eigenvectors

T T T T

1 2 3 41 0 0 2 , 3 2 1 6 , 0 1 0 0 , 1 0 0 0 v v v v ,

where 1 2 3, ,v v v is a length 3 chain and 4v is an ordinary eigenvector. The general so-

lution tx defined as in Problem 21 has scalar components

2

1 1 2 4 2 3 3

2 2 3 3

3 2 3

24 1 2 2 3 3

3 3

2 2

2 6

2

2 6

t

t

t

t

tx t e c c c c t c t c

x t e c c c t

x t e c c t

x t e c c c t c t c t

In Problems 23 and 24 there are only two distinct eigenvalues 1 and 2 . However, the eigen-

vector equation A I v 0 yields the three linearly independent eigenvectors 1v , 2v , and 3v

that are given. We list the scalar components of the corresponding general solution 1 2 2

1 1 2 2 3 3t t tt c e c e c e x v v v .

23. 1 1 : 1v with T

1 1 1 2 v ;

2 3 : 2v with T

2 4 0 9v and 3v with T

3 0 2 1v .

Scalar components:

31 1 2

32 1 3

3 33 1 2 3

4

2

2 9

t t

t t

t t t

x t c e c e

x t c e c e

x t c e c e c e

24. 1 2 : 1v with T

1 5 3 3 v ;

2 3 : 2v with T

2 4 0 1 v and 3v with T

3 2 1 0 v .

Scalar components:



2 3 31 1 2 3

2 32 1 3

2 33 1 2

5 4 2

3

3

t t t

t t

t t

x t c e c e c e

x t c e c e

x t c e c e

In Problems 25, 26, and 28 there is given a single eigenvalue of multiplicity 3. We find that

2 A I 0 but 3 A I 0 . We therefore start with T

3 1 0 0v and define

2 3 v A I v and 1 2 v A I v 0 , thereby obtaining the length 3 chain 1 2 3, ,v v v of

generalized eigenvectors based on the ordinary eigenvector 1v . We list the scalar components of

the corresponding general solution

2

1 1 2 1 2 3 1 2 3 2

t t ttc e c t e c t et

x v v v v v v .

25. 1 2 3, ,v v v with

T T T

1 2 31 0 1 , 4 1 0 , 1 0 0 v v v

Scalar components:

22

1 1 2 3 2 3 3

22 2 3

22

3 1 2 3

4 42

2

t

t

t


x t e c c t

tx t e c c t c

26. 1 2 3, ,v v v with

T T T

1 2 30 2 2 , 2 1 3 , 1 0 0 v v v ;

General solution:

31 2 3 3

3 22 1 2 2 3 3

3 23 1 2 2 3 3

2 2

2 2

2 3 2 3

t

t

t

x t e c c c t



27. We find that the triple eigenvalue 2 has the two linearly independent eigenvectors

T1 1 0 and T

1 0 1 . Next we find that A I 0 but 2 A I 0 . We

therefore start with T

2 1 0 0v and define

T1 2 5 3 8 v A I v 0 ,

Section 5.5 341


thereby obtaining the length 2 chain 1 2,v v of generalized eigenvectors based on the

ordinary eigenvector 1v . If we take T

3 1 1 0v , then the general solution

21 1 2 1 2 3 3 tt e c c t c x v v v v


21 1 2 3 2

22 1 2

23 1 2

5 5

3 3

8 8

t

t

t

x t e c c c c t

x t e c c t

x t e c c t

28. 1 2 3, ,v v v with

T T T

1 2 3119 289 0 , 17 34 17 , 1 0 0 v v v ;

General solution

22

1 1 2 3 2 3 3

22

2 1 2 2 3 3

23 2 3

119 17 119 17 1192

289 34 289 34 2892

17 17

t

t

t


tx t e c c c t c t c

x t e c c t

In Problems 29 and 30 the matrix A has two distinct eigenvalues 1 and 2 each having multi-

plicity 2 and defect 1. First, we select 2v so that 1 1 2 v A I v 0 but 1 1 A I v 0 , so

1 2,v v is a length 2 chain based on 1v . Next, we select 2u so that 1 1 2 u A I u 0 but

1 1 A I u 0 , so 1 2,u u is a length 2 chain based on 1u . We give the scalar components of

the corresponding general solution

211 1 2 1 2 3 1 4 1 2 ttt e c c t ce c t x v v v u u u .

