CH06 - ruby.fgcu.edu

CHAPTER 6 Linear Systems ofDifferential Equations

6.1 Theory of Linear DE Systems

!!!! Nullcline Sketching

1. ′ =′ =

x xy y

Equilibrium (unstable) at (0, 0)h y

x− =− =

nullcline nullcline

00υ

(See figure.)

2–2

–2

2y

xh -nullcline

υ -nullcline

2. ′ = +′ =

x x yy x

3 22

Equilibrium (unstable) at (0, 0)h x

x y− =− + =

nullcline nullcline

03 2 0υ

(See figure.)

2–2

–2

2y

x

3. ′ =′ = − −

x yy x y5 2

Equilibrium (stable) at (0, 0)h x y

y− + =− =

nullcline nullcline

5 2 00υ

(See figure.)

2–2

–2

2y

x

435

436 CHAPTER 6 Linear Systems of Differential Equations

4. ′ =′ = − − +

x yy x y

22 2

Equilibrium (stable) at 1 0, ( )

h x yy

− + =− =

nullcline nullcline

2 20υ

(See figure.)

3–1

–2

2y

x

5. ′ = + −′ = −

x x yy y

21

Equilibrium (unstable) at 1 1, ( )

h yx y

− =− + =

nullcline nullcline

12υ

(See figure.)

2–1

–1

2y

x

6. ′ = −′ = − − +

x yy x y

15 2 2

Equilibrium (stable) at 0 1, ( )

h x yy

− + =− =

nullcline nullcline

5 2 21υ

(See figure.)

–2

2y

2–1x

!!!! Sketching Second-Order DEs

7. ′′ + ′ + =x x x 0

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

′ =′ = − −

x yy x y .

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

SECTION 6.1 Theory of Linear DE Systems 437

(c) h x yy

− + =− =

nullcline nullcline

00υ

(See figure.)

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

(e) A mass-spring system with this equationshows damped oscillatory motion aboutx t( ) ≡ 0 is stable.

2–2

–2

2y

x

8. ′′ − ′ + =x x x 0


′ =′ = − +

x yy x y .


(c) h x yy

− − =− =

nullcline nullcline

00υ

(See figure.)

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

(e) A mass-spring system with this equationtends to fly apart. Hence, x t( ) ≡ 0 is

unstable.

2–2

–2

2y

x

9. ′′ + =x x 1


′ =′ = − +

x yy x 1.



(c) h xy

− =− =

nullcline nullcline

10υ

(See figure.)

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

(e) A mass-spring system with this equationshows no damping and steady forcing;hence, periodic motion about an equilib-rium is to the right of the origin. Hence,x t( ) ≡ 1 is stable.

3–1

–2

2y

x

10. ′′ + ′ + =x x x2 2


′ =′ = − − +

x yy x y2 2.

(b) The equilibrium point is x y, ( ) = ( )2, 0 .

(c) h x yy

− + =− =

nullcline nullcline

2 20υ

(See figure.)

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

(e) A mass-spring system with this equationshows heavy damping. The force movesthe equilibrium two units to the right ofthe origin. Hence, x t( ) ≡ 2 is stable.

3–1

–2

2y

x

!!!! Breaking Out Systems

11. ′ = +′ = −

x x xx x x

1 1 2

2 1 2

24

12. ′ =′ = − +

x xx x

1 1

2 2 1

13. ′ = + +′ = − −

−x x x ex x x

t1 1 2

2 1 2

4 3 14. ′ =′ =′ = − + + +

x xx xx x x x t

1 2

2 3

3 1 2 32 3 sin


!!!! Checking It Out

15. ′ = LNMOQPx x

1 33 1

Plugging

u tee

t

t( ) =LNMOQP

4

4 and v tee

t

t( ) =

−LNMOQP

−

−

2

2

into the given system easily verifies:

44

1 33 1

4

4

4

4ee

ee

t

t

t

tLNMOQP =LNMOQPLNMOQP

and

−LNMOQP =LNMOQP −LNMOQP

−

−

−

−

22

1 33 1

2

2

2

2ee

ee

t

t

t

t .

The fundamental matrix is

e ee e

t t

t t

4 2

4 2

−

−−LNM

OQP .

The general solution of this 2 2× system is !x Ax= is

x t cee

cee

t

t

t

t( ) =LNMOQP + −LNMOQP

−

−1

4

4 2

2

2 .

16. ′ =−LNMOQPx x

4 12 1

By substitution, we verify

u tee

t

t( ) =LNMOQP

3

3 and v tee

t

t( ) =LNMOQP

2

22

satisfy the system. The fundamental matrix is

e ee e

t t

t t

3 2

3 22LNM

OQP .

The general solution is

x t cee

cee

t

t

t

t( ) =LNMOQP +LNMOQP1

3

3 2

2

22.


17. ′ = LNMOQPx x

1 14 1


u t ee

t

tb g =−

LNMOQP

−

−2 and v t e

e

t

tb g = LNMOQP

3

32


e ee e

t t

t t

−

−−

LNM

OQP

3

32 2.


x t c ee

c ee

t

t

t

tb g =−

LNMOQP+LNMOQP

−

−1 2

3

32 2.


0 11 0


u ttt

( ) = LNMOQP

sincos

and v ttt

b g =−LNMOQP

cossin


sin coscos sin

t tt t−

LNM

OQP .


x t ctt

ctt

( ) = LNMOQP + −LNMOQP1 2

sincos

cossin

.

!!!! Uniqueness in the Phase Plane

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

We have drawn three distinct trajectories for thesix initial conditions

x y0 0 1 0( ) ( ) =, ,a f a f ,2, 0a f , 3 0,a f, 0 1,a f, 0 2,a f , 0 3,a f.

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

4–4

–4

4y

x


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.

–1

1

t

2

–2

4 8

x

102

–3

3

6

x x t= ( )

y

–1

1t

2

–2

4 8 102

–3

3

6

y y t= ( )

!!!! Verification

20. Plugging

v =−LNMOQP

−

−

ee

t

t

into

′′LNMOQP =LNMOQPLNMOQP

xx

xx

1

2

1

2

1 22 1

yields

−LNMOQP =LNMOQP −LNMOQP

−

−

−

−

ee

ee

t

t

t

t

1 22 1

or expanding this, we may write

− = −

= −

− − −

− − −

e e ee e e

t t t

t t t

22

which verifies the solution.

!!!! Third-Order Verification

21. To verify u, v, w , you should follow the procedure carried out in Problem 20. To show that thevector functions u, v, w are linear independent, set

c c c c ee

cee c

tetee

t

t

t

t

t

t

t1 2 3 1 2

2

23

2

2

2

0

0

000

u v w+ + =L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP=L

NMMM

O

QPPP

or if this were to be written out in scalar form


c e c tec e c e c tec e c e

t t

t t t

t t

22

32

1 22

32

1 32

000

+ =+ + =

+ =.

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

cc cc c

2

1 2

1 3

000

=+ =

+ =

or c c c1 2 3 0= = = .

!!!! Euler’s Method Numerics

22. (a) The IVP′′ + =x x01 0. , x 0 1( ) = , ′( ) =x 0 0

studied in Example 3 can be solved numerically with a spreadsheet using the followingcoding:

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

Doing this results in the following values on 0 1≤ ≤t .

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


If the range is continued to t = 40, then the graphs that correspond to Figure 6.1.3 looklike the following.

x

y

xy component graph

t

y

yt component graph

t

x

xt component graph

t

x y

3D txy view

(b) The IVP

′′ + ′ + =x x x0 05 01 0. . , x −( ) = −5 01. , ′ −( ) =x 5 05.


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


Doing this results in 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


x

y

xy phase portrait

t

y

yt component graph


t

x

xt component graph

t

x y

3D txy view

(c) The IVP

′′ + =x x t01 05. . cos , x 0 1( ) = , ′( ) =x 0 0


A B C D E

1 t x y dxdt

dydt

2 0 1 0 = C2 = –0.1 * B2 + 0.05 * cos(A2)

3 = A2 + 0.1 = B2 + 0.1 * D2 = C2 + 0.1 * E2 = C3 = –0.1 * B3 + 0.05 * cos(A3)

Doing this results in the following values on 0 1≤ ≤t .

t x y dxdt

dydt

0.0 1.0000 0.0000 0.0000 0.4000

0.1 1.0000 0.0400 0.0400 0.3975

0.2 1.0040 0.0798 0.0798 0.3896

0.3 1.0120 0.1187 0.1187 0.3765

0.4 1.0238 0.1564 0.1564 0.3581

0.5 1.0395 0.1922 0.1922 0.3348

0.6 1.0587 0.2257 0.2257 0.3068

0.7 1.0813 0.2563 0.2563 0.2743

0.8 1.1069 0.2838 0.2838 0.2377

0.9 1.1353 0.3075 0.3075 0.1973

1.0 1.1660 0.3273 0.3273 0.1535



x

y

xy phase portrait

t

y

yt component graph

t

x

xt component graph

t

x y

3D txy view

!!!! Matching Games

23. A 24. C 25. D 26. B

!!!! Finding Trajectories

27. ′ =x x , ′ =y y . Write

dydx

yx

yx

= ′′= .

Separating variables, yields

dyy

dxx

=

2–2

–2

2y

x


or

ln lnln

y x c

y e e

y e xy C x

x c

c

= +

=

= ±

=

where C is an arbitrary constant. Hence, the trajectories consist of a family of semi-infinite linesoriginating 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. (Compare with the solution to Problem 1.) Allsolutions are further and further from the origin and go faster and faster the further away theyare.

28. ′ =x y , ′ = −y x . We write these equations as

dydx

yx

xy

=′′= − .

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

12

12

2 2y x c= − + , or x y C2 2+ =

3–3

–3

3y

x

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 pathsaround the origin have the same period, but thepaths with the larger radius move at a faster rate.

3–3

–3

3y

x


!!!! Computer Check

29. The computer phase portrait for Problem 27 and 28 are shown.

2–2

–2

2y

x

Computer Phase Portrait for Problem 27

3–3

–3

3y

x

Computer Phase Portrait for Problem 28

!!!! Computer Lab: Skew-Symmetric Matrices

30. (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 = ( )x y, is a constant.

3–3

–3

3y

x

(b) ′ =x ky , ′ = −y kx

Write this system as the single equation

′′ + =x k x2 0

which has a general solution of

x R kt= −( )cos δ .

Then find

yk

x R kt= ′ = − −( )1 sin δ .

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

x t y t R kt R kt R2 2 2 2 2( ) + ( ) = −( ) + −( ) =cos sinδ δ .


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

frequency ωπ

=k

2 and period 2π

k.

An open-ended graphic solver can be used to verify these facts.

!!!! The Wronskian

When the Wronskian is not zero, the vectors are linearly independent and form a fundamental set. (If theWronskian of two solutions is nonzero on I it will always be nonzero.)

31. We ee

et t

ttx x1 2

2

232

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

32. We ee e

et t

t ttx x1 2

3

322

35 0, =

−= − ≠

−

−, so the vectors form a fundamental set.

33. We ee

et t

ttx x1 2

220

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

34. We ee e

et t

t ttx x1 2

4 4

4 483

2 0, = = ≠ , so the vectors form a fundamental set.

35. We t e te t e t

e t t et t

t tt tx x1 2

2 2 2 2 0,cos sinsin cos

cos sin =−

= + = ≠b g , so the vectors form a fundamental set.

36. Wt tt t

t tx x1 22 23 3

3 33 3 1 0,

cos sinsin cos

cos sin =−

= + = ≠ , so the vectors form a fundamental set.

