Chapter 6 diff eq sm

564

CHAPTER 6 Linear Systems ofDifferential Equations

6.1 Theory of Linear DE Systems

Breaking Out Systems

1. 1 1 2

2 1 2

24

x x xx x x′ = +′ = −

2. 1 1

2 2 1x xx x′ =′ = − +

3. 1 1 2

2 1 2

4 3 tx x x ex x x

−′ = + +′ = − −

4. 1 2

2 3

3 1 2 32 3 sin

x xx xx x x x t

′ =′ =′ = − + + +

Checking It Out

5. 1 33 1⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

Substituting ( )4

4

t

t

et

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

u and ( )2

2

t

t

et

e

−

−

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦v into the given system easily verifies:

4 4

4 4

4 1 33 14

t t

t t

e e

e e

⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

and 2 2

2 2

2 1 33 12

t t

t t

e e

e e

− −

− −

⎡ ⎤ ⎡ ⎤− ⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥

−⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦.

The fundamental matrix ( )4 2

4 2

t t

t t

e et

e e

−

−

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦X .

The general solution of this 2 2× system ( )4 2

1 24 2

t t

t t

e et c c

e e

−

−

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦x .

SECTION 6.1 Theory of Linear DE Systems 565

6. 4 12 1

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

By substitution, we verify that ( )3

3

t

t

et

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

u and ( )2

22

t

t

et

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

v satisfy the system.

The fundamental matrix ( )3 2

3 22

t t

t t

e et

e e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

X .

The general solution ( )3 2

1 23 22

t t

t t

e et c c

e e

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦x .

7. 1 14 1⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

By substitution, we verify that ( )2

t

t

et

e

−

−

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦u and ( )

3

32

t

t

et

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

v satisfy the system.

The fundamental matrix ( )3

32 2

t t

t t

e et

e e

−

−

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦X .

The general solution ( )3

1 2 32 2

t t

t t

e et c c

e e

−

−

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦x .

8. 0 11 0

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

By substitution, we verify that ( ) sincos

tt

t⎡ ⎤

= ⎢ ⎥⎣ ⎦

u and ( ) cossin

tt

t⎡ ⎤

= ⎢ ⎥−⎣ ⎦v satisfy the system.

The fundamental matrix ( ) sin coscos sin

t tt

t t⎡ ⎤

= ⎢ ⎥−⎣ ⎦X .

The general solution ( ) 1 2sin coscos sin

t tt c c

t t⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .

566 CHAPTER 6 Linear Systems of Differential Equations

Uniqueness in the Phase Plane

9. The direction field of x y′ = , y x′ = − is shown.

We have drawn three distinct trajectories for thesix initial conditions ( ) ( )( )0 , 0x y =

( )1, 0 , ( )2, 0 , ( )3, 0 , ( )0,1 , ( )0, 2 , ( )0, 3 .

Note that although the trajectories may (and do)coincide if one starts at a point lying on another,they never cross each other.

However, if we plot coordinate ( )x x t= or ( )y y t= for these same six initial conditions we get

the six intersecting curves shown in the tx and ty planes.

Intersecting solutions ( )x x t= Intersecting solutions ( )y y t=


Verification

10. Substituting t

t

e

e

−

−

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦v into 1 1

2 2

1 22 1

x xx x′⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥′ ⎣ ⎦⎣ ⎦ ⎣ ⎦

yields

1 22 1

t t

t t

e e

e e

− −

− −

⎡ ⎤ ⎡ ⎤− ⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥

−⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦, or 2

2

t t t

t t t

e e e

e e e

− − −

− − −

− = −

= −,

which verifies that v is the solution.

Third-Order Verification

11. To verify , , ,u v w , you should follow the procedure carried out in Problem 10. To show that the vector functions , , u v w are linearly independent, set

2 2

2 21 2 3 1 2 3

2

0 00

0 0

t t

t t t

t t

e te

c c c c e c e c te

e e

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥+ + = + + =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

u v w

or, in scalar form,

2 22 3

2 21 2 3

21 3

0

0

0

t t

t t t

t t

c e c te

c e c e c te

c e c e

+ =

+ + =

+ =

.

Because it was assumed that these equations are true for all t, they must be true for 0t = :

2

1 2

1 3

000

cc cc c

=+ =

+ =

The unique solution of this system is 1 2 3 0c c c= = = . Hence the vector functions are linearly

independent.


Euler’s Method Numerics

12. (a) The IVP studied in Example 5, 0.1 0x x′′ + = , ( )0 1x = , ( )0 0x′ = ,

can be solved numerically with a spreadsheet using the following coding:

A B C D E

1 t x y dxdt

dydt

2 0 1 0 = C2 = –0.1 * B2

3 = A2 + 0.1 = B2 + 0.1 * D2 = C2 + 0.1 * E2 = C3 = –0.1 * B3

Row 3 can now be dragged down to produce the following values on 0 1t≤ ≤ .

t x y dxdt

dydt

0.0 1.0000 0.0000 0.0000 –0.1000

0.1 1.0000 –0.0100 –0.0100 –0.1000

0.2 0.9990 –0.0200 –0.0200 –0.0999

0.3 0.9970 –0.0300 –0.0300 –0.0997

0.4 0.9940 –0.0400 –0.0400 –0.0994

0.5 0.9900 –0.0499 –0.0499 –0.0990

0.6 0.9850 –0.0598 –0.0598 –0.0985

0.7 0.9790 –0.0697 –0.0697 –0.0979

0.8 0.9721 –0.0794 –0.0794 –0.0972

0.9 0.9641 –0.0892 –0.0892 –0.0964

1.0 0.9552 –0.0988 –0.0988 –0.0955

continued on the next page


If the domain is continued to 40t = , then the graphs that correspond to Figure 6.1.1 look

like the following.

Note that the xy trajectory does not close as in Figure 6.1.1; with Euler’s method, a smaller step size would do better.

(b) The IVP studied in Example 6

0.05 0.1 0x x x′′ ′+ + = , ( )5 0.1x − = − , ( )5 0.5x′ − = ,

can be solved numerically with a spreadsheet using the following coding:

A B C D E

1 t x y dxdt

dydt

2 0 1 0 = C2 = –0.1 * B2 – 0.05 * C2

3 = A2 + 0.1 = B2 + 0.1 * D2 = C2 + 0.1 * E2 = C3 = –0.1 * B3 – 0.05 * C3


Row 3 can now be dragged down to produce the following values on 5 4t− ≤ ≤ − .

t x y dxdt

dydt

–5.0 –0.1000 0.5000 0.5000 –0.0150

–4.9 –0.0500 0.4985 0.4985 –0.0199

–4.8 –0.0001 0.4965 0.4965 –0.0248

–4.7 0.0495 0.4940 0.4940 –0.0297

–4.6 0.0989 0.4911 0.4911 –0.0344

–4.5 0.1480 0.4876 0.4876 –0.0392

–4.4 0.1968 0.4837 0.4837 –0.0439

–4.3 0.2451 0.4793 0.4793 –0.0485

–4.2 0.2931 0.4745 0.4745 –0.0530

–4.1 0.3405 0.4692 0.4692 –0.0575

–4.0 0.3874 0.4634 0.4634 –0.0619

If the domain is continued to 25t = , then the graphs that correspond to Figure 6.1.2 look

like the following.

continued on the next page


(c) The system of Example 9,

2

2

3

cos0 0 1sin0 1 2 1 ,

1 1 3 00 1 1 t

tt tt

t

et

⎡ ⎤ ⎡ ⎤+⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

x x + or

2

21 1 4

2 2 3 4

33 1 2 3

4 2 3 4

1 cos

2 sin

3t

tx t x t

x x x t

x x x t

x x tx e

= + + +

= + − +

= − + +

= + + +

x

x

x

x

,

can be solved numerically with a spreadsheet using the following coding. We choose stepsize h = 0.01, in hope of a good approximation, and enter the initial conditions in boldface. There are so many columns for a 4D-system that we have to break our display into two lines.

Dragging down Row 3 results in the following value for the first 100 steps, or −1 ≤ t ≤ 0.


Choosing to plot columns B, C, D, E as “lines” gives the following “chart” for component graphs, or time series.

These curves agree with the graphs shown in Example 9, Figure 6.1.7, for the behaviors between t0 = −1 and tf = 0.


Finding Trajectories

13. x x′ = , y y′ = . Write

dy y ydx x x

′= =

′.

Separating variables, yields

dy dxy x=

or

ln

ln lnx c

c

y x c

y e e

y e x

y C x

= +

=

= ±

=

where C is an arbitrary constant. Hence, the trajectories consist of a family of semi-infinite lines originating at the origin. The equations x x′ = , y y′ = show that solutions move along these lines

away from the origin as indicated in the figure. All solutions go faster and faster the further away they are from the origin.

14. x y′ = , y x′ = − . We write these equations as dy y xdx x y

′= = −

′.

Separating variables, yields the equation in thedifferential form ydy xdx= − . Integrating, yields

2 21 12 2

y x c= − + , or 2 2x y C+ =

where C is an arbitrary nonnegative constant.

Hence, the trajectories consist of a family of cir-

cles centered at the origin. The equations x y′ = ,y x′ = − show that solutions move along the tra-

jectories in the clockwise direction as illustrated.

Keep in mind that solutions do not allmove at the same speed. All circular paths aroundthe origin have the same period, but the paths with the larger radius move at a faster rate.


Computer Check

15. See the computer phase portraits shown in the solutions for Problem 13 and 14.

Computer Lab: Skew-Symmetric Matrices

16. (a) x y′ = , y x′ = −

Trajectories of this skew symmetric sys-tem are given in the figure. Note thattrajectories are circles centered aroundthe origin, and, hence, the length of thevector ( ), x y=x is a constant.

(The figure does not have the properaspect ratio to show circles, but you can note that both axes go from −3 to 3.)

(b) x ky′ = , y kx′ = −

Write this system as the single equation

2 0x k x′′ + =

which has a general solution of

( )cosx A kt δ= − .

Then find

( )1 siny x A ktk

δ′= = − − .

Hence, the length of any solution vector ( ), x y=x is

( ) ( ) ( ) ( )2 2 2 2 2 2cos sinx t y t A kt A kt Aδ δ+ = − + − = .

Therefore, the trajectories of the system are circles centered around the origin with

frequency 2kωπ

= and period 2kπ . We see that does not affect the size of the circles.

An open-ended graphic solver can be used to verify these facts for 1k ≠ .


The Wronskian

When the Wronskian is not zero, the vectors are linearly independent and form a fundamental set.

(If the Wronskian of two solutions is nonzero on any interval it will always be nonzero on that interval.)

17. [ ]2

31 2 2

2, 0

0

t tt

t

e eW e

e= = − ≠x x , so the vectors form a fundamental set.

18. [ ]3

21 2 3

2, 5 0

3

t tt

t t

e eW e

e e

−

−= = − ≠

−x x , so the vectors form a fundamental set.

19. [ ] 21 2

2, 0

0

t tt

t

e eW e

e= = − ≠x x , so the vectors form a fundamental set.

20. [ ]4 4

81 2 4 4

3, 2 0

t tt

t t

e eW e

e e= = ≠x x , so the vectors form a fundamental set.

21. [ ] ( )2 2 2 21 2

cos sin, cos sin 0

sin cos

t tt t

t t

e t e tW e t t e

e t e t= = + = ≠−

x x ; the vectors form a fundamental set.

22. [ ] 2 21 2

cos3 sin3, cos 3 sin 3 1 0

sin 3 cos3t t

W t tt t

= = + = ≠−

x x , so the vectors form a fundamental set.

Suggested Journal Entry I

23. Student Project

Suggested Journal Entry II

24. Student Project


6.2 Linear Systems with Real Eigenvalues

Sketching Second-Order DEs

1. 0x x x′′ ′+ + =

(a) Letting y x′= , we write the equation as the first-order system

.

x yy x y′ =′ = − −

(b) The equilibrium point is ( ) ( ), 0, 0x y = .

(c) nullcline 0nullcline 0

h x yyυ

− + =− =

(See figure.)

(d) From the direction field, the equilibrium point ( ) ( ), 0, 0x y = is stable.

(e) A mass-spring system with this equationshows damped oscillatory motion.

2. 0x x x′′ ′− + =


.

x yy x y′ =′ = − +



h x yyυ

− − =− =

(See figure.)

(d) From the direction field, the equilibriumpoint ( ) ( ), 0, 0x y = is unstable.

(e) A mass-spring system with this equationtends to fly apart.

SECTION 6.2 Linear Systems with Real Eigenvalues 577

3. 1x x′′ + =


1.

x yy x′ =′ = − +



h xyυ

− =− =

(See figure.)

(d) From the direction field, the equilibriumpoint ( ) ( ), 1, 0x y = is stable.

(e) A mass-spring system with this equationshows no damping and steady forcing;hence, periodic motion about an equilib-rium to the right of the origin.

4. 2 2x x x′′ ′+ + =


2 2.

x yy x y′ =′ = − − +


(c) nullcline 2 2nullcline 0

h x yyυ

− + =− =

(See figure.)

(d) From the direction field, the equilibriumpoint ( ) ( ), 2, 0x y = is stable.

(e) A mass-spring system with this equation shows heavy damping. The force movesthe equilibrium two units to the right ofthe origin.

Matching Games

5. A 6. C 7. D 8. B


Solutions in General

9. 4 22 1−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The characteristic equation of the system is ( ) 24 25 0

2 1p

λλ λ λ

λ− −

= = + =− −

,

which has solutions 1 0λ = , 2 5λ = − . Finding the eigenvectors corresponding to each eigenvalue

yields

1 1 2 21 2

0 , 5 .2 1

λ λ−⎡ ⎤ ⎡ ⎤

= ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v

Hence, the general solution is ( ) 51 2

1 22 1

tt c c e−−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

10. 2 13 6

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The characteristic equation of the system is ( ) 22 18 15 0

3 6p

λλ λ λ

λ−

= = − + =− −

,

which has solutions 1 3λ = , 2 5λ = .

Finding the eigenvectors corresponding to each eigenvalue yields

1 1 2 21 1

3 , 5 .1 3

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v

Hence, the general solution is ( ) 3 51 2

1 11 3

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

11. 1 12 4

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

The characteristic equation of the system is

( ) 21 15 6 0

2 4p

λλ λ λ

λ− −

= = − + =−

,


The eigenvectors corresponding to each eigenvalue are

1 1 2 21 1

2 , 3 .1 2

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦v v


1 11 2

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦x .


12. 10 5

8 12−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The characteristic equation of the system is

( ) 210 52 80 0

8 12p

λλ λ λ

λ− −

= = + − =− −

,

which has solutions 1 10λ = − , 2 8λ = .


1 1 2 21 5

10 , 8 .4 2

λ λ⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 54 2

t tt c e c e− ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .

13. 5 13 1

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x


3 1p

λλ λ λ

λ− −

= = − + =−

,



1 1 2 21 1

2 , 4 .3 1

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 13 1

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

14. 1 24 3⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x


4 3p

λλ λ λ

λ−

= = − − =−

,



1 1 2 21 1

1 , 5 .1 2

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 11 2

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


15. 1 02 2

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The characteristic equation of the system is ( ) ( )( )1 01 2 0

2 2p

λλ λ λ

λ−

= = − − =− −

,

which has solutions 1 1λ = , 2 2λ = . The eigenvectors corresponding to each eigenvalue are

1 1 2 21 0

1 , 2 .2 1

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 02 1

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

16. 3 31 1

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x


1 1p

λλ λ λ

λ−

= = − =− − −

,



1 1 2 21 3

0 , 2 .1 1

λ λ− −⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 31 1

tt c c e− −⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

17. 3 22 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


2 2p

λλ λ λ

λ− −

= = − − =− −

,



1 1 2 21 2

1 , 2 .2 1

λ λ⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 22 1

t tt c e c e− ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


18. 4 34 4

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x


4 4p

λλ λ

λ−

= = − =− − −

,

which has solutions 1 2λ = , 2 2λ = − .


1 1 2 23 1

2 , 2 .2 2

λ λ−⎡ ⎤ ⎡ ⎤

= ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦v v


3 12 2

t tt c e c e−−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .

19. 1 23 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


3 4p

λλ λ λ

λ− −

= = + + =− −

,

which has solutions 1 2λ = − , 2 1λ = − . The eigenvectors corresponding to each eigenvalue are

1 1 2 22 1

2 , 1 .3 1

λ λ⎡ ⎤ ⎡ ⎤

= − ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


2 13 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .

20. 5 22 8

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


2 8p

λλ λ λ

λ− −

= = − + =− −

,



1 1 2 22 1

4 , 9 .1 2

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦v v


2 11 2

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .


21. 4 38 6

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


8 6p

λλ λ λ

λ− −

= = + =− −

,

which has solutions 1 0λ = , 2 2λ = − .

The eigenvectors corresponding to each eigenvalue are

1 1 2 23 1

0 , 2 .4 2

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


3 14 2

tt c c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

22. 5 31 1

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


1 1p

λλ λ λ

λ−

= = − + =− −

,



1 1 2 21 3

2 , 4 .1 1

λ λ− −⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 31 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

Repeated Eigenvalues

23. 1 14 3−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x


4 3p

λλ λ λ

λ− −

= = − + =− −

,

which has solutions 1, 1λ = with one linearly independent eigenvector 12⎡ ⎤

= ⎢ ⎥⎣ ⎦

v .

The general solution is, therefore, ( ) 11 2

2

1 12 2

t t t ut c e c te e

u⎧ ⎫⎡ ⎤⎡ ⎤ ⎡ ⎤⎪ ⎪= + +⎨ ⎬⎢ ⎥⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭

x ,

where u is a generalized eigenvector satisfying (A-λ I) u = v , or

1

2

2 1 14 2 2

uu

− ⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦

.


This condition gives one linearly independent equation, 1 22 1u u− + = . Hence,

1

2

0 11 2 1 2

u kk

u k⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= = = +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦u

and

( ) 1 21 1 1 0

.2 2 2 1

t t t tt c e c te ke e⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎪ ⎪= + + +⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭

x

Because the term involving k is a scalar multiple of the first term, we combine them to get a final expression

( ) 21 1 02 2 1

t t tt ce c te e⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎪ ⎪= + +⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭

x .

24. 3 28 5

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x


8 5p

λλ λ λ

λ−

= = + + =− − −

,

which has solutions 1, 1λ = − − with one linearly independent eigenvector, 12

⎡ ⎤= ⎢ ⎥−⎣ ⎦

v .

The general solution is, therefore,

( ) 11 2

2

1 12 2

t t t ut c e c te e

u− − −⎧ ⎫⎡ ⎤⎡ ⎤ ⎡ ⎤⎪ ⎪= + +⎨ ⎬⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭

x

where u is a generalized eigenvector satisfying (A-λ I) u = v , or

1

2

4 2 18 4 2

uu⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎣ ⎦,

which has one linearly independent equation, 1 24 2 1u u+ = . Hence,

( )

1

2

1 2

0 11 1 222 2

01 1 1.12 2 2

2

t t t t

kuk

u k

t c e c te ke e− − − −

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= = = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ −− ⎣ ⎦⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎧ ⎫⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎪ ⎪⎢ ⎥= + + +⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎪ ⎪⎢ ⎥⎣ ⎦⎩ ⎭

u

x

Because the term involving k is a multiple of the first term, we have

( ) 2

01 112 22

t tt c e c e t− −⎧ ⎫⎡ ⎤⎡ ⎤ ⎡ ⎤⎪ ⎪⎢ ⎥= + +⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎪ ⎪⎢ ⎥⎣ ⎦⎩ ⎭

x .


Solutions in Particular

25. 2 15 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x , ( ) 1

03⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


5 4p

λλ λ λ

λ− −

= = − − =− −

,

which has the solutions 1 3λ = and 2 1λ = − .


1 1 2 21 1

3 , 1 .5 1

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v

The general solution is ( ) 31 2

1 15 1

t tt c e c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

Substituting the initial conditions ( ) 10

3⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

15 3c cc c+ =+ =

which gives 112

c = , 212

c = . The solution of the IVP is ( ) 3 1 11 15 12 2

t tt e e−⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦x .

26. 1 32 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x , ( ) 1

01

⎡ ⎤= ⎢ ⎥−⎣ ⎦

x


2 2p

λλ λ λ

λ− −

= = − − =− −

,

which has the solutions 1 1λ = − and 2 4λ = .


1 1 2 23 1

1 , 42 1

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


3 12 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


1⎡ ⎤

= ⎢ ⎥−⎣ ⎦x yields

1 2

1 2

3 12 1

c cc c− =+ = −

which gives 1 0c = , 2 1c = − . The solution of the IVP is ( ) 4 11

tt e−⎡ ⎤

= − ⎢ ⎥⎣ ⎦

x .


27. 2 00 3⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 50

4⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The characteristic equation of the system is ( ) ( )( )2 02 3 0

0 3p

λλ λ λ

λ−

= = − − =−

,

which has the solutions 1 2λ = and 2 3λ = .


1 1 2 21 0

2 , 3 .0 1

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v

The general solution is ( ) 2 31 2

1 00 1

t tt c e c e⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .


4⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields 1 5c = and 2 4c = .

The solution of the IVP is ( )2

2 33

1 0 55 4

0 1 4

tt t

t

et e e

e

⎡ ⎤⎡ ⎤ ⎡ ⎤= + = ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦x .

