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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
(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 .
(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
(a) Letting y x= ′ , we write the equation as the first-order system
′ =′ = − +
x yy x 1.
(b) The equilibrium point is x y, , ( ) = ( )1 0 .
438 CHAPTER 6 Linear Systems of Differential Equations
(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
(a) Letting y x= ′ , we write the equation as the first-order system
′ =′ = − − +
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
SECTION 6.1 Theory of Linear DE Systems 439
!!!! 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.
440 CHAPTER 6 Linear Systems of Differential Equations
17. ′ = LNMOQPx x
1 14 1
By substitution, we verify
u t ee
t
tb g =−
LNMOQP
−
−2 and v t e
e
t
tb g = LNMOQP
3
32
satisfy the system. The fundamental matrix is
e ee e
t t
t t
−
−−
LNM
OQP
3
32 2.
The general solution is
x t c ee
c ee
t
t
t
tb g =−
LNMOQP+LNMOQP
−
−1 2
3
32 2.
18. ′ =−LNMOQPx x
0 11 0
By substitution, we verify
u ttt
( ) = LNMOQP
sincos
and v ttt
b g =−LNMOQP
cossin
satisfy the system. The fundamental matrix is
sin coscos sin
t tt t−
LNM
OQP .
The general solution is
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
SECTION 6.1 Theory of Linear DE Systems 441
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
442 CHAPTER 6 Linear Systems of Differential Equations
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
SECTION 6.1 Theory of Linear DE Systems 443
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.
studied in Example 4 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 – 0.05 * C2
3 = A2 + 0.1 = B2 + 0.1 * D2 = C2 + 0.1 * E2 = C3 = –0.1 * B3 – 0.05 * C3
444 CHAPTER 6 Linear Systems of Differential Equations
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
If the range is continued to t = 25, then the graphs that correspond to Figure 6.1.4 looklike the following.
x
y
xy phase portrait
t
y
yt component graph
SECTION 6.1 Theory of Linear DE Systems 445
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
studied in Example 5 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 + 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
446 CHAPTER 6 Linear Systems of Differential Equations
If the range is continued to t = 40, then the graphs that correspond to Figure 6.1.5 looklike the following.
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
SECTION 6.1 Theory of Linear DE Systems 447
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
448 CHAPTER 6 Linear Systems of Differential Equations
!!!! 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 .
SECTION 6.1 Theory of Linear DE Systems 449
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
450 CHAPTER 6 Linear Systems of Differential Equations
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. ′ =−LNMOQPx x
2 13 6
The characteristic equation of the system is
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 .
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP1
32
511
13
.
SECTION 6.2 Linear Systems with Real Eigenvalues 451
3. ′ =−LNMOQPx x
1 12 4
The characteristic equation of the system is
p t( ) =− −
−= − + =
1 12 4
5 6 02λλ
λ λ ,
which has solutions λ1 2= , λ2 3= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
211
312
= ⇒ =−LNMOQP
= ⇒ =−LNMOQP
v
v .
Hence, the general solution is
x t c e c et t( ) =−LNMOQP + −
LNMOQP1
22
311
12
.
4. ′ =−−
LNM
OQPx x
10 58 12
The characteristic equation of the system is
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 .
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP
−1
102
814
52
.
5. ′ =−LNMOQPx x
5 13 1
The characteristic equation of the system is
p t( ) =− −
−= − + =
5 13 1
6 8 02λλ
λ λ ,
452 CHAPTER 6 Linear Systems of Differential Equations
which has solutions λ1 2= , λ2 4= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
213
411
= ⇒ = LNMOQP
= ⇒ = LNMOQP
v
v .
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP1
22
413
11
.
6. ′ = LNMOQPx x
1 24 3
The characteristic equation of the system is
p tb g = −−
= − − =1 2
4 34 5 02λ
λλ λ ,
which has solutions λ1 1= − , λ2 5= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
111
512
= − ⇒ =−LNMOQP
= ⇒ = LNMOQP
v
v .
Hence, the general solution is
x t c e c et t( ) =−LNMOQP +
LNMOQP
−1 2
511
12
.