29. 1 : 1 2,v v with T

1 1 3 1 2 v and T

2 0 1 0 0v ;

2 : 1 2,u u with T

1 0 1 1 0 u and T

2 0 0 2 1u ;

Scalar components

1 1 2

22 1 2 2 3 4

23 1 2 3 4 4

24 1 2 4

3 3

2

2 2

t

t t

t t

t t

x t e c c t

x t e c c c t e c c t

x t e c c t e c c c t

x t e c c t e c



30. 1 : 1 2,v v with T

1 0 1 1 3 v and T

2 0 0 1 2v ;

2 : 1 2,u u with T

1 1 0 0 0 u and T

2 0 0 3 5u ;

Scalar components

21 3 4

2 1 2

23 1 2 2 4

24 1 2 2 4

3

3 2 3 5

t

t

t t

t t

x t e c c t

x t e c c t

x t e c c c t e c

x t e c c c t e c

31. We have the single eigenvalue 1 of multiplicity 4. Starting with T

3 1 0 0 0v ,

we calculate 2 3 v A I v and 1 2 v A I v 0 , and find that 1 A I v 0 .

Therefore 1 2 3, ,v v v is a length 3 chain based on the ordinary eigenvector 1v . Next, the

eigenvector equation A I v 0 yields the second linearly independent eigenvector

T

4 0 1 3 0v . With

T T

1 2

T T

3 4

42 7 21 42 , 34 22 10 27 ,

1 0 0 0 , 0 1 3 0

v v

v v

the general solution

2

1 1 2 1 2 3 1 2 3 4 4 2

t tt e c c t c t c

x v v v v v v v


21 1 2 3 2 3 3

2

2 1 2 4 2 3 3

2

3 1 2 4 2 3 3

24 1 2 2 3 3

42 34 42 34 21

7 22 7 22 72

21 10 3 21 10 212

42 27 42 27 21

t

t

t

t

x t e c c c c t c t c t




32. Here we find that the matrix A has five linearly independent eigenvectors:

2 : Eigenvectors T1 8 0 3 1 0 v and T

2 1 0 0 0 3v ;

3 : Eigenvectors T

3 3 2 1 0 0 v , T4 2 2 0 3 0 v , and

T

5 1 1 0 0 3 v .

The general solution

Section 5.5 343


2 31 1 2 2 3 3 4 4 5 5 t tt e c c e c c c x v v v v v


2 31 1 2 3 4 5

32 3 4 5

2 33 1 3

2 34 1 4

2 35 2 5

8 3 2

2 2

3

3

3 3

t t

t

t t

t t

t t

x t e c c e c c c

x t e c c c

x t e c e c

x t e c e c

x t e c e c

33. The chain 1 2,v v was found using the matrices

4 4 1 0 1 0 0

4 4 0 1 0 0 1 0

0 0 4 4 0 0 0 1

0 0 4 4 0 0 0 0

i i

i

i

i

A I

and

2

32 32 8 8 1 0 0

32 32 8 8 0 0 1( )

0 0 32 32 0 0 0 0

0 0 32 32 0 0 0 0

i i i

i i i

i

i

A I ,

where signifies reduction to row-echelon form. The resulting real-valued solution vectors are

T31

T32

T33

T34

cos4 sin 4 0 0

sin 4 cos4 0 0

cos4 sin 4 cos4 sin 4

sin 4 cos4 sin 4 cos4

t

t

t

t

t e t t

t e t t

t e t t t t t t

t e t t t t t t

x

x

x

x

34. The chain 1 2,v v was found using the matrices

3 0 8 3 1 0 0 0

18 3 3 0 0 0 1 0 3 3

9 3 27 3 9 0 0 1 0

33 10 90 30 3 0 0 0 0

i

i i

i

i

A I

and



2

36 6 54 48 18 18 1 0 3

54 108 18 144 54 0 1 9 10 3 3

54 18 18 162 54 0 0 0 0

198 60 6 540 18 180 0 0 0 0

i i i i

i i i i

i i i i

i i i i

A I ,

where signifies reduction to row-echelon form. The resulting real-valued solution vectors are

T21

T22

T23

T24

sin 3 3cos3 3sin 3 0 sin 3

cos3 3sin 3 3cos3 0 cos3

3cos3 sin 3 3 10 cos3 3 9 sin 3 sin 3 sin 3

cos3 3sin 3 3 9 cos3 3 10 sin 3 cos3 cos3

t

t

t

t

t e t t t t

t e t t t t

t e t t t t t t t t t t

t e t t t t t t t t t t

x

x

x

x


0 0 1 0

0 0 0 1

1 1 2 1

1 1 1 2

A

has the following eigenvalues and corresponding eigenvectors:

Corresponding eigenvector(s)

0 T

1 1 1 0 0v

1 T

2 1 0 1 0 v and T

3 0 1 0 1 v

2 T

4 1 1 2 2 v

When we impose the given initial conditions on the general solution

21 1 2 2 3 3 4 4 t t tt c c e c e c e x v v v v

we find that 1 0c v , 2 3 0c c v , 4 0c . Hence the position functions of the two mass-

es are given by

01 2 1 tx t x t v e .

Each mass travels a distance 0v before stopping.