!!!! Suggested Journal Entry

37. Student Project


6.2 Linear Systems with Real Eigenvalues

!!!! Solutions in General

1. ′ =−

−LNM

OQPx x

4 22 1

The characteristic equation of the system is

p t( ) =− −

− −= + =

4 22 1

5 02λλ

λ λ ,

which has solutions λ1 0= , λ2 5= − . Finding the eigenvectors corresponding to each eigenvalue

yields

λ

λ

1 1

2 2

012

521

= ⇒ = LNMOQP

= − ⇒ =−LNMOQP

v

v .

Hence, the general solution is

x t c c e t( ) = LNMOQP +

−LNMOQP

−1 2

512

21

.


2 13 6


p t( ) =−− −

= − + =2 1

3 68 15 02λ

λλ λ ,

which has solutions λ1 3= , λ2 5= . Finding the eigenvectors corresponding to each eigenvalue

yields

λ

λ

1 1

2 2

311

513

= ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .


x t c e c et t( ) = LNMOQP +

LNMOQP1

32

511

13

.

SECTION 6.2 Linear Systems with Real Eigenvalues 451


1 12 4


p t( ) =− −

−= − + =

1 12 4

5 6 02λλ

λ λ ,


yields

λ

λ

1 1

2 2

211

312

= ⇒ =−LNMOQP

= ⇒ =−LNMOQP

v

v .


x t c e c et t( ) =−LNMOQP + −

LNMOQP1

22

311

12

.

4. ′ =−−

LNM

OQPx x

10 58 12


p t( ) =− −

− −= + − =

10 58 12

2 80 02λλ

λ λ ,

which has solutions λ1 10= − , λ2 8= . Finding the eigenvectors corresponding to each eigenvalue

yields

λ

λ

1 1

2 2

1014

852

= − ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .



LNMOQP

−1

102

814

52

.


5 13 1


p t( ) =− −

−= − + =

5 13 1

6 8 02λλ

λ λ ,



yields

λ

λ

1 1

2 2

213

411

= ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .



LNMOQP1

22

413

11

.

6. ′ = LNMOQPx x

1 24 3


p tb g = −−

= − − =1 2

4 34 5 02λ

λλ λ ,


yields

λ

λ

1 1

2 2

111

512


= ⇒ = LNMOQP

v

v .


x t c e c et t( ) =−LNMOQP +

LNMOQP

−1 2

511

12

.


1 02 2


p t( ) =−− −

= −( ) −( ) =1 0

2 21 2 0

λλ

λ λ ,


yields

λ

λ

1 1

2 2

112

201

= ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .




LNMOQP1 2

212

01

.

8. ′ =− −LNMOQPx x

3 31 1


p t( ) =−− − −

= − =3 3

1 12 02λ

λλ λ ,


yields

λ

λ

1 1

2 2

011

231

= ⇒ =−LNMOQP

= ⇒ =−LNMOQP

v

v .


x t c c e t( ) =−LNMOQP +

−LNMOQP1 2

211

31

.

9. ′ =−−LNMOQPx x

3 22 2


p t( ) =− −

− −= − − =

3 22 2

2 02λλ

λ λ ,


yields

λ

λ

1 1

2 2

112

221

= − ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .



LNMOQP

−1 2

212

21

.


10. ′ =− −LNM

OQPx x

4 34 4


p t( ) =−− − −

= − =4 3

4 44 02λ

λλ ,


yields

λ

λ

1 1

2 2

232

212

= ⇒ =−LNMOQP


v

v .



LNMOQP

−1

22

232

12

.


1 23 4


p t( ) =− −

− −= + + =

1 23 4

3 2 02λλ

λ λ ,

which has solutions λ1 2= − , λ2 1= − . Finding the eigenvectors corresponding to each eigenvalue

yields

λ

λ

1 1

2 2

223

111

= − ⇒ = LNMOQP

= − ⇒ = LNMOQP

v

v .



LNMOQP

− −1

22

23

11

.

12. ′ =−

−LNM

OQPx x

5 22 8


p t( ) =− −− −

= − + =5 2

2 813 36 02λ

λλ λ ,



yields

λ

λ

1 1

2 2

421

912

= ⇒ = LNMOQP

= ⇒ =−LNMOQP

v

v .


x t c e c et tb g = LNMOQP + −

LNMOQP1

42

921

12

.


4 38 6


p t( ) =− −

− −= + =

4 38 6

2 02λλ

λ λ ,


yields

λ

λ

1 1

2 2

034

212

= ⇒ = LNMOQP

= − ⇒ = LNMOQP

v

v .


x t c c e t( ) = LNMOQP +

LNMOQP

−1 2

234

12

.


5 31 1


p t( ) =−− −

= − + =5 3

1 16 8 02λ

λλ λ ,


yields

λ

λ

1 1

2 2

211

431

= ⇒ =−LNMOQP

= ⇒ =−LNMOQP

v

v .




−LNMOQP1

22

411

31

.

!!!! Repeated Eigenvalues

15. The characteristic equation of the system is

p t( ) =− −

− −= − + =

1 14 3

2 1 02λλ

λ λ ,

which has solutions λ1 1= and λ2 1= with one linearly independent eigenvector

v = LNMOQP

12

.

The general solution is, therefore,

x t c e c te et t t( ) = LNMOQP +

LNMOQP +LNMOQP

RSTUVW1 2

1

2

12

12

uu

where u1 and u2 satisfy A I u−( ) = LNMOQP

12

, or

−−LNMOQPLNMOQP =LNMOQP

2 14 2

12

1

2

uu

,

which has one linearly independent equation, − + =2 11 2u u . Hence,

uuu

x

= LNMOQP = +LNMOQP =LNMOQP +LNMOQP

( ) = LNMOQP +

LNMOQP +LNMOQP +LNMOQP

RSTUVW

1

2

1 2

1 201

12

12

12

12

01

kk

k

t c e c te ke et t t t .

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

x t c e c te et t t( ) = LNMOQP +

LNMOQP +LNMOQP

RSTUVW1 2

12

12

01

.


p t( ) =−− − −

= + + =3 2

8 52 1 02λ

λλ λ ,

which has solutions λ1 1= − and λ2 1= − with one linearly independent eigenvector

v =−LNMOQP

12

.



x t c e c te et t tb g =−LNMOQP + −

LNMOQP +LNMOQP

RSTUVW

− − −1 2

1

2

12

12

uu

where u1 and u2 satisfy

A I u+( ) =−LNMOQP

12

or

4 28 4

12

1

2− −LNMOQPLNMOQP = −LNMOQP

uu

,

which has one linearly independent equation, 4 2 11 2u u+ = . Hence,

uuu

x

=LNMOQP = −LNMMOQPP =LNMMOQPP + −LNMOQP

=−LNMOQP + −

LNMOQP + −LNMOQP +LNMMOQPP

RS|T|

UV|W|

− − − −

1

2

1 2

12

2012

12

12

12

12

012

kk k

t c e c te ke et t t tb g .

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

x t c e c e tt t( ) =−LNMOQP + −

LNMOQP +LNMMOQPP

RS|T|UV|W|

− −1 2

12

12

012

.

!!!! Solutions in Particular


p t( ) =− −

− −= − − =

2 15 4

2 3 02λλ

λ λ ,

which has the solutions λ1 3= and λ2 1= − . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

1 1

2 2

315

111

= ⇒ = LNMOQP

= − ⇒ = LNMOQP

v

v .



LNMOQP

−1

32

15

11

.


Plugging into the initial conditions

x 013

( ) = LNMOQP

yields

c cc c1 2

1 2

15 3

+ =+ =

which gives c112

= and c212

= . The solution of the IVP is

x t e et t( ) = FHIKLNMOQP +FHIKLNMOQP

−12

15

12

11

3 .


p t( ) =− −− −

= − − =1 3

2 23 4 02λ

λλ λ ,

which has the solutions λ1 1= − and λ2 4= . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

1 1

2 2

132

411

= − ⇒ = LNMOQP

= ⇒ =−LNMOQP

v

v



−LNMOQP

−1 2

432

11

.


x 011

( ) =−LNMOQP

yields

3 12 1

1 2

1 2

c cc c− =+ = −

which gives c1 0= and c2 1= − . The solution of the IVP is

x t e t( ) = −−LNMOQP

4 11

.



p t( ) =−

−= −( ) −( ) =

2 00 3

2 3 0λ

λλ λ ,

which has the solutions λ1 2= and λ2 3= . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

1 1

2 2

210

301

= ⇒ = LNMOQP

= ⇒ = LNMOQP

v

v .



LNMOQP1

22

310

01

.


x 054

( ) = LNMOQP

yields c1 5= and c2 4= . The solution of the IVP is

x t e eee

t tt

t( ) = L

NMOQP +LNMOQP =LNMOQP5

10

401

54

2 32

3 .


p t( ) =− −

−= + − =

2 41 1

6 02λλ

λ λ ,


eigenvalue, yields

λ

λ

1 1

2 2

341

211


= ⇒ = LNMOQP

v

v .



LNMOQP

−1

32

241

11

.


x 011

( ) =−LNMOQP


yields

− + = −+ =

4 11

1 2

1 2

c cc c

which gives c125

= and c235


x t e et t( ) = FHIK

−LNMOQP +FHIKLNMOQP

−25

41

35

11

3 2 .


p t( ) =−

−= − =

1 11 1

2 02λλ

λ λ ,


eigenvalue, yields

λ

λ

1 1

2 2

011

211

= ⇒ =−LNMOQP

= ⇒ = LNMOQP

v

v .


x t c c e t( ) =−LNMOQP +

LNMOQP1 2

211

11

.


x 023

( ) = LNMOQP

yields

− + =+ =

c cc c1 2

1 2

23

which gives c112

= and c252


x t e t( ) =−LNMOQP +FH IKLNMOQP

12

11

52

11

2 .


p t( ) =− −

− −= + + =

3 21 2

5 4 02λλ

λ λ ,


which has the solutions λ1 4= − and λ2 1= − . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

1 1

2 2

421

111


= − ⇒ = LNMOQP

v

v .



LNMOQP

− −1

42

21

11

.


x 016

( ) =−LNMOQP

yields

− + = −+ =

2 16

1 2

1 2

c cc c

which gives c173

= and c2113


x t e et t( ) = FHIK

−LNMOQP +FHIKLNMOQP

− −73

21

113

11

4 .


p t( ) =− −

− −= + =

2 14 2

4 02λλ

λ λ ,

which has the solutions λ1 0= and λ2 4= − . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

1 1

2 2

012

412

= ⇒ = LNMOQP


v

v .


x t c c e t( ) = LNMOQP + −

LNMOQP

−1 2

412

12

.



x 024

( ) = LNMOQP

yields

c cc c

1 2

1 2

22 2 4

+ =− =

which gives c1 2= and c2 0= . The solution of the IVP is

x t( ) = LNMOQP

24

.


p t( ) =−

−= − − =

1 123 1

2 35 02λλ

λ λ ,


eigenvalue, yields

λ

λ

1 1

2 2

521

721


= ⇒ = LNMOQP

v

v .



LNMOQP

−1

52

721

21

.


x 001

( ) = LNMOQP

yields

− + =+ =

2 2 01

1 2

1 2

c cc c

which gives c112

= and c212


x t e et t( ) = FH IK−LNMOQP +FH IKLNMOQP

−12

21

12

21

5 7 .



p t( ) =− −

−= − + =

1 12 4

5 6 02λλ

λ λ ,


eigenvalue, yields

λ

λ

1 1

2 2

211

312

= ⇒ =−LNMOQP

= ⇒ =−LNMOQP

v

v .