7. ′ =−LNMOQPx x
1 02 2
The characteristic equation of the system is
p t( ) =−− −
= −( ) −( ) =1 0
2 21 2 0
λλ
λ λ ,
which has solutions λ1 1= , λ2 2= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
112
201
= ⇒ = LNMOQP
= ⇒ = LNMOQP
v
v .
SECTION 6.2 Linear Systems with Real Eigenvalues 453
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP1 2
212
01
.
8. ′ =− −LNMOQPx x
3 31 1
The characteristic equation of the system is
p t( ) =−− − −
= − =3 3
1 12 02λ
λλ λ ,
which has solutions λ1 0= , λ2 2= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
011
231
= ⇒ =−LNMOQP
= ⇒ =−LNMOQP
v
v .
Hence, the general solution is
x t c c e t( ) =−LNMOQP +
−LNMOQP1 2
211
31
.
9. ′ =−−LNMOQPx x
3 22 2
The characteristic equation of the system is
p t( ) =− −
− −= − − =
3 22 2
2 02λλ
λ λ ,
which has solutions λ1 1= − , λ2 2= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
112
221
= − ⇒ = LNMOQP
= ⇒ = LNMOQP
v
v .
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP
−1 2
212
21
.
454 CHAPTER 6 Linear Systems of Differential Equations
10. ′ =− −LNM
OQPx x
4 34 4
The characteristic equation of the system is
p t( ) =−− − −
= − =4 3
4 44 02λ
λλ ,
which has solutions λ1 2= , λ2 2= − . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
232
212
= ⇒ =−LNMOQP
= − ⇒ =−LNMOQP
v
v .
Hence, the general solution is
x t c e c et t( ) =−LNMOQP + −
LNMOQP
−1
22
232
12
.
11. ′ =−−LNMOQPx x
1 23 4
The characteristic equation of the system is
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 .
Hence, the general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP
− −1
22
23
11
.
12. ′ =−
−LNM
OQPx x
5 22 8
The characteristic equation of the system is
p t( ) =− −− −
= − + =5 2
2 813 36 02λ
λλ λ ,
SECTION 6.2 Linear Systems with Real Eigenvalues 455
which has solutions λ1 4= , λ2 9= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
421
912
= ⇒ = LNMOQP
= ⇒ =−LNMOQP
v
v .
Hence, the general solution is
x t c e c et tb g = LNMOQP + −
LNMOQP1
42
921
12
.
13. ′ =−−LNMOQPx x
4 38 6
The characteristic equation of the system is
p t( ) =− −
− −= + =
4 38 6
2 02λλ
λ λ ,
which has solutions λ1 0= , λ2 2= − . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
034
212
= ⇒ = LNMOQP
= − ⇒ = LNMOQP
v
v .
Hence, the general solution is
x t c c e t( ) = LNMOQP +
LNMOQP
−1 2
234
12
.
14. ′ =−LNMOQPx x
5 31 1
The characteristic equation of the system is
p t( ) =−− −
= − + =5 3
1 16 8 02λ
λλ λ ,
which has solutions λ1 2= , λ2 4= . Finding the eigenvectors corresponding to each eigenvalue
yields
λ
λ
1 1
2 2
211
431
= ⇒ =−LNMOQP
= ⇒ =−LNMOQP
v
v .
456 CHAPTER 6 Linear Systems of Differential Equations
Hence, the general solution is
x t c e c et t( ) =−LNMOQP +
−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
.
16. The characteristic equation of the system is
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
.
SECTION 6.2 Linear Systems with Real Eigenvalues 457
The general solution is, therefore,
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
17. The characteristic equation of the system is
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 .
The general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP
−1
32
15
11
.
458 CHAPTER 6 Linear Systems of Differential Equations
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 .
18. The characteristic equation of the system is
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
The general solution is
x t c e c et t( ) = LNMOQP +
−LNMOQP
−1 2
432
11
.
Plugging into the initial conditions
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
.
SECTION 6.2 Linear Systems with Real Eigenvalues 459
19. The characteristic equation of the system is
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 .