36. The coefficient matrix is the same as in Problem 35 except that 44 1a . Now the ma-

trix A has the eigenvalue 0 with eigenvector T

0 1 1 0 0v , and the triple ei-

Section 5.5 345


genvalue 1 with associated length 2 chain 1 2 3, ,v v v consisting of the generalized

eigenvectors

T T T

1 2 30 1 0 1 , 1 0 1 1 , 1 0 0 0 v v v

When we impose the given initial conditions on the general solution

0 0 1 1 2 1 2 3 1 2 3

2

2tt c e c c t c

tt

x v v v v v v v

we find that 0 02c v , 1 02c v , and 2 3 0c c v . Hence the position functions of the

two masses are given by

2

1 0 2 0 2 2 , 2 22

t t t t ttx t v e te x t v e te e

Each travels a distance 02v before stopping.

SECTION 5.6

MATRIX EXPONENTIALS AND LINEAR SYSTEMS

In Problems 1–8 we first use the eigenvalues and eigenvectors of the coefficient matrix A to find first a fundamental matrix t for the homogeneous system x Ax . Then we apply the for-

mula 1

00t t x x to find the solution vector tx that satisfies the initial condition

00 x x . Formulas (11) and (12) in the text provide inverses of 2 2 and 3 3 matrices.

1. Eigensystem: 1 1 , T

1 1 1 v ; 2 3 , T

1 1 1v

1 2

3

1 2 3( )

t tt t

t t

e et e e

e e

v v

3 3

3 3

1 1 3 51 1

1 1 22 2 5

t t t t

t t t t

e e e et

e e e e

x

2. Eigensystem: 1 0 , T

1 1 2v ; 2 4 , 2 4 , T2 1 2 v

1 2

4

1 2 4

1

2 2

tt t

t

et e e

e

v v

4 4

4 4

2 1 21 3 51 1

2 1 14 42 2 6 10

t t

t t

e et

e e

x

346 MATRIX EXPONENTIALS AND LINEAR SYSTEMS


3. Eigensystem: 4i , T1 2 2i v ;

cos4 2sin 4 2cos4 sin 4( ) Re Im

2cos4 2sin 4t t t t t t

t e et t

v v

cos4 2sin 4 2cos4 sin 4 0 2 0 5sin 41 1

2cos4 2sin 4 2 1 1 4cos4 2sin 44 4

t t t t tt

t t t t

x

4. Eigensystem: 2,2 , 1 2{ , }v v with T

1 1 1v , T2 1 0v ;

21 1 2

1 1

1t t t t

t e t e et

v v v

2 21 1 0 1 1 1

1 1 1 0t tt t

t e et t

x

5. Eigensystem: 3i , T1 3i v ;

cos3 sin 3 cos3 sin 3Re Im

3cos3 3sin 3t t t t t t

t e et t

v v

cos3 sin 3 cos3 sin 3 0 1 1 3cos3 sin 31 1

3cos3 3sin 3 3 1 1 3cos3 6sin 33 3

t t t t t tt

t t t t

x

6. Eigensystem: 5 4i , T1 2 2i v ;

5 cos4 2sin 4 2cos4 2sin 4Re Im

2cos4 2sin 4t t t t t t t

t e e et t

v v

5 5cos4 2sin 4 2cos4 2sin 4 0 2 2 cos4 sin 412

2cos4 2sin 4 2 4 0 sin 44t tt t t t t t

t e et t t

x

7. Eigensystem:

T T T

1 1 2 2 3 30, 6 2 5 ; 1, 3 1 2 ; 1, 2 1 2 v v v ;

31 21 2 3

6 3 2

2

5 2 2

t t

tt t t t

t t

e e

t e e e e e

e e

v v v

Section 5.6 347


6 3 2 0 2 1 2 12 12 2

2 1 2 2 1 4 4

5 2 2 1 3 0 0 10 8 2

t t t t

t t t t

t t t t

e e e e

t e e e e

e e e e

x

8. Eigensystem:

T T T

1 1 2 2 3 32, 0 1 1 ; 1, 1 1 0 ; 3, 1 1 1 v v v ;

31 2

3

2 31 2 3

2 3

0

0

t t

tt t t t t

t t

e e

t e e e e e e

e e

v v v

3

2 3 2

2 3 2

0 1 1 0 1

0 1 1 0

0 1 1 1 1

t t t

t t t t t

t t t

e e e

t e e e e e

e e e

x

In each of Problems 9-20 we first solve the given linear system to find two linearly independent

solutions 1x and 2x , then set up the fundamental matrix 1 2t t t x x , and finally cal-

culate the matrix exponential 10te t

A .