28. 2 41 1

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 10

1−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


1 1p

λλ λ λ

λ− −

= = + − =−

,



1 1 2 24 1

3 , 2 .1 1

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


4 11 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


1−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

4 11

c cc c

− + = −+ =

which gives 125

c = , 235


t tt e e− −⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

x .


29. 1 11 1⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 20

3⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


1 1p

λλ λ λ

λ−

= = − =−

,



1 1 2 21 1

0 , 2 .1 1

λ λ−⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 11 1

tt c c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .


3⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

23

c cc c

− + =+ =

which gives 112

c = , 252

c = . The solution of the IVP is ( ) 21 11 51 12 2

tt e−⎡ ⎤ ⎡ ⎤⎛ ⎞= +⎢ ⎥ ⎜ ⎟ ⎢ ⎥

⎝ ⎠⎣ ⎦ ⎣ ⎦x .

30. 3 21 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x , ( ) 1

06−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


1 2p

λλ λ λ

λ− −

= = + + =− −

,

which has the solutions 1 4λ = − and 2 1λ = − .


1 1 2 22 1

4 , 1 .1 1

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


2 11 1

t tt c e c e− −−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


6−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

2 16

c cc c

− + = −+ =

which gives 173

c = , 2113


t tt e e− −−⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

x .


31. 2 14 2−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x , ( ) 20

4⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


4 2p

λλ λ λ

λ− −

= = + =− −

,

which has the solutions 1 0λ = and 2 4λ = − .


1 1 2 21 1

0 , 4 .2 2

λ λ⎡ ⎤ ⎡ ⎤

= ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦v v


1 12 2

tt c c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .


4⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

22 2 4

c cc c

+ =− =

which gives 1 2c = and 2 0c = . The solution of the IVP is ( ) 24

t⎡ ⎤

= ⎢ ⎥⎣ ⎦

x .

32. 1 123 1⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 00

1⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


3 1p

λλ λ λ

λ−

= = − − =−

,



1 1 2 22 2

5 , 7 .1 1

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


2 21 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


1⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

2 2 01

c cc c

− + =+ =

which gives 112

c = , 212


t tt e e− −⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

x .


33. 1 12 4

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 10

0⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


2 4p

λλ λ λ

λ− −

= = − + =−

,



1 1 2 21 1

2 , 3 .1 2

λ λ−⎡ ⎤ ⎡ ⎤

= ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦v v


1 11 2

t tt c e c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .


0⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

12 0

c cc c− + =− =

which gives 1 2c = − and 2 1c = − . The solution of the IVP is ( ) 2 31 12

1 2t tt e e−⎡ ⎤ ⎡ ⎤

= − −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .

34. 1 22 1⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 10

3⎡ ⎤

= ⎢ ⎥⎣ ⎦

x


2 1p

λλ λ λ

λ−

= = − − =−

,



1 1 2 21 1

1 , 3 .1 1

λ λ−⎡ ⎤ ⎡ ⎤

= − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v v


1 11 1

t tt c e c e− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .


3⎡ ⎤

= ⎢ ⎥⎣ ⎦

x yields

1 2

1 2

13

c cc c

− + =+ =

which gives 1 1c = , 2 2c = . The solution of the IVP is ( ) 31 12

1 1t tt e e− −⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .


Creating New Problems

35. (a) An example is

1 0

0 00 0

aa

b

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

A .

The characteristic equation of this matrix is ( ) ( ) ( )2p a bλ λ λ= − − , giving a double root

of a and a single root of b for the eigenvalues. However, the eigenvector corresponding to a is found by solving for x, y, z in the equation

1 0

0 00 0

a x xa y a y

b z z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, or, ax y ax

ay aybz az

+ ===

,

which implies 0z = , x α= , 0y = . In other words, it has only one (linearly independent) eigenvector [ ]1, 0, 0 . The eigenvector corresponding to the single eigenvalue b is found

by solving for x, y, z in the equation

1 0

0 00 0

a x xa y b y

b z z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, or, ax y bx

ay bybz bz

+ ===

,

which implies 0, 0x y= = , z α= . In other words, the eigenvector is [ ]0, 0, 1 .

(b) An example is

1 0

0 00 0

aa

a

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

A .

The characteristic equation of this matrix is ( ) ( )3p aλ λ= − , giving a triple root of a for

the eigenvalues. To find the eigenvector, solve for x, y, z in the equation

1 0

0 00 0

a x xa y a y

a z z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, or ax y ax

ay ayaz az

+ ===

,

which implies 0y = , x α= , z β= , α, β arbitrary. The two (linearly independent) eigenvectors are [ ]1, 0, 0 and [ ]0, 0, 1 .


Repeated Eigenvalue Theory

36. a bc d⎡ ⎤

= ⎢ ⎥⎣ ⎦

A has characteristic equation

( ) ( )2 0a d ad bcλ λ− + + − = .

2( ) ( ) 4( )

2a d a d ad bc

λ+ ± + − −

=

There is a double eigenvalue if and only if the discriminant

( ) ( )2 4 0a d ad bc+ − − = (a − d)2 + 4bc = 0

in which case, 1 ( )2

a dλ = + .

(b) If a = d, then bc = 0 which implies that either b or c = 0, possibly both.

The double eigenvalue .a dλ = = To find the eigenvectors,

1

2

0 0( )

0 0vb

c vλ

⎡ ⎤ ⎡ ⎤⎡ ⎤− = =⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦A I v ⇒ bv2 = 0 and cv1 = 0.

Case 1: b = 0, c ≠ 0 Then v2 is free and v1 = 0 and 01

⎧ ⎫⎡ ⎤⎪ ⎪⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

is a basis.

Case 2: b ≠ 0, c = 0 Then v1 is free and v2 = 0 and 10

⎧ ⎫⎡ ⎤⎪ ⎪⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

is a basis.

Case 3: b = c = 0 Then both v1 and v2 are free and 1 0

,0 1

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

is a basis.

Therefore the requirement that both the off-diagonal elements be zero, i.e., when ,a dλ = = implies that A must be a multiple of

the identity matrix.

(c) Assume a ≠ d: With zero discriminant, the double eigenvalue is now 1 ( ).2

a d+

To find the eigenvectors, set

1

2

1 ( ) 02( )1 0( )2

a a d b vvc d a d

λ

⎡ ⎤− +⎢ ⎥ ⎡ ⎤ ⎡ ⎤− = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎣ ⎦⎣ ⎦− +⎢ ⎥⎣ ⎦

A I v .


The resulting system,

1 21 ( ) 02

a d v bv− + =

1 21 ( ) 02

cv d a v+ − =

gives 1 22bv v

d a=

− so

2bd a

⎧ ⎫⎡ ⎤⎪ ⎪⎨ ⎬⎢ ⎥−⎪ ⎪⎣ ⎦⎩ ⎭

is a basis for the eigenspace.

An alternate basis vector is .2

d ac−⎡ ⎤

⎢ ⎥⎣ ⎦

(d) Assume a ≠ d.

Because there is only one eigenvector for 1 ( ),2

a dλ = + we need a generalized

eigenvector u .

To find u , set ( )λ− =A I u v

1 2

1 2

1 ( ) 22

1 ( )2

a d u bu b

cu d a u d a

⎫− + = ⎪⎪⎬⎪+ − = −⎪⎭

⇒ u1 = 0, u2 = 2.

Therefore u = 02⎡ ⎤⎢ ⎥⎣ ⎦

is a generalized eigenvector.

The general solution of the system is 1 22 2 0

( ) .2

t tb bt c e c e t

d a d aλ λ ⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= + +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠x

Quick Sketch

37. Student Project

Generalized Eigenvectors

38. (a) (i) 1 1( ) ( ) ( )t t t te e e eλ λ λ λλ λ ′= = = = =Ax A v Av v v x

(ii) 2 1 1 1( ) ( ) ( )t t tt e e t e tλ λ λ λ λ= + = + = + +Ax A v u Av Au v v u

= 1 2t t tte e eλ λ λλ λ ′+ + =v v u x .


(iii) 2 23 1 2 1 2

1 12 2

t tt t e e t tλ λ⎛ ⎞ ⎛ ⎞= + + = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Ax A v u u Av Au Au

21 1 2

21 1 2

21 1 2

1 ( )212

1 .2

t

t

t t t t t

e t t

e t t t

t e te te e e

λ

λ

λ λ λ λ λ

λ λ λ

λ λ λ

λ λ λ

⎛ ⎞= + + + +⎜ ⎟⎝ ⎠⎛ ⎞= + + + +⎜ ⎟⎝ ⎠

′= + + + + = 3

v v u u u

v v u u u

v v u u u x

(b) We first show that the set 1 2{ , , }v u u is linearly independent.

(i) Observe that, from the definitions of eigenvector and generalized eigenvectors,

( )λ− =A I v 0 , 21( )λ− =A I u v , 2 1( ) ,λ− =A I u u

it follows that 2( )λ− =A I v 0 , 2

1( )λ− =A I u 0 , 22( ) .λ− =A I u v

Now, let us suppose that 1 2 1 3 3 ,c c c+ + =v u u 0 which implies

21 2 1 3 2( ) ( ) .c c cλ− + + =A I v u u 0

Expanding the left side gives 2 2 2

1 2 1 3 2( ) ( ) ( ) ,c c cλ λ λ− + − + −A I v A I u A I uv0 0

so we have 3c v = 0 which means that c3 = 0.

Thus our supposition reduces to 1 2 1c cv + u = 0, which implies

1 2 1( )( ) .c cλ− + =A I v u 0

By expanding the left side

1 2 1( ) ( )c cλ λ− + −A I v A I u

0 0

we have 2c =v 0 which means that c2 = 0.

Because c3 = c2 = 0, we are left with 1c =v 0 which implies c3 = 0.

Therefore, we conclude that the set 1 2{ , , }v u u is linearly independent.


(ii) We are now ready to show that the vectors 1 2 1, ( )t te t eλ λ= = +x v x v u and

23 1 2

12

tt t eλ⎛ ⎞= + +⎜ ⎟⎝ ⎠

x v u u are linearly independent.

Suppose, for all t, that

21 1 2 2 3 3 1 2 1 3 1 2

1( ) ( ) ( ) { ( )2

tc t c t c t e c c t c t tλ ⎛ ⎞+ + = + + + + + =⎜ ⎟⎝ ⎠

x x x v v u v u u 0 .

Because this equation must hold for t = 0, it follows that 1 2 1 3 2 .c c c+ + =v u u 0

We proved in (i) that the set 1 2{ , , }v u u is linearly independent, so we must have

c1 = c2 = c3 = 0.

Thus we have proved that the vectors 1 2 3, , and x x x are linearly independent.

(c) Since 1 1 10 1 10 0 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

A

is upper triangular, it has eigenvalue 1, 1, 1λ = .

The eigenspace is one dimensional and an eigenvector for λ is 10 .0

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v

A solution of 1( )λ− =A I u v is 1

01 ,0

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

u and a solution of

2 1( )λ− =A I u u is 2

011

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

u

Using parts (a) and (b), the general solution of ′ =x Ax is

21 2 3

1 1 0 1 0 01( ) 0 0 1 0 1 1 .2

0 0 0 0 0 1

t t tt c e c e t c e t t⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + + + + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠

x


One Independent Eigenvector

39. 0 0 11 0 30 1 3

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

A

(a) The eigenvalue is 1λ = , with an algebraic multiplicity of 3. We find the eigenvector(s) by substituting 1λ = into the equation λ=Av v and solving for the vector v . Doing this yields the single eigenvector [ ]1, 2, 1c − .

(b) From the eigenvalue and eigenvector, one solution has been found

( )1

121

tt ce⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

x .

(c) Now we solve for a second solution of the form ( )2t tt te e= +x v u , where [ ]1, 2, 1= −v

is the first eigenvector, and [ ]1 2 3, , u u u=u is an unknown vector. Substituting ( )2 tx

into the system ′ =x Ax and comparing coefficients of tte and te yields equations for 1u ,

2u , 3u , giving 1 1u = − , 2 1u = , 3 0u = . Hence, we obtain as a second solution

( )2

1 12 11 0

t tt te e−⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥= − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

(d) To find a third (linearly independent) solution, we try the specific form

( ) 23

12

t t tt t e te e= + +x v u w

where v and u are vectors previously found and w is the unknown vector. Substituting ( )3 tx into the system results in the system of equations ( )− =A I w u . We then find

[ ]1 2 3, , w w w=w . Solving this system yields 1 1w = , 2 0w = , 3 0w = . Hence, we obtain

as a third solution

( ) 23

1 1 11 2 1 02

1 0 0

t t tt t e te e−⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .


Solutions in Space

40. 3 2 21 4 12 4 1

⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥− − −⎣ ⎦

x x

The characteristic equation of the system is ( ) 3 2

3 2 21 4 1 6 11 6 02 4 1

pλ

λ λ λ λ λλ

−= − = − + − + =

− − − −,

which has solutions 1 1λ = , 2 2λ = , and 3 3λ = .


1 1 2 2 3 3

1 2 01 0 , 2 1 , 3 1 .

1 0 1λ λ λ

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⇒ = = ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

v v v

Hence, the general solution is ( ) 2 31 2 3

1 2 00 1 11 0 1

t t tt c e c e c e−⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

41. 1 1 01 2 10 3 1

−⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥−⎣ ⎦

x x

The characteristic equation of the system is ( ) 3

1 1 01 2 1 7 6 00 3 1

pλ

λ λ λ λλ

− −= − = − + + =

− −,

which has solutions 1 1λ = − , 2 3λ = , and 3 2λ = − .


1 1 2 2 3 3

1 1 11 0 , 3 4 , 2 1 .

1 3 3λ λ λ

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − ⇒ = = ⇒ = = − ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

v v v


1 1 10 4 11 3 3

t t tt c e c e c e− −

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .


Spatial Particulars

42. 1 1 00 1 31 1 0

−⎡ ⎤⎢ ⎥′ = −⎢ ⎥⎢ ⎥−⎣ ⎦

x x , ( )0

0 01

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x

We find the eigenvalues and eigenvectors of the coefficient matrix by the usual procedure, obtaining

1 1 2 2 3 3

3 1 10 3 , 2 1 , 2 3 .

1 1 1λ λ λ

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⇒ = = ⇒ = = − ⇒ = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

v v v


3 1 13 1 31 1 1

t tt c c e c e−− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

Substituting this vector into the initial condition ( ) [ ]0 0, 0, 1=x yields the three equations

1 2 3

1 2 3

1 2 3

3 03 3 0

0

c c cc c cc c c

− − =+ − =+ + =

with the solution 114

c = , 238

c = , 338

c = .

Hence, the IVP has the solution ( ) 2 2

3 1 11 3 33 1 34 8 8

1 1 1

t tt e e−− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

43. 1 1 01 1 00 0 1

⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥−⎣ ⎦

x x , ( )2

0 42

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x

We find the eigenvalues and eigenvectors of the coefficient matrix by the usual procedure, obtaining

1 1 2 2 3 3

1 0 10 1 , 1 0 , 2 1 .

0 1 0λ λ λ

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⇒ = = − ⇒ = = ⇒ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

v v v


1 0 11 0 10 1 0

t tt c c e c e−

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .


Substituting this vector into the initial condition ( ) [ ]0 2, 4, 2=x yields the three equations

1 3

1 3

2

242

c cc c

c

− + =+ =

=

with has the solution 1 1c = , 2 2c = , 3 3c = .

Hence, the IVP has the solution ( ) 2

1 0 11 2 0 3 10 1 0

t tt e e−

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

Verification of Independence

44. To show that the vectors

( ) ( )4 41 2

1 and

2 2 1t t t

t e t et

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

x x

are linearly independent, we must show that

4 41 2

1 02 2 1 0

t t tc e c e

t⎡ ⎤ ⎡ ⎤ ⎡ ⎤

+ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

for all. If this must hold for all t, it must hold for 0t = , which yields the equations

1 1 20, 2 0c c c= − − = ; their solution is 1 2 0c c= = .

Adjoint Systems

45. 0 11 0⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

(a) The negative transpose of the given matrix is simply the matrix with –1s in the place of 1s, hence the adjoint system is

T 0 11 0

−⎡ ⎤′ = − = ⎢ ⎥−⎣ ⎦w A w w .

(b) The first equality is simply the product rule for matrix derivatives. Using the adjoint system, yields

( )TT T T′ = − = −w A w w A ,

and hence, T T T T 0′ ′ ′ ′+ = − + =w x w x w x w x .


(c) The characteristic equation of the matrix is simply 2 1 0λ − = , and hence, the eigenvalues are +1, –1. The eigenvector corresponding to +1 can easily be found and is [1, 1]. Likewise, the eigenvector for –1 is [1, -1]. Hence,

( ) 1 21 11 1

t tt c e c e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦x .

Substituting in the initial condition (0)x = [1, 0], yields 1 212

c c= = .

The IVP solution is ( ) 1 11 1 11 12 2 2

t tt t

t t

e et e e

e e

−−

−

⎡ ⎤+⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ −⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦x .

(d) In the adjoint system the eigenvalues are also 1 and –1, but the eigenvectors are reversed, [1,-1] and [1,1], respectively. Hence

( ) 1 21 11 1

t tt k e k e−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦w .

If the initial conditions are ( )0w = [0, 1], then

112

k = , 212

k = − . So the solution of this IVP is

1( )2

t t

t t

e et

e e

−

−

⎡ ⎤− += ⎢ ⎥

+⎢ ⎥⎣ ⎦w .

(e) Part (b) shows that the vectors ( )tw and ( )tx have a constant dot product.

-2

-2

2

2

x,w

x,w

w(t)

x(t)

Note that the initial conditions ( )0x = [1, 0] and ( )0w = [0, 1] are orthogonal vectors, so this constant is zero. Hence the two resulting trajectories will be orthogonal for all 0t > . As trajectories evolve, the vector ( )tw for the adjoint system is always orthogonal to the vector ( )tx for the original system, as shown for a typical t value.

Cauchy-Euler Systems

46. t ′ =x Ax

(a) Let ( )t tλ=x v , where λ is an eigenvalue of A and v is a corresponding eigenvector.

Then 1tλλ −′ =x v or t tλλ′ =x v .

On the other hand,

t t tλ λ λλ= = =Ax A v Av v ,

because v is an eigenvector of A. Therefore, t ′ =x Ax .

w(t)x(t)


(b) We have

3 22 2

t−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x , 0t > .

The characteristic equation is ( ) ( )( )3 2 4 0,p λ λ λ= − − − + =

which yields eigenvalues 1 1λ = − and 2 2.λ = Corresponding eigenvectors are

[ ]1 1, 2=v and [ ]2 2, 1=v .

From part (a), the general solution is then ( ) 1 21 2

1 22 1

t c t c t− ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .

Computer Labs: Predicting Phase Portraits

For each of the linear systems (47−50) a few trajectories in the phase plane have been drawn. The analytic solutions are then computed.

47. x x′ = , y y′ = −

Solve each of these equations individu-ally, obtaining

1tx c e= and 2

ty c e−= .

Eliminating t yields the trajectories

cyx

= ,

which is the family of hyperbolas shown.

48. 0x′ = , y y′ = −

Solve each of these equations individu-ally, obtaining

1x c= and 2ty c e−= .

Eliminating t yields the trajectories

,x c=

which is the family of vertical lines. For any starting point ( )0 0, x y the solution moves

asymptotically towards ( )0 , 0x . The x-axis is composed entirely of stable equilibrium

points.


49. x x y′ = + , y x y′ = +

Because x y′ ′= , solutions are a familyof straight lines x y c= + in the phase

plane, with slope 1 and y-intercept ( ), 0c .

The line x y= − consists entirely of

unstable equilibrium points.

50. x y′ = , y x′ =

We write these equations as the singleequation

0x x′′ − = ,

which has solution

1 2t tx c e c e−= + .

Hence, 1 2( ) t ty t c e c e−= − .

Now we add and subtract these equations, yielding

1

2

2

2 .

t

t

x y c e

x y c e−+ =

− =

Combining these gives

21

2 12 , or, ,cx y c x y kx y x y

⎛ ⎞+ = + =⎜ ⎟− −⎝ ⎠

which is a family of hyperbolas with axes y x= and y x= − . (See figure.)

Radioactive Decay Chain

51. (a) The amount of iodine is decreasing via radioactive decay; hence, 1dI k Idt

= − , where 1k is

the decay constant of iodine. Work in Chapter 2 showed that the decay constant is ln 2 divided by the half-life of the material; hence,

1ln 2 0.10345486.7

k = ≈ .


The amount of xenon x(t) is increasing with the decay of iodine, but decreasing with its own radioactive decay, hence, the equation

1 2 ,dx k I k xdt

= − where 2ln 2 0.07534219.2

k = ≈ .

(b) In matrix form, the equations become

1

1 2

0kI Ik kx x

′ −⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥′ −⎣ ⎦ ⎣ ⎦⎣ ⎦

.

The eigenvalues of this triangular matrix can easily be seen and their eigenvectors calculated as

2 11 1 1 2 2 2

1

0, ; , .

1k k

k kk

λ λ−⎡ ⎤ ⎡ ⎤

= − = = − =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

v v

Hence, the solution is ( ) 1 22 11 2

1

0 1

k t k tk kt c e c e

k− −−⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

x .

Multiple Compartment Mixing I

52. (a) Let x1, x2 represent the amount of salt (in lbs) in Tank A and Tank B, respectively.