LNMOQP1

22

311

12

.


x 010

( ) = LNMOQP

yields− + =− =

c cc c

1 2

1 2

12 0

which gives c1 2= − and c2 1= − . The solution of the IVP is

x t e et t( ) = −−LNMOQP − −LNMOQP2

11

12

2 3 .


p t( ) =−

−= − − =

1 22 1

2 3 02λλ

λ λ ,


eigenvalue, yields

λ

λ

1 1

2 2

111

311


= ⇒ = LNMOQP

v

v .



LNMOQP

−1 2

311

11

.



x 013

( ) = LNMOQP

yields− + =

+ =c cc c1 2

1 2

13

which gives c1 1= and c2 2= . The solution of the IVP is

x t e et t( ) =−LNMOQP +LNMOQP

− 11

211

3 .

!!!! Creating New Problems

27. (a) An example is

A =L

NMMM

O

QPPP

aa

b

1 00 00 0

.

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

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

aa

b

xyz

axyz

1 00 00 0

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

orax y ax

ay aybz az

+ ===

which implies z = 0, x = α , y = 0. 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 equationa

ab

xyz

bxyz

1 00 00 0

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

orax y bx

ay bybz bz

+ ===

which implies x y= =0 0, , z = α . In other words, the eigen-vector 0 1, , 0 .


(b) An example is

A =L

NMMM

O

QPPP

aa

a

1 00 00 0

.

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

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

aa

a

xyz

axyz

1 00 00 0

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

orax y ax

ay ayaz az

+ ===

which implies y = 0, x = α , z = β , α, β arbitrary. In other words, the two (linearlyindependent) eigenvectors are 1 0 0, , and 0 0 1, , .

!!!! One Independent Eigenvector

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

(b) From the eigenvalue and eigenvector, one solution

x1

121

t cet( ) = −L

NMMM

O

QPPP

has been found.

(c) Now we solve for a second solution of the form x v u2 t te et t( ) = + , where v = −( )1 2, 1,

is the first eigenvector, and u u u u= 1 2 3, , a f is an unknown vector. Plugging x2 t( ) intothe system ′ =x Ax and comparing coefficients of tet and et yields equations for u1 ,u2 , u3 , giving u1 1= − , u2 1= , u3 0= . Hence, we obtain a second solution of

x2

121

110

t te et t( ) = −L

NMMM

O

QPPP+

−L

NMMM

O

QPPP

.


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

x v u w321

2t t e te et t tb g = + +

where v and u are vectors previously found and w is the unknown vector. Plugging thisinto the system results in the system of equations A I w u−( ) = . We then findw w w w= 1 2 3, , a f . Solving this system yields w1 1= , w2 0= , w3 0= . Hence, we

obtain a third solution of

x321

2

121

110

100

t t e te et t tb g = −L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+L

NMMM

O

QPPP

.

!!!! Solutions in Space


p t( ) =−

−− − − −

= − + − + =3 2 2

1 4 12 4 1

6 11 6 03 2λ

λλ

λ λ λ ,

which has solutions λ1 1= , λ2 2= , and λ3 3= . Finding the eigenvectors corresponding to each

eigenvalue, yields

λ

λ

λ

1 1

2 2

3 3

1101

2210

3011

= ⇒ =−

L

NMMM

O

QPPP

= ⇒ =−L

NMMM

O

QPPP

= ⇒ =−

L

NMMM

O

QPPP

v

v

v .


x t c e c e c et t t( ) =−

L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+

−

L

NMMM

O

QPPP

1 22

33

101

210

011

.


p tb g =− −

−− −

= − + + =1 1 0

1 2 10 3 1

7 6 03

λλ

λλ λ ,


which has solutions λ1 1= − , λ2 3= , and λ3 2= − . Finding the eigenvectors corresponding to

each eigenvalue, yields

λ

λ

λ

1 1

2 2

3 3

1101

3143

2113

= − ⇒ =−

L

NMMM

O

QPPP

= ⇒ =L

NMMM

O

QPPP

= − ⇒ =−

−

L

NMMM

O

QPPP

v

v

v .


x t c e c e c et t tb g =−

L

NMMM

O

QPPP+L

NMMM

O

QPPP+

−

−

L

NMMM

O

QPPP

− −1 2

33

2

101

143

113

.

!!!! Spatial Particulars

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

λ

λ

λ

1 1

2 2

3 3

0331

2111

2131

= ⇒ =L

NMMM

O

QPPP

= ⇒ =−L

NMMM

O

QPPP

= − ⇒ =−−L

NMMM

O

QPPP

v

v

v .


x t c c e c et t( ) =L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+

−−L

NMMM

O

QPPP

−1 2

23

2331

111

131

.

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

3 03 3 0

0

1 2 3

1 2 3

1 2 3

c c cc c cc c c

− − =+ − =+ + =


which has the solution c114

= , c238

= , c338

= . Hence, the IVP has the solution

x t e et t( ) =L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+

−−L

NMMM

O

QPPP

−14

331

38

111

38

131

2 2 .

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

λ

λ

λ

1 1

2 2

3 3

0110

1001

2110

= ⇒ =−L

NMMM

O

QPPP

= − ⇒ =L

NMMM

O

QPPP

= ⇒ =L

NMMM

O

QPPP

v

v

v .


x t c c e c et t( ) =−L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP

−1 2 3

2110

001

110

.

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

− + =+ =

=

c cc c

c

1 3

1 3

2

242

which has the solution c1 1= , c2 2= , c3 3= . Hence, the IVP has the solution

x t e et tb g =−L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP

−

110

2001

3110

2 .

!!!! Repeated Eigenvalue Theory

33. (a) The characteristic equation of the matrix is a d bc−( ) −( ) − =λ λ 0, or

λ λ2 0− +( ) + −( ) =a d ad bc .

This has repeated roots if and only if the discriminate is zero (i.e., if and only if

a d ad bc+( ) − −( ) =2 4 0,

which simplifies to a d bc−( ) + =2 4 0.)


(b) Suppose a d≠ and a d bc−( ) + =2 4 0. Then the single eigenvalue is a d+2

. To find the

corresponding eigenvector, we obtain the equations

a d c b

c d a b

−+ =

+−

=2

0

20

1 2

1 2

v v

v v.

Solving these equations, yields

v v v= = −( )1 2 2, , a f b d a .

(c) With one linearly independent eigenvector we obtain one linearly independent solution

x12 2

t ceb

d aa d tb g b g=

−LNMOQP

+ .

A second solution has the form of

xuu2

2 2 1

2

2t te

bd a

ea d t a d tb g b g b g=−LNMOQP +

LNMOQP

+ + .

We find u by solving A I u v−( ) =λ , we obtain u1 0= and u2 1= .

!!!! Phase Portraits


2 13 6

The eigenvalues and vectors are

λ1 3= , v1 1 1= ( ),

λ2 5= , v2 1 3= ( ), .2–2

–2

2y

x

v2 v1


1 12 4


λ1 2= , v1 1 1= −( ),

λ2 3= , v2 1 2= −( ), .2–2

–2

2y

x

v2v1



5 13 1


λ1 2= , v1 1 3= ( ),

λ2 4= , v2 1 1= ( ), .

2–2

–2

2y

x

v2v1


1 23 4


λ1 2= − , v1 2, 3= ( )

λ2 1= − , v2 1 1= ( ), .

2–2

–2

2y

x

v2v1

!!!! Verification of Independence

38. To show

x t e

x t et

t

t

t

14

24

12

2 1

( ) =−LNMOQP

( ) =− −LNMOQP

are linearly independent, we show that the constants c1 and c2 for which

c e c et

tt t

14

241

2 2 100−

LNMOQP + − −

LNMOQP =LNMOQP

for all t are c c1 2 0= = . If this must hold for all t, it must hold for t = 0, which yields the equa-

tions

cc c

1

1 2

02 0

=− − =

whose solution is c c1 2 0= = .

!!!! Adjoint Systems

39. (a) The negative transpose of the given matrix is simply the matrix with –1s in the place of1s, hence the adjoint system is

′ = − =−

−LNMOQPw A w wT 0 1

1 0.


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

′ = − = −w A w w AT T T Tb g ,

and hence,

′ + ′ = − ′ + ′ =w x w x w x w xT T T T 0.

(c) The characteristic equation of the matrix is simply λ2 1 0− = , and hence, the eigenvaluesare +1, –1. The eigenvector corresponding to +1 can easily be found and is 1 1, ( ).Likewise, the eigenvector for –1 is 1 1, −( ) . Hence,

x t c e c et t( ) = LNMOQP + −

LNMOQP

−1 2

11

11

.

Plugging in the initial condition x 0 1 0( ) = ( ), , yields c c1 212

= = . So the solution of the

IVP is

x t e ee ee e

t tt t

t t( ) = FHIKLNMOQP +FHIK −LNMOQP =FHIK

+−LNM

OQP

−−

−12

11

12

11

12

.

(d) If the initial conditions are w t( ) = ( )0 1, , then c112

= , c212

= − . Note that the initial

conditions x 0 1 0( ) = ( ), and w 0 0 1( ) = ( ), are orthogonal vectors, and by the result in

part (b) the two resulting solutions will always be orthogonal for all t > 0 . So thesolution of the IVP is

w =−+LNM

OQP

−

−12

e ee e

t t

t t .

(e) Trajectories are orthogonal.

!!!! Cauchy-Euler Systems

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

′ = −x vλ λt 1

or

t t′ =x vλ λ .

On the other hand,

Ax A v Av v= = =t t tλ λ λλ

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


(b) We have

t ′ =−−

LNMOQPx x

3 22 2

, t > 0 .

The characteristic equation is

p λ λ λ( ) = −( ) − −( ) + =3 2 4 0

whose eigenvalues are λ1 1= − and λ2 2= and corresponding eigenvectors are

v1 1 2= ( ), and v2 2, 1= ( ) .

From part (a), the general solution is then

x t c t c t( ) = LNMOQP +LNMOQP

−1

12

212

21

.

!!!! Computer Labs: Predicting Phase Portraits

41. For each of the linear systems (a)–(d) a few trajectories in the phase plane have been drawn. Theanalytic solutions are then computed.

(a) ′ =x x , ′ = −y y

Solve each of these equations individu-ally, obtaining

x c et= 1 and y c e t= −2 .

Eliminating t yields the trajectories

y cx

= ,

which is the family of hyperbolas shown.

2–2

–2

2y

x

(b) ′ =x 0, ′ = −y y

Solve each of these equations individu-ally, obtaining

x c= 1 and y c e t= −2 .

Eliminating t yields the trajectories

x c= ,

2–2

–2

2y

x

which is the family of vertical lines. For any starting point x y0 0, a f the solution movesasymptotically towards x0 0, a f .

The x-axis is composed entirely of stable equilibrium points.


(c) ′ = +x x y , ′ = +y x y x y c= + , because

′ = ′x y ;

which is a family of straight lines in thephase plane with slope 1 and y-interceptc, 0( ) . x y= − is a line of unstable equi-

librium points.

2–2

–2

2y

x

(d) ′ =x y , ′ =y x

We write these equations as the singleequation

′′ − =x x 0,

which has solution

x c e c et t= + −1 2 .

2–2

–2

2y

x

Hence,

y c e c et t= − −1 2 .

Now we add and subtract these equations, yielding

x y c ex y c e

t

t

+ =

− = −

22

1

2

or

x y kx y

+ =−1 ,

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

!!!! Radioactive Decay Chain

42. (a) The amount of iodine is simply decreasing via radioactive decay; hence, dIdt

k I= − 1 ,

where k1 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,

k l12

6 701034548= ≈

ln.