9. Eigensystem: T T

1 1 2 21, 1 1 ; 3, 2 1 v v ;

1 2

3

1 2 3

2t tt t

t t

e et e e

e e

v v

3 3 3

3 3 3

1 22 2 2 2

1 1 2

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A

10. Eigensystem: T T

1 1 2 20, 1 1 ; 2, 3 2 v v ;

1 2

2

1 2 2

1 3

1 2

tt t

t

et e e

e

v v

2 2 2

2 2 2

2 31 3 2 3 3 3

1 11 2 2 2 3 2

t t tt

t t t

e e ee

e e e

A


1 1 2 22, 1 1 ; 3, 3 2 v v ;

1 2

2 3

1 2 2 3

3

2

t tt t

t t

e et e e

e e

v v



2 3 2 3 2 3

2 3 2 3 2 3

2 33 2 3 3 3

1 12 2 2 3 2

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 21, 1 1 ; 2, 4 3 v v ;

1 2

2

1 2 2

4

3

t tt t

t t

e et e e

e e

v v

2 2 2

2 2 2

3 44 3 4 4 4

1 13 3 3 4 3

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 21, 1 1 ; 3, 4 3 v v

1 2

3

1 2 3

4

3

t tt t

t t

e et e e

e e

v v

3 3 3

3 3 3

3 44 3 4 4 4

1 13 3 3 4 3

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 21, 2 3 ; 3, 3 4 v v ;

1 2

2

1 2 2

2 3

3 4

t tt t

t t

e et e e

e e

v v

2 2 2

2 2 2

4 32 3 8 9 6 6

3 23 4 12 12 9 8

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 21, 2 1 ; 2, 5 2 v v ;

1 2

2

1 2 2

2 5

2

t tt t

t t

e et e e

e e

v v

2 2 2

2 2 2

2 52 5 4 5 10 10

1 22 2 2 5 4

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 21, 3 2 ; 2, 5 3 v v ;

1 2

2

1 2 2

3 5

2 3

t tt t

t t

e et e e

e e

v v

Section 5.6 349


2 2 2

2 2 2

3 53 5 9 10 15 15

2 32 3 6 6 10 9

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 22, 1 1 ; 4, 1 1 v v ;

1 2

2 4

1 2 2 4

t tt t

t t

e et e e

e e

v v

2 4 2 4 2 4

2 4 2 4 2 4

1 11 1

1 12 2

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 22, 1 1 ; 6, 1 1 v v ;

1 2

2 6

1 2 2 6

t tt t

t t

e et e e

e e

v v

2 6 2 6 2 6

2 6 2 6 2 6

1 11 1

1 12 2

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 25, 1 2 ; 10, 2 1 v v ;

1 2

5 10

1 2 5 10

2

2

t tt t

t t

e et e e

e e

v v

5 10 5 10 5 10

5 10 5 10 5 10

1 22 4 2 21 1

2 15 52 2 2 4

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A


1 1 2 25, 1 2 ; 15, 2 1 v v ;

1 2

5 15

1 2 5 15

2

2

t tt t

t t

e et e e

e e

v v

5 15 5 15 5 15

5 15 5 15 5 15

1 22 4 2 21 1

2 15 52 2 2 4

t t t t t tt

t t t t t t

e e e e e ee

e e e e e e

A

21. 2 A 0 , so 1

1t t t

e tt t

A I A .

22. 2 A 0 , so 1 6 4

9 1 6t t t

e tt t

A I A .



23. 3 A 0 , so

2

2 2 2

11

12

0 0 1

t

t t t t

e t t t t t t

A I A A .

24. 3 A 0 , so 2 2 2 2

1 3 0 31

5 18 1 7 182

3 0 1 3

t

t t

e t t t t t t

t t

A I A A .

25. 2 A I B , where 2 B 0 , so 2 2t t t te e e e t A I B I I B . Hence

2 2

2

2

4 4 355,

7 70

t tt t t

t

te tee t e e

e

A Ax .

26. 7 A I B , where 2 B 0 , so 7 7t t t te e e e t A I B I I B . Hence

7

7

7 7

5 50,

10 10 5511

tt t t

t t

ee t e e

tte e

A Ax .

27. A I B , where 3 B 0 , so 2 21

2t t t te e e e t t

A I B I I B . Hence

2 22 3 2 4 4 28 12

0 2 , 5 5 12

0 0 6 6

t t t

t t t t t

t

e te t t e t t

e e te t e e t

e

A Ax .

28. 5 A I B , where 3 B 0 , so 5 5 2 21

2t t t te e e e t t

A I B I I B B . Hence

5

5 5 5

22 5 5 5

0 0 40 40

10 0 , 50 50 400

60 60 2300 600020 150 30

t

t t t t t

t t t

e

e te e t e e t

t tt t e te e

A Ax .

29. A I B , where 4 B 0 , so 2 2 3 31 1

2 6t t t te e e e t t t

A I B I I B B B . Hence

Section 5.6 351


2 2 3 2 3

2 2

11 2 3 6 4 6 4 1 9 12 4

10 1 6 3 6 1 9 6,

10 0 1 2 1 2

10 0 0 1 1

t t t t

t t t t t t t t t

t t t t te e t e e

t t

A Ax .