The general solution is
x t c e c et t( ) = LNMOQP +
LNMOQP1
22
310
01
.
Plugging into the initial conditions
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 .
20. The characteristic equation of the system is
p t( ) =− −
−= + − =
2 41 1
6 02λλ
λ λ ,
which has the solutions λ1 3= − and λ2 2= . Finding the eigenvectors corresponding to each
eigenvalue, yields
λ
λ
1 1
2 2
341
211
= − ⇒ =−LNMOQP
= ⇒ = LNMOQP
v
v .
The general solution is
x t c e c et t( ) =−LNMOQP +
LNMOQP
−1
32
241
11
.
Plugging into the initial conditions
x 011
( ) =−LNMOQP
460 CHAPTER 6 Linear Systems of Differential Equations
yields
− + = −+ =
4 11
1 2
1 2
c cc c
which gives c125
= and c235
= . The solution of the IVP is
x t e et t( ) = FHIK
−LNMOQP +FHIKLNMOQP
−25
41
35
11
3 2 .
21. The characteristic equation of the system is
p t( ) =−
−= − =
1 11 1
2 02λλ
λ λ ,
which has the solutions λ1 0= and λ2 2= . Finding the eigenvectors corresponding to each
eigenvalue, yields
λ
λ
1 1
2 2
011
211
= ⇒ =−LNMOQP
= ⇒ = LNMOQP
v
v .
The general solution is
x t c c e t( ) =−LNMOQP +
LNMOQP1 2
211
11
.
Plugging into the initial conditions
x 023
( ) = LNMOQP
yields
− + =+ =
c cc c1 2
1 2
23
which gives c112
= and c252
= . The solution of the IVP is
x t e t( ) =−LNMOQP +FH IKLNMOQP
12
11
52
11
2 .
22. The characteristic equation of the system is
p t( ) =− −
− −= + + =
3 21 2
5 4 02λλ
λ λ ,
SECTION 6.2 Linear Systems with Real Eigenvalues 461
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
= − ⇒ = LNMOQP
v
v .
The general solution is
x t c e c et t( ) =−LNMOQP +
LNMOQP
− −1
42
21
11
.
Plugging into the initial conditions
x 016
( ) =−LNMOQP
yields
− + = −+ =
2 16
1 2
1 2
c cc c
which gives c173
= and c2113
= . The solution of the IVP is
x t e et t( ) = FHIK
−LNMOQP +FHIKLNMOQP
− −73
21
113
11
4 .
23. The characteristic equation of the system is
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
= − ⇒ =−LNMOQP
v
v .
The general solution is
x t c c e t( ) = LNMOQP + −
LNMOQP
−1 2
412
12
.
462 CHAPTER 6 Linear Systems of Differential Equations
Plugging into the initial conditions
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
.
24. The characteristic equation of the system is
p t( ) =−
−= − − =
1 123 1
2 35 02λλ
λ λ ,
which has the solutions λ1 5= − and λ2 7= . Finding the eigenvectors corresponding to each
eigenvalue, yields
λ
λ
1 1
2 2
521
721
= − ⇒ =−LNMOQP
= ⇒ = LNMOQP
v
v .
The general solution is
x t c e c et t( ) =−LNMOQP +
LNMOQP
−1
52
721
21
.
Plugging into the initial conditions
x 001
( ) = LNMOQP
yields
− + =+ =
2 2 01
1 2
1 2
c cc c
which gives c112
= and c212
= . The solution of the IVP is
x t e et t( ) = FH IK−LNMOQP +FH IKLNMOQP
−12
21
12
21
5 7 .
SECTION 6.2 Linear Systems with Real Eigenvalues 463
25. The characteristic equation of the system is
p t( ) =− −
−= − + =
1 12 4
5 6 02λλ
λ λ ,
which has the solutions λ1 2= and λ2 3= . Finding the eigenvectors corresponding to each
eigenvalue, yields
λ
λ
1 1
2 2
211
312
= ⇒ =−LNMOQP
= ⇒ =−LNMOQP
v
v .
The general solution is
x t c e c et t( ) =−LNMOQP + −
LNMOQP1
22
311
12
.