1x ′ = 1 26 2

100 100x x− + , x1(0) = 25

1 22

6 6100 100

x xx ′ = − , x2(0) = 0

6 2100 1006 6

100 100

⎡ ⎤−⎢ ⎥′ = =⎢ ⎥

⎢ ⎥−⎢ ⎥⎣ ⎦

x x Ax

We use the fact that if ,λ=Av v then kAv = kλv , so the eigenvalues for A are 1100

of

the eigenvalues for 100A.

Furthermore, the eigenvectors are precisely the same, so we can use6 2

6 6−⎡ ⎤⎢ ⎥−⎣ ⎦

.

Then 2 12 24 0, and 6 2 3λ λ λ+ + = = − ± .

For 6 2 3λ = − + :

2 3 2 0

06 2 3

ab

⎡ ⎤− ⎡ ⎤ ⎡ ⎤= ⇒⎢ ⎥ ⎢ ⎥ ⎢ ⎥

− ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

2 3 2 0

6 2 3 0

a b

a b

− + =

− = 1

1

3

⎡ ⎤⇒ = ⎢ ⎥

⎣ ⎦v .


For 6 2 3λ = − − , by similar calculation, 2

1

3

⎡ ⎤= ⎢ ⎥

−⎣ ⎦v .

Hence the general solution 1 1( 6 2 3) ( 6 2 3)

100 1001 2

0.025 0.0951 2

1 1( )

3 3

1 1

3 3

t t

t

t c e c e

c e c e

− + − −

− −

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

−⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x

Substituting the initial conditions x1(0) = 25, x2(0) = 0, we obtain

25 = c1 + c2

0 = 1 23 3c c−

which yields the IVP solution 0.025 0.0951 1( ) 12.5 12.5 .

3 3t tt e e− −⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x

(b)

(c) Yes. At the time when the graphs of 1( )tx and 2 ( )tx intersect, the amount of salt in Tank

2 equals and then exceeds the amount of salt in Tank 1.

(d) The amount of salt in each tank approaches zero as time becomes large.


Multiple Compartment Mixing II

53. Let x1, x2 represent the amount of salt in Tank A and Tank B, respectively.

VA = 150 gal in which 25 lb of salt is dissolved.

VB = 100 gal of pure water.

1 21

6 2150 100

x xx ′ = − + , x1(0) = 25.

1 22

6 6150 100

x xx ′ = − , x1(0) = 0.

′ =x Ax =

1 125 501 325 50

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

2 5 2 2 10, so ,50 1250 25 50

λ λ λ − −+ + = = .

For 2 ,25

λ = − 1v = 12⎡ ⎤⎢ ⎥⎣ ⎦

; for 1 ,50

λ = − 2v = 11⎡ ⎤⎢ ⎥⎣ ⎦

.

Hence, the general solution

2 125 50

21 1

( )2 1

t tt ce c e

− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x .

Substituting the initial conditions x1(0) = 25, x2(0) = 0, we obtain

25 = c1 + c2

0 = 2c1 + c2,

which yields the IVP solution 2 125 50

1 1( ) 25 50

2 1t t

t e e− −⎡ ⎤ ⎡ ⎤

= − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

(b)


(c) No. The amount of salt in Tank 1 never exceeds the amount of salt in Tank 2, i.e.,

2 1( ) ( )t t≤x x for a ll t ≥ 0.

(d) The amount of salt in each tank approaches zero as time becomes large.

Mixing and Homogeneity

54. Instead of pouring pure water into Tank A, pour in a brine solution of 12

lb/gal. Then the

equations would be

1 21

6 22100 100

x xx ′ = − + , 1 22

6 6 ,100 100

x xx ′ = −

6 22100 100

6 6 0100 100

⎡ ⎤−⎢ ⎥ ⎡ ⎤′ = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦−⎢ ⎥⎣ ⎦

x x .

Aquatic Compartment Model

55. 1 1 2

2 1 2 3

3 1 2 3

0.10 0.10

0.06 0.11 0.05

0.04 0.01 0.05 ,

x x x

x x x x

x x x x

⎫′ = − +⎪⎪′ = − + ⎬⎪′ = + − ⎪⎭

which can be written as .10 .10 0

.06 .11 0.5

.04 .01 .05

−⎡ ⎤⎢ ⎥′ = −⎢ ⎥⎢ ⎥−⎣ ⎦

x x .

Electrical Circuits

56. R1 = R2 = R3 = 4 ohms, L1 = L2 = 2 henries.

Noting that I2 = I1 − I3, we obtain from Kirchoff’s 2nd law,

(Loop 1) 4I1 + 4(I1 − I3) + 12I ′ = 0

(Loop 2) 4I3 + 32I ′ − (I1 − I3)2 = 0

1 1 3

3 1 3

4 2

3

I I I

I I I

′ = − +

′ = − or

4 21 3−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

I I , where 1

3

II⎡ ⎤

= ⎢ ⎥⎣ ⎦

I .

The eigenvalues and eigenvectors are

1 2,λ = − 111⎡ ⎤

= ⎢ ⎥⎣ ⎦

v ; 2 5,λ = − 22

1⎡ ⎤

= ⎢ ⎥−⎣ ⎦v .

Thus, the general solution is

1 2 51 2

3

1 2( )

1 1t tI

t c e c eI

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦

I and I2(t) = I1(t) − I3(t) = 3c2e−5t.


57. R1 = 4 ohms, R3 = 6 ohms, L1 = 1 henry, L2 = 2 henries.

Using the fact that I3 = I1 − I2, we obtain from Kirchoff’s 2nd Law

(Loop 1) 1 1 1 24 6( ) 0I I I I′ + + − =

(Loop 2) 2 1 22 6( ) 0I I I′ − − =

so we have

1 1 2

2 1 2

10 6

3 3

I I I

I I I

′ = − +

′ = − or 1 1

2 2

10 63 3

I II I

′ −⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦⎣ ⎦ ⎣ ⎦

.

The eigenvalues and eigenvectors are

λ1 = −1, 123⎡ ⎤

= ⎢ ⎥⎣ ⎦

v ; λ2 = −12, 231

⎡ ⎤= ⎢ ⎥−⎣ ⎦

v .

The general solution of our system is 1 121 2

2

2 33 1

t tIc e c e

I− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦, so

I1(t) = 2c1e−t + 3c2e−12t

I2(t) = 3c1e−t − c2e−12t

I3(t) = −c1e−t + 4c2e−12t

Suggested Journal Entry

58. Student Project


6.3 Linear Systems with Nonreal Eigenvalues

For all problems in 6.3, iλ α β= ± and .i= ±v p q

Solutions in General

1. 0 11 0

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The characteristic equation for the matrix is 2 1 0λ + = , which has complex eigenvalues λ = ±i. Substituting i into λ=Av v for λ, yields the single equation 2 1v iv= . so v = [ ]1, i± . Therefore,

0α = , 1β = , [ ]1, 0=p , [ ]0, 1=q .

Two linearly independent solutions result:

( )

( )

1

2

1 0cos sin cos sin ,

0 1

1 0sin cos sin cos .

0 1

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q

The general solution is

( ) 1 1 2 2 1 2

cos sin( ) ( ) ,

sin cost t

t c t c t c ct t

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x x or 1 2

1 2

( ) cos sin( ) sin cos .

x t c t c ty t c t c t

= += − +

2. 1 21 3−⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦

x x

The characteristic equation for the matrix is 2 4 5 0λ λ+ + = , which yields complex eigenvalues 2 iλ = − ± . Corresponding eigenvectors are [ ]1 , 1i= −v ∓ . Therefore,

2α = − , 1β = , [ ]1, 1= −p , [ ]1, 0= −q .

SECTION 6.3 Linear Systems with Nonreal Eigenvalues 607


( )

( )

2 21

2 22


1 0


1 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

− −⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦− −⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q

The general solution is ( ) 2 21 1 2 2 1 22

cos sin sin cos( ) ( )

cos sint t

t

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

e t t− −

−

− + − −⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .

3. 1 22 1

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The characteristic equation for the matrix is 2 2 5 0λ λ− + = , which yields complex eigenvalues 1 2iλ = ± . Substituting these values into λ=Av v , respective eigenvectors [ ]1, i= ±v .

Therefore,

1α = , 2β = , [ ]1, 0=p , [ ]0, 1=q .


( )

( )

1

2

1 0cos sin cos 2 sin 2 ,

0 1

1 0sin cos sin 2 cos 2 .

0 1

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


The general solution is ( ) 1 1 2 2 1 2

cos2 sin 2( ) ( )

sin 2 cos2t tt t

t c t c t c e c et t

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x x .

4. 6 15 2

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

The characteristic equation for the matrix is 2 8 17 0λ λ− + = , which yields complex eigenvalues 4 iλ = ± . Corresponding eigenvectors are [ ]2 , 5i= ±v . Therefore,

4α = , 1β = , [ ]2, 5=p , [ ]1, 0=q .


( )

( )

4 41

4 42


5 0


5 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


2cos sin 2sin cos( ) ( )

5cos 5sint tt t t t


− +⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .


5. 1 12 1

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x

The eigenvalues are iλ = ± , with corresponding [ ]1, 1 i= − ±v . Therefore,

0α = , 1β = , [ ]1, 1= −p , [ ]0, 1=q .


( )

( )

1

2


1 1


1 1

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x p q

x p q


cos sin( ) ( )

cos sin sin cost t

t c t c t c ct t t t

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦

x x x .

6. 2 42 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The eigenvalues are 2iλ = ± with corresponding eigenvectors [ ]1 , 1i= ±v . Therefore,

0α = , 2β = , [ ]1, 1=p , [ ]1, 0=q .


( )

( )

1

2

1 1cos sin cos 2 sin 2

1 0


1 0

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q



( ) 1 1 2 2 1 2

cos2 sin 2 cos2 sin 2( ) ( )

cos 2 sin 2t t t t

t c t c t c ct t

− +⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .

7. 3 24 1

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x The eigenvalues are 1 2iλ = ± , with corresponding eigenvectors [ ]1, 1 i=v ∓ .

Therefore,

1α = , 2β = , [ ]1, 1=p , [ ]0, 1= −q .


( )

( )

1

2

1 0cos sin cos 2 sin 2

1 1


1 1

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x p q

x p q


( ) 1 1 2 2 1 2

cos2 sin 2( ) ( )

cos2 sin 2 cos2 sin 2t tt t

t c t c t c e c et t t t

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥+ − +⎣ ⎦ ⎣ ⎦

x x x .


8. 2 51 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x The eigenvalues are iλ = ± , with complex eigenvectors [ ]2 , 1i= ±v .

Therefore,

0α = , 1β = , ( )2, 1=p , [ ]1, 0=q .


( )

( )

1

2


1 0


1 0

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


2cos sin cos 2sin( ) ( )

cos sint t t t

t c t c t c ct t

− +⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .

9. 1 15 3

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The eigenvalues are 1 iλ = − ± , with complex eigenvectors [ ]2 , 5i= ±v . Therefore,

1α = − , 1β = , [ ]2, 5=p , [ ]1, 0=q .


( )

( )

1

2


5 0


5 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q



( ) 1 1 2 2 1 2

2cos sin cos 2sin( ) ( )

5cos 5sint tt t t t


− −− +⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .

10. 2 33 2

− −⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x The eigenvalues are 2 3iλ = − ± , with complex eigenvectors [ ], 1i= ±v .

Therefore,

2α = − , 3β = , [ ]0, 1=p , [ ]1, 0=q .


( )

( )

2 21

2 22

0 1cos sin cos3 sin 3

1 0

0 1sin cos sin3 cos3 .

1 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


sin3 cos3( ) ( )

cos3 sin 3t tt t


− −−⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x .


11. 3 12 1− −⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The eigenvalues are 2 iλ = − ± , with complex eigenvectors [ ]1, 1 i= − ±v . Therefore,

2α = − , 1β = , [ ]1, 1= −p , [ ]0, 1=q .


( )

( )

2 21

2 22


1 1


1 1

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

−⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦−⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


cos sin( ) ( )

cos sin sin cost tt t

t c t c t c e c et t t t

− −− −⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦

x x x .

12. 2 42 2

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x The eigenvalues are 2iλ = ± , with complex eigenvectors [ ]2, 1 i= − ±v .

Therefore,

0α = , 2β = , [ ]2, 1= −p , [ ]0, 1=q .


( )

( )

1

2


1 1

2 0sin cos sin 2 cos2 .

1 1

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x p q

x p q


2cos2 2sin 2( ) ( )

cos2 sin 2 sin 2 cos2t t

t c t c t c ct t t t

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦

x x x .


Solutions in Particular

13. 1 11 1

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 10

1−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The coefficient matrix has eigenvalues 1 iλ = ± and corresponding eigenvectors [ ], 1i= ±v .

Hence, two linearly independent solutions obtained are

( )

( )

1

2


1 0


1 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q

Substituting the initial conditions into 1 1 2 2( ) ( ) ( ) :t c t c t= +x x x

( ) ( ) ( )1 1 2 2 1 2

0 1 10 0 0

1 0 1c c c c

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦x x x ⇒ 1

2

11

cc== −

The solution is, therefore, ( ) ( ) ( )1 2

sin coscos sin

t t tt t t e

t t− −⎡ ⎤

= − = ⎢ ⎥−⎣ ⎦x x x .


14. 0 41 0

−⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x , ( ) 10

1⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The coefficient matrix has eigenvalues 2iλ = ± and corresponding eigenvectors [ ]2, i=v ∓ .

Hence, two linearly independent solutions are

( )

( )

1

2


0 1

2 0sin cos sin 2 cos2 .

0 1

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x p q

x p q


( ) ( ) ( )1 1 2 2 1 2

2 0 10 0 0

0 1 1c c c c

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x x x

yields 112

c = and 2 1c = − .

The solution is, therefore, ( ) ( ) ( )1 2

cos2 2sin 21

12 sin 2 cos22

t tt t t

t t

−⎡ ⎤⎢ ⎥= − =⎢ ⎥+⎢ ⎥⎣ ⎦

x x x .

15. 3 21 1

−⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x , ( ) 1

01⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The coefficient matrix has eigenvalues 2 iλ = − ± , and corresponding eigenvectors [ ]1 , 1i=v ∓ .


( )

( )

2 21

2 22


1 0

1 1sin cos sin cos2 .

1 0

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

−⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦−⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q



( ) ( ) ( )1 1 2 2 1 2

1 1 10 0 0

1 0 1c c c c

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦x x x

yields 1 1c = and 2 0c = .

The solution is, therefore, ( ) ( ) 21

cos sincos

t t tt t e

t− +⎡ ⎤

= = ⎢ ⎥⎣ ⎦

x x .

16. 1 51 3

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x , ( ) 5

04⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The coefficient matrix has eigenvalues 1 iλ = − ± and corresponding eigenvectors [ ]5, 2 i=v ∓ .


( )

( )

1

2


2 1


2 1

t t t t

t t t t

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x p q

x p q


( ) ( ) ( )1 1 2 2 1 2

5 0 50 0 0

2 1 4c c c c

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x x x

yields 1 1c = and 2 2c = − . The solution is, therefore,

( ) ( ) ( )1 2

5cos 5sin 5cos 10sin2 2

2cos sin 2sin cos 5sint t tt t t t

t t t e e et t t t t

− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x x x .


Nonreal Conditions

17. A = a bc d⎡ ⎤⎢ ⎥⎣ ⎦

(a) The characteristic equation is 2 det 0trλ λ− + =A A , with solution

2

2

det2

( ) ( ) 4( )2

( ) 42 2

tr tr

a d a d ad bc

a d bca d

λ ± −=

+ ± + − −=

− ++= ±

A A A

The discrimant is 2( ) 4 must be always negative forpositive nonreal

a d bc

λ

− +

Hence, for bc to be negative, either b or c, but not both, must be negative.

(b) If trA = 0, then the solution to the characteristic equation reduces to

detλ = ± − A

If the eigenvalues are imaginary, there can be no real part of the eigenvalues λ, so 2

a d+

must be zero.


Rotation Direction

18. For =x Ax with A = a bc d⎡ ⎤⎢ ⎥⎣ ⎦

and nonreal eigenvalues, the off-diagonal elements b and c must

be nonzero and of opposite sign. (See Problem 17(a).) We also know that nonreal eigenvalues give solutions with a rotation factor (see text equation (13)), so it will be sufficient to have a qualitative look at the vector field, determined by

,x ax byy cx dy′ = +′ = +

for some sample points.

For example, if b is negative and c is positive,

• Along the positive y-axis (where x = 0), x′ points left, not right (regardless of

whether y′ points up or down).

• Along the positive x-axis (where y = 0), y′ points up, not down (regardless of whether x′ points right or left).

Some sample possible phase-plane vectors are drawn in the first figure, and they show that the rotation is counterclockwise.

Counter clockwise rotation Clockwise rotation for negative b, positive c. for positive v, negative c

By similar reasoning, if b is positive and c is negative, rotation is clockwise, as shown by second figure.


Complexities of Complex Eigenvectors

19. A = a bc d⎡ ⎤⎢ ⎥⎣ ⎦

, 2, iλ β β= ± = A .

(a)

2 2

( )

0.

0det det

a i b bc d i a i

ab bi ab bibc ad di ai i

tr i

βλ

β β

β β

β β ββ

− −⎡ ⎤⎡ ⎤− = ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦

− + + −⎡ ⎤⎡ ⎤⎢ ⎥= =− + − + + ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥−−⎣ ⎦

A I v

A AA

(b)

2 2 2

( ) *

0,

0

a i b a bic d i c

a i bcac ci cd ci

βλ

β

ββ β

− − −⎡ ⎤⎡ ⎤− = ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦

⎡ ⎤− + − ⎡ ⎤= =⎢ ⎥ ⎢ ⎥− − − + ⎣ ⎦⎣ ⎦

A I v

because for imaginary eigenvalues d = −a

(c) The most obvious difference between v and v * is in the imaginary parts

0β

⎡ ⎤= ⎢ ⎥−⎣ ⎦

q points down along the vertical axis;

*0β−⎡ ⎤

= ⎢ ⎥⎣ ⎦

q points left along the horzontal axis.

The real parts p and *p are even less related, a good example of the caution that

complex numbers and vectors can lead to vastly different expressions and pictures, though the actual solutions to the DE will have the same trajectories.

Elliptical Shape and Tilt

20. For ′ =x Ax with purely imaginary eigenvalues, by Problem 17(a), it follows that trA = 0 so

a = −d and α = 0, so λ = ±βi.

We have from text equations (6) and (7) the solution

1 Re 2 Im( )t c c= +x x x , with

Re

Im

cos sinsin cos

t tt t

β ββ β

−⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

x pqx

Let us choose the initial conditions so that c1 = 1, c2 = 0.


(a) Then we have 0

, ,b

a β−⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦p q and

( ) cos sin ,( ) sin cos .t t tt t t

β ββ β β β

= −′ = − −

x p qx p q

(b) Substitution in the solution equations from part (a) gives position and velocity vectors at four points on an elliptical trajectory:

(0) , (0) β′= = −x p x q (initial position),

, π π ββ β

⎛ ⎞ ⎛ ⎞′= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x p x q (halfway around),

and

, 2 2π π ββ β

⎛ ⎞ ⎛ ⎞′= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x q x p (quarter of the way around)

3 3, 2 2π π ββ β

⎛ ⎞ ⎛ ⎞′= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x q x p (three quarters of the way around).

Plotting these vectors, as shown for an example in 6.3.6, determines the shape and tilt of an elliptical trajectory that results from iλ β= ± .

“Boxing” the Ellipse

From Problem 20, , , . a b b

ic d a

λ α βλ

−⎡ ⎤⎡ ⎤′ = = ± = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦x x v

21. 4 55 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

53 ,

4 3i

iλ

⎡ ⎤= ± = ⎢ ⎥

⎣ ⎦v

∓.

Therefore, α = 0, β = 3, [ ]5,4=p , [ ]0, 3= −q .

See graph.


22. 1 15 1− −⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x 1

2 , 1 2

ii

λ⎡ ⎤

= ± = ⎢ ⎥−⎣ ⎦v

∓.

Therefore, α = 0, β = 2, [ ]1, 1= −p , [ ]0, 2= −q .

See graph.

23. 1 12 1

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

1,

1i

iλ

⎡ ⎤= ± = ⎢ ⎥

⎣ ⎦v

∓.

Therefore, α = 0, β = 1, [ ]1,1=p , [ ]0, 1= −q .

See graph.

24. 1 12 1−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x 1

, 1

ii

λ−⎡ ⎤

= ± = ⎢ ⎥−⎣ ⎦v

∓.

Therefore, α = 0, β = 1, [ ]1, 1= − −p , [ ]0, 1= −q .

See graph.


Tilt with Precision

25. From Problem 20(a) we have

( ) cos sin ( ) sin cos t t tt t t

β ββ β β β

= −′ = − −

x p qx p q

( )2 2

2 2 2 2

( ) ( ) ( cos sin ) cos sin cos sin

cos sin (cos sin ) .1 cos2sin 22

t t t t t t t t

t t t ttt

β β β β β β β β β β

β β β β β βββ

′⋅ = − ⋅ − ⋅ + ⋅ + ⋅

= − − − ⋅

x x p p p q q p q q

q p p q

This product must be zero at the endpoints of the major and minor axes of an elliptical trajectory, which occurs when

2 2

2 tan 2 .tβ ⋅=

−

p qq p

Axes for Ellipses

From Problem 25, , , . a b b

ic d a

λ α βλ

−⎡ ⎤⎡ ⎤′ = = ± = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦x x v

26. 4 55 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

53 ,

4 3i

iλ

⎡ ⎤= ± = ⎢ ⎥

⎣ ⎦v

∓.