. .

The amount of xenon is increasing with the decay of iodine, but decreasing with its ownradioactive decay, hence, the equation


dxdt

k I k xee= −1 2

where

k22

9 200753421= ≈

ln.

. .

(b) In matrix form, the equations become

′′LNMOQP =

−−

LNM

OQPLNMOQP

Ix

kk k

Ixe e

1

1 2

0.

The eigenvalues and vectors of this matrix can easily be seen as

λ

λ

1 1 12 1

1

2 2 201


= − ⇒ = LNMOQP

kk k

k

k

v

v .

Hence, the solution

x t c ek k

kc ek t k t( ) =

−LNMOQP +

LNMOQP

− −1

2 1

121 2

01

.

!!!! Multiple Compartment Mixing

43. (a) Calling x t1( ) , x t2( ) the number of pounds of salt in tank A and B, respectively, at time t,

the IVP that describes these amounts is

′ = − FH IK + FH IK ( ) =

′ = FH IK − FH IK ( ) =

x x x x

x x x x

11 2

1

21 2

2

6100

2100

0 25

6100

2100

0 0

,

, .

To solve this linear homogeneous system, we find the eigenvalues and vectors of thecoefficient matrix

A =−

−LNMOQP

6 22 6

,

which are

λ

λ

1 1

2 2

811

411


= − ⇒ = LNMOQP

v

v .



xx

c e c et t1

21

82

411

11

LNMOQP = −

LNMOQP +

LNMOQP

− − .

Plugging the initial conditions into this general solution yields

250

11

111 2

LNMOQP = −LNMOQP +LNMOQPc c

or c1 12 5= . and c2 12 5= − . . Hence,

x t e et t( ) =−LNMOQP +

LNMOQP

− −12511

12511

8 4. . .

(b) See figure for graph of the amount ofsalt.

(c) The figure in part (b) indicates that theamount of salt in tank B is never as largeas the amount of salt in tank A; i.e.,

xB never exceeds xA at any t.

(d) The amount of salt in each tank goes tozero as expected, i.e., xB → 0; xA → 0.

–1

t

2

–2

0.4 0.8

x

0.2

3

0.6

xA

xB

Time, minutes


44. Student Project


6.3 Linear Systems with Nonreal Eigenvalues

!!!! Solutions in General


0 11 0

The characteristic equation for the matrix is λ2 1 0+ = , which has complex eigenvalues ±i.Plugging i into Av v= λ for λ, yields the single equation v v2 1= i . Setting v1 1= yields v2 = i .

Therefore,

α = 0 , β = 1 , p = 1 0, , q = 0 1, .

Two linearly independent solutions are then obtained.

x p q

x p q

1

2

10

01

10

01

t e t e t t t

t e t e t t t

t t

t t

( ) = − = LNMOQP −LNMOQP

( ) = + = LNMOQP +LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c ctt

ctt

( ) = + =−LNMOQP +LNMOQP1 1 2 2 1 2

cossin

sincos

or written out in component form,

x c t c ty c t c t= += − +

1 2

1 2

cos sinsin cos .

2. ′ =−− −LNMOQPx x

1 21 3

The characteristic equation for the matrix is λ λ2 4 5 0+ + = , which has complex solutions λ1 ,and λ2 2= − ± i . Plugging these values into Av v= λ yields the respective eigenvectors−( )1 1∓ i, . Therefore,

α = −2 , β = 1 , p = −( )1 1, , q = −1 0, b g .Two linearly independent solutions are then obtained.

x p q

x p q

12 2

22 2

11

10

11

10

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

b g

b g

= − =−LNMOQP −

−LNMOQP

= + =−LNMOQP +

−LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

SECTION 6.3 Linear Systems with Nonreal Eigenvalues 477


x x xt c c c et t

e tc e

t tt

tt

tb g = + =− +LNM

OQP +

− −LNM

OQP

−−

−1 1 2 2 1

22 2

2cos sincos

sin cossin

.


1 22 1

The characteristic equation for the matrix is λ λ2 2 5 0− + = , which has complex solutions λ1

and λ2 1 2= ± i . Plugging these values into Av v= λ yields the respective eigenvectors 1, ±( )i .

Therefore,

α = 1, β = 2 , p = ( )1 0, , q = ( )0 1, .


x p q

x p q

1

2

210

201

210

201

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

b g

b g

= − =LNMOQP −

LNMOQP

= + =LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c c ett

c ett

t t( ) = + =−LNM

OQP +LNMOQP1 1 2 2 1 2

22

22

cossin

sincos

.


6 15 2

The characteristic equation for the matrix is λ λ2 8 17 0− + = , which has complex solutions λ1

and λ2 4= ± i . Plugging these values into Av v= λ yields the respective eigenvectors 2 ±( )i, 5 .

Therefore,

α = 4 , β = 1 , p = ( )2, 5 , q = ( )1 0, .


x p q

x p q

14 4

24 4

25

10

25

10

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c c et t

tc e

t tt

t t( ) = + =−L

NMOQP +

+LNM

OQP1 1 2 2 1

42

42

52

5cos sin

cossin cos

sin.



1 12 1

The eigenvalues are λ1 and λ2 = ±i . Plugging these values into Av v= λ yields the respectiveeigenvectors 1 1, − ±( )i . Therefore,

α = 0 , β = 1 , p = −( )1 1, , q = ( )0 1, .


x p q

x p q

1

2

11

01

11

01

t e t e t t t

t e t e t t t

t t

t t

( ) = − =−LNMOQP −LNMOQP

( ) = + =−LNMOQP +LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c ct

t tc

tt t

b g = + =− −LNM

OQP + − +LNM

OQP1 1 2 2 1 2

coscos sin

sinsin cos

.


2 42 2

The eigenvalues are λ1 and λ2 2= ± i with corresponding eigenvectors 1 1±( )i, . Therefore,

α = 0 , β = 2 , p = ( )1 1, , q = ( )1 0, .


x p q

x p q

1

2

211

210

211

210

t e t e t t t

t e t e t t t

t t

t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c ct t

tc

t tt

( ) = + =−L

NMOQP +

+LNM

OQP1 1 2 2 1 2

2 22

2 22

cos sincos

cos sinsin

.

7. ′ =−−

LNMOQPx x

3 24 1

The eigenvalues are λ1 and λ2 1 2= ± i . Plugging these values into Av v= λ yields complexeigenvectors 1 1, ∓ i( ) . Therefore,

α = 1, β = 2 , p = ( )1 1, , q = −( )0 1, .



x p q

x p q

1

2

211

201

211

201

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

b g

b g

= − =LNMOQP − −

LNMOQP

= + =LNMOQP + −

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c c et

t tc e

tt t

t tb g = + =+

LNM

OQP + − +LNM

OQP1 1 2 2 1 2

22 2

22 2

coscos sin

sincos sin

.


2 51 2

The eigenvalues are λ1 and λ2 = ±i . Plugging these values into Av v= λ yields complexeigenvectors 2 1±( )i, . Therefore,

α = 0 , β = 1 , p = ( )2, 1 , q = ( )1 0, .

Two linearly independent solutions are the obtained.

x p q

x p q

1

2

21

10

21

10

t e t e t t t

t e t e t t t

t t

t t

( ) = − = LNMOQP −LNMOQP

( ) = + = LNMOQP +LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c ct t

tc

t tt

( ) = + =−L

NMOQP +

+LNM

OQP1 1 2 2 1 2

2 2cos sincos

cos sinsin

.


1 15 3

The eigenvalues are λ1 and λ2 1= − ± i .Plugging these values into Av v= λ yields complexeigenvectors 2 ±( )i, 5 . Therefore,

α = −1, β = 1 , p = ( )2, 5 , q = ( )1 0, .


x p q

x p q

1

2

25

10

25

10

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .



x x xt c c c et t

tc e

t tt

t t( ) = + =−L

NMOQP +

+LNM

OQP

− −1 1 2 2 1 2

25

25

cos sincos

cos sinsin

.

10. ′ =− −

−LNMOQPx x

2 33 2

The eigenvalues are λ1 and λ2 2 3= − ± i . Plugging these values into Av v= λ yields complexeigenvectors ±( )i, 1 . Therefore,

α = −2 , β = 3 , p = ( )0 1, , q = ( )1 0, .


x p q

x p q

12 2

22 2

301

310

301

310

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c c ett

c ett

t t( ) = + =−LNMOQP +

LNMOQP

− −1 1 2 2 1

22

233

33

sincos

cossin

.

11. ′ =− −

−LNMOQPx x

3 12 1

The eigenvalues are λ1 and λ2 2= − ± i . Plugging these values into Av v= λ yields complexeigenvectors − ±( )1 1, i . Therefore,

α = −2 , β = 1 , p = −( )1 1, , q = ( )0 1, .


x p q

x p q

12 2

22 2

11

01

11

01

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

( ) = − =−LNMOQP −

LNMOQP

( ) = + =−LNMOQP +

LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c c et

t tc e

tt t

t t( ) = + =−−

LNM

OQP +

−+

LNM

OQP

− −1 1 2 2 1

22

2cos

cos sinsin

sin cos.



2 42 2

The eigenvalues are λ1 and λ2 2= ± i .Plugging these values into Av v= λ yields complexeigenvectors 2, 1 − ±( )i . Therefore,

α = 0 , β = 2 , p = −( )2, 1 , q = ( )0 1, .


x p q

x p q

1

2

221

201

221

201

t e t e t t t

t e t e t t t

t t

t t

( ) = − =−LNMOQP −

LNMOQP

( ) = + =−LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


x x xt c c ct

t tc

tt t

( ) = + =− −LNM

OQP + − +LNM

OQP1 1 2 2 1 2

2 22 2

2 22 2

coscos sin

sinsin cos

.

!!!! Solutions in Particular


1 11 1

, x 011

( ) =−LNMOQP

The coefficient matrix A has eigenvalues λ λ1 2 1= = ± i and corresponding eigenvectors ±( )i, 1 .

Hence, the two linearly independent solutions obtained are

x p q

x p q

1

2

01

10

01

10

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in

x x x0 0 001

10

111 1 2 2 1 2( ) = ( ) + ( ) = LNM

OQP +LNMOQP =

−LNMOQPc c c c

yields c1 1= and c2 1= − . The solution is, therefore,

x x xt t t et tt t

t( ) = ( ) − ( ) =− −

−LNM

OQP1 2

sin coscos sin

.



0 41 0

, x 011

( ) = LNMOQP

The coefficient matrix A has eigenvalues λ λ1 2 2= = ± i and corresponding eigenvectors 2, ∓ i( ) .

Hence, the two linearly independent solutions obtained are

x p q

x p q

1

2

220

201

220

201

t e t e t t t

t e t e t t t

t t

t t

b g

b g


LNMOQP

= + =LNMOQP + −

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in

x x x0 0 020

01

111 1 2 2 1 2b g b g b g= + =

LNMOQP + −LNMOQP =LNMOQPc c c c

yields c112

= and c2 1= − . The solution is, therefore,

x x xt t tt tt tb g b g b g= − =−

+LNMM

OQPP

12

2 2 212

2 21 2

cos sinsin cos .