Plugging into the initial conditions
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 .
26. The characteristic equation of the system is
p t( ) =−
−= − − =
1 22 1
2 3 02λλ
λ λ ,
which has the solutions λ1 1= − and λ2 3= . Finding the eigenvectors corresponding to each
eigenvalue, yields
λ
λ
1 1
2 2
111
311
= − ⇒ =−LNMOQP
= ⇒ = LNMOQP
v
v .
The general solution is
x t c e c et t( ) =−LNMOQP +
LNMOQP
−1 2
311
11
.
Plugging into the initial conditions
464 CHAPTER 6 Linear Systems of Differential Equations
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 .
SECTION 6.2 Linear Systems with Real Eigenvalues 465
(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
.
466 CHAPTER 6 Linear Systems of Differential Equations
(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
29. The characteristic equation of the system is
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 .
Hence, the general solution is
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
.
30. The characteristic equation of the system is
p tb g =− −
−− −
= − + + =1 1 0
1 2 10 3 1
7 6 03
λλ
λλ λ ,
SECTION 6.2 Linear Systems with Real Eigenvalues 467
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 .
Hence, the general solution is
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 .
Hence, the general solution is
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
− − =+ − =+ + =
468 CHAPTER 6 Linear Systems of Differential Equations
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 .
Hence, the general solution is
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.)
SECTION 6.2 Linear Systems with Real Eigenvalues 469
(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
34. ′ =−LNMOQPx x
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
35. ′ =−LNMOQPx x
1 12 4
The eigenvalues and vectors are
λ1 2= , v1 1 1= −( ),
λ2 3= , v2 1 2= −( ), .2–2
–2
2y
x
v2v1
470 CHAPTER 6 Linear Systems of Differential Equations
36. ′ =−LNMOQPx x
5 13 1
The eigenvalues and vectors are
λ1 2= , v1 1 3= ( ),
λ2 4= , v2 1 1= ( ), .
2–2
–2
2y
x
v2v1
37. ′ =−−LNMOQPx x
1 23 4
The eigenvalues and vectors are
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.
SECTION 6.2 Linear Systems with Real Eigenvalues 471
(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 .
472 CHAPTER 6 Linear Systems of Differential Equations
(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.
SECTION 6.2 Linear Systems with Real Eigenvalues 473
(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
474 CHAPTER 6 Linear Systems of Differential Equations
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
= − ⇒ = 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
= − ⇒ = LNMOQP
v
v .
SECTION 6.2 Linear Systems with Real Eigenvalues 475
Hence, the general solution is
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
!!!! Suggested Journal Entry
44. Student Project
476 CHAPTER 6 Linear Systems of Differential Equations
6.3 Linear Systems with Nonreal Eigenvalues
!!!! Solutions in General
1. ′ =−LNMOQPx x
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 .
The general solution is, therefore,
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
The general solution is, therefore,
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
.
3. ′ =−LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
x x xt c c c ett
c ett
t t( ) = + =−LNM
OQP +LNMOQP1 1 2 2 1 2
22
22
cossin
sincos
.
4. ′ =−LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
Hence, the general solution is
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.
478 CHAPTER 6 Linear Systems of Differential Equations
5. ′ =− −LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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
.
6. ′ =−−LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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, .
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 479
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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
.
8. ′ =−−LNMOQPx x
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 .
The general solution is, therefore,
x x xt c c ct t
tc
t tt
( ) = + =−L
NMOQP +
+LNM
OQP1 1 2 2 1 2
2 2cos sincos
cos sinsin
.
9. ′ =−−LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
480 CHAPTER 6 Linear Systems of Differential Equations
The general solution is, therefore,
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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.
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 481
12. ′ =− −LNMOQPx x
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, .
Two linearly independent solutions are then obtained.
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 .
The general solution is, therefore,
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
13. ′ =−LNMOQPx x
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
.
482 CHAPTER 6 Linear Systems of Differential Equations
14. ′ =−LNMOQPx x
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 + −
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
.
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 483
16. ′ =−−LNMOQPx x
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 + −
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 .