30. 3 A I B , where 4 B 0 , so 3 3 2 2 3 31 1

2 6t t t te e e e t t t

A I B I I B B B . Hence

3 32 2

2 3 2 2 3

1 0 0 0 1 1

6 1 0 0 1 1 6,

9 18 6 1 0 1 1 15 18

12 54 36 9 18 6 1 1 1 27 72 36

t t t tt te e t e e

t t t t t

t t t t t t t t t

A Ax

33. cosh sinh

cosh sinhsinh cosh

t t te t t

t t

A I A , so the general solution of x Ax is

1 2

1 2

cosh sinh

sinh cosht c t c t

t ec t c t

Ax c .

34. Direct calculation gives 2 4 A I , and it follows that 3 4 A A and 4 16A I . There-fore

2 3 4 5

2 4 3 5

4 4 16 16

2! 3! 4! 5!

2 2 2 211 2

2! 4! 2 3! 5!

1cos2 sin 2 .

2

t t t t te t

t t t tt

t t

A I A I AI A

I A

I A

In Problems 35–40 we give first the linearly independent generalized eigenvectors 1 2, , , nu u u

of the matrix A and the corresponding solution vectors 1 2, , , nt t tx x x defined by Eq. (34)

in the text, then the fundamental matrix 1 2 nt t t t x x x . Finally we calcu-

late the exponential matrix 10te t

A .

35. 3 : T

1 4 0u , T

2 0 1u ;

1 2,u u is a length 2 chain based on the ordinary (rank 1) eigenvector 1u , so 2u is a

generalized eigenvector of rank 2.



1 1 2 2 2,t tt e t e t x u x u A I u

31 2

4 4

0 1t t

t t t e

x x

3 34 4 1 0 1 41

0 1 0 4 0 14t t tt t

e e e

A

36. 1 : T

1 8 0 0u , T

2 5 4 0u , T

3 0 1 1u ;

1 2 3, ,u u u is a length 3 chain based on the ordinary (rank 1) eigenvector 1u , so 2u and

3u are generalized eigenvectors of ranks 2 and 3 (respectively).

2

2

1 1 2 2 2 3 3 3 3, ,2

t t t tt e t e t t e t

x u x u A I u x u A I u A I u

2

1 2 3

8 5 8 5 4

0 4 1 4

0 0 1

t

t t t

t t t t e t

x x x

2 28 5 8 5 4 4 5 5 1 2 3 41

0 4 1 4 0 8 8 0 1 432

0 0 1 0 0 32 0 0 1

t t t

t t t t t t

e e t e t

A

37. 1 2 : T

1 1 0 0u , 11 1

tt ex u ;

2 1 : T

2 9 3 0 u , T

3 10 1 1 u ;



2 22 2 3 3 2 3,t tt e t e t x u x u A I u

2

1 2 3

9 10 9

0 3 1 3

0 0

t t t

t t

t

e e t e

t t t t e t e

e

x x x

2 9 10 9 3 9 391

0 3 1 3 0 1 13

0 0 0 0 3

t t t

t t t

t

e e t e

e e t e

e

A

Section 5.6 353


2 2 23 3 13 9 13

0 3

0 0

t t t t t

t t

t

e e e t e e

e te

e

2 2 23 3 13 9 13

0 3

0 0

t t t t t

t t

t

e e e t e e

e te

e

38. 1 10 : T

1 4 1 0u , 11 1

tt ex u ;

2 5 : T

2 50 0 0u , T

3 0 4 1 u ;



2 22 2 3 3 2 3,t tt e t e t x u x u A I u

10 5 5

10 51 2 3

5

4 50 50

0 4

0 0

t t t

t t

t

e e te

t t t t e e

e

x x x

10 5 5 5 10 5 10 5

10 5 10 10 5

5 5

4 50 50 0 50 200 4 4 16 16 501

0 4 1 4 16 0 4 450

0 0 0 0 50 0 0

t t t t t t t t

t t t t t t

t t

e e te e e e e t e

e e e e e e

e e

A

39. 2 1 : T

1 3 0 0 0u , T

2 0 1 0 0u ;

1 2,u u is a length 2 chain based on the ordinary (rank 1) eigenvector 1u , so 2u is a gen-

eralized eigenvector of rank 2.