15. ′ =−− −LNMOQPx x

3 21 1

, x 011

( ) = LNMOQP

The coefficient matrix A has eigenvalues λ λ1 2 2= = − ± i and corresponding eigenvectors1 1∓ i, ( ) . Hence, the two linearly independent solutions obtained are

x p q

x p q

12 2

22 2

11

10

11

210

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

b g

b g

= − =LNMOQP −

−LNMOQP

= + =LNMOQP +

−LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in

x x x0 0 011

10

111 1 2 2 1 2b g b g b g= + =

LNMOQP +

−LNMOQP =LNMOQPc c c c

yields c1 1= and c2 0= . The solution is, therefore,

x xt t et t

ttb g b g= =

+LNM

OQP

−1

2 cos sincos

.



1 51 3

, x 054

( ) = LNMOQP

The coefficient matrix A has eigenvalues λ λ1 2 1= = − ± i and corresponding eigenvectors5 2, ∓ i( ) . The two linearly independent solutions obtained are

x p q

x p q

1

2

52

01

52

01

t e t e t e t e t

t e t e t e t e t

t t t t

t t t t

b g

b g


LNMOQP

= + =LNMOQP + −

LNMOQP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in

x x x0 0 052

01

541 1 2 2 1 2b g b g b g= + =

LNMOQP + −LNMOQP =LNMOQPc c c c

yields c1 1= and c2 2= − . The solution is, therefore,

x x xt t t et

t te

tt t

et t

tt t t( ) = ( ) − ( ) =

−LNM

OQP − −LNM

OQP =

−−

LNM

OQP

− − −1 22

52

25

25 10

5cos

cos sinsin

sin coscos sin

sin.

!!!! 3 3× System

17. ′ =−

−

L

NMMM

O

QPPP

x x1 0 00 0 20 2 0

(a) Solving the characteristic equation

− −−− −

L

NMMM

O

QPPP= − +( ) + =

1 0 00 20 2

1 4 02λ

λλ

λ λb g ,

which has roots λ1 1= − , λ2 , λ3 2= ± i .

(b) Solving for x, y, z in the equation

−

−

L

NMMM

O

QPPP

L

NMMM

O

QPPP= −L

NMMM

O

QPPP

1 0 00 0 20 2 0

1xyz

xyz

yields x = α , y = 0 , z = 0 , α arbitrary, so the eigenvector corresponding to λ1 1= − is

x1

100

=L

NMMM

O

QPPP

−e t .


(c) Solving for x, y, z in the system

−

−

L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP

1 0 00 0 20 2 0

2xyz

ixyz

yields the eigenvector 0 1, , i( ) . Hence, we identify

α = 0 , β = 2 , p = ( )0 1 0, , , q = ( )0 0 1, , .

The two linearly independent solutions obtained are

x p q

x p q

2

3

2010

2001

2010

2001

t e t e t t t

t e t e t t t

t t

t t

( ) = − =L

NMMM

O

QPPP−L

NMMM

O

QPPP

( ) = + =L

NMMM

O

QPPP+L

NMMM

O

QPPP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

(d) Writing out the general solution

x x x xt c c c( ) = + +1 1 2 2 3 3

in scalar form, yields

x t c ey t c t c tz t c t c t

t( ) =( ) = +( ) = −

−1

2 3

3 2

2 22 2

cos sincos sin .

(e) Substituting the IC

x x x x0 0 0 0100

010

001

101

1 1 2 2 3 3 1 2 3( ) = ( ) + ( ) + ( ) =L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP=L

NMMM

O

QPPP

c c c c c c

yields c1 1= , c2 0= , and c3 1= . The solution of the IVP is, therefore,

x x xt t t e tt

t( ) = ( ) + ( ) =L

NMMM

O

QPPP+L

NMMM

O

QPPP

−1 3

100

022

sincos

or in coordinate form

x t ey t tz t t

t( ) =( ) =( ) =

−

sincos .

22


(f) The trajectory of

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

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

x

y

z

!!!! Threefold Solutions

18. ′ =−L

NMMM

O

QPPP

x x1 0 10 2 01 0 1

The characteristic polynomial is given by

− + − + = − −( ) − +λ λ λ λ λ λ3 2 24 6 4 2 2 2b g .Hence, the eigenvalues are λ1 2= , λ2 and λ3 1= ± i . Substituting 2 and 1+ i into the equation

Av v= λ yields the eigenvectors associated with each eigenvalue. Doing this, yields

λλ

1 1

2 2

2 0 1 01 0 1

= ⇒ == + ⇒ =

vv

, ,, , .

i i

Therefore,

α = 1 , β = 1, p = ( )0 0 1, , , q = ( )1 0 0, , .

The three independent solutions are

x

x p q

x p q

12

2

3

010

001

100

001

100

t e

t e t e t e t e t

t e t e t e t e t

t

t t t t

t t t t

( ) =L

NMMM

O

QPPP

( ) = − =L

NMMM

O

QPPP−

L

NMMM

O

QPPP

( ) = + =L

NMMM

O

QPPP+

L

NMMM

O

QPPP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .



x t c e c et

tc e

t

t

e c t c tc e

e c t c t

t t t

t

t

t( ) =

L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+L

NMMM

O

QPPP=

−

+

L

NMMM

O

QPPP

12

2 3

3 2

12

2 3

010

0 0sin

cos

cos

sin

cos sin

cos sin

a f

a f

19. ′ =−

L

NMMM

O

QPPP

x x0 1 00 0 11 0 0

The characteristic polynomial of this system is

−−

− −

L

NMMM

O

QPPP= − + =

λλ

λλ

1 00 11 0

1 03b g

with eigenvalues λ1 1= − , λ2 and λ312

32

= ± i . The eigenvectors corresponding to these eigen-

values are

λ

λ

1 1

2 2

1 1 1 1

12

32

1 3 1 3 2

= − ⇒ = −

= + ⇒ = − − −

v

v

, ,

, , .

i i i

Therefore,

α =12

, β = 32

, p = −1 1, , 2b g , q = − −3 3 0, , e j .

Hence, the general solution

x t c e c e

t t

t t

t

c e

t t

t t

t

t t t( ) = −L

NMMM

O

QPPP+

− +

+

L

N

MMMMMMM

O

Q

PPPPPPP

+

− −

−

L

N

MMMMMMM

O

Q

PPPPPPP

−1 2

23

2111

32

3 32

32

3 32

2 32

32

3 32

32

3 32

2 32

cos sin

cos sin

cos

sin cos

sin cos

sin

.

20. ′ = −L

NMMM

O

QPPP

x x1 0 02 1 23 2 1

The characteristic polynomial is given by

− + − + = − −( ) − +λ λ λ λ λ λ3 2 23 7 5 1 2 5b g .


Hence, the eigenvalues are λ1 1= and λ λ2 3 1 2= = ± i . Substituting 1 and 1 2+ i into the

equation Av v= λ yields the eigenvectors associated with each eigenvalue. Doing this, yields

λλ

1 1

2 2

1 2, 3 21 2 0 1

= ⇒ = −= + ⇒ = −

vv

,, , .i i

Therefore,

α = 1 , β = 2 , p = ( )0 1 0, , , q = −( )0 0, , 1 .

Hence the general solution is

x t c e c e tt

c e tt

t t t( ) = −L

NMMM

O

QPPP+L

NMMM

O

QPPP+

−

L

NMMM

O

QPPP

1 2 3

232

022

022

cossin

sincos

.

21. ′ =− −

− −L

NMMM

O

QPPP

x x3 1 20 1 12 0 0

The eigenvalues are

λ1 2= − and λ λ2 3 1 2= = − ± i .

Substituting –2 and − +1 2i into the equation Av v= λ yields the eigenvectors associ-

ated with each eigenvalue. Doing this, yields

λ

λ1 1

2 2

2 1 1 1

1 2 2 2 2, 2 2

= − ⇒ = −

= − + ⇒ = + −

v

v

, ,

, .

i i i

Therefore,

α = −1 , β = 2 , p = ( )2, 2, 0 , q = −2 0 2, , 2d hand so three independent solutions are

x

x p q

x p q

12

2

3

111

2220

2 2102

2220

2 2102

t e

t e t e t e t e t

t e t e t e t e t

t

t t t t

t t t t

( ) =−L

NMMM

O

QPPP

( ) = − =L

NMMM

O

QPPP−

−

L

NMMM

O

QPPP

( ) = + =L

NMMM

O

QPPP+

−

L

NMMM

O

QPPP

−

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .


Hence,

x t c e c et t

tt

c et t

tt

t t t( ) =−L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+

+

−

L

NMMM

O

QPPP

− − −1

22 3

111

2 2 2 22 2

2 2 2

2 2 2 22 2

2 2 2

cos sincossin

sin cossincos

.

!!!! Triple IVPs

22. ′ =−

− −−

L

NMMM

O

QPPP

x x3 0 10 3 10 2 1

, x 05

1326

( ) =−

−

L

NMMM

O

QPPP

The eigenvalues and vectors of the coefficient matrix are

λ1 13 1 0 0= ⇒ =v , ,

and

λ λ λ2 3 2 22 5 13 13, ; , ,= − ± ⇒ = + +i i i - 26v

with α = −2 , β = 1, #p = −5 13, , 26 , #q = 1, , 13 0 . The three independent solutions are

x

x p q

x p q

13

22 2

32 2

100

51326

1130

51326

1130

t e

t e t e t e t e t

t e t e t e t e t

t

t t t t

t t t t

( ) =L

NMMM

O

QPPP

( ) = − = −L

NMMM

O

QPPP−

L

NMMM

O

QPPP

( ) = + = −L

NMMM

O

QPPP+

L

NMMM

O

QPPP

− −

− −

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in the initial conditions

x 0100

51326

1130

51326

1 2 3( ) =L

NMMM

O

QPPP+ −L

NMMM

O

QPPP+L

NMMM

O

QPPP=

−

−

L

NMMM

O

QPPP

c c c

yields c1 0= , c2 1= − , and c3 0= . The solution of the IVP is

x x= − = −−

− −L

NMMM

O

QPPP

−2

2

513 13

26e

t tt t

t

tcos sin

cos sincos


23. ′ = − −L

NMMM

O

QPPP

x x0 1 01 0 10 1 0

, x 0011

( ) =L

NMMM

O

QPPP

The eigenvalues and vectors of the coefficient matrix are

λ1 10 1 0 1= ⇒ = −v , ,

and

λ λ2 3 2 32 1 2 1, , ,,= ± ⇒ = ±i iv ,

therefore, three independent solutions are

x

x p q

x p q

1

2

3

101

2101

2 2010

2101

2 2010

t

t e t e t t t

t e t e t t t

t t

t t

( ) =−L

NMMM

O

QPPP

( ) = − =L

NMMM

O

QPPP−

L

NMMM

O

QPPP

( ) = + =L

NMMM

O

QPPP+

L

NMMM

O

QPPP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos .

Plugging in the initial conditions

x 0101

101

020

011

1 2 3( ) =−L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP=L

NMMM

O

QPPP

c c c

yields c c1 212

= = and c32

2= . Hence, the solution of the IVP is

x x x x= + + =−L

NMMM

O

QPPP+ −

L

NMMM

O

QPPP+

L

NMMM

O

QPPP

=−L

NMMM

O

QPPP+

L

NMMM

O

QPPP+ −

L

NMMM

O

QPPP

12

12

22

12

101

12

22 2

2

22

22 2

2

12

101

12

2121

22

2111

1 2 3

cossincos

sincossin

cos sin .

ttt

ttt

t t

!!!! Matter of Independence

24. The Wronskian of two vector functions is defined as the determinant of the matrix formed byplacing the vectors as columns in the matrix. If the vector functions are also solutions of a linearsystem 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 twovector solutions

x

x

11

2

1

2

21

2

1

2

t e taa

e tbb

t e taa

e tbb

t t

t t

( ) = LNMOQP −

LNMOQP

( ) = LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin

sin cos

formed from the eigenvalues α β± i and eigenvectors p = a a1 2, a f, q = b b1 2, a f of a matrix. Weevaluate x1 t( ) , x2 t( ) when t = 0, yielding

x1 1 2t a a( ) = , a f, x2 1 2t b b( ) = , a f .Hence, the Wronskian of x1 t( ) and x2 t( ) at t = 0 is

W x x1 21 1

2 20, ( ) =

a ba b

.