Therefore, α = 0, β = 3, [ ]5, 4=p , [ ]0, 3= −q .

(a) tan 2β t* = 2 2

2 2( 12) 0.759 41

⋅ −= =

−−

p qq p

.

2β t* = 1tan (0.75)− = 0.64 radians, or, 0.64 + π = 3.78 radians.

Thus, the parameter β t* = 0.32 radians or 1.89 radians.

(b) For an endpoint of one axis of the ellipse, the value β t* = 0.32 gives coordinates

5 0 4.74 4.74cos.32 sin.32

4 3 3.8 0.94 4.74⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦.

For an endpoint of the other axis of the ellipse, the value β t* = 1.89 gives coordinates

5 0 1.566 1.6cos1.89 sin1.89

4 3 1.253 2.85 1.6− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ ≈⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦


(c) See figure.

27. 1 15 1− −⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x 1

2 , 1 2

ii

λ⎡ ⎤

= ± = ⎢ ⎥−⎣ ⎦v

∓.

Therefore, α = 0, β = 2, [ ]1, 1= −p , [ ]0, 2= −q .

(a) tan 2β t* = 2 2

2 2(2) 24 2

⋅= =

−−

p qq p

.

2β t* = 1tan (2)− ≈ 1.11 radians or 1.11 + π = 4.25 radians.


(b) For an endpoint of one axis of the ellipse, the value β t* = .55 gives coordinates

1 0 .85 .85cos.55 sin.55

1 2 .85 1.04 .19⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦.


1 0 .52 .52cos 2.12 sin 2.12

1 2 .52 1.70 2.22− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ ≈⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(c) See figure.

1.61.6−⎡ ⎤⎢ ⎥⎣ ⎦

4.704.70

⎡ ⎤⎢ ⎥⎣ ⎦

5

4

⎡ ⎤⎢ ⎥⎣ ⎦

.522.22−⎡ ⎤⎢ ⎥⎣ ⎦

0.850.19

⎡ ⎤⎢ ⎥⎣ ⎦

11−

⎡ ⎤⎢ ⎥⎣ ⎦


28. 1 12 1

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

1,

1i

iλ

⎡ ⎤= ± = ⎢ ⎥

⎣ ⎦v

∓.

Therefore, α = 0, β = 1, [ ]1,1=p , [ ]0, 1= −q .

(a) tan 2β t* = 2 2

2 2( 1) 21 2

⋅ −= =

−−

p qq p

.

2β t* = 1tan (2)− ≈ 1.11 radians or 1.11 + π = 4.25 radians.


(b) For an endpoint of one axis of the ellipse, the value β t* = .55 gives coordinates

1 0 .85 .85cos.55 sin.55

1 1 .85 .52 1.37⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦.


1 0 .52 .52cos2.12 sin 2.12

1 1 .52 .85 .33− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ ≈⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(c) See figure.

29. 1 12 1−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x 1

, 1

ii

λ⎡ ⎤

= ± = ⎢ ⎥−⎣ ⎦v

∓.

Therefore, α = 0, β = 1, [ ]1, 1= − −p , [ ]0, 1= −q .

(a) tan 2β t* = 2 2

2 2( 1) 21 2

⋅ −= = −

−−

p qq p

.

2β t* = 1tan ( 2)− − ≈ −1.11 radians or −1.11 + π = 2.03 radians.

Thus, the parameter β t* = −.55 radians or 1.02 radians.

5.26.324−

+

⎡ ⎤⎢ ⎥⎣ ⎦

.851.375

⎡ ⎤⎢ ⎥⎣ ⎦

11

⎡ ⎤⎢ ⎥⎣ ⎦


(b) For an endpoint of one axis of the ellipse, the value β t* = −.55 gives coordinates

1 0 .85 .85cos( .55) sin( .55)

1 1 .85 .52 1.37− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− − − ≈ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦.


1 0 .52 .52cos1.02 sin1.02

1 1 .52 .85 .33− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− ≈ ≈⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦.

(c) See figure.

×3 3 System

30. 1 0 00 0 20 2 0

−⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥−⎣ ⎦

x x

(a) The characteristic equation, ( )( )2

1 0 00 2 1 4 00 2

λλ λ λ

λ

− −⎡ ⎤⎢ ⎥− = − + + =⎢ ⎥⎢ ⎥− −⎣ ⎦

,

has roots 1 1λ = − , 2λ , 3 2iλ = ± .

(b) For 1 1λ = − , solving for x, y, z in the equation

1 0 00 0 2 1 0 2 0

x xy yz z

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

yields x α= , 0y = , 0z = , α arbitrary, so the corresponding eigenvector is 1

100

te−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x .

.52

.33−

+

⎡ ⎤⎢ ⎥⎣ ⎦

.851.37−

+

⎡ ⎤⎢ ⎥⎣ ⎦

11−

−

⎡ ⎤⎢ ⎥⎣ ⎦


(c) For 2 2 ,iλ = solving for x, y, z in the system

1 0 00 0 2 20 2 0

x xy i yz z

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⇒ 2

01i

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v

Hence, we identify

0α = , 2β = , [ ]0, 1, 0=p , [ ]0, 0, 1=q .

Using complex conjugates 2 3,λ λ and 2 3, ,v v two linearly independent solutions are

( )

( )

2

3

00 0cos sin cos2 1 sin 2 0 cos2

0 1 sin 2

00 0sin cos sin 2 1 cos2 0 sin 2 .

0 1 cos 2

t t

t t

t e t e t t t tt

t e t e t t t tt

α α

α α

β β

β β

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= − = − = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= + = + = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x p q

x p q

(d) The general solution, from (b) and (c), is

( ) 1 2 3

1 0 00 cos2 sin 2 ,0 sin 2 cos 2

tt c e c t c tt t

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x

or, ( )( )( )

1

2 3

3 2

cos2 sin 2

cos2 sin 2 .

tx t c e

y t c t c t

z t c t c t

−=

= +

= −

(e) Substituting the IC:

( ) ( ) ( ) ( )1 1 2 2 3 3 1 2 3

1 0 0 10 0 0 0 0 1 0 0

0 0 1 1c c c c c c

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + = + + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x x x x ⇒ 1

2

3

101

ccc

==−

.

The solution of the IVP is, therefore,

( ) ( ) ( )1 3

1 00 sin 20 cos 2

tt t t e tt

−

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x x x , or, in coordinate form, ( )( )( )

sin 2

cos2 .

tx t e

y t t

z t t

−=

=

=


(f) The trajectory of

( ) ( ) ( )( ), , x t y t z t

in 3D space is a helix (i.e., it rotatesaround the x-axis but approaches the yz-plane.)

x

y

z

Threefold Solutions

31. 1 0 10 2 01 0 1

−⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥⎣ ⎦

x x

The characteristic polynomial is given by ( )( )3 2 24 6 4 2 2 2λ λ λ λ λ λ− + − + = − − − + .

Hence, the eigenvalues and eigenvectors are:

[ ][ ]

1 1

2 3 2 3

2 0, 1, 0 ,

, 1 , , 0, 1i i

λ

λ λ

= ⇒ =

= ± ⇒ = ±

v

v v

Therefore, from 2 2, ,λ v we have

1α = , 1β = , [ ]0, 0, 1=p , [ ]1, 0, 0=q .

and three independent solutions are

( )

( )

( )

21

2

3

01 ,0


1 0


1 0

t

t t t t

t t t t

t e

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x p q

x p q

Hence, the general solution is

( )( )

( )

3 22 2

1 2 3 1

2 3

0 sin cos cos sin1 0 00 cos sin cos sin

t

t t t t

t

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

t t e c t c t

⎡ ⎤− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + = ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x


32. 0 1 00 0 11 0 0

⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥−⎣ ⎦

x x

The characteristic equation of this system is ( )3

1 00 1 1 01 0

λλ λ

λ

−⎡ ⎤⎢ ⎥− = − + =⎢ ⎥⎢ ⎥− −⎣ ⎦

.

The eigenvalues and corresponding eigenvectors are

[ ]1 1

2 3 2

1 1, 1, 1

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

i i i

λ

λ λ

= − ⇒ = −

⎡ ⎤= ± ⇒ = −⎣ ⎦

v

v ∓ ∓

Therefore, from 2 2, λ v we have

12

α = , 32

β = , [ ]1, 1, 2= −p , 3, 3, 0⎡ ⎤= − −⎣ ⎦q .

The general solution can be written as

( ) 2 21 2 3

3 3 3 3cos 3 sin sin 3 cos2 2 2 213 3 3 31 cos 3 sin sin 3 cos

2 2 2 21

3 32cos 2sin2 2

t t t

t t t t

t c e c e t t c e t t

t t

−

⎡ ⎤ ⎡ ⎤− + − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= − + + + −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

33. 1 0 02 1 23 2 1

⎡ ⎤⎢ ⎥′ = −⎢ ⎥⎢ ⎥⎣ ⎦

x x

The characteristic polynomial is given by ( )( )3 2 23 7 5 1 2 5λ λ λ λ λ λ− + − + = − − − + .

Hence, the eigenvalues and corresponding eigenvectors are

[ ][ ]

1 1

2 3 2 3

1 2, 3, 2 ,

, 1 2 , 0, 1, .i i

λ

λ λ

= ⇒ = −

= ± ⇒ =

v

v v ∓


1α = , 2β = , [ ]0, 1, 0=p , [ ]0, 0, 1= −q .

Hence the general solution is

( ) 1 2 3

2 0 03 cos2 sin 22 sin 2 cos2

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

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .


34. 3 1 20 1 12 0 0

− −⎡ ⎤⎢ ⎥′ = − −⎢ ⎥⎢ ⎥⎣ ⎦

x x

The characteristic equation is 3 24 7 6 0.λ λ λ+ + + =

Hence the eigenvalues and corresponding eigenvectors are

[ ]1 1

2 3 2 3

2 1, 1, 1 ,

, 1 2 , 2 2, 2, 2 2 .i i i

λ

λ λ

= − ⇒ = −

⎡ ⎤= − ± ⇒ = ±⎣ ⎦

v

v v ∓


1α = − , 2β = , [ ]2, 2, 0=p , 2, 0, 2 2⎡ ⎤= −⎣ ⎦q ,

and three independent solutions are

( )

( )

( )

21

2

3

11 ,1

2 1cos sin cos 2 2 2 sin 2 0 ,

0 2

2 1sin cos sin 2 2 2 cos 2 0 .

0 2

t

t t t t

t t t t

t e

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

−

− −

− −

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x

x p q

x p q

Hence, the general solution can be written as

( ) 21 2 3

2cos 2 2 sin 2 2sin 2 2 cos 211 2cos 2 2sin 21 2 2 sin 2 2 2 cos 2

t t t

t t t t

t c e c e t c e t

t t

− − −

⎡ ⎤ ⎡ ⎤− +−⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= + +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

Triple IVPs

35. 3 0 10 3 10 2 1

−⎡ ⎤⎢ ⎥′ = − −⎢ ⎥⎢ ⎥−⎣ ⎦

x x , ( )5

0 1326

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

x

The eigenvalues and eigenvectors of the coefficient matrix are

[ ]1 13 1, 0, 0λ = ⇒ =v ,

[ ]2 3 2 3, 2 , 5 , 13 13 , 26i i iλ λ = − ± ⇒ = ± − ±v v


with 2α = − , 1β = , [ ]5, 13, 26= −p , [ ]1, 13, 0=q . Hence, the three independent solutions are

( )

( )

( )

31

2 22

2 23

100


26 0


26 0

t

t t t t

t t t t

t e

t e t e t e t e t

t e t e t e t e t

α α

α α

β β

β β

− −

− −

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + = − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x p q

x p q

Substituting the initial conditions:

( ) 1 2 3

1 5 1 50 0 13 13 13

0 26 0 26c c c

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x ⇒ 1

2

3

01

0.

ccc

== −=

The solution of the IVP is 22

5cos sin( ) ( ) 13cos 13sin .

26cos

t

t tt t e t t

t

−

−⎡ ⎤⎢ ⎥= − = − − −⎢ ⎥⎢ ⎥⎣ ⎦

x x

36. 0 1 01 0 10 1 0

⎡ ⎤⎢ ⎥′ = − −⎢ ⎥⎢ ⎥⎣ ⎦

x x , ( )0

0 11

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x


[ ]1 10 1, 0, 1λ = ⇒ = −v ,

2 3 2,3, 2 1, 2, 1i iλ λ ⎡ ⎤= ± ⇒ = ±⎣ ⎦v ,

therefore, three independent solutions are

( )

( )

( )

1

2

3

101

1 0cos sin cos 2 0 2 sin 2 1

1 0

1 0sin cos sin 2 0 2 cos 2 1 .

1 0

t t

t t

t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x p q

x p q


Substituting the initial conditions

( ) 1 2 3

01 1 00 0 0 2 1

1 1 0 1c c c

⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

x

yields 1 212

c c= = and 32

2c = . Hence, the solution of the IVP is

1 2 31 1 22 2 2

cos 2 sin 211 1 20 2 sin 2 2 cos 22 2 2

1 cos 2 sin 2

1 111 1 20 cos 2 2 sin 2 2 .2 2 2

1 1 1

t t

t t

t t

t t

= + +

⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= + − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤−⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= + + −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x x x x

Matter of Independence

37. The Wronskian of two vector functions is defined as the determinant of the matrix formed by placing the vectors as columns in the matrix. If the vector functions are also solutions of a linear system of differential equations, then the vectors are linearly independent if and only if the Wronskian is nonzero for any t in the interval of interest. In this problem, we obtain the two vector solutions

( )

( )

1 11

2 2

1 12

2 2

cos sin

sin cos

t t

t t

a bt e t e t

a b

a bt e t e t

a b

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x

formed from the eigenvalues iα β± and eigenvectors [ ]1 2, a a=p , [ ]1 2, b b=q of a matrix. We evaluate ( )1 tx , ( )2 tx when 0t = , yielding

( ) [ ]1 1 20 , a a=x , ( ) [ ]2 1 20 , b b=x .

Hence, the Wronskian of ( )1 tx and ( )2 tx at 0t = is

[ ]( ) 1 12

2 2

0a b

Wa b

=1x , x .

But the columns of this matrix are linearly independent and thus the Wronskian is nonzero. Hence, the vectors ( )1 tx and ( )2 tx are linearly independent vector functions.


Skew-Symmetric Systems

38. 0

0k

k⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The characteristic equation of the coefficient matrix is

( ) 2 2 0k

p kkλ

λ λλ

−⎡ ⎤= = + =⎢ ⎥− −⎣ ⎦

,

which has roots ikλ = ± , with corresponding eigenvectors

1 1 0

0 1i

i⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= = ±⎢ ⎥ ⎢ ⎥ ⎢ ⎥±⎣ ⎦ ⎣ ⎦ ⎣ ⎦v .

We then identify

0α = , kβ = , [ ]1, 0=p , [ ]0, 1=q .

Two linearly independent vector solutions are then

( )

( )

1

2

1 0cos sin cos sin

0 1

1 0sin cos sin cos

0 1

t t

t t

t e t e t kt kt

t e t e t kt kt

α α

α α

β β

β β

⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x p q

x p q


1 1 2 2 1 2

cos sin( ) ( ) ( ) ,

sin coskt kt

t c t c t c ckt kt

⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x x

or, in component form,

1 2

1 2

cos sinsin cos .

x c kt c kty c kt c kt= += − +

To verify that the length of the solution vector is a constant for all t, we write the system as the single equation

2 0x k x′′ + = whose general solution is

( )cosx C k t δ= − .

We then find

( )1 siny k C ktk

δ= − − .

The length of any solution vector [ ], x y=x is

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 2cos sin cos sinx t y t C kt C kt C kt kt Cδ δ δ δ⎡ ⎤+ = − + − = − + − =⎣ ⎦ .


Coupled Mass-Spring System

39. The coupled mass-spring matrix

1 2 2

1 1

2 32

2 2

0 1 0 0

0 0

0 0 0 1

0 0

k k km m

k kkm m

⎡ ⎤⎢ ⎥+⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥+

−⎢ ⎥⎣ ⎦

simplifies, with 1 2 3 1 2 1k k k m m= = = = = to

0 1 0 02 0 1 00 0 0 11 0 2 0

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥

−⎣ ⎦

.

We find purely complex eigenvalues and their corresponding eigenvectors to be

[ ]1,2 1,2

3,4 3 4

, , 1 , , 1

3, 1, 3, 1, 3 .

i i i

i i i

λ

λ

= ± =

⎡ ⎤= ± = − ±⎣ ⎦

∓ ∓

∓,

v

v

For 1,2λ we have

0α = , 1β = and 1,2 [0,1,0,1]=p , 1,2 [ 1,0, 1,0]= − −q

For 3,4λ we have

0α = , 3β = and 3,4 [ 1,0,1,0]= −p , 3,4 [0, 3,0, 3]= −q .

Then four linearly independent solutions are

1 1,2 1,2

2 1,2 1,2

0 11 0

( ) cos sin cos sin ,0 11 0

0 11 0

( ) sin cos sin cos ,0 11 0

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= + = +⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x

p q

p q


3 3,4 3,4

4 3,4 3,4

010 3

( ) cos sin cos 3 sin 3 ,1 00 3

010 3

( ) sin cos sin 3 cos 3 .1 00 3

t t

t t

t e t e t t t

t e t e t t t

α α

α α

β β

β β

⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= − = − ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= + = + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

x

x

p q

p q


( ) 1 1 2 2 3 3 4 4t c c c c= + + +x x x x x .

Substituting the initial conditions

( )1 0 0x = , ( )2 0 0x = , ( )3 0 2x = , and ( )4 0 0x =

we get

1 0c = , 2 1c = − , 3 1c = , and 4 0c = .

Finally, because 1x x= , and 3y x= , we have the desired result

( )( )

cos cos 3

cos cos 3 .

x t t t

y t t t

= −

= +

Computer Lab: Phase Portrait

40. 0 15 2

⎡ ⎤′ = ⎢ ⎥− −⎣ ⎦x x , ( ) 2

02⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

The trajectory of the IVP solution is shown in thephase plane. See figures for plot of the

x-coordinate and the y-coordinate as a function oft. Note that these graphs are consistent with thesolution in the phase plane.

Phase plane trajectory


41. 4 55 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x , ( ) 2

02

⎡ ⎤= ⎢ ⎥−⎣ ⎦

x .

The trajectory of the IVP solution in the phaseplane is shown. Note that the graphs of ( )x t and

( )y t versus t are consistent with the phase plane

graph.

Phase plane trajectory


42. Student Project

y(t) x(t)


6.4 Stability and Linear Classification

Classification Verification

1. (saddle point) 1 14 2⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

⎥

⎥x

The matrix has eigenvalues –3 and 2. Because it has at least one positive eigenvalue, it is unstable. As the eigenvalues are real and have opposite signs, the origin is a saddle point.

2. (center) 0 11 0

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

The matrix has eigenvalues ±i. Because the real part is zero, the origin is stable, but not asymptotically stable at equilibrium point. The origin is a center.

3. (star node) 2 00 2−⎡ ⎤′ = ⎢ −⎣ ⎦

x x

The matrix has eigenvalues –2 and –2. Because both eigenvalues are negative, the origin is an asymptotically stable equilibrium point. Also the matrix has two linearly independent eigenvectors (in fact every vector in the plane is an eigenvector), and hence, the origin is a star node.

4. (degenerate node) 2 10 2−⎡ ⎤′ = ⎢ −⎣ ⎦

x

The matrix has eigenvalues –2 and –2. Because both eigenvalues are negative, the origin is an asymptotically stable equilibrium point. Also there exists only one linearly independent eigenvector corresponding to the eigenvalue; hence, the origin is a degenerate node.

5. (node) 2 13 4⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

The matrix has eigenvalues 1 and 5, which means the origin is an unstable equilibrium point. The fact that the roots are real and unequal means the origin is a nondegenerate node.

SECTION 6.4 Stability and Linear Classification 637

6. (spiral sink) 0 11 1

⎡′ = ⎢− −⎣ ⎦x

⎤⎥ x

The matrix has eigenvalues 12 2

i− ±3 . Because the real part of the eigenvalues is negative, the

origin is an asymptotically stable equilibrium point. The fact that the eigenvalues are complex with negative real parts also means the origin is a spiral sink.

Undamped Spring

7. 20 0x xω+ =

Denote 1x x= and 2x x= ; the equation becomes

1 12

2 20

0 1

0x xx xω

⎡ ⎤⎡ ⎤ ⎡= ⎢ ⎥

⎤⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣⎣ ⎦ ⎦

.

The coefficient matrix has eigenvalues 0iω± , so the origin ( )0, 0 is a center point and thus

classified as neutrally stable.

Damped Spring

8. 0mx bx kx+ + =

Let x y= . The second-order equation can be written as the linear system

0 1x xk by ym m

⎡ ⎤⎡ ⎤ ⎡⎢ ⎥=⎤

⎢ ⎥ ⎢⎢ ⎥− − ⎥⎣ ⎦ ⎣⎢ ⎥⎣ ⎦ ⎦

.