1 11 1 2 2 1 2,t tt e t e t x u x u A I u

2 2 : T

3 144 36 12 0u , T

4 0 27 17 4u ;



2 23 3 4 4 2 4,t tt e t e t x u x u A I u

2 2

2 2

1 2 3 4 2 2

2

3 3 144 144

0 36 27 36

0 0 12 17 12

0 0 0 4

t t t t

t t t

t t

t

e te e te

e e t et t t t t

e t e

e

x x x x



2 2

2 2

2 2

2

16 0 192 8163 3 144 144

0 48 144 2880 36 27 36 1

0 0 4 17480 0 12 17 12

0 0 0 120 0 0 4

t t t t

t t tt

t t

t

e te e te

e e t ee

e t e

e

A

2 2

2 2

2 2

2

3 12 9 12 51 18 51 36

0 3 3 6 6 9

0 0 3

0 0 0

t t t t t t

t t t t t

t t

t

e te t e te t e t e

e e e e t e

e te

e

40. 1 3 : T

1 100 20 4 1u , 11 1

tt ex u ;

2 : T

2 16 0 0 0u , T

3 0 4 0 0u , T

4 0 1 1 0 u ;

2 3 4, ,u u u is a length 3 chain based on the ordinary (rank 1) eigenvector 2u , so 3u and

4u are generalized eigenvectors of ranks 2 and 3 (respectively).

2 2

2

2 2 3 3 2 3

22

4 4 2 4 2 4

, ,

2

t t

t

t e t e t

tt e t

x u x u A I u

x u A I u A I u

3 2 2 2 2

3 2 2

1 2 3 4 3 2

3

100 16 16 8

20 0 4 1 4

4 0 0

0 0 0

t t t t

t t t

t t

t

e e te t e

e e t et t t t t

e e

e

x x x x

3 2 2 2 2

3 2 2

3 2

3

0 0 0 16100 16 16 8

1 0 0 10020 0 4 1 4 1

0 4 4 96164 0 0

0 0 16 640 0 0

t t t t

t t tt

t t

t

e e te t e

e e t ee

e e

e

A

2 2 2 2 3 2 2

2 2 3 2

2 3 2

3

4 4 8 100 100 96 32

0 4 20 20 16

0 0 4 4

0 0 0

t t t t t

t t t t

t t t

t

e te t t e e t t e

e te e t e

e e e

e

Section 5.7 355


SECTION 5.7

NONHOMOGENEOUS LINEAR SYSTEMS

1. Substitution of the trial solution px t a , py t b yields the equations

2 3 0, 2 2 0a b a b

with solution 7

3a ,

8

3b . Thus we obtain the particular solution

7 8,

3 3x t y t .

2. When we substitute the trial solution 1 1px t a b t , 2 2py t a b t and collect coeffi-

cients, we get the equations

1 2

1 2

2 2

2 3 0

b b

b b

and 1 2 2

1 2 1

2

2 3 5

a a b

a a b

.

We solve the first pair for 1

3

2b and 2 1b Then we can solve the second pair for

1

1

8a and 2

5

4a . This gives the particular solution

1 11 12 , 5 4

8 4x t t y t t .

3. When we substitute the trial solution 2 2

1 1 1 2 2 2,p px a b t c t y a b t c t

and collect coefficients, we get the equations

1 2 1 1 2 1 1 2

1 2 2 1 2 2 1 2

3 4 3 4 2 3 4 0

3 2 3 2 2 3 2 1 0

a a b b b c c c

a a b b b c c c

Working backwards, we solve first for 1

2

3c and 2

1

2c , then for 1

10

9b and

2

7

6b , and finally for 1

31

27a and 2

41

36a . This determines the particular solution

px t , py t . Next, the coefficient matrix of the associated homogeneous system has

eigenvalues 1 1 and 2 6 , with eigenvectors T

1 1 1 v and T

2 4 3v , re-

spectively, so the complementary solution is given by

6 61 2 1 24 , 3t t t t

c cx t c e c e y t c e c e .

356 NONHOMOGENEOUS LINEAR SYSTEMS


When we impose the initial conditions 0 0x , 0 0y on the general solution

c px t x t x t , c py t y t y t we find that 1

8

7c and 2

1

756c . This final-

ly gives the desired particular solution

6 2

6 2

1864 4 868 840 504 ,

7561

864 3 861 882 378 .756

t t

t t

x t e e t t

y t e e t t

4. The coefficient matrix of the associated homogeneous system has eigenvalues 1 5

and 2 2 , with eigenvectors T

1 1 1 v and T

2 1 6 v , respectively, so the

complementary solution is given by

5 2 5 21 2 1 2, 6t t t t


Then we try tpx t ae , t

py t be and find readily the particular solution

1

12t

px t e , 3

4t

py t e . Thus the general solution is

5 2 5 21 2 1 2

1 3, 6

12 4t t t t t tx t c e c e e y t c e c e e .

Finally we apply the initial conditions 0 0 1x y to determine 1

33

28c and

2

2

21c . The resulting particular solution is given by

5 2 5 21 199 8 7 , 99 48 63

84 84t t t t t tx t e e e y t e e e .

5. The coefficient matrix of the associated homogeneous system has eigenvalues 1 1

and 2 5 , so the nonhomogeneous term te duplicates part of the complementary solu-

tion. We therefore try the particular solution

1 1 1 2 2 2,t t t tp px t a b e c te y t a b e c te .