But the columns of this matrix are linearly independent and thus the Wronskian is nonzero.Hence, the vectors x1 t( ) and x2 t( ) are linearly independent vector functions.

!!!! Matrix Exponential

25.ddt

e ddt

t t t t t t t

t t e

t

t

A

A

I A A A A A A A A

A I A A A

= + + + + +FHG

IKJ = + + + +

= + + +FHG

IKJ =

22

33

44 2

23

34

22

2 3 4 2 3

2

! ! ! ! !

!

$ $

$

26. We differentiate x cAt et( ) = , obtaining ′( ) =x A cAt et . Plugging x x, ′ into ′ =x Ax , yields the

identity A c A cA Ae et t= .

27. Given

A = LNMOQP

0 11 0

and by computing the powers, yields

A2 1 00 1

= LNMOQP , A A3 = , A I4 = , .$

In general, we use the rule

A I2n = , A A2 1n+ = .


Hence, the matrix exponential is

e t t tt t t t t

t t t t t

t tt t

tA I= + LNMOQP +LNMOQP +LNMOQP+ =

+ + + + + +

+ + + + + +

L

N

MMM

O

Q

PPP

= LNMOQP

0 11 0 2

1 00 1 3

0 11 0

12 4 3 5

3 51

2 4

2 32 4 3 5

3 5 2 4$$ $

$ $! ! !

! ! !cosh sinhsinh cosh

.

The general solution to x Ax= can then be written as

x tt tt t

cc

ctt

ctt

( ) = LNMOQPLNMOQP =LNMOQP +LNMOQP

cosh sinhsinh cosh

coshsinh

sinhcosh

1

21 2 .

This could be written as a linear combination of vectors involving et and e t− because

cosh t e et t=

+ −

2, sinh t e et t

=− −

2.

28. Given

A =−LNMOQP

0 11 0

and by computing the powers, yields

A I

A AA I

2

3

4

1 00 1

=−

−LNM

OQP = −

= −

=

and so on. Hence, the matrix exponential is

e t t tt t t t t

t t t t t

t tt t

tA I= +−LNMOQP +

−−

LNM

OQP +

−LNMOQP+ =

− + − − + −

− + − + − + −

L

N

MMMM

O

Q

PPPP=

−LNM

OQP

0 11 0 2

1 00 1 3

0 11 0

12 4 3 5

3 51

2 4

2 3

2 4 3 5

3 5 2 4!! ! !

! ! !cos sinsin cos

.

$$ $

$ $

The general solution can then be written as

x tt tt t

cc

ctt

ctt

b g =−LNM

OQPLNMOQP = −LNMOQP +LNMOQP

cos sinsin cos

cossin

sincos

1

21 2 .


29. A =− −LNM

OQP

3 52 4

The characteristic equation is given by λ λ2 2 0+ − = . Thus, by the Cayley-Hamilton theorem,

A A I2 2 0+ − = .

Multiplying by A yields the equation

A A A3 2 2 0+ − = .

Hence, we solve for A2 and A3 to get

A A IA A A

2

3 2

22

= − +

= − + .

Using these equations, we find

A

A

2

3

3 52 4

21 00 1

1 52 6

1 52 6

23 52 4

7 156 14

= −− −LNM

OQP +LNMOQP =

− −LNM

OQP

= −− −LNM

OQP + − −LNM

OQP = − −LNM

OQP.

Hence, we approximate the matrix exponential as

e t t ttA I= +− −LNM

OQP +

− −LNM

OQP + − −LNM

OQP+

3 52 4 2

1 52 6 3

7 156 14

2 3

!….

!!!! Skew-Symmetric Systems


00k

k

We solve this using eigenvalues and vectors. The characteristic equation of the coefficient matrixis

pk

kkλ

λλ

λ( ) =−− −LNM

OQP = + =2 2 0,

which has roots λ λ1 2= = ±ik . Plugging in iv into the equation

00

1

2

1

2

kk

ik−LNMOQPLNMOQP =LNMOQP

vv

vv

gives k ikv v2 1= . Setting v1 1= yields v2 = i . Hence, one of the conjugate eigenvectors is

v = LNMOQP =LNMOQP +LNMOQP

1 10

01i

i .


We then identify

α = 0, β = k , p = ( )1 0, , q = ( )0, 1 .

The two linearly independent vector solutions are then

x p q

x p q

1

2

10

01

10

01

t e t e t kt kt

t e t e t kt kt

t t

t t

( ) = − = LNMOQP −

LNMOQP

( ) = + = LNMOQP +

LNMOQP

α α

α α

β β

β β

cos sin cos sin

sin cos sin cos


x x x= + =−LNMOQP +LNMOQPc c c

ktkt

cktkt1 1 2 2 1 2

cossin

sincos

.

In component form

x c kt c kty c kt c kt= += − +

1 2

1 2

cos sinsin cos .

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

′′ + =x k x2 0

whose general solution is

x C xt= −( )cos δ .

We then find

yk

x C kt= ′ = − −( )1 sin δ .

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

x t y t C kt C kt C kt kt C2 2 2 2 2 2 2 2 2 2( ) + ( ) = −( ) + −( ) = −( ) + −( ) =cos sin cos sinδ δ δ δ .


!!!! Computer Lab: Phase Portrait


0 15 2

, x 022

( ) = LNMOQP

The solution of the IVP is shown in the phaseplane.

See figures for plot of the x-coordinateand the y-coordinate as a function of t. Note thatthese graphs are consistent with the solution inthe phase plane.

3–3x

–3

3y

(2, 2)

Phase plane solution

–1

1

t

2

–2

32 4 5

x

1

–3

3

–1

1

t

2

–2

32 4 5

y

1

–3

3


4 55 4

, x 022

( ) =−LNMOQP .

The solution of the IVP in the phase plane isshown.

Note that the graphs of x t( ) and y t( ) ver-

sus t are consistent with the phase plane graph.

8–8

–8

8y

x

–4

4

t

8

–8

32 4 51

x t( )

–4

4

t

8

–8

32 4 51

y t( )


!!!! Coupled Mass-Spring System

33. We find the eigenvalues and vectors of the matrix

0 1 0 00 0

0 0 0 10 0

1 2

1

2

1

2

2

2 3

2

−+

−+

L

N

MMMMM

O

Q

PPPPP

k km

km

km

k km

in terms of the k1, k2 , k3 , m1, m2 , but this is extremely involved so we let the parameter equalk k k m m1 2 3 1 2 1= = = = = to get

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

−

−

L

N

MMMM

O

Q

PPPP.

Finding the eigenvalues using the computer algebra system Maple, yields the purely complexnumbers λ λ1 2= = ±i and λ λ3 4 3= = ±i with corresponding vectors

λ

λ1 1

2 2

1 1

3 1 3 1 3

= ⇒ = − −

= ⇒ = − −

i i i

i i i

v

v

, , ,

, , , .

Hence, the eigenvalues are α β± i , where α = 0, β = 1 and α = 0, β = 3 . The corresponding

eigenvectors are

α β

α β

+ =

=

−

−

L

N

MMMM

O

Q

PPPP= + =

L

N

MMMM

O

Q

PPPP+

−

−

L

N

MMMM

O

Q

PPPP+ =

=

−−L

N

MMMM

O

Q

PPPP= + =

−L

N

MMMM

O

Q

PPPP+

−L

N

MMMM

O

Q

PPPP

i ii

ii i

i i

i

i

i i

v p q

v p q

1

1

0101

1010

31313

1010

0303

.


The four linearly independent solutions are then

x p q

x p q

x p q

1

2

3

0101

1010

0101

1010

3

1010

3

0303

t e t e t t t

t e t e t t t

t e t e t t t

t t

t t

t t

b g

b g

b g

= − =

L

N

MMMM

O

Q

PPPP−

−

−

L

N

MMMM

O

Q

PPPP

= + =

L

N

MMMM

O

Q

PPPP+

−

−

L

N

MMMM

O

Q

PPPP

= − =

−

−

L

N

MMMM

O

Q

PPPP−

−L

N

MMM

α α

α α

α α

β β

β β

β β

cos sin cos sin

sin cos sin cos

cos sin cos sinM

O

Q

PPPP

=

−L

N

MMMM

O

Q

PPPP+

−L

N

MMMM

O

Q

PPPPx4 3

1010

3

0303

t t tb g sin cos


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

Plugging this into the initial conditions x1 0 0( ) = , x2 0 0( ) = , x3 0( ) = d , and x4 0 0( ) = we getc1 0= , c2 1= − , c3 1= , and c4 0= . Finally, because x x= 1, and y = x3, we have the desired result

x t t t

y t t t

( ) = −

( ) = +

cos cos

cos cos .

3

3


34. Student Project

SECTION 6.4 Uncoupling a Linear DE System 497

6.4 Uncoupling a Linear DE System

!!!! Uncoupling Homogeneous Linear Systems

1. ′ =− −−LNM

OQPx x

1 22 2

The coefficient matrix has eigenvalue and eigenvectors

λλ

1 1

2 2

3 1 22 2, 1

= ⇒ = −= − ⇒ =

vv

,.

Matrices of eigenvectors are

P =−LNMOQP

1 22 1

and P− =−L

NMOQP

1 15

1 22 1

.

Therefore,

P AP− =−L

NMOQP− −−LNM

OQP −LNMOQP = −LNMOQP

1 15

1 22 1

1 22 2

1 22 1

3 00 2

.

Hence, transforming from x to the new variables

w P x= −1

yields the uncoupled system

′ =′ = −

w ww w

1 1

2 2

32 .

Solving this uncoupled system yields w1 13t c e t( ) = and w2 2

2t c e t( ) = − . The solution of the

original system is

x Pwt tc e

c ec e c e

t

tt t( ) = ( ) =

−LNMOQPLNMOQP = −

LNMOQP +

LNMOQP−

−1 22 1

12

21

13

22 1

32

2 .

2. ′ =−

−LNMOQPx x

0 13 2


λλ

1 1

2 2

3 1 31 1 1

= ⇒ = −= − ⇒ =

vv

,, .


P =−LNMOQP

1 13 1

and P− =−L

NMOQP

1 14

1 13 1

.


Therefore,

P AP− =−L

NMOQP

−−LNM

OQP −LNMOQP = −LNMOQP

1 14

1 13 1

0 13 2

1 13 1

3 00 1

.

Hence, transforming from x to the new variables w P x= −1 yields the uncoupled system ′ =w w1 13and ′ = −w w2 2 . Solving this uncoupled system yields w1 1

3t c e t( ) = and w2 2t c e t( ) = − . The solution

of the original system is

x Pwtc ec e

c e c et

tt t( ) = =

−LNMOQPLNMOQP = −LNMOQP +

LNMOQP−

−1 13 1

13

11

13

21

32 .

3. ′ =−

−LNMOQPx x

0 11 0


λλ

1 1

2 2

1 1 11 1 1

= ⇒ = −= − ⇒ =

vv

,, .


P =−LNMOQP

1 11 1

and P− =−LNMOQP

1 12

1 11 1

.