The determinant of the coefficient matrix is km

, which is assumed positive. Hence, the matrix is

nonsingular and is an isolated equilibrium point. The eigenvalues of this system are the

roots of

0x y= =

( )21

1 0m b kk b mm m

λλ λ

λ

−= + + =

− − −,

which are

2

14

2b b mk

mλ − + −= and

2

24

2b b mk

mλ − − −

= .


From these roots, we see that when , regardless of the values of , and , the roots

will either be real and negative or complex with negative real parts. In either case, the origin is asymptotically stable.

0b > 0m > 0k >

When the three parameters m, k, and b are positive, the origin will always be asymptotically stable, which is the nature of real systems with friction.

One Zero Eigenvalue

9. (a) If 1 0λ = and 2 0λ ≠ , then A is a singular matrix because 1 0λ= − =A A I . Hence, the

rank of A is less than 2. But the rank of A is not 0 because if it were it would be the matrix of all zeros, which would have both eigenvalues 0. The rank of A is 1, which means the kernel of A consists of a one-dimensional subspace of , a line through the origin. But the kernel of A is simply the set of solutions of

2R=Ax 0 , which are the

equilibrium points of . We use the solution of the form ′ =x Ax

( ) 21 2

tx a ct c c e

y bλ

d⎡ ⎤ ⎡ ⎤ ⎡

= = +⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣

x⎦

to find the equilibrium points. We compute the derivatives and set them to zero. Setting , yields the equation 0x y= =

( ) 22 2

00

t ct c e

dλλ⎡ ⎤ ⎡ ⎤′ = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x ,

which implies . The points that satisfy 2 0c = 0x y= = are the points

( ) 1x a

t cy b⎡ ⎤ ⎡

= =⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

x⎦

,

which consists of all multiples of a given vector (i.e., a line through the origin.)

(b) If a solution starts off the line of equilibrium points, then 2 0c ≠ . If 2 0λ > , the second

term

22

t cc e

dλ ⎡ ⎤⎢ ⎥⎣ ⎦

becomes larger and larger. Hence, the solution moves farther and farther away from the line of equilibrium points. On the other hand, if 2 0λ < , the second term becomes smaller

and smaller, the solution moves towards the line.


Zero Eigenvalue Example

10. 0 01 1

x xy y′⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(a) The characteristic equation ( )1λ λ 0− − = yields eigenvalues 1 0λ = and 2 1λ = .

The corresponding eigenvectors are [ ]1 1, 1=v and [ ]2 0, 1=v .

(b) Setting , we see that all points on the line 0x y′ ′= = x y= are equilibrium points, and thus ( )0, 0 is not an isolated equilibrium point.

(c) We set 0x and ′ = y x y′ = − + , yielding ( ) 1x t c= and 1y c y′ = − + . Hence,

( ) 2 1ty t c e c−= +

where c and c are arbitrary constants. In vector form, this is 1 2

( ) 1 2

011 t

xt c c

y e−⎡ ⎤⎡ ⎤ ⎡ ⎤

= = + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

(d) Because ( )x t c= , the solutions move along vertical lines (or don’t move at all). To

examine this further, assume we start at an initial point ( ) ( )( ) ( )0 00 , 0 , x y x= y . Finding

constants, and , yields the solution 1c 2c

( )( ) ( )

0

0 0 0t

x t x

y t x y x e−=

= + −

which says that starting at any point ( )0 0, x y , the solution moves vertically approaching the 45-degree line and the point ( )0 0, x x .

Both Eigenvalues Zero

11. 0 10 0⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x 0,0λ =

For λ = 0, there is only one linearly independent

eigenvector: 10⎡ ⎤

= ⎢ ⎥⎣ ⎦

v

We can check the direction of the solutions through other

points by checking 0 10 0

xy⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦ for a given point (x, y).


12. 2 14 2

⎡′ = ⎢− −⎣ ⎦x

⎤⎥x 0,0λ =


eigenvector: 12

⎡ ⎤= ⎢ ⎥−⎣ ⎦

v

13. 3 91 3

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x 0,0λ =



= ⎢ ⎥⎣ ⎦

v

14. 4 28 4

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x 0,0λ =



= ⎢ ⎥⎣ ⎦

v


Zero Again

15. 1 21 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

(a) The characteristic equation of this system is 2 0λ λ+ = , yielding 1 0λ = and 2 1λ = − . The corresponding eigenvectors can be seen to be [ ]1 2, 1=v , . [ ]2 1, 1=v

(b) Setting

, 0

0

x y′ ′= =

we see that all points on the line

2x y− =

are equilibrium points, and thus ( )0, 0 is

not an isolated equilibrium point. Also from the differential equations,

2x y x y′ ′= = − ,

Sample trajectories of a singular system

we see that solutions move along trajectories on 45-degree lines. Above the line

2 0x y = 2 0x y x y, ′ ′= = − < −

and the movement is downward and to the left. Below the line , movement is

upward and to the right. This outcome is shown in the phase plane. (See the figure.) Note that the solutions below the equilibrium line approach the line because the trajectories move along the 45-degree lines, but the equilibrium line goes up by less than 45 degrees, and the solutions above the equilibrium line move down towards the line.

2 0x y− =

All Zero

16. 0 00 0⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

Nothing moves; all trajectories are points.


Stability

17. 0

0 1k⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The characteristic equation of this system is

( )( )01 0

0 1k

kλ

λ λλ

−,= − + =

− −

hence the roots are 1 kλ = , 2 1λ = − .

(a) ( ), 1k∈ −∞ − implies that the origin ( )0, 0 is an asymptotically stable nondegenerate

node.

(b) implies that the origin 1k = − ( )0, 0 is an asymptotically stable star node.

(c) ( )1, 0k∈ − implies that the origin ( )0, 0 is an asymptotically stable nondegenerate

node.

(d) implies that the matrix is singular; hence, the origin is not an isolated equilibrium point (all trajectories of this system move vertically towards the

0k =

1x axis).

(e) ( )0, k∈ ∞ implies the origin is an unstable saddle point.

k = −2 k = −1 k = −0.5

k = 0 k = 2


Bifurcation Point

18. The characteristic equation of 0 11 k

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x is 2 1 0kλ λ− + = , which has roots

( )21 2

1 42

k kλ λ= = + − .

When 2k < , the roots are complex and the solutions oscillate. When 2k ≥ the solutions ema-

nate from an unstable node. Hence, the bifurcation values are 2k = ± .

Interesting Relationships

19. a bc d⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x

The characteristic equation is

( ) ( ) ( )( ) ( )2 21 2 1 2 1 20 0a d ad bc r r r r r rλ λ λ λ λ λ− + + − = = − − = − + + =

If the characteristic roots are and , we factor the quadratic on the left. We see by equating

the coefficients that 1r 2r

(a) (the coefficient of λ) is always the negative of the sum of the roots (i.e., Tr− A( )1 2Tr r r= − +A ).

(b) A (the constant term) is always the product of the roots (i.e., 1 2r r=A ).

Interpreting the Trace-Determinant Graph

In these problems we use the basic fact that the eigenvalues can be written in terms of the trace and determinant of A using the basic formula

1λ , ( )2

2

42

Tr Trλ

± −=

A A A.

20. 0>A , ( )2 4 0Tr − >A A

Using the basic formula, the eigenvalues are real, unequal, and of the same sign; hence, the equilibrium point ( )0, 0 is a node. Whether it is an attracting or repelling node depends on the

trace.

21. 0<A

Using the basic formula, the determinant of A is negative then ( )2 4Tr − >A A 0 and 0Tr >A ,

so the eigenvalues must be positive and have opposite signs. Hence, the origin is a saddle point and an unstable equilibrium.


22. , 0Tr ≠A ( )2 4 0Tr − <A A

Using the basic formula, the eigenvalues are complex with a nonzero real part. Hence, the origin is a spiral equilibrium point. Whether it is an attracting or repelling spiral depends on whether the trace is positive or negative. If it is negative the origin is attracting, so that it is a spiral sink. If

is positive, the origin is repelling so that it is a spiral source. TrA

23. , 0Tr =A 0>A

Using the basic formula, the eigenvalues are purely complex. Hence, the origin is a center point and neutrally stable.

24. ( )2 4 0Tr − =A A , 0Tr ≠A

Using the basic formula, (real) nonzero eigenvalues are repeated. Hence, the origin is a degenerate or star node.

25. or 0Tr >A 0<A

Using the basic formula, if the trace is positive, then either the roots are complex with positive part or the roots are real with at least one positive root. In either case the origin is an unstable equilibrium point. In the case when det 0<A , then, from the basic formula the roots are real and

at least one root is positive, again showing that the origin is unstable.

26. 0>A and 0Tr =A

Using the basic formula, the eigenvalues are purely imaginary. Hence, the origin is a center point and neutrally stable.

27. and 0Tr <A 0>A

Using the basic formula, the eigenvalues are real and both negative. Hence, the origin is asymptotically stable.


28. Student Project

SECTION 6.5 Decoupling a Linear DE System 645

6.5 Decoupling a Linear DE System

Decoupling Homogeneous Linear Systems

1. 1 22 2− −⎡ ⎤′ = ⎢ ⎥−⎣ ⎦

x x

The coefficient matrix has eigenvalue and eigenvectors

[ ][ ]

1 1

2 2

3, 1, 2

2, 2, 1 .

λ

λ

= = −

= − =

v

v

The matrix of eigenvectors is

1 22 1

⎡ ⎤= ⎢ ⎥−⎣ ⎦

P , and 1 1 212 15

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

Therefore,

1 1 2 1 2 1 2 3 012 1 2 2 2 1 0 25

− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

P AP .

Hence, transforming from to the new variable x 1−=w P x

yields the decoupled system

1 1

2 2

32 .

w ww w′ =′ = −

Solving this decoupled system yields ( ) 31 1

tw t c e= and ( ) 22 2

tw t c e−= .

The solution of the original system is

( ) ( )3

1 3 21 22

2

1 2 1 22 1 2 1

tt t

t

c et t c e c e

c e−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.

2. 0 13 2

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


[ ][ ]

1 1

2 2

3, 1, 3

1, 1, 1 .

λ

λ

= = −

= − =

v

v



1 13 1

⎡ ⎤= ⎢ ⎥−⎣ ⎦

P , and 1 1 113 14

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

Therefore,

1 1 1 0 1 1 1 3 013 1 3 2 3 1 0 14

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

P AP .

Hence, transforming from to the new variables x 1−=w P x yields the uncoupled system 1 13w w′ = and . Solving this decoupled system yields 2w w′ = − 2 ( ) 3

1 1tw t c e= and ( )2 2

tw t c e−= . The solution

of the original system is

( )3

1 31 2

2

1 1 1 13 1 3 1

tt t

t

c et c e

c e−

−

⎡ ⎤c e

⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.

3. 0 11 0

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


[ ][ ]

1 1

2 2

1, 1, 1

1, 1, 1 .

λ

λ

= = −

= − =

v

v


1 11 1−⎡ ⎤

= ⎢ ⎥⎣ ⎦

P , and 1 1 111 12

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

Therefore,

1 1 1 0 1 1 1 1 011 1 1 0 1 1 0 12

− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

P AP .

Hence, transforming from to the new variable x 1−=w P x yields the decoupled system 1 1w w′ = and . Solving this decoupled system yields 2w w′ = − 2 ( )1 1

tw t c e= and ( )2 2tw t c e−= . The solution

of the original system is

( ) ( ) 11 2

2

1 1 1 11 1 1 1

tt t

t

c et t c e c e

c e−

−

⎡ ⎤− −⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.


4. 2 31 4⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x


[ ][ ]

1 1

2 2

1, 3, 1

5, 1, 1 .

λ

λ

= = −

= =

v

v


3 11 1

−⎡ ⎤= ⎢ ⎥⎣ ⎦

P , and 1 1 111 34

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

Therefore,

1 1 1 2 3 3 1 1 011 3 1 4 1 1 0 54

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦P AP .

Hence, transforming from to the new variable x 1−=w P x yields the uncoupled system 1 1w w′ = and . Solving this decoupled system yields 2 5w w′ = 2 ( )1 1

tw t c e= and ( ) 52 2

tw t c e= . Hence, the

solution of the original system is

( ) ( ) 1 51 25

2

3 1 3 11 1 1 1

tt t

t

c et t c e c e

c e

⎡ ⎤− −⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.

5. 2 32 5

−⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x


[ ][ ]

1 1

2 2

1, 3, 1

4, 1, 2 .

λ

λ

= =

= − =

v

v


3 11 2⎡ ⎤

= ⎢ ⎥⎣ ⎦

P , and 1 2 111 35

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦

P .

Therefore,

1 2 1 2 3 3 1 1 011 3 2 5 1 2 0 45

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

P AP .


Hence, transforming from x to the new variable 1−=w P x yields the decoupled system and . Solving this decoupled system yields

1 1w w′ =

2 4w′ = − 2w ( )1 1tw t c e= and ( ) 4

2 2tw t c e−= . Hence, the


( ) ( ) 1 41 24

2

3 1 3 11 2 1 2

tt t

t

c et t c e c e

c e−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.

6. 0 11 0⎡ ⎤′ = ⎢ ⎥⎣ ⎦

x x


[ ][ ]

1 1

2 2

1, 1, 1

1, 1, 1 .

λ

λ

= − = −

= =

v

v


1 11 1−⎡ ⎤

= ⎢ ⎥⎣ ⎦

P , and 1 1 111 12

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

Therefore,

1 1 1 0 1 1 1 1 011 1 1 0 1 1 0 12

− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦P AP .

Hence, transforming from to the new variable x 1−=w P x yields the decoupled system 1 1w w′ = − and . Solving this decoupled system yields 2w w′ = 2 ( )1 1

tw t c e−= and ( )2 2tw t c e= . Hence, the


( ) ( ) 11 2

2

1 1 1 11 1 1 1

tt t

t

c et t c e c e

c e

−−

⎡ ⎤− −⎡ ⎤ ⎡ ⎤= = = +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.

7. 1 1 11 1 11 1 1

⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥⎣ ⎦

x x


[ ][ ][ ]

1 1

2 2

3 3

3, 1, 1, 1

0, 1, 1, 0

0, 1, 0, 1 .

λ

λ

λ

= =

= = −

= = −

v

v

v



1 1 11 1 01 0 1

− −⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P , and 1

1 1 11 1 2 13

1 1 2

−

⎡ ⎤⎢ ⎥= = − −⎢ ⎥⎢ ⎥− −⎣ ⎦

P .

Therefore,

1

1 1 1 1 1 1 1 1 1 3 0 01 1 2 1 1 1 1 1 1 0 0 0 03

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

−

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢= = − − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

P AP⎤⎥⎥⎥⎦

.

Hence, transforming from to the new variable x 1−=w P x yields the decoupled system 1 13w w′ = , , and . Solving this decoupled system yields 2 0w′ = 3 0w′ = ( ) 3

1 1tw t c e= , ( )2 2w t c= , and

( )3w t c= 3 . The solution of the original system is

( ) ( )3

13

2 1 2 3

3

1 1 1 1 1 11 1 0 1 1 01 0 1 1 0 1

t

t

c et t c c e c c

c

⎡ ⎤− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = + +⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

x Pw

⎣ ⎦

3

.

In scalar form

31 2

31 2

31 3 .

t

t

t

x c e c c

y c e c

z c e c

= − −

= +

= +

8. 0 0 00 1 01 0 1

⎡ ⎤⎢ ⎥′ = ⎢ ⎥⎢ ⎥⎣ ⎦

x x


[ ][ ][ ]

1 1

2 2

3 3

0, 1, 0, 1

1, 0, 0, 1

1, 0, 1, 0 .

λ

λ

λ

= = −

= =

= =

v

v

v


1 0 00 0 11 1 0

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P , and 1

1 0 01 0 10 1 0

−

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P .


Therefore,

1

1 0 0 0 0 0 1 0 0 0 0 01 1 0 1 0 1 0 0 0 1 0 1 03

0 1 0 1 0 1 1 1 0 0 0 1

−

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

P AP⎤⎥⎥⎥⎦

.

Hence, transforming from to the new variable x 1−=w P x yields the decoupled system 1 0w′ = , , and . Solving this decoupled system yields 2w w′ = 2 33w w′ = ( )1 1w t c= , ( )2 2

tw t c e= , and

( )3 3tw t c e= . Hence, the solution of the original system is

. ( ) ( )1

2 1 2 3

3

1 0 0 1 0 00 0 1 0 0 11 1 0 1 1 0

t t

t

c

t t c e c c e c e

c e

⎡ ⎤− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = + +⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

x Pw t

1 1x c= − , 2 3tx c e= , and 3 1 2

tx c c e= + .

9. (See Problem 43 in Section 5.4) 1 0 04 3 04 2 1

⎡ ⎤⎢′ = −⎢⎢ ⎥−⎣ ⎦

x ⎥⎥ x

The eigenvalues are λ1 = 1, 1 and λ2 = 3, with eigenvectors

1

1 02 , 00 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

v and 2

011

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v ,

so that

1 0 00 1 00 0 3

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

D and 1 0 02 0 10 1 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P .

We change the variable to , to find 1−=w P x ′ =w Dw :

1 1

2 2

3 3

1 0 0( ) 0 1 0

0 0 3 3

w wt w w w

w w

′⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥′ ′= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥′ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

w1

2

3

w

w.

Solving we obtain ( )1 1 ,tw t c e= ( )2 2 ,tw t c e= ( ) 33 3

tw t c e= .

Thus ( ) ( )11

32 1 3

3 33 2 3

1 0 02 0 1 20 1 1

tt

t t

t t t

c ec e

t t c e c e c e

c e c e c e

t

⎡ ⎤⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥= = = +⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x Pw .


10. (See Problem 44 in Section 5.4) 3 2 01 0 01 1 3

−⎡ ⎤⎢′ = ⎢⎢ ⎥−⎣ ⎦

x ⎥⎥ x

The eigenvalues are λ = 1, 2, 3 with respective eigenvectors 1 2 01 , 1 , 0 , 0 1 1

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

so that

1 0 00 2 00 0 3

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

D and 1 2 01 1 00 1 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P .

We change the variable to 1−=w P x , to find ′ =w Dw :

1 1

2 2

3 3

1 0 0( ) 0 2 0 2

0 0 3 3

w wt w w w

w w

′ 1

2

3

w

w

⎡ ⎤ ⎡ ⎤ ⎡⎡ ⎤ ⎤⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥′ ′= = = ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥′ ⎣ ⎦ ⎥⎣ ⎦ ⎣ ⎦ ⎣

w

⎦

t

t

Solving the individual linear DEs 2 31 1 2 2 3 3( ) , ( ) , ( )t tw t c e w t c e w t c e= = =

Thus

( )

21 21

2 22 1 2

3 2 33 2 3

21 2 0( ) 1 1 0

0 1 1

t tt

t t

t t t

c e c ec e

t t c e c e c e

c e c e c e

⎡ ⎤⎡ ⎤ +⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥= = = +⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x Pw .

Decoupling Nonhomogeneous Linear Systems

11. 0 1 11 0 1⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦

] The eigenvalues are 1 and –1, and their two independent eigenvectors are and [1, 1 [ ]1, 1− . We

form the matrices

1 11 1

−⎡ ⎤= ⎢ ⎥⎣ ⎦

P and 1 1 111 12

− ⎡ ⎤= ⎢ ⎥−⎣ ⎦

P .

We change to the variable 1−=w P x , to yield the decoupled system

1 0 1 1 110 1 1 1 12⎡ ⎤ ⎡ ⎤ ⎡′ = +

⎤⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣

w w⎦

2

or and . Solving these, yields 1 1 1w w′ = + 2w w′ = − ( )1 1 1tw t c e= − and ( )2 2tw t c e−= . Thus

( ) ( ) 11 2

2

11 1 1 1 11 1 1 1 1

tt t

t

c et t c e c e

c e−

−

⎡ ⎤−− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = + +⎢ ⎥

⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pw⎣ ⎦

.


12. 3 1 sin1 3 0

t−⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢−⎣ ⎦ ⎣x x

⎤⎥⎦

The eigenvalues are –2 and –4, and their two independent eigenvectors are [ ]1, 1 and [ . We

form the matrices

]1, 1−

1 11 1

−⎡ ⎤= ⎢ ⎥⎣ ⎦

P and 1 1 111 12

− ⎡ ⎤= ⎢ ⎥−⎣ ⎦

P .

We change to the variable , to yield the decoupled system 1−=w P x

2 0 1 1 sin10 4 1 1 02

t−⎡ ⎤ ⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢ ⎥ ⎢− −⎣ ⎦ ⎣ ⎦ ⎣w w

⎤⎥⎦

t, or 1 1

2 2

2 sin4 .

w ww w′ = − +′ = −

Solving these yields ( ) ( )2 41 1 2 2

1 2cos sin , .5 5

t tw t c e t t w t c e− −= − + =

Thus

( ) ( )

21

42

2 41 2

1 2cos sin1 15 5

1 1

1 1 cos 2s1 .1 1 cos 2s5

t

t

t t

c e t tt t

c e

t tc e c e

t t

−

−

− − inin

⎡ ⎤− +−⎡ ⎤ ⎢ ⎥= = ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦− − +⎡ ⎤ ⎡ ⎤ ⎡

= + +⎤

⎢ ⎥ ⎢ ⎥ ⎢− + ⎥⎣ ⎦ ⎣ ⎦ ⎣

x Pw

⎦

⎤⎥⎦

13. . 1 11 1 1

t⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x

The eigenvalues are 0 and 2, and their two independent eigenvectors are [ ]1, 1− and [ . We

form the matrices

]1, 1

1 11 1

⎡ ⎤= ⎢ ⎥−⎣ ⎦

P and 1 1 111 12

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦P .

We change to the variable , to yield the decoupled system 1−=w P x

0 0 1 110 2 1 1 12

t−⎡ ⎤ ⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

w w⎤⎥⎦

, or ( ) ( )1 2 21 11 , 2 1 .2 2

w t w w t′ ′= − = + +

Solving these, yields ( ) ( )2 21 1 2 2

1 3, .4 2 4 8

tt tw t t c w t c e= − + = − −

Thus

( ) ( )

22

12

1 2 22

2

3 31 1 1 1 4 4 84 21 1 1 13 3

4 8 4 4 8

t

t

t tt t ct t c c e

t t tc e

⎡ ⎤⎡ ⎤− −− + ⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥= = = + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦− − − + −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

x Pw .


14. 5 4 51 2 0

t⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦

] The eigenvalues are 6 and 1, and their two independent eigenvectors are [ and 4, 1 [ ]1, 1− . We

form the matrices

4 11 1

−⎡ ⎤= ⎢ ⎥⎣ ⎦

P and 1 1 111 45

− ⎡ ⎤= ⎢ ⎥−⎣ ⎦

P .

We change to the variable 1−=w P x to yield the decoupled system

6 0 1 1 510 1 1 4 05

t⎡ ⎤ ⎡ ⎤ ⎡′ = +⎤

⎢ ⎥ ⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣ ⎦ ⎣

w w⎦

t, or 1 1

2 2

6.

w ww w t′ = +′ = −

Solving these yields ( ) ( )61 1 2 2

1 , 1.6 36

t ttw t c e w t c e t= − − = + +

Thus the general solution is

( ) ( )6

1 61 2

2

5 1014 1 4 1 3 96 361 1 1 1 5 35

16 36

tt t

t

ttc et t c e c e

tc e t

⎡ ⎤⎡ ⎤ − −⎢ ⎥− −− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥= = = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ++ +⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

x Pw .

15. 1 42 3 2

tt

⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦

1 5,λ = eigenvector , 11⎡ ⎤⎢ ⎥⎣ ⎦

2 1,λ = − eigenvector 21

⎡ ⎤⎢ ⎥−⎣ ⎦

,

, 5 00 1⎡ ⎤

= ⎢ ⎥−⎣ ⎦D

1 21 1⎡ ⎤

= ⎢ ⎥−⎣ ⎦P , 1 1 2 1 21 1

1 1 1 13 3− − −⎡ ⎤ ⎡= − =

⎤⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣

P⎦

.

We change to the variable 1−=w P x , so that 1 ( ).t−′ = +w Dw P f

1

2

5 0 1 21( )0 1 1 1 23

w tt

tw⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡′ = +⎢ ⎥

⎤⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦

w⎦

= 1

2

55313

w t

w t

⎡ ⎤+⎢ ⎥. ⎢ ⎥

⎢ ⎥− −⎢ ⎥⎣ ⎦

Solving these linear DEs gives

5 553( )

13

t t

t t

e te dtt

e te d

−

− t

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

∫

∫w (integration by parts)

51

2

13 15

13 3

t

t

t c e

t c e−

⎡ ⎤− − +⎢ ⎥= ⎢ ⎥⎢ ⎥− + +⎢ ⎥⎣ ⎦

.


Thus we obtain

( )5 5

1 1 2

52 1 2

1 321 2 3 15 5( )1 1 1 2

3 3 5

t t t

t t t

t c e c e c e tt t

t c e c e c e

−

− −

⎡ ⎤ ⎡− − + + − +⎢ ⎥ ⎢⎡ ⎤= = =⎢ ⎥ ⎢⎢ ⎥− ⎢ ⎥ ⎢⎣ ⎦ − + + − −⎢ ⎥ ⎢⎣ ⎦ ⎣

x Pw

⎤⎥⎥⎥⎥⎦

⎥

.

16. 1 44 11

t

t

e

e

⎡ ⎤⎡ ⎤′ = + ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎢⎣ ⎦x x

Eigenvalues are 3, 9λ = , with eigenvectors 2 1

, ,1 2⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

so that

3 00 9⎡ ⎤

= ⎢ ⎥⎣ ⎦

D , 2 11 2⎡ ⎤

= ⎢ ⎥⎣ ⎦

P , 1 2 111 23

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦

P .

We change the variable to , so that 1−=w P x 1 ( ) :t−′ = +w Dw P f

1 1

2 2

3 0 2 110 9 1 23

t

t

w w ew w e

⎡ ⎤′ −⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤′ = = + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥′ −⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦w =

1

2

133193

t

t

w e

w e

⎡ ⎤+⎢ ⎥⎢ ⎥⎢ ⎥+⎢ ⎥⎣ ⎦

.

Solving these linear DEs gives

3 2

9 8

13( )13

t t

t t

e e dtt

e e dt

−

−

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∫

∫w

31

92

16124

t t

t t

e c e

e c e

⎡ ⎤− +⎢ ⎥= ⎢ ⎥⎢ ⎥− +⎢ ⎥⎣ ⎦

.

To find ( )tx

( )3 3

1 1

9 32 1

1 3 22 1 6 8( )1 2 1 1 2

24 4

t t t t

t t t t

e c e e c e c et t

e c e e c e c e

⎡ ⎤ ⎡− + − + +⎢ ⎥ ⎢⎡ ⎤= = =⎢ ⎥ ⎢⎢ ⎥

⎢ ⎥ ⎢⎣ ⎦ − + − + +⎢ ⎥ ⎢⎣ ⎦ ⎣

x Pw

92

92

t

t

⎤⎥⎥⎥⎥⎦

.

17. 11 0 0

4 3 0 04 2 1 1

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥′ = − + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x

Double eigenvalue 1 1,λ = ; eigenvectors 1 02 , 00 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

, 2 3,λ = eigenvector 011

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

.

1 0 02 0 10 1 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P , 1 0 00 1 00 0 3

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

D , 1

1 0 02 1 12 1 0

−

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P .


= 1 ( )t−′ = +w Dw P f1

2

3

1 0 0 1 0 00 1 0 2 1 10 0 3 2 1 0

www

⎡ ⎤⎡ ⎤ ⎡ ⎤⎥⎥⎥⎦

1

2

3

110 31 3 2

www

+⎢ ⎥⎢ ⎥ ⎢+ −⎢ ⎥⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢−⎣ ⎦ ⎣⎣ ⎦

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ = +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦

⇒ 1 1

2 2

33 3

1

323

t

t

t

w c e

w c e

w c e

= −

= −

= +

( )1

13

2 1 3

333

2 3

111 0 04( ) 2 0 1 3 23

0 1 1 2 73 3

tt

t t t

tt t

c ec e

t t c e c e c e

c e c e c e

⎡ ⎤⎡ ⎤ ⎢ ⎥−−⎢ ⎥⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= = − = + −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥+ ⎢ ⎥+ −⎣ ⎦ ⎢ ⎥⎣ ⎦

x Pw .

18. 43 2 0

1 0 0 61 1 3 1

− ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥′ = + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x

1 1,λ = ; 1

110

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v 2 2,λ = 2

211

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v ; 2 3,λ = 3

001

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

v .

1 2 01 1 00 1 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P , 1 0 00 2 00 0 3

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

D , 1

1 2 01 1 01 1 1

−

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P .

= 1 ( )t−′ = +w Dw P f1

2

3

41 0 0 1 2 00 2 0 1 1 0 60 0 3 1 1 1 1

www

−⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥+ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

= 1

2

3

82 23 3

www

+⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥+⎣ ⎦

⇒ 1 1

2 23

3 3

8

1

1

t

t

t

w c e

w c e

w c e

= −

= +

= −

( )

21 1 2

2 22 1 2

3 23 2 3

8 21 2 0( ) 1 1 0 1 7

0 1 1 1

t t t

t t t

t t

c e c e c e

t t c e c e c e

c e c e c e3

6

t

⎡ ⎤ ⎡ ⎤− + −⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= = + = + −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ − +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x Pw .

19. We first find the eigenvalues and eigenvectors. 2

4 1 1 12 5 21 1 2

t

t

⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥′ = − + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

x x

(Maple or Matlab can help):

[ ] [ ] [1 1 2 2 3 35, 1, 2, 1 , 3, 0, 1, 1 , 3, 1, 0, 1 .λ λ λ= = = = = =v v v ]

Hence,

1 0 12 1 01 1 1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

P , and 1

1 1 11 2 0 22

1 1 1

−

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P ,


so

, and 1

5 0 00 3 00 0 3

−

⎡ ⎤⎢= = ⎢⎢ ⎥⎣ ⎦

D P AP ⎥⎥

2

1 2

2 2

111 2 22

1

t t

t t

t t t

−

⎡ ⎤− + +⎡ ⎤⎢ ⎥⎢ ⎥

= −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ − +⎣ ⎦ ⎢ ⎥⎣ ⎦

P .

The decoupled system is 1−′ = +w Dw P f , or

21 1

22 2

23 3

15 0 010 3 0 2 22

0 0 3 1

t tw ww ww w t t

t

⎡ ⎤− + +′⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ = + −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ − +⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

.

Solving these three equations individually yields

( )

( )

( )

25

1 1

23

2 2

23

3 3

3 1410 50 125

2 73 9 27

4 .6 18 27

t

t

t

t tw t c e

t tw t c e

t tw t c e

⎛ ⎞= + − −⎜ ⎟

⎝ ⎠⎛ ⎞

= + − − +⎜ ⎟⎝ ⎠⎛ ⎞

= + − + −⎜ ⎟⎝ ⎠

Transforming back ( ) ( )t t=x Pw = 1 1

2 2

3 3 1

1 0 12 1 0 21 1 1

1 3

1 2

2 3

x w w wx w w wx w w w w

+⎡ ⎤ ⎡ ⎤ ⎡⎡ ⎤ ⎤⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥= = ⎥+⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ + +⎣ ⎦ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, or

( )

25 3

1 1 3

25 3

2 1 2

25 3

3 1 2 3

878( ) ,15 225 3375

2 77 119( ) 2 ,15 225 3375

2 17 1( ) .5 75 1125

t t

t t

t t

t tx t c e c e

t tx t c e c e

t tx t c e c c e

⎛ ⎞= + − + +⎜ ⎟

⎝ ⎠⎛ ⎞

= + − + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + − + +⎜ ⎟

⎝ ⎠

20.

0 0 1 00 0 0 1 01 0 0 00 1 0 0 1

t

t

⎡ ⎤⎢ ⎥⎢ ⎥′ = +⎢ ⎥ −⎢ ⎥⎣ ⎦

x x

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

We first find the eigenvalues and eigenvectors. (Note: Maple or Matlab can help.)

⇒

[ ][ ][ ][ ]

1 1

2 2

3 3

4 4

1, 0, 1, 0, 1

1, 1, 0, 1, 0

1, 0, 1, 0, 1

1, 1, 0, 1, 0 .

λ

λ

λ

λ

= =

= =

= − = −

= − = −

v

v

v

v

0 1 0 11 0 1 00 1 0 11 0 1 0

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

P .


We can now obtain

, 1

0 1 0 11 0 1 00 1 0 11 0 1 0

−

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥−⎢ ⎥−⎣ ⎦

P 1

1 0 0 00 1 0 00 0 1 00 0 0 1

−

⎡ ⎤⎢ ⎥⎢ ⎥= =⎢ ⎥−⎢ ⎥

−⎣ ⎦

D P AP , 1

10 01

121 2

t

tt

−

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥

−⎣ ⎦ ⎣ ⎦

P .

Hence, the decoupled system is 1−′ = +w Dw P f , or

1 1

2 2

3 3

4 4

1 0 0 0 10 1 0 0 010 0 1 0 120 0 0 1 2

w ww ww ww w t

′⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥

⎤⎢ ⎥ ⎢′⎢ ⎥ ⎢ ⎥

⎥⎢ ⎥ ⎢= +

⎢ ⎥ ⎢ ⎥⎥

⎢ ⎥ ⎢′ −⎢ ⎥ ⎢ ⎥

⎥⎢ ⎥ ⎢′ ⎥

− −⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦ ⎦

.

Solving these four equations individually yields

( ) ( ) ( ) ( )1 1 2 2 3 3 4 41 1, , , 1.2 2

t t t tw t c e w t c e w t c e w t c e t− −= − = = + = − +

Transforming back yields the solution ( ) ( )t =x Pw t , which turns out to be

2 41 1

1 32 2

2 43 3

1 34 4

0 1 0 11 0 1 00 1 0 11 0 1 0

w wx ww wx ww wx ww wx w

−− ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ −− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ +⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ +⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

,

1 2 4

2 1 3

3 2 4

4 1 3

( ) 1

( ) 1

( ) 1

( ) .

t t

t t

t t

t t

x t c e c e t

x t c e c e

x t c e c e t

x t c e c e

−

−

−

−

= − + −

= − −

= + − +

= +

Working Backwards

21. Given eigenvalues are 1 and –1 and respective eigenvectors are [ ]1, 1 and [ , we form the

matrices

]

⎤⎥

1, 2

and 1 11 2⎡

= ⎢⎣ ⎦

P 1 2 11 1

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦

P , and then the diagonal matrix , 1 00 1⎡ ⎤

= ⎢ ⎥−⎣ ⎦D

whose diagonal elements are the eigenvalues. Using the relation 1−=D P AP , we premultiply by P, and postmultiply by , yielding 1−P

1 1 1 1 0 2 1 3 21 2 0 1 1 2 4 3

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

A PDP .


Jordan Form

22. (a) The system 2 11 4

⎡ ⎤′ = ⎢ ⎥−⎣ ⎦x x

has a double eigenvalue of 3 and only one independent eigenvector [ ]1, 1=v .

We must find the generalized eigenvector w that satisfies the equations ( )3− =A I w v ,

or

1

2

1 1 11 1 1

ww

− ⎡ ⎤⎡ ⎤ ⎡=⎢ ⎥

⎤⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣⎣ ⎦ ⎦

.

This dependent system reduces to 1 2 1w w− + = , which has as one solution and . Hence, .

1 1w =

2 2w = [ ]1, 2=w

We now form the matrix

1 11 2⎡

⎡ ⎤= = ⎢⎣ ⎦⎣ ⎦

P v w⎤⎥ , and compute 1 2 1

, so that1 1

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦

P

1 2 1 2 1 1 1 3 1.

1 1 1 4 1 2 0 3− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦P AP

(b) Transforming from x to the new variables 1−=u P x yields the new system

and . Solving this system yields 1 13u u u′ = + 2

22 3u u′ =

( ) 3 31 1 2

t tu t c e c te= + , ( ) 32 2

tu t c e= .

The solution of the original system is

( ) ( ) ( ) 31 2 1 2 23

31 2 22

1 121 2

tt

t

c tc e c c tct t e

c c tcc e

⎡ ⎤+ + +⎡ ⎤⎡ ⎤= = =⎢ ⎥ ⎢ ⎥⎢ ⎥ + +⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x Pu .

Complex Decoupling

23. Student Project


24. Student Project

SECTION 6.6 Matrix Exponential 659

6.6 Matrix Exponential

Matrix Exponential Functions

1. A = A is a diagonal matrix, so 1 00 1⎡⎢ −⎣ ⎦

⎤⎥

0

0

tt

t

ee

e−⎡ ⎤

= ⎢ ⎥⎢ ⎥⎣ ⎦

A .

2. A = 0

0i

iπ

π−⎡ ⎤⎢ ⎥⎣ ⎦

A is a diagonal matrix, so 0 cos sin 0

0 cos sin0

i tt

i t

e t i te

t i te

π

π

π ππ π

−⎡ ⎤ −⎡ ⎤= =⎢ ⎥ ⎢ ⎥+⎣ ⎦⎢ ⎥⎣ ⎦

A .

3. A = A has eigenvalues 0 and 1, with eigenvectors 1 01 0⎡⎢⎣ ⎦

⎤⎥

01⎡ ⎤⎢ ⎥⎣ ⎦

and , respectively. 11⎡ ⎤⎢ ⎥⎣ ⎦

Therefore a fundamental matrix is 0

( )1

t

t

et

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

X , and

1 0 1 1( ) (0)

1 01 1

t tt

t t

e ee t

e e− ⎡ ⎤ ⎡−⎡ ⎤

= = =⎢ ⎥ ⎢⎢ ⎥−⎣ ⎦⎢ ⎥ ⎢⎣ ⎦ ⎣

A X X0

1

⎤⎥⎥⎦

⎤⎥

⎤⎥⎦

⎥⎥

4. A = Note that A0 10 0⎡⎢⎣ ⎦

2 = 0, so that

1 0 0 1

.0 1 0 0 0 1

t t te t

⎡ ⎤ ⎡ ⎤ ⎡= + = + =⎢ ⎥ ⎢ ⎥ ⎢

⎣ ⎦ ⎣ ⎦ ⎣A I A

5. A = is a diagonal matrix, so 1 0 00 2 00 0 3

⎡ ⎤⎢⎢⎢ ⎥⎣ ⎦

2

3

0 0

0 0

0 0

t

t t

t

e

e e

e

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A .

6. A = Note that A0 1 10 0 10 0 0

⎡ ⎤⎢⎢⎢ ⎥⎣ ⎦

⎥⎥

3 = 0, so that

22 21 0 0 0 1 1 0 0 1 1 / 2

( ) 0 1 0 0 0 1 0 0 0 = 0 12 2

0 0 1 0 0 0 0 0 0 0 0 1

t

t t tt te t t t

⎡ ⎤+⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

A AI A .


DE Solutions using Matrix Exponentials

7. x xy y′ =′ =

for , which gives ′ =x Ax⎤⎥

1 00 1⎡

= ⎢⎣ ⎦

A0

0

tt

t

ee

e

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

A ,

which yields the general solution, 1 1

2 2

0( )

0

tt

t t

c c eet

ce c e

⎡ ⎤⎡ ⎤ ⎡ ⎤= = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

x .

8. x yy x′ =′ =

for ′ =x Ax0 11 0⎡ ⎤

= ⎢ ⎥⎣ ⎦

A

The matrix A has eigenvalues ±1 with eigenvectors 1

1±⎡ ⎤⎢ ⎥⎣ ⎦

.

A fundamental matrix is X(t) = , with X(0)t t

t t

e e

e e

−

−

⎡ ⎤−⎢⎢ ⎥⎣ ⎦

⎥ −1 = 1 111 12

⎡ ⎤⎢ ⎥−⎣ ⎦

, so

= 1( ) (0)te t −=A X X1 11 11 12 2

t t t t t

t t t t t t

e e e e e e

e e e e e e

− − t

− −

⎡ ⎤ ⎡− +⎡ ⎤=⎢ ⎥ ⎢⎢ ⎥− − +⎣ ⎦⎢ ⎥ ⎢⎣ ⎦ ⎣

cosh sinhsinh cosh

t tt t

−

−

⎤−⎥⎥⎦

= ⎡ ⎤⎢ ⎥⎣ ⎦

tt

The general solution is 1 1 2

2 1 2

cosh sinhcosh sinh( )

sinh cosh sinh coshc c t ct t

tt t c c t c

+⎡ ⎤ ⎡ ⎤⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎢ ⎥ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x .

9. x x yy y′ = +′ =

Note that A = 1 10 1⎡ ⎤⎢ ⎥⎣ ⎦

is not diagonalizable.

So we must use the definition of matrix exponential.

We find that A = , A1 10 1⎡⎢ ⎥⎣ ⎦

⎤ ⎤⎥

2 = A1 2

,0 1⎡⎢⎣ ⎦

3 = 1 3

,0 1⎡ ⎤⎢ ⎥⎣ ⎦

…, An = 1

,0 1

n⎡ ⎤⎢ ⎥⎣ ⎦

…,

and so 2 3

0

1 1 1 2 1 3 1 1... ...

0 1 0 1 0 1 0 1 0 12! 3! ! !

k kt

k

k kt t t te tk k

∞

=

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡= + + + + + + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦∑A I

⎤⎥

0 1

0

! ( 1)!.

00!

k k

t tk k

k t

k

t te tek k

t ek

∞ ∞

= =

∞

=

⎡ ⎤⎢ ⎥ ⎡ ⎤−⎢ ⎥= = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

∑ ∑

∑

Note: we have used the fact that tet = 1

0 0 10

! ! (

k k k

k k k

t t ttk k k

+∞ ∞ ∞

= = =

= = +1)!−∑ ∑ ∑ .

Hence, . 1

2

( )0

t t

t

ce tet

ce⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

x


10. 0

x y zy zz

′ = +′ =′ =

A = 0 1 10 0 10 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⇒

21 /0 10 0 1

t

t t te

2t

⎡ ⎤+⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

A (from Problem 6)


221 2 31

2 2 3

3 3

( / 21 / 2( ) 0 1

0 0 1

c c t c t tct t tt t c c c t

c c

)⎡ ⎤⎡ ⎤ + + ++ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= = +⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

x .