Upon solving the six linear equations we get by collecting coefficients after substitution of this trial solution into the given nonhomogeneous system, we obtain the particular so-lution

1 112 7 , 6 7

3 3t t tx t e te y t te .

Section 5.7 357


6. The coefficient matrix of the associated homogeneous system has eigenvalues

17 89

2 , so there is no duplication. We therefore try the particular solution

1 1 2 2,t t t tp pt tx b e c te y b e c te .

Upon solving the four linear equations we get by collecting coefficients after substitution of this trial solution into the given nonhomogeneous system, we obtain the particular so-lution

1 191 16 , 25 16

256 32t tx t t e y t t e .

7. First we try the particular solution

1 1 2 2sin cos , sin cosp px t a t b t y t a t b t .

Upon solving the four linear equations we get by collecting coefficients after substitution

of this trial solution into the given nonhomogeneous system, we find that 1

21

82a ,

1

25

82b , 2

15

41a , and 2

12

41b . The coefficient matrix of the associated homoge-

neous system has eigenvalues 1 1 and 2 9 , with eigenvectors T

1 1 1v and

T

2 2 3 v , respectively, so the complementary solution is given by

9 91 2 1 22 , 3t t t t


When we impose the initial conditions 0 1x , 0 0y we find that 1

9

10c and

2

83

410c . It follows that the desired particular solution c px x x , c py y y is given

by

9

9

1369 166 125cos 105sin ,

4101

369 249 120cos 150sin .410

t t

t t

x t e e t t

y t e e t t

8. The coefficient matrix of the associated homogeneous system has eigenvalues 2i , so the complementary function involves cos2t and sin 2t . There being therefore no du-plication, we substitute the trial solution

1 1 2 2sin cos , sin cosp px t a t b t y t a t b t

into the given nonhomogeneous system. Upon solving the four linear equations that re-sult upon collection of coefficients, we obtain the particular solution



1 117cos 2sin , 3cos 5sin

3 3x t t t y t t t .

9. Here the associated homogeneous system is the same as in Problem 8, so the nonhomo-geneous term cos2t duplicates the complementary function. We therefore substitute the trial solution

1 1 1 1

2 2 2 2



p

p

x t a t b t c t t d t t

y t a t b t c t t d t t

and use a computer algebra system to solve the system of 8 linear equations that results when we collect coefficients in the usual way. This gives the particular solution

1 1sin 2 2 cos2 sin 2 , sin 2

4 4x t t t t t t y t t t .

10. The coefficient matrix of the associated homogeneous system has eigenvalues 3i , so there is no duplication. Substitution of the trial solution

1 1 2 2cos sin , cos sint t t tp px t a e t b e t y t a e t b e t

yields the equations

2 1

2 1

2 0

2 0

a b

b a

and 1 2 2

2 1 2

2 2 0

2 2 1

a a b

a b b

.

The first two equations enable us to eliminate two of the variables immediately, and we

readily solve for the values 1

4

13a , 2

3

13a , 1

6

13b , and 2

2

13b that give the partic-

ular solution

1 14cos 6sin , 3cos 2sin

13 13t tx t e t t y t e t t .

11. The coefficient matrix of the associated homogeneous system has eigenvalues 1 0 and

2 4, so there is duplication of constant terms. We therefore substitute the particular

solution

1 1 2 2,p px t a b t y t a b t

and solve the resulting equations for 1 2a , 2 0a , 1 2b , and 2 1b . The eigenvec-

tors of the coefficient matrix associated with the eigenvalues 1 0 and 2 4 are

T

1 2 1 v and T

2 2 1v , respectively, so the general solution of the given non-

homogeneous system is given by

4 41 2 1 22 2 2 2 ,t tx t c c e t y t c c e t .

Section 5.7 359


When we impose the initial conditions 0 1x , 0 1y we find readily that 1

5

4c ,

2

1

4c . This gives the desired particular solution

4 41 11 4 , 5 4

2 4t tx t t e y t t e .


2 2, so there is duplication of constant terms in the first natural attempt. We must

multiply the t-terms by t and include all lower-degree terms in the trial solution. Thus we substitute the trial solution

2 21 1 1 2 2 2,p px t a b t c t y t a b t c t .

The resulting six equations in the coefficients are satisfied by 1 1 2 2 0a b a b ,

1 1c , 2 1c . This gives the particular solution

2 2,x t t y t t .


2 3, so there is duplication of te terms. We therefore substitute the trial solution

1 1 2 2,t tp px t a b t e y t a b t e

This leads readily to the particular solution

1 51 5 ,

2 2t tx t t e y t te .


2 4, so there is duplication of both constant terms and 4te terms. We therefore substi-

tute the particular solution

4 4 4 41 1 1 1 2 2 2 2,t t t t

p pt tx a b t c e d te y a b t c e d te .