Therefore,

P AP− =−LNMOQP

−−LNM

OQP−LNMOQP = −LNMOQP

1 12

1 11 1

0 11 0

1 11 1

1 00 1

.

Hence, transforming from x to the new variables w P x= −1 yields the decoupled system ′ =w w1 1

and ′ = −w w2 2 . Solving this decoupled system yields w1 1t c et( ) = and w2 2t c e t( ) = − . The solution

of the original system is

x Pwt tc e

c ec e c e

t

tt t( ) = ( ) =

−LNMOQPLNMOQP =

−LNMOQP +

LNMOQP−

−1 11 1

11

11

1

21 2 .

4. ′ = LNMOQPx x

2 31 4


λλ

1 1

2 2

1 3 15 1 1

= ⇒ = −= ⇒ =

vv

,, .


P =−LNMOQP

3 11 1

and P− =−LNMOQP

1 14

1 11 3

.


Therefore,

P AP− =−LNMOQPLNMOQP−LNMOQP =LNMOQP

1 14

1 11 3

2 31 4

3 11 1

1 00 5

.

Hence, transforming from x to the new variables w P x= −1 yields the uncoupled system ′ =w w1 1

and ′ =w w2 25 . Solving this uncoupled system yields w1 1t c et( ) = and w2 25t c e t( ) = . Hence, the

solution of the original system is

x Pwt tc e

c ec e c e

t

tt t( ) = ( ) =

−LNMOQPLNMOQP =

−LNMOQP +LNMOQP

3 11 1

31

11

1

25 1 2

5 .


2 32 5


λλ

1 1

2 2

1 3 14 1 2

= ⇒ == − ⇒ =

vv

,, .


P = LNMOQP

3 11 2

and P− =−

−LNM

OQP

1 15

2 11 3

.

Therefore,

P AP− =−

−LNM

OQP

−−

LNMOQPLNMOQP = −LNMOQP

1 15

2 11 3

2 32 5

3 11 2

1 00 4

.

Hence, transforming from x to the new variables w P x= −1 yields the decoupled system ′ =w w1 1

and ′ = −w w2 24 . Solving this decoupled system yields w1 1t c et( ) = and w2 24t c e t( ) = − . Hence, the

solution of the original system is

x Pwt tc e

c ec e c e

t

tt t( ) = ( ) = LNM

OQPLNMOQP =LNMOQP +

LNMOQP−

−3 11 2

31

12

1

24 1 2

4 .

6. ′ = LNMOQPx x

0 11 0


λλ

1 1

2 2

1 1 11 1 1

= − ⇒ = −= ⇒ =

vv

,, .


P =−LNMOQP

1 11 1

and P− =−LNMOQP

1 12

1 11 1

.


Therefore,

P AP− =−LNMOQPLNMOQP−LNMOQP =

−LNMOQP

1 12

1 11 1

0 11 0

1 11 1

1 00 1

.

Hence, transforming from x to the new variables w P x= −1 yields the uncoupled system′ = −w w1 1 and ′ =w w2 2. Solving this uncoupled system yields w1 1t c e t( ) = − and w2 2t c et( ) = .

Hence, the solution of the original system is

x Pwt tc ec e

c e c et

tt t( ) = ( ) =

−LNMOQPLNMOQP =

−LNMOQP +LNMOQP

−−

1 11 1

11

11

1

21 2 .

7. ′ =L

NMMM

O

QPPP

x x1 1 11 1 11 1 1


λλλ

1 1

2 2

3 3

3 1 1 10 1 1 00 1 0 1

= ⇒ == ⇒ = −= ⇒ = −

vvv

, ,, ,, , .


P =− −L

NMMM

O

QPPP

1 1 11 1 01 0 1

and P− = = − −− −

L

NMMM

O

QPPP

1 13

1 1 11 2 11 1 2

.

Therefore,

P AP− = = − −− −

L

NMMM

O

QPPP

L

NMMM

O

QPPP

− −L

NMMM

O

QPPP=L

NMMM

O

QPPP

1 13

1 1 11 2 11 1 2

1 1 11 1 11 1 1

1 1 11 1 01 0 1

3 0 00 0 00 0 0

.

Hence, transforming from x to the new variables w P x= −1 yields the decoupled system′ =w w1 13 , ′ =w2 0, and ′ =w3 0. Solving this uncoupled system yields w1 1

3t c e t( ) = , w2 2t c( ) = ,and w3 3t c( ) = . The solution of the original system is

x Pwt tc e

cc

c e c c

t

t( ) = ( ) =− −L

NMMM

O

QPPP

L

NMMM

O

QPPP=L

NMMM

O

QPPP+

−L

NMMM

O

QPPP+

−L

NMMM

O

QPPP

1 1 11 1 01 0 1

111

110

101

13

2

3

13

2 3 .

In scalar form

x = − −

= +

= +

c e c cy c e cz c e c

t

t

t

13

2 3

13

2

13

3 .


8. ′ =L

NMMM

O

QPPP

x x0 0 00 1 01 0 1


λλλ

1 1

2 2

3 3

0 1 0 11 0 0 11 0 1 0

= ⇒ = −= ⇒ == ⇒ =

vvv

, ,, ,, , .


P =−L

NMMM

O

QPPP

1 0 00 0 11 1 0

and P− =−L

NMMM

O

QPPP

11 0 01 0 10 1 0

.

Therefore,

P AP− = =−L

NMMM

O

QPPP

L

NMMM

O

QPPP

−L

NMMM

O

QPPP=L

NMMM

O

QPPP

1 13

1 0 01 0 10 1 0

0 0 00 1 01 0 1

1 0 00 0 11 1 0

0 0 00 1 00 0 1

.

Hence, transforming from x to the new variables w P x= −1 yields the decoupled system ′ =w1 0,′ =w w2 2, and ′ =w w3 3. Solving this decoupled system yields w1 1t c( ) = , w2 2t c et( ) = , and

w3 3t c et( ) = . Hence, the solution of the original system is

x Pwt tc

c ec e

c c e c et

t

t t( ) = ( ) =−L

NMMM

O

QPPP

L

NMMM

O

QPPP=

−L

NMMM

O

QPPP+L

NMMM

O

QPPP+L

NMMM

O

QPPP

1 0 00 0 11 1 0

101

001

010

1

2

3

1 2 3 .

x1 1= −c , x2 3= c et , and x3 1 2= +c c et .

!!!! Uncoupling Nonhomogeneous Linear Systems

9. ′ = LNMOQP +LNMOQPx x

0 11 0

11

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

form the matrices

P =−LNMOQP

1 11 1

and P− =−LNMOQP

1 12

1 11 1

.

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

′ =−

LNMOQP +

−LNMOQPLNMOQPw w

1 00 1

12

1 11 1

11


or ′ = +w w1 1 1 and ′ = −w w2 2 . Solving these, yields w1 1 1t c et( ) = − and w2 2t c e t( ) = − . Thus

x Pwt tc e

c ec e c e

t

tt t( ) = ( ) =

−LNMOQP

−LNMOQP =LNMOQP +

−LNMOQP +

−−LNMOQP−

−1 11 1

1 11

11

11

1

21 2 .

10. ′ =−

−LNMOQP +LNMOQPx x

3 11 3 0

sin t

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

form the matrices

P =−LNMOQP

1 11 1

and P− =−LNMOQP

1 12

1 11 1

.


′ =−

−LNM

OQP +

−LNMOQPLNMOQPw w

2 00 4

12

1 11 1 0

sin t

or

′ = − +′ = −

w ww w

1 1

2 2

24

sin.

t

Solving these, yields

w

w

1 12

2 24

15

25

t c e t t

t c e

t

t

( ) = − +

( ) =

−

−

cos sin

.

Thus

x Pwt t c e t t

c e

c e c et tt t

t

t

t t

( ) = ( ) =−L

NMOQP

− +LNMM

OQPP

= LNMOQP +

−LNMOQP +

− +− +LNM

OQP

−

−

− −

1 11 1

15

25

11

11

15

22

12

24

12

24

cos sin

cos sincos sin

.


1 11 1 1

t.

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

form the matrices

P =−LNMOQP

1 11 1

and P− =−L

NMOQP

1 12

1 11 1

.


′ = LNMOQP +

−LNMOQPLNMOQPw w

0 00 2

12

1 11 1 1

t


or

′ = −( )

′ = + +( )

w

w w

1

2 2

12

1

2 12

1

t

t .


w

w

12

1

2 22

14 2

438

t t t c

t c e tt

( ) = − +

( ) = − − .

Thus

x Pwt t

t t c

c e t c c e

t t

t tt

t( ) = ( ) =−LNMOQP

− +

− −

L

N

MMM

O

Q

PPP=

−LNMOQP +

LNMOQP +

− −

− + −

L

N

MMM

O

Q

PPP1 11 1

4 2

438

11

11

434

38

4 438

2

1

22

1 22

2

2 .


5 41 2

50t

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

form the matrices

P =−LNMOQP

4 11 1

and P− =−LNMOQP

1 15

1 11 4

.


′ = LNMOQP +

−LNMOQPLNMOQPw w

6 00 1

15

1 11 4

50t

or

′ = +′ = −

w ww w

1 1

2 2

6 tt .


w

w

1 16

2 2

6136

1

t c e t

t c e t

t

t

( ) = − −

( ) = + + .


Thus

x Pwt t c e t

c e tc e c e

t

tt

tt t( ) = ( ) =

−LNMOQP

− −

+ +

LNMM

OQPP =

LNMOQP +

−LNMOQP +

− −

+

L

NMMM

O

QPPP

4 11 1 6

136

1

41

11

53

109

56

3536

16

21

62 .

13. ′ =−−

L

NMMM

O

QPPP+L

NMMM

O

QPPP

x x4 1 12 5 21 1 2

1

2t

t

We use Maple to first find the eigenvalues and eigenvectors, yielding

λλλ

1 1

2 2

3 3

5 1 2, 13 0 1 13 1 0 1

= ⇒ == ⇒ == ⇒ =

vvv

,, ,, , .

Hence,

P =L

NMMM

O

QPPP

1 0 12 1 01 1 1

.

We then find

P− =−

−−

L

NMMM

O

QPPP

1 12

1 1 12 0 21 1 1

so

P AP− =L

NMMM

O

QPPP

15 0 00 3 00 0 3

and

P−L

NMMM

O

QPPP=

− + +−

− +

L

NMMM

O

QPPP

1

2

2

2

2

112

12 2

1t

t

t tt

t t.

The decoupled system is

′ = + −w Dw P f1


or

′′′

L

NMMM

O

QPPP=L

NMMM

O

QPPP

L

NMMM

O

QPPP+

− + +−

− +

L

NMMM

O

QPPP

www

www

1

2

3

1

2

3

2

2

2

5 0 00 3 00 0 3

12

12 2

1

t tt

t t.

Solving these three equations individually yields

w

w

w

1 15

2

2 23

2

3 33

2

10350

14125

329

727

6 184

27

t c e t t

t c e t t

t c e t t

t

t

t

( ) = + − −FHG

IKJ

( ) = + − − +FHG

IKJ

( ) = + − + −FHG

IKJ .

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

xxx

www

w ww w

w w w

1

2

3

1

2

3

1 3

1 2

1 2 3

1 0 12 1 01 1 1

2L

NMMM

O

QPPP=L

NMMM

O

QPPP

L

NMMM

O

QPPP=

++

+ +

L

NMMM

O

QPPP

,

x

x

x

1 15

33

2

2 15

23

2

3 15

2 33

2

15 2258783375

2 215

77225

1193375

25

1775

11125

= + − + +FHG

IKJ

= + − + −FHG

IKJ

= + + − + +FHG

IKJ

c e c e t t

c e c e t t

c e c c e t t

t t

t t

t ta f .