11. Because A is diagonal, 1

2

1 0 10 2 0

xx

− ⎡ ⎤ ⎡ ⎤⎡ ⎤′ = ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣⎣ ⎦

x +⎦ 2

0

0

tt

t

ee

e

−⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

A , so we have

12 2 2

2 0

12 2

2 0

21

2 22

0 0 0( )

00 0 0

0 000 0

0 100 0

0

0

t t st

t t s

t t t s

t t

t t t t

t t

t

ce e et d

ce e e

ce e e dsce e

ce e e ece e

e

e

− −

−

− −

− −

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ 1s

⎡ ⎤= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤

= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ −

= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

=

∫

∫

x

12

2

1 .0

t

t

c ec

−⎡ ⎤ ⎡ ⎤⎡ ⎤ −+⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

12. Because A is diagonal, we have 1

2

2 0 00 3 6

xx⎡ ⎤ ⎡ ⎤⎡ ⎤′ = ⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣⎣ ⎦x +

⎦ 3

0

0

tt

t

ee

e

2⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

A , so

2 2 21

3 3 32 0

2 21

33 32 0

2 21

33 32

0 0 0( )

60 0 0

00 060 0

00 02 20 0

t t st

t t s

t t t

st t

t t

tt t

ce e et d

ce e e

ce eds

c ee e

ce ec ee e

−

−

−

−

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ 0s

⎡ ⎤= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤

= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤

= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

∫

∫

x

13. From Problem 8, 1

2

0 1 11 0 1

xx⎡ ⎤ ⎡ ⎤⎡ ⎤′ = ⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦x +

cosh sinhsinh cosh

t t te

t t⎡ ⎤

= ⎢ ⎥⎣ ⎦

A , so

1

2 0

cosh sinh cosh sinh cosh sinh 1( )

sinh cosh sinh cosh sinh cosh 1

tct t t t s st d

t t t t s sc−⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦∫x s

The integral becomes

0 0

cosh sinh 1sinh cosh 1

s tt t

s t

s s e eds ds

s s e e

− −

− −

⎡ ⎤ ⎡ ⎤− − +⎡ ⎤= =⎢ ⎥ ⎢⎢ ⎥− +

⎥− +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

∫ ∫⎦

.


When we use the fact that cosh s − sinh s = 2 2

s s s sse e e e e

− −−+ −

− = , we get

1

2

1

2

cosh sinh 11( )sinh cosh 2 1

cosh sinh 1.

sinh cosh 1

t t t t t

t t t t t

t

t

ct t e e e e et

t t c e e e e e

ct t et t c e

− − −

− − −

⎡ ⎤⎡ ⎤+ − − +⎡ ⎤⎡ ⎤= + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

− + − +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤− +⎡ ⎤⎡ ⎤

= + ⎢ ⎥⎢ ⎥⎢ ⎥− +⎣ ⎦ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x

14. 1 1

2 2

(0)0 1 1 1,

1 0 0 (0)x xx x⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤′ = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

x1

=

The eigenvalues are ±i and a fundamental matrix is X(t) = cos sinsin cos

t tt t

⎡ ⎤⎢ ⎥−⎣ ⎦

.

Note that X(0) = I = X(0)−1, so cos sin

( )sin cos

t t te t

t t⎡ ⎤

= = ⎢ ⎥−⎣ ⎦A X .


1

2 0

1

2 0

cos sin cos sin cos sin 1( )

sin cos sin cos sin cos 0

cos sin cos sin cossin cos sin cos sin

cos sinsin cos

t

t

ct t t t s st d

t t t t s sc

ct t t t sds

t t t tc s

t tt t

−⎡ ⎤s

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤= ⎢ ⎥−⎣ ⎦

∫

∫

x

1

2

1

2

cos sin sinsin cos cos 1

cos sin sin.

sin cos 1 cos

c t t tt tc t

ct t tt t c t

⎡ ⎤ ⎡ ⎤⎡ ⎤+⎢ ⎥ ⎢ ⎥⎢ ⎥− − +⎣ ⎦ ⎣ ⎦⎣ ⎦

⎡ ⎤ ⎡ ⎤⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥− − +⎣ ⎦ ⎣ ⎦⎣ ⎦

Substituting into the initial conditions, 1

2

1 1 0 0(0)

0 11 0cc⎡ ⎤⎡ ⎤ ⎡ ⎤

= = +⎢ ⎥⎡ ⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥⎣ ⎦ ⎣⎣ ⎦

x⎦

−

, we find c1 = 1, c2 = 1.

Thus, . cos sin 1 sin cos 2sin

( )sin cos 1 1 cos sin 2cos 1

t t t t tt

t t t t t+⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤

= + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥− − + − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦x


Products of Matrix Exponentials

15. A = and B = 0 10 0

−⎡⎢⎣ ⎦

⎤⎥

0 01 0⎡ ⎤⎢ ⎥⎣ ⎦

(a) Note that A2 = 0 = B2, so that te t= + =A I A10 1

t⎡ ⎤⎢ ⎥⎣ ⎦

and . te t= + =B I B1 0

1t⎡ ⎤⎢ ⎥⎣ ⎦

(b) To find , we note that A + B = ( )te +A B 0 11 0

−⎡ ⎤⎢ ⎥⎣ ⎦

has eigenvalues iλ = ± ,

and a fundamental matrix cos sin

( )sin cos

t tt

t t−⎡ ⎤

= ⎢ ⎥⎣ ⎦

X .

Note that X(0) = I = X(0)−1. Then ( ) cos sin( )

sin cost t t

e tt t

+ −⎡ ⎤= = ⎢ ⎥

⎣ ⎦A B X I

(c) No, because ( ) cos sinsin cos

t t te

t t+ −⎡ ⎤

= ⎢ ⎥⎣ ⎦

A B

21 1 0 1

0 1 1 1t t t t te e

t t⎡ ⎤⎡ ⎤ ⎡ ⎤ +

≠ = = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

A B .

Properties of Matrix Exponentials

16. 2 3 2 3

2 22

1 1 1 1... ...2! 3! 2! 3!

( ) ... , because all other terms will cancel.2! 2!

e e− ⎡ ⎤ ⎡ ⎤= + + + + − + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎛ ⎞

= + − + + − + =⎜ ⎟⎝ ⎠

A A I A A A I A A A

A AI A A A I

Thus, . 1( )e e− −=A A

17. 2 31 1( ) ( ) ( )2! 3!

e + = + + + + + + +A B I A B A B A B ...

2 2

3 2 2 2 2 3

2 2 3 2 3

2 2 3 2 2 3

2 3

1 ( )2!

1 ( ) ...3!

1 1( 2 ) ( 3 3 )2! 3!

... (because is given.)

...2! 2! 3! 2! 2! 3!

1 1( ...2! 3!

= + + + + + +

+ + + + + + + + +

= + + + + + + + + +

+ =

= + + + + + + + + + +

= + + + +

2

I A B A AB BA B

A A B ABA AB BA BAB B A B

I A B A AB B A A B AB B

AB BA

A B A A B AB BI A B AB

I A A A 2 31 1)( ...)2! 3!

.e e

+ + + +

= A B

I B B B


Nilpotent Example

18. (a) A = . 2 3

1 1 1 1 0 1 0 0 01 0 1 , 0 0 0 , 0 0 01 1 1 1 0 1 0 0 0

− −⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢− = =⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢− −⎣ ⎦ ⎣ ⎦ ⎣

A A⎤⎥⎥⎥⎦

(b) Since A3 = 0, we have 2

2

2

2 22

2 2

2!1 0 0 1 1 1 1 0 10 1 0 1 0 1 0 0 0

2!0 0 1 1 1 1 1 0 1

12! 2!

1 .

12! 2!

t te t t

tt

t tt t t

t t

t tt t t

= + +

− −⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢= + − +⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢

⎤⎥⎥⎥− −⎣ ⎦ ⎣ ⎦ ⎣

⎡ ⎤+ + − −⎢ ⎥

⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥+ − −⎢ ⎥⎣ ⎦

A I A A

⎦

The general solution of is ′ =x Ax teA c , or

2 22

1

22 2

3

12! 2!

( ) 1 .

12! 2!

t tt t t ct t t

ct tt t t

⎡ ⎤+ + − −⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥+ − −⎢ ⎥⎣ ⎦

x c

An Exponential Pattern

19. (a) A2 = I, A3 = A, A4 = I, A5 = A, …

So A2n = I and A2n+1 = A for positive integers n.

(b) 2 3 4 5

2 3 4 5 ...2! 3! 4! 5!

t t t t te t= + + + + + +A I A A A A A

2 3 4 5

2 3 4 5

2 4 6 3 5

3 5 2 4 6

...2! 3! 4! 5!

1 0 0 1 1 0 0 1 1 0 0 1...

0 1 1 0 0 1 1 0 0 1 1 02! 3! 4! 5!

1 ... ...2! 4! 6! 3! 5!

... 1 ...3! 5! 2! 4! 6!

t t t tt

t t t tt

t t t t tt

t t t t tt

= + + + + + +

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + + + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤+ + + + + + +⎢ ⎥

⎢=⎢

+ + + + + + +⎢⎣ ⎦

I A I A I A

cosh sinhsinh cosh

t tt t

⎡ ⎤⎥ = ⎢ ⎥⎥ ⎣ ⎦⎥

+

(c) The general solution of is ′ =x Ax 1

2

cosh sinh( ) .

sinh cosht ct t

t et t c

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦Ax c


Nilpotent Criterion

20. An n × n matrix A is nilpotent if and only if its eigenvalues are zero.

(⇒) Suppose An = 0. Let λ be an eigenvalue of A. Then λn is an eigenvalue of An, that is,

Akv = λkv for a non-zero vector v. Because An = 0, we have λkv = 0 and hence λk = 0. Thus λ = 0.

(⇐) If all eigenvalues of A are 0, then the characteristic equation is 0 = .nIλ λ− =A

By the Cayley-Hamilton Theorem, An = 0, hence A is nilpotent.

Fundamental Matrices

21. For A = , λ = 1, 1 but there is only one linearly independent eigenvector . 1 20 1⎡⎢⎣ ⎦

⎤⎥

10⎡ ⎤⎢ ⎥⎣ ⎦

A generalized eigenvector is 01⎡ ⎤⎢ ⎥⎣ ⎦

.

A fundamental matrix because X(0) = I = X( ) ,0

t tt

t

e tet

e

⎡ ⎤= =⎢ ⎥⎢ ⎥⎣ ⎦

AX e −1(0).

Because A is not diagonalizable, the second method 1t te e −=A DP P is not applicable.

22. A = 1 1 10 2 10 0 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

The matrix A has eigenvalues λ1 = 1, λ2 = 2, λ3 = 3,

with corresponding eigenvectors 1v = 10 ,0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2v = 110

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

, 3v = 111

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

.

Using A fundamental matrix 1( ) (0) :te t −=A X X

X(t) =

2 3

2 3 1

3

1 1 00 , with (0) 0 1 1

0 0 10 0

t t t

t t

t

e e e

e e

e

−

⎡ ⎤ −⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

X .

Therefore, 2 3 2 3 2

2 3 2 3 2

3 3

1 1 00 0 1 1 0

0 0 10 0 0 0

t t t t t t t t

t t t t t

t t

e e e e e e e e

e e e e e

e e

⎡ ⎤ ⎡ − −−⎡ ⎤⎢ ⎥ ⎢⎢ ⎥= − =⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢⎢ ⎥⎣ ⎦⎢ ⎥ ⎢⎣ ⎦ ⎣

A te

⎤⎥

− ⎥⎥⎥⎦

.


Using The matrix P = 1 :t te e −=A DP P1 1 10 1 10 0 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

diagonalizes A, so

1

1 2

3

2

3

2 3

0 01 1 1 1 1 10 1 1 0 0 0 1 10 0 1 0 0 10 0

0 01 1 1 1 1 0 0 1 1 0 0 0 1 1

0 0 1 0 0 10 0

t

t t t

t

t

t

t

t t t

e

e e e

e

e

e

e

e e e e

−

−

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥= = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤ −⎡ ⎤ ⎡⎢ ⎥⎢ ⎥ ⎢= −⎢ ⎥⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢⎣ ⎦ ⎣⎢ ⎥⎣ ⎦

−

=

A DP P

2

2 3 2

3

0 .

0 0

t t

t t t

t

e

e e e

e

⎡ ⎤−⎢ ⎥

−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⎤⎥⎥⎥⎦

23. 1 14 1⎡ ⎤⎢ ⎥⎣ ⎦

The method using was shown in Example 5. 1( ) (0)te t −=A X X

The method using is applicable because the matrix has distinct eigenvalues −1, 3,

with corresponding eigenvectors and

1t te e −=A DP P12

⎡ ⎤⎢ ⎥−⎣ ⎦

12⎡ ⎤⎢ ⎥⎣ ⎦

, respectively,

1 12 2

⎡= ⎢−⎣ ⎦

P⎤⎥ , D =

1 00 3−⎡ ⎤⎢ ⎥⎣ ⎦

and PP

−1 = 2 112 14

−⎡ ⎤⎢ ⎥⎣ ⎦

3

3

3

3 3

3 3

1 1 0 2 112 2 2 140

2 112 14 2 2

2 214 4 4 2 2

tt

t

t t

t t

t t t t

t t t t

ee

e

e e

e e

e e e e

e e e e

−

−

− −

− −

⎡ ⎤ ⎛ ⎞−⎡ ⎤ ⎡= ⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎢⎜ ⎟−⎣ ⎦ ⎣⎢ ⎥ ⎝ ⎠⎣ ⎦

⎡ ⎤ −⎡ ⎤= ⎢ ⎥ ⎢ ⎥

− ⎣ ⎦⎢ ⎥⎣ ⎦⎡ ⎤+ − +

= ⎢ ⎥− + +⎢ ⎥⎣ ⎦

A ⎤⎥⎦

Computer Lab

24. A = ; .

0 0 0 10 0 1 00 1 0 01 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

cosh 0 0 sinh0 cosh sinh 00 sinh cosh 0

sinh 0 0 cosh

t

t tt t

et t

t t

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

A


25. A = ;

0 0 0 10 0 1 00 1 0 01 0 0 0

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

cos 0 0 sin0 cos sin 00 sin cos 0sin 0 0 cos

t

t tt t

et t

t t

⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

A .

Computer DE Solutions

26. A = has characteristic equation 3 22 2

−⎡⎢ −⎣ ⎦

⎤⎥

2 2 0λ λ− − =

1 22; 1λ λ= = −

1 2

2 11 2⎡ ⎤ ⎡

= =⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

v v⎦

⎤⎥

X(t) = 2

2

22

t t

t t

e ee e

−

−

⎡ ⎤⎢ ⎥⎣ ⎦

X(0) = , X2 11 2⎡⎢⎣ ⎦

−1(0) = 2 111 23

−⎡ ⎤⎢ ⎥−⎣ ⎦

2

2

2 2

2 2

2 121( ) (0)1 23 2

4 21 =3 2 2 4

t tt

t t

t t t t

t t t t

e ee t

e e

e e e ee e e e

−

−

− −

− −

−⎡ ⎤ ⎡ ⎤= = ⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

⎡ ⎤− − +⎢ ⎥− − +⎣ ⎦

A X X

2

1

2

2 21 2

2 21 2

( )

1 4 2 23 3 3 32 2 4 13 3 3 3

t

t t t t

t t t t

ct e

c

e e c e e c

e e c e e c

− −

− −

⎡ ⎤= ⎢ ⎥

⎣ ⎦⎡ ⎤⎛ ⎞ ⎛− + + −⎜ ⎟ ⎜⎢ ⎥⎝ ⎠ ⎝⎢ ⎥=⎢ ⎥⎛ ⎞ ⎛− + + −⎢ ⎥⎜ ⎟ ⎜⎝ ⎠ ⎝⎣ ⎦

Ax

⎞⎟⎠⎞⎟⎠

⎤⎥

27. A = has characteristic equation 1 52 1

⎡⎢− −⎣ ⎦

2 9 0λ + = , so . 1

53 ;

1 3i

iλ

−⎡ ⎤= ± = ⎢ ⎥

⎣ ⎦v

∓

Re

5 0( ) cos3 sin 3

1 3t t t

−⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x

Im

5 0( ) sin3 cos3

1 3t t t

−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

x

X(t) = [ ] Re Im

5cos3 5sincos3 3sin 3 sin 3 3cos

t tt t t

− −⎡ ⎤= ⎢ ⎥+ −⎣ ⎦

x xt


X(0) = X5 0

1 3−⎡⎢ −⎣ ⎦

⎤⎥

−1(0) = 3 011 515

−⎡ ⎤⎢ ⎥− −⎣ ⎦

5cos3 5sin 3 3 01cos3 3sin 3 sin 3 3cos3 1 515

15cos3 5sin3 25sin3110sin 3 5sin 3 15cos315

t t te

t t t t

t t tt t

− − −⎡ ⎤= ⎢ ⎥+ − −⎣ ⎦

+⎡ ⎤= ⎢ ⎥− − +⎣ ⎦

A

t

⎡ ⎤⎢ ⎥−⎣ ⎦

1 cos3 2sin 3( )

1 cos3 sin 3t t t

t et t+⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦Ax

28. A = 1 1 12 1 18 5 3

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− − −⎣ ⎦

2 2 2

12 2 2 2 2 2

2

32 2 2 2 2 2

3 23 5 13 4 5 1 4 5( ) 42 2 12 3 4 12 3 43 7 13 2 7 1 2 722 2 12 3 4 12 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 t t

e e e e e e ct e e e e e e e e e e c

ce e e e e e e e e

− − − − − −

− − − − − −

− − − − − −

⎡ ⎤⎢ ⎥− − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= = − + − + − + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎢ ⎥− − + − − + − − +⎢ ⎥⎣ ⎦

Ax c

29. A = , 3 1 00 3 10 0 3

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1(0) 0

0

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x

3 3 2 33

3 3

3

12

0 ( ) (0)0 0 0

t t tt

t t t t

t

e te t e ee e te t e

e

⎡ ⎤⎢ ⎥

0⎡ ⎤

⎢ ⎥ ⎢ ⎥= ⇒ =⎢ ⎥ = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

A Ax x .

30. A = 6 3 24 1 2

13 9 3

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎣ ⎦

,

− − −

− −

⎡ ⎤− + − − + − − +⎢ ⎥= − + + + − −⎢ ⎥⎢ ⎥− + + − −⎣ ⎦

A

2

2

2

5 3

( ) (0) 5 2 3

5 6

t t t

t t t t

t t t

e e e

t e e e e

e e e

−

−

−

1(0) 0

0

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

x

⇒

2 2

2 2

2 2

3 5 3 25 3 2 2 2 35 6 3 4 2

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

e e e e e e e ee e e e e e e e e

e e e e e e e e

− − −

−

⎡ ⎤− −⎢ ⎥

= = − + +⎢ ⎥⎢ ⎥+ −⎢ ⎥⎣ ⎦

Ax x .


31. Student Project


6.7 Theory of Linear DE Systems

Superposition for Systems

1. ( ) 1 2 20 1 2

teL⎡ ⎤⎡ ⎤ +′= − = ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦x x x

We know that

20

1 1.

1 1

t t

t

e eLe

L

⎡ ⎤ ⎡ ⎤−=⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎣ ⎦⎣ ⎦⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Multiplying the first equation by 12

− , and the second equation by 2, then adding, yields

1 12 21 12 21 12 2 0 2

t t t

t

e e eL Le

⎡ ⎤ ⎡ ⎤ ⎡⎡ ⎤ ⎡ ⎤ ⎤− +− + = − + =⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥

⎥⎣ ⎦ ⎣⎣ ⎦ ⎦

.

Using properties of linear transformations we get

1 2 221 222

tt

t

e eLe

⎡ ⎤− +⎢ ⎥ ⎡ ⎤+=⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎣ ⎦− −⎢ ⎥⎣ ⎦

.

We, therefore, have a particular solution

1 221 22

t

t

e

e

⎡ ⎤− +⎢ ⎥= ⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

px .

Superposition for Systems Once More

2. We know that

11 1 3

1 1.

2 5

t tL

t

L

+⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

Adding 2 times the first equation to 3 times the second equation yields

1 1 1 2 5

2 3 2 31 2 1 3 5 6 1t t

L Lt t

+ +7

t⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡+ = + =

⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − − − − ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.


Using properties of linear transformations, we find

2 3 2 5

8 6t t

Lt

+ +17

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

.

Hence, the solution of the given equation is

2 3

8pt +⎡ ⎤

= ⎢ ⎥⎣ ⎦

x .

Nonhomogeneous Illustration

3. As seen in Section 6.2 of the text, the general solution of the homogeneous linear system

1 14 1⎡ ⎤′ = = ⎢ ⎥⎣ ⎦

x Ax x

is

( ) 31 2

1 12 2

t th t c e c e−

⎡ ⎤ ⎡= +

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

x⎦

.

Hence, by the principle of superposition, the nonhomogenous system

2

4 1 4

t

t

t e

t e

⎡ ⎤− +′ = + ⎢ ⎥

− −⎢ ⎥⎣ ⎦x Ax

has the general solution of

( ) 31 2

1 1( ) ( )

2 2 1

tt t

h p t

e tt t t c e c e

e− ⎡ ⎤−⎡ ⎤ ⎡ ⎤

= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦x x x .

Systematic Prediction

4. . 1 4 31 1 0⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦

The homogeneous solution is

( ) 31 2

2 21 1

t tt c e c e−⎡ ⎤ ⎡

= +⎤

⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣x

⎦.