When we use a computer algebra system to solve the resulting system of 8 equations in 8 unknowns, we find that 2a and 2c can be chosen arbitrarily. With both zero we get the

particular solution

4 4 412 4 2 , 2

8 2t t tt

x t t e te y t e .

In Problems 15 and 16 the amounts 1x t and 2x t in the two tanks satisfy the equations

1 0 1 1 2 1 1 2 2,x rc k x x k x k x ,



where ii

rk

V , in terms of the flow rate r, the inflowing concentration 0c , and the volumes 1V

and 2V of the two tanks.

15. (a) We solve the initial value problem

11 1

1 22 2

20 , 0 010

, 0 010 20

xx x

x xx x

for 101 200 1 tx t e , 10 20

2 400 1 2t tx t e e .

(b) Evidently 1 200galx t and 2 400galx t as t .

(c) It takes about 6 min 56 sec for tank 1 to reach a salt concentration of 1 lb/gal, and about 24 min 34 sec for tank 2 to reach this concentration.

16. (a) We solve the initial value problem

11 1

1 22 2

30 , 0 020

, 0 020 10

xx x

x xx x

for 201 600 1 tx t e , 10 20

2 300 1 2t tx t e e .

(b) Evidently 1 600galx t and 2 300galx t as t .

(c) It takes about 8 min 7 sec for tank 1 to reach a salt concentration of 1 lb/gal, and about 17 min 13 sec for tank 2 to reach this concentration.

In Problems 17–34 we apply the variation of parameters formula in Eq. (28) of Section 5.6. The answers shown below were actually calculated using the Mathematica code listed in the compu-ting project for Section 5.7. For instance, for Problem 17 we first enter the coefficient matrix

A = {{6, -7}, {1, -2}};

the initial vector

x0 = {{0}, {0}};

and the vector

f[t_] := {{60}, {90}};

of nonhomogeneous terms. It simplifies the notation to rename Mathematica’s exponential ma-trix function by defining

exp[A_] := MatrixExp[A]

Then the integral in the variation of parameters formula is given by

Section 5.7 361


integral = Integrate[exp[-A*s] . f[s], {s, 0, t}] // Simplify

and yields the output

5

5

102 7 95

96 95

t t

t t

e e

e e

.

Finally the desired particular solution is given by

solution = exp[A*t] . (x0 + integral) // Simplify

which yields

5

5

102 7 95

96 95

t t

t t

e e

e e

.

(Maple and MATLAB versions of this computation are provided in the applications manual that accompanies the textbook.)

In each succeeding problem, we need only substitute the given coefficient matrix A, initial vector x0, and the vector f of nonhomogeneous terms in the above commands, and then re-execute them in turn. We give below only the component functions of the final results.

17. 51 102 95 7t tx t e e , 5

2 96 95 t tx t e e

18. 51 68 110 75 7t tx t t e e , 5

2 74 80 75 t tx t t e e

19. 3 21 70 60 16 54t tx t t e e , 3 2

2 5 60 32 27t tx t t e e

20. 2 2 31 3 60 3t t tx t e te e , 2 2 3

2 6 30 6t t tx t e te e

21. 2 31 14 15t t tx t e e e , 2 3

2 5 10 15t t tx t e e e

22. 3 31 10 7 10 5t t t tx t e te e te , 3 3

2 15 35 15 5t t t tx t e te e te

23. 21 3 11 8x t t t , 2

2 5 17 24x t t t

24. 1 2 lnx t t t , 2

15 3 3lnx t t t

t

25. 1 1 8 cos 8sinx t t t t , 2 2 4 2cos 3sinx t t t t

26. 1 3cos 32sin 17 cos 4 sinx t t t t t t t , 2 5cos 13sin 6 cos 5 sinx t t t t t t t



27. 3 41 8 6x t t t , 2 3 4

2 3 2 3x t t t t

28. 2 21 7 14 6 4 lnx t t t t t , 2 2

2 7 9 3 ln 2 ln 2 lnx t t t t t t t t

29. 1 cos ln cos sinx t t t t t , 2 sin ln cos cosx t t t t t

30. 21

1cos2

2x t t t , 2

2

1sin 2

2x t t t

31. 2 31 9 4 tx t t t e , 2

2 6 tx t t e , 3 6 tx t te

32. 21 44 18 44 26t tx t t e t e , 2

2 6 6 6t tx t e t e , 23 2 tx t te

33. 2 3 4 51 15 60 95 12x t t t t t , 2 3 4

2 15 55 15x t t t t , 2 33 15 20x t t t ,

24 15x t t

34. 3 2 21 4 4 16 8 tx t t t t e , 2 2

2 3 2 4 tx t t t e , 2 23 2 4 2 tx t t t e ,

24 1 tx t t e

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