14. ′ =

L

N

MMMM

O

Q

PPPP+−

L

N

MMMM

O

Q

PPPPx x

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

0

1

t

t

We use Maple to first find the eigenvalues and eigenvectors, yielding

λλλλ

1 1

2 2

3 3

4 4

1 0 1 0 11 1 0 1 0

1 0 1 0 11 1 0 1 0

= ⇒ == ⇒ == − ⇒ = −= − ⇒ = −

vv

vv

, , ,, , ,

, , ,, , , .


Hence,

P =

−−

L

N

MMMM

O

Q

PPPP

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

.

Then find

P PT− = =−

−

L

N

MMMM

O

Q

PPPP1 1

2

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

and so

P AP− =−

−

L

N

MMMM

O

Q

PPPP1

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

and

P−−

L

N

MMMM

O

Q

PPPP=

−

L

N

MMMM

O

Q

PPPP1 0

1

12

101

2

t

tt

.

Hence, the decoupled system is

′ = + −w Dw P f1

or

′′′′

L

N

MMMM

O

Q

PPPP=

−−

L

N

MMMM

O

Q

PPPP

L

N

MMMM

O

Q

PPPP+

−

L

N

MMMM

O

Q

PPPP

wwww

wwww

1

2

3

4

1

2

3

4

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

12

101

2t

.

Solving these three equations individually yields

w

w

w

w

1 1

2 2

3 3

4 4

12

12

1

t c e

t c e

t c e

t c e t

t

t

t

t

( ) = −

( ) =

( ) = +

( ) = − +

−

− .


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

xxxx

wwww

w ww ww ww w

1

2

3

4

1

2

3

4

2 4

1 3

2 4

1 3

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

L

N

MMMM

O

Q

PPPP=

−−

L

N

MMMM

O

Q

PPPP

L

N

MMMM

O

Q

PPPP=

−−++

L

N

MMMM

O

Q

PPPP,

xxxx

1 2 4

2 1 3

3 2 4

4 1 3

11

1

= − + −

= − −

= + − +

= +

−

−

−

−

c e c e tc e c ec e c e tc e c e

t t

t t

t t

t t .

!!!! Working Backwards

15. Given eigenvalues are 1 and –1 and respective eigenvectors are 1 1, ( ) and 1 2, ( ) , we form the

matrices

P = LNMOQP

1 11 2

and P− =−

−LNMOQP

1 2 11 1

and then form the diagonal matrix

D =−LNMOQP

1 00 1

,

whose diagonal elements are the eigenvalues. We then use the relation D P AP= −1 , premultiplyby P, and postmultiply by P−1, yielding

A PDP= = LNMOQP −LNMOQP

−LNMOQP =

−−

LNMOQP

−1 1 11 2

1 00 1

2 11 2

3 24 3

.

!!!! Jordan Form

16. (a) The system

′ =−LNMOQPx x

2 11 4

has a soluble eigenvalue of 3 and only one independent eigenvector v = 1 1, . We findthe generalized eigenvector w that satisfies the equations A I w v−( ) =3 , or

−−LNMOQPLNMOQP =LNMOQP

1 11 1

11

1

2

ww

or − + =w w1 2 1, which yields w1 1= and w2 2= . Hence, w = 1 2, . We now form the

matrix


P v w= = LNMOQP

1 11 2

and compute

P

P AP

−

−

=−

−LNM

OQP

=−

−LNM

OQP −LNMOQPLNMOQP =LNMOQP

1

1

2 11 12 11 1

2 11 4

1 11 2

3 10 3

.

(b) Transforming from x to the new variables u P x= −1 yields the new system u u u21 13′ = +

and u u2 23′ = . Solving this system yields

u2 23t c e t( ) = , u1 1

32

3t c e c tet tb g = + .

The solution of the original system is

x Put tc tc e

c ee

c c tcc c tc

t

tt( ) = ( ) = LNM

OQP

+LNM

OQP =

+ ++ +LNM

OQP

1 11 2 2

1 23

23

3 1 2 2

1 2 2

a f.


17. Student Project

SECTION 6.5 Stability and Linear Classification 509

6.5 Stability and Linear Classification

!!!! Classification Verification


1 14 2

(saddle point)

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


0 11 0

(center)

The matrix has eigenvalues ±i. Because the real part is zero, the origin is stable, but notasymptotically stable at equilibrium point. The origin is a center.

3. ′ =−

−LNM

OQPx x

2 00 2

(star node)

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

4. ′ =−

−LNM

OQPx x

2 10 2

(degenerate node)

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

5. ′ = LNMOQPx x

2 13 4

(node)

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


0 11 1

(spiral sink)

The matrix has eigenvalues − ±12

32

i . Because the real part of the eigenvalues is negative, the

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


!!!! Undamped Spring

7. x x+ =ω 02 0

Denote x x1 = and x x2 = ! ; the equation becomes

!!xx

xx

1

2 02

1

2

0 10

LNMOQP = −LNM

OQPLNMOQPω

.

The coefficient matrix has eigenvalues ±iω0, so the origin 0 0, ( ) is a center point and thus

classified as neutrally stable.

!!!! Damped Spring

8. m b k!! !x x x+ + = 0

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

!!xy

km

bm

xy

LNMOQP = − −LNMM

OQPPLNMOQP

0 1.

The determinant of the coefficient matrix is km

, which is assumed positive. Hence, the matrix is

nonsingular and x y= = 0 is an isolated equilibrium point. The eigenvalues of this system are the

roots of

−− − − = + + =λ

λ λ λ1 1 02k

mbm m

m b kb g ,

which are

λ1

2 42

=− + −b b mk

m and λ2

2 42

=− − −b b mk

m.

From these roots, we see that when b > 0, regardless of the values of m > 0, and k > 0, the rootswill either be real and negative or complex with negative real parts. In either case, the origin isasymptotically stable.

Because only when the three parameters m, k, and b are positive is considered, the originwill always be asymptotically stable, which is the nature of real systems with friction.

!!!! One Zero Eigenvalue

9. (a) If λ1 0= and λ2 0≠ , then A is a singular matrix because A A I= − =λ1 0. Hence, the

rank of A is less than 2. But the rank of A is not 0 because if it were it would be thematrix of all zeros, which would have both eigenvalues 0. The rank of A is 1, whichmeans the kernel of A consists of a one-dimensional subspace of R2 , a line through the


origin. But the kernel of A is simply the act of solutions of Ax = 0, which are theequilibrium points of !x Ax= . We use the solution of the form

x txy

cab

c ecd

t( ) = LNMOQP =LNMOQP +

LNMOQP1 2 2λ

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

!x t c ecd

t( ) = LNMOQP =LNMOQP2 2 2

00

λ λ ,

which implies c2 0= . The points that satisfy ! !x = =y 0 are the points

x txy

cab

( ) = LNMOQP =LNMOQP1 ,

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 c2 0≠ . If λ 2 0> , the second

term

c ecd

t2 2λ LNMOQP

becomes larger and larger. Hence, the solution moves farther and farther away from theline of equilibrium points. On the other hand, if λ2 0< , the second term becomes smaller

and smaller, the solution moves towards the line.

!!!! Zero Eigenvalue Example

10.′′LNMOQP = −LNMOQPLNMOQP

xy

xy

0 01 1

(a) The characteristic equation of this system is − − =λ λ1 0b g ; it yields eigenvalues

λ1 0= and λ2 1= .

The corresponding eigenvectors are

υ1 1 1= , and υ2 0 1= , .

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

(c) We set ′ =x 0 and !y x y= − + , yielding x t c( ) = 1 and ′ = − +y c y1 . Hence,

y t c e ct( ) = +−2 1


where c1 and c2 are arbitrary constants. In vector form, this is

x txy

c ce t( ) = LNM

OQP =LNMOQP +LNMOQP−1 2

11

0.

(d) Because x t c( ) = , the solutions move along vertical lines (or don’t move at all). Toexamine this further, assume we start at an initial point x y x y0 0 0 0( ) ( )( ) =, , a f . Findingconstants, c1 and c2 , yields the solution

x t xy t x y x e t

( ) =

( ) = + − −0

0 0 0a fwhich says that starting at any point x y0 0, a f, the solution moves vertically approachingthe 45-degree line and the point x x0 0, a f.

!!!! Both Eigenvalues Zero

11. When both eigenvalues of A are zero, such as in the matrices

0 00 0LNMOQP or

0 10 0LNMOQP ,

the solution is

x t c eab

c tecd

cab

c tcd

t t( ) = LNMOQP +

LNMOQP =LNMOQP +LNMOQP1

02

01 2 .

!!!! Strange System I

12. (a) The characteristic equation of this system is λ λ2 0+ = , yielding λ1 0= and λ2 1= − .

The corresponding eigenvectors can be seen to be

υ1 2, 1= , υ2 1 1= , .

(b) Setting

′ = ′ =x y 0 ,

we see that all points on the line

x y− =2 0

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

not an isolated equilibrium point. Alsofrom the differential equations,

′ = ′x y ,

2–2

–2

2y

x

Sample trajectories of a singular system


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

x y− =2 0 , ′ = ′ = − <x y x y2 0

and the movement is downward and to the left. Below the line x y− =2 0 , movement is

upward and to the right. This outcome is shown in the phase plane. (See the figure.) Notethat the solutions below the equilibrium line approach the line because the trajectoriesmove 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.

(c) It is normally assumed that the matrix A is nonsingular so normally 0 0, ( ) is an isolated

equilibrium point.

!!!! Strange System II

13. ′ = LNMOQPx x

0 00 0

Nothing moves; all trajectories are points.

!!!! Strange System III

14. ′ =−

LNMOQPx x

k 00 1

The characteristic equation of this system is

kk

−− −

= −( ) +( ) =λ

λλ λ

00 1

1 0

and, hence, the roots are λ1 = k , λ2 1= − .

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

node.

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

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

(d) k = 0 implies that the matrix is singular; hence, the origin is not an isolated equilibriumpoint (all trajectories of this system move vertically towards the x1 axis).

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


!!!! Bifurcation Point

15. The characteristic equation of

′ =−LNMOQPx x

0 11 k

is λ λ2 1 0− + =k , which has roots

λ λ1 221

24= = + −k ke j .

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

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

!!!! Interesting Relationships

16. ′ = LNMOQPx x

a bc d

The characteristic equation is

λ λ λ λ λ λ21 2

21 2 1 20 0− +( ) + −( ) = = − − = − + + =a d ad bc r r r r r ra fa f a f

If the characteristic roots are r1 and r2 , we factor the quadratic on the left. We see by equating

the coefficients that

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

(b) A (the constant term) is always the product of the roots (i.e., A = r r1 2).

!!!! Interpreting the Trace-Determinant Graph

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

λ1, λ2

2 42

=± ( ) −Tr TrA A A .

17. A > 0, TrA A( ) − >2 4 0

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

trace.


18. A < 0

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

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

19. TrA ≠ 0, TrA A( ) − <2 4 0

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

20. TrA = 0, A > 0

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

21. TrA A( ) − =2 4 0, TrA ≠ 0

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

22. TrA > 0 or A < 0

Using the basic formula, if the trace is positive, then either the roots are complex with positivepart or the roots are real with at least one positive root. In either case the origin is an unstableequilibrium point. In the case when det A < 0, then, from the basic formula the roots are real andat least one root is positive, again showing that the origin is unstable.

23. A > 0 and TrA = 0

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

24. TrA < 0 and A > 0

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

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