We seek a particular solution of the form ( ) [ ], p t A B=x . Substituting this expression into the

system yields , , which has the solution 4 3 0A B+ + = 0A B+ = 1A = , and 1B = − . Hence, we

obtain the general solution

( ) 31 2

2 2( ) ( )

1 1t t

h pt t t c e c e−11

⎡ ⎤ ⎡ ⎤ ⎡= + = + +

⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣

x x x⎦

.


5. . 1 4 01 1 9t⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦

The homogeneous solution is ( ) 31 2

2 21 1

t th t c e c e−

⎡ ⎤ ⎡= +

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

x⎦

.

We seek a particular solution of the form ( )pA C

t tB D⎡ ⎤ ⎡ ⎤

= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

Substituting this expression into the nonhomogeneous system yields

1 4 01 1 9

A At CB Bt D t

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, or ( ) ( )( ) ( )

4

9 .

A At C Bt D

B At C Bt D t

= + + +

= + + + +

Expanding these equations yields

( ) ( )( ) ( )

4 4

9 0

A B t A C D

A B t C D B

0

.

+ + − + + =

+ + + + − =

Equating coefficients of like terms we obtain four equations

4 04

9 00

A BA C D

A BC D B

0+ =

− + + =+ + =+ − =

⇒

1238

5

ABCD

= −=== −

Hence, we have the general solution

( ) 31 2

2 2 12( ) ( )

1 1 3t t

h pt t t c e c e t− − 85

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡= + = + + +

⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

x x x⎦

.

6. . 01 4

1 1 te⎡ ⎤⎡ ⎤′ = + ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦x x


2 21 1

t th t c e c e−

⎡ ⎤ ⎡= +

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

x⎦

.

We seek a particular solution of the form ( ) tp

At e

B⎡ ⎤

= ⎢ ⎥⎣ ⎦

x .

Substituting this expression into the nonhomogeneous system yields equations in A and B, that give , and . Hence, the general solution of the nonhomogeneous system is 1A = − 0B =

( ) 31 2

2 2( ) ( )

1 1t t t

h pt t t c e c e e− −10

⎡ ⎤ ⎡ ⎤ ⎡= + = + +

⎤⎢ ⎥ ⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣ ⎦ ⎣

x x x⎦

.


7. . 1 4 01 1 10sin t⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦


2 21 1

t th t c e c e−

⎡ ⎤ ⎡= +

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

x⎦

.

We seek a particular solution of the form ( ) cos sinpA B

t t tC D⎡ ⎤ ⎡

= +⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

x⎦

.

Substituting this expression into the nonhomogeneous system yields the equations

( ) ( )

( ) ( )4 sin 4 cos

10 sin cos 0.

B D A t A C B t

B C D t A C D t

+ + + + − =

+ + + + + − =

0

Equating coefficients of like terms yields the four equations

4 04 0

10 00.

B D AA C B

B C DA C D

+ + =+ − =

+ + + =+ − =

⇒

483

1

ABCD

== −= −=

Hence, it yields the general solution

( ) 31 2

2 2 4( ) ( ) cos sin

1 1 3t t

h pt t t c e c e t t− −81

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡= + = + + +

⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

x x x⎦

.

System Superposition

8. Assume that x is a solution to the nonhomogeneous system ( )L =x f , so ( )i iL x f=

for each i = 1, …, n.. Multiplying by , yields . Using the fact that L is a linear operator, we write ic ( )i i ic L x c f= i

( )i i i iL c x c f= . Adding these equations for 1, i n= we get

( ) ( ) ( )1 1 2 2 1 1 2 2n n n nL c x L c x L c x c f c f c f+ + + = + + + .

Again using the fact that L is a linear transformation, yields

( )1 1 2 2 1 1 2 2n n n nL c x c x c x c f c f c f+ + = + + + ,

which proves the desired result.


Variation of Parameters

9. 1 1 34 1 9

−⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢−⎣ ⎦ ⎣x x

⎤⎥⎦

The eigenvalues and vectors of the coefficient matrix are

1 1 2 2

1 13, ; 1,

2 2λ λ

⎡ ⎤ ⎡= = = − =

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

v v⎦

.

Thus, the homogeneous solution is given by

( ) 31 2

1 12 2

t th t c e c e−⎡ ⎤ ⎡

= +⎤

⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣x

⎦.

We seek a particular solution of the form ( ) 1

2p

kt

k⎡ ⎤

= ⎢ ⎥⎣ ⎦

x . Substituting this into the system yields

1 14 1

39

1

2

LNMOQPLNMOQP =LNMOQP

kk

,

which gives k and k . Hence, we have the particular solution , 1 2= 2 1= ( ) 21p t⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

and the general solution ( ) 31 2

1 1( ) ( )

2 2t t

h pt t t c e c e− 21

⎡ ⎤ ⎡ ⎤ ⎡= + = + +

⎤⎢ ⎥ ⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣ ⎦ ⎣

x x x⎦

.

10. 1 14 1 4

t

t

ee

⎡ ⎤⎡ ⎤′ = + ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎣ ⎦x x


1 1 2 2

1 13, ; 1,

2 2λ λ

⎡ ⎤ ⎡= = = − =

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

v v⎦

Thus, the homogeneous solution is given by ( ) 31 2

1 12 2

t th t c e c e−⎡ ⎤ ⎡

= +⎤

⎢ ⎥ ⎢− ⎥⎣ ⎦ ⎣

x⎦

.

We seek a particular solution of the form ( ) 1

2

tp

kt e

k⎡ ⎤

= ⎢ ⎥⎣ ⎦

x .

Substituting this into the nonhomogeneous system yields

k ek e

k ek e

ee

t

t

t

t

t

t1

2

1

2

1 14 1 4

LNMOQP =LNMOQPLNMOQP + −LNMOQP ,

which gives k and k1 1= 2 1= − . Hence, we have the particular solution , ( )t

p t

et

e⎡ ⎤

= ⎢ ⎥−⎣ ⎦x

and the general solution . ( ) 31 2

1 1( ) ( )

2 2

tt t

h p t

et t t c e c e

e− ⎡ ⎤⎡ ⎤ ⎡ ⎤

= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦x x x


11. 0 1 33 4 9

t−⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢⎣ ⎦ ⎣

x x⎤⎥⎦


1 1 2 2

1 11, ; 3,

1 3λ λ

⎡ ⎤ ⎡= = = =

⎤⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣

v v⎦

Thus the homogeneous solution is given by ( ) 31 2

1 11 3

t th t c e c e

⎡ ⎤ ⎡= +

⎤⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣

x⎦

.

We seek a particular solution of the form ( ) 1 1

2 2p

a bt t

a b⎡ ⎤ ⎡

= +⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

x⎦

.

Substituting this into the nonhomogeneous system yields

aa

a t ba t b

t1

2

1 1

2 2

0 13 4

39

LNMOQP =

−LNMOQP

++

LNM

OQP +LNMOQP , or

a a t b ta a t b a t b

1 2 2

2 1 1 2 2

33 4 9

= − − += + + + +a f a f .

Equating coefficients of like terms yields four equations

1 2

2

2 1 2

1 2

3 03 4

3 4 0.

a ba

a b ba a

9

= −− + =

= + ++ =

Solving yields a , 1 4= − b1223

= − , a2 3= , and b2 4= .

Hence, 224

,33 4

p t−⎡ ⎤−⎡ ⎤ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

x so that

( ) 31 2

221 1 4( ) ( ) 3

1 3 3 4

t th pt t t c e c e t

⎡ ⎤− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥= + = + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

x x x .

12. 31 1 2

4 1 0

te⎡ ⎤⎡ ⎤′ = + ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

x x


1 1 2 2

1 13, ; 1,

2 2λ λ

⎡ ⎤ ⎡= = = − =

⎤⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣

v v⎦

.

The fundamental matrix and its inverse are

( ) ( )3 3 32

13 3 3

2 21; .4 42 2 2 2

t t t t t tt

t t t t t t

e e e e e eet te e e e e e

− − −−−

−

⎡ ⎤ ⎡ ⎤ ⎡− −= = =⎢ ⎥ ⎢ ⎥ ⎢−− −⎣ ⎦ ⎣ ⎦ ⎣

X X− − ⎤

⎥− ⎦


We first find a particular solution by computing

( ) ( )3 3 3

14

12 214 2 0

t t t

tt t

e e et t

ee e

− −− ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦ ⎣ ⎦X f .

Integrating yields

( ) ( )141

4t

tt t dt

e−

⎡ ⎤⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

∫X f .

Finally, multiplying by X gives a particular solution t( )

( ) ( ) ( )3

1 343

14

1 12 2 24 2

t tt

p tt t

t te et t t dt

ee e t

−−

−e

⎡ ⎤+⎡ ⎤ ⎢ ⎥⎡ ⎤ ⎢ ⎥= = = ⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦ −⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

∫x X X f .

Thus the general solution of the system is

( ) ( ) 3 31 2

11 1 4( )2 2 12

2

t tp

tt t t c e c e

t

− te

⎡ ⎤+⎢ ⎥⎡ ⎤ ⎡ ⎤= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦ ⎣ ⎦ −⎢ ⎥⎣ ⎦

x X c x .

13. 2 2 11 3 t⎡ ⎤ ⎡′ = +⎢ ⎥ ⎢−⎣ ⎦ ⎣

x x⎤⎥⎦


1 1 2 2

2 11, ; 4,

1 1λ λ

−⎡ ⎤ ⎡= = = =

⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

v v⎦

.


( ) ( )4

14 4

2 1; .3 2

t t t t

t t t t

e e e et t

e e e e

− −−

− −4

⎡ ⎤ ⎡− −= =

⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

X X⎦

We find a particular solution by computing

( ) ( )14 4 4 4

11 13 32 2

t t t t

t t t

e e e tet t

te e e te

− − − −−

− − − − t

⎡ ⎤ ⎡− ⎡ ⎤ − −= =

⎤⎢ ⎥ ⎢⎢ ⎥− −⎣ ⎦

⎥⎣ ⎦ ⎣

X f⎦

.

Integrating yields

( ) ( )14 4

211 138 2

t t

t t

te et t dt

e te

− −

−− −

⎡ ⎤+⎢ ⎥= ⎢ ⎥− +⎢ ⎥⎣ ⎦

∫X f .



( ) ( ) ( )4

14 4 4

1 11221 2 8

1 1 1 538 2 2 8

t tt t

p t t t t

tte ee et t t dt

e e e te t

− −

−− −

⎡ ⎤⎡ ⎤ − −+ ⎢ ⎥⎡ ⎤− ⎢ ⎥= = = ⎢ ⎥⎢ ⎥ ⎢ ⎥− + ⎢ ⎥⎣ ⎦ +⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

∫x X X f .


( ) 41 2

1 112 1 2 8( )1 1 1

2 8

t tp

tt t c e c e

t 5

⎡ ⎤− −⎢ ⎥−⎡ ⎤ ⎡ ⎤= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦ +⎢ ⎥⎣ ⎦

x X c x .

14. 1

1

4 22 1 2 4

tt

−

−

− ⎡⎡ ⎤′ = + ⎢ ⎥⎢ ⎥− +⎣ ⎦ ⎣ ⎦x x

⎤


1 1 2 2

1 20, ; 5,

2 1λ λ

−⎡ ⎤ ⎡= = = − =

⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

v v⎦

.


( ) ( )5

15 55

1 21 2 1, .252

t

t tt

et t

e ee

−−

−

⎡ ⎤− ⎡ ⎤= =⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

X X

We find a particular solution by computing

( ) ( )1

15 5 1

5

51 2 81 125 52 4 4

t tt

tt t t

e e t e

−−

−

⎡ ⎤⎡ ⎤ +⎡ ⎤ ⎢ ⎥= =⎢ ⎥⎢ ⎥ ⎢ ⎥− +⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

X f .

Integrating yields

( ) ( )1

5

8ln5

425

t

ttt t dt

e

−

⎡ ⎤+⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∫X f .


( ) ( ) ( )5

15

5

8 8ln ln1 2 5 54 162 2ln25 5 25

t

p tt

t t t tet t t dt

e e t t

−−

−

825

4

⎡ ⎤ ⎡ ⎤+ + −⎢ ⎥ ⎢ ⎥⎡ ⎤−= = =⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫x X X f .


( ) 51 2

8 8ln1 2 5 25( )2 1 162ln

5 25

tp

t tt t c c e

t t

−

4

⎡ ⎤+ −⎢ ⎥−⎡ ⎤ ⎡ ⎤= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦ + +⎢ ⎥⎣ ⎦

x X c x .


15. 3

2

4 28 4

tt

−

−

− ⎡⎡ ⎤′ = + ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦x x

⎤

The characteristic equation is λ2 0= , which yields eigenvalues λ λ1 2 0= = , but only a single

eigenvector , which yields one solution for the homogeneous system: 1

12⎡ ⎤

= ⎢ ⎥⎣ ⎦

v

01

1 12 2

te⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x .

We seek a second linearly independent solution 2x of the form

0 02 2

1 12 2

t tte e t⎡ ⎤ ⎡ ⎤

2= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

x v v ,

where is the generalized eigenvector that satisfies 2

ab⎡ ⎤

= ⎢ ⎥⎣ ⎦

v ( ) 2

10 .

2⎡ ⎤

− = ⎢ ⎥⎣ ⎦

A I v

Solving

4 28 4

12

−−LNMOQPLNMOQP =LNMOQP

ab

, or 4 2 1a b− = , 8 4 2a b− = ,

2

014 1 12

2 2

aaaab

⎡ ⎤ ⎡⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢= = = +−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢

⎤⎥⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢⎣ ⎦ ⎣

v⎥⎦

, and 2

01 112 22

t a⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

x .

For convenience, we choose a = 0, and write the general solution of the homogeneous system as

1 1 2 2 1 2

01 112 22

h c c c c t⎛ ⎞⎡ ⎤⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥= + = + +⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥−⎣ ⎦ ⎣ ⎦⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

x x x .

Hence,

( ) ( )11 1 4 2

, and .1 4 22 22

t t tt t

t−

⎡ ⎤ −⎡ ⎤⎢ ⎥= = ⎢ ⎥⎢ ⎥ −− ⎣ ⎦⎢ ⎥⎣ ⎦

X X

We compute

( ) ( )

2

3 31

2

3

2 4 11 4 2

4 2 2 4

t tt t t tt t

t tt

−−

−

⎡ ⎤− − +⎢ ⎥− ⎡ ⎤⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥− − +⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

X f .


and integrate,

( ) ( )2

1

2

1 4 2ln2

2 2

tt tt t dt

t t

−

⎡ ⎤− + −⎢ ⎥= ⎢ ⎥⎢ ⎥− −⎢ ⎥⎣ ⎦

∫X f ,

to get the particular solution

( ) ( ) ( )

( )( )

2 2

2 21

2 2

2 2

2

22

1 4 1 4 4 ln 41 2ln2 2

1 2 22 2 5 4 ln 42

1 4 4 ln 11 .10 8 ln 12

p

t t t tt tt t tt t t dt

t t t t tt t t

t t tt t tt

−

⎡ ⎤− + − −⎡ ⎤− + −⎡ ⎤ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= = =⎢ ⎥⎢ ⎥− ⎢ ⎥− −⎢ ⎥− −⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤− + − +

= ⎢ ⎥+ − −⎢ ⎥⎣ ⎦

∫x X X f

Finally, we have the general solution,

( ) ( )( )

2

1 1 2 2 1 2 22

1 1 4 4112 10 8 ln 1222

p

t t t tt c c c c

t t ttt

⎡ ⎤ ln 1⎡ ⎤− + − +⎡ ⎤ ⎢ ⎥= + + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥ − +− ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

x x x x .

16. 0 1 01 0 tan x

− ⎡ ⎤⎡ ⎤′ = ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

x x +

11 2

2

1

cos sin cos sin( )

sin cossin cos

cos sin cos sin 0( )

sin cos sin cos tan

cos sin cos sin 0

sin cos sin cos tan

h

p

Ct t t tt C C

t tt t C

t t t tt dt

t t t t t

t t t tt t t t

−

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−− ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

∫

∫

x

x

( )( ) 2

cos sin sin tan

sin cos sin

(cos ) sin ln sec tan sin cossin ln sec tancos sinsin cos cos (sin ) sin ln sec tan cos

cos ln sec tan

1 sin

t t t tdt dt

t tt t

t t t t tt t tt tt t t t t t t t

t t t

⎡ ⎤ ⎡ ⎤⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦⎣ ⎦ ⎣ ⎦

t⎡ ⎤− + + +⎡ ⎤− + +⎡ ⎤ ⎢ ⎥= =⎢ ⎥⎢ ⎥− ⎢ ⎥− + + −⎣ ⎦ ⎣ ⎦ ⎣ ⎦+

=− +

∫

.ln sec tant t t

⎡ ⎤⎢ ⎥

+⎢ ⎥⎣ ⎦

Hence, the general solution is

1 2

cos ln sec tancos sin( ) ( ) ( )

sin cos 1 sin ln sec tanh p

t t tt tt t t C C

t t t t t

⎡ ⎤+⎡ ⎤ ⎡ ⎤= + = + + ⎢ ⎥⎢ ⎥ ⎢ ⎥− − + +⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦

x x x


Two-Tank Mixing Problem

17. (a) 11

6 24100 100

2x xx ′ = − + , with x1(0) = x2(0) = 0

1 22

6 6100 100

x xx ′ = −

or

0.06 0.02 4,

0.06 0.06 0− ⎡ ⎤⎡ ⎤′ = + ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

x x 0

(0)0⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

(b) 0.025 0.0950.50 0.50 100( ) 157.47 42.53

0.87 0.87 100t tt e e− − −⎡ ⎤ ⎡ ⎤

= − + +⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

⎤⎥−⎣ ⎦

x (given)

x 157.47(0.50) 42.53( 0.50) 100 0

(0)157.47(0.87) 42.53(0.87) 100 0− + − +⎡ ⎤

= ≈⎢ ⎥− + +⎣ ⎦

From the given solution we identify λ1 = −0.025 and λ2 = −0.095.

From the matrix ⎡⎢ , we obtain a reasonable check:

0.06 0.020.06 0.06−

λ2 + .12 λ + .0024 = 0

λ = .12 .0144 4(.0024)

.095, .02542

− ± −≈ − −

(c) The equilibrium solution

⎡⎢x is approached as t becomes large. 100

( )100

t⎤

= ⎥⎣ ⎦

x2(t)

x1(t)


Two-Loop Circuit

18. RAB = 2 ohms, REF = 1 ohm, LKL = 1henry, LDG = 5 henries.

From Kirchoff’s Laws we obtain

Loop 1: 1 1I ′ 1 22 ( ) 60I I I+ + − =

) 0I I I′

Loop 2: 5 (2 1 2− = −

Hence the IVP is

1 1

22

3 1 601/ 5 1/ 5 0

I III

⎡ ⎤′ − ⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ = +⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥′ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦, 1

2

(0) 0(0) 0

II⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

.

and

2 16 2 0 3.070, 0.13035 5

λ λ λ+ + = ⇒ = − − .

Solving (with Maple) gives

1 3.07 0.13

2

( ) 1 .3319 33.33

( ) 0.07 .94 30t tI t

e eI t

− −− −⎡ ⎤ 30⎡ ⎤ ⎡ ⎤= +⎢ ⎥

⎡ ⎤+⎢ ⎥ ⎢ ⎥− ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

Check:

19 33.3( 0.33) 30 0(0)

19(0.07) 33.3( 0.94) 30 0− + − +⎡ ⎤′ = ≈⎢ ⎥+ − +⎣ ⎦

I .

I1(t)

I2(t)

When the current reaches steady-state, the voltage drop across the 2 henry inductor is zero, so the 1 ohm resistor is shorted out. Loop 1 then has only 2 ohms of resistance.

Thus the current I1 = I2 = 30 amps.


Multiple Loop RL Circuit with AC Input

19. RAB = 4 ohms, RDG = 6 ohms, LLK = 1 henry, LEH = 2 henries. From Kirchoff’s Law we obtain Loop 1: 1 1 1 24 6( ) 220sinI I I I′ + + − = t

0 Loop 2: 2 1 22 6( )I I I′ − − =

Hence the IVP is

1 1 1

2 22

(0)10 6 220sin 0,

3 3 0 (0)I I It

I II

⎡ ⎤′ − ⎡ ⎤ ⎡ ⎤0

⎡ ⎤ ⎡⎡ ⎤⎢ ⎥ = +⎢ ⎥ ⎢ ⎥⎤

=⎢ ⎥ ⎢⎢ ⎥−⎢ ⎥′ ⎣ ⎦⎥

⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎦

Solving, with Maple, gives

1 12

2

2 3 34.9 21.21210 sin cos3 1 25 29.629

t tIe e t t

I− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡

= + + +⎢ ⎥⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎦

I2(t)

I1(t)

I1ss(t) = 34.9 sin t − 21.2 cos t

2 2

2 2 2 2

34.9 21.234.9 22.2 sin cos34.9 22.2 34.9 22.2

41.36sin( )

t t

t δ

⎛ ⎞= + −⎜ ⎟⎜ ⎟+ +⎝ ⎠= +

Note that δ is unimportant to the amplitude, which is 41.36 Amps.


20. Student Project

Documents

Chapter 6 diff eq sm