484 CHAPTER 6 Linear Systems of Differential Equations
(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
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 485
(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 .
486 CHAPTER 6 Linear Systems of Differential Equations
Hence, the general solution is
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 .
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 487
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 .
488 CHAPTER 6 Linear Systems of Differential Equations
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
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 489
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
490 CHAPTER 6 Linear Systems of Differential Equations
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+ = .
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 491
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 .
492 CHAPTER 6 Linear Systems of Differential Equations
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
30. ′ =−LNMOQPx x
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 .
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 493
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
The general solution is
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 .
494 CHAPTER 6 Linear Systems of Differential Equations
!!!! Computer Lab: Phase Portrait
31. ′ =− −LNMOQPx x
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
32. ′ =−−LNMOQPx x
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( )
SECTION 6.3 Linear Systems with Nonreal Eigenvalues 495
!!!! 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
.
496 CHAPTER 6 Linear Systems of Differential Equations
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
The general solution is
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
!!!! Suggested Journal Entry
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
The coefficient matrix has eigenvalue and eigenvectors
λλ
1 1
2 2
3 1 31 1 1
= ⇒ = −= − ⇒ =
vv
,, .
Matrices of eigenvectors are
P =−LNMOQP
1 13 1
and P− =−L
NMOQP
1 14
1 13 1
.
498 CHAPTER 6 Linear Systems of Differential Equations
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
The coefficient matrix has eigenvalue and eigenvectors
λλ
1 1
2 2
1 1 11 1 1
= ⇒ = −= − ⇒ =
vv
,, .
Matrices of eigenvectors are
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
The coefficient matrix has eigenvalue and eigenvectors
λλ
1 1
2 2
1 3 15 1 1
= ⇒ = −= ⇒ =
vv
,, .
Matrices of eigenvectors are
P =−LNMOQP
3 11 1
and P− =−LNMOQP
1 14
1 11 3
.
SECTION 6.4 Uncoupling a Linear DE System 499
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 .
5. ′ =−−LNMOQPx x
2 32 5
The coefficient matrix has eigenvalue and eigenvectors
λλ
1 1
2 2
1 3 14 1 2
= ⇒ == − ⇒ =
vv
,, .
Matrices of eigenvectors are
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
The coefficient matrix has eigenvalue and eigenvectors
λλ
1 1
2 2
1 1 11 1 1
= − ⇒ = −= ⇒ =
vv
,, .
Matrices of eigenvectors are
P =−LNMOQP
1 11 1
and P− =−LNMOQP
1 12
1 11 1
.
500 CHAPTER 6 Linear Systems of Differential Equations
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
The coefficient matrix has eigenvalue and eigenvectors
λλλ
1 1
2 2
3 3
3 1 1 10 1 1 00 1 0 1
= ⇒ == ⇒ = −= ⇒ = −
vvv
, ,, ,, , .
Matrices of eigenvectors are
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 .
SECTION 6.4 Uncoupling a Linear DE System 501
8. ′ =L
NMMM
O
QPPP
x x0 0 00 1 01 0 1
The coefficient matrix has eigenvalue and eigenvectors
λλλ
1 1
2 2
3 3
0 1 0 11 0 0 11 0 1 0
= ⇒ = −= ⇒ == ⇒ =
vvv
, ,, ,, , .
Matrices of eigenvectors are
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
502 CHAPTER 6 Linear Systems of Differential Equations
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
.
We change the variable w P x= −1 , to yield the decoupled system
′ =−
−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
.
11. ′ = LNMOQP +LNMOQPx x
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
.
We change the variable w P x= −1 , to yield the decoupled system
′ = LNMOQP +
−LNMOQPLNMOQPw w
0 00 2
12
1 11 1 1
t
SECTION 6.4 Uncoupling a Linear DE System 503
or
′ = −( )
′ = + +( )
w
w w
1
2 2
12
1
2 12
1
t
t .
Solving these, yields
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 .
12. ′ = LNMOQP +LNMOQPx x
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
.
We change the variable w P x= −1 , to yield the decoupled system
′ = LNMOQP +
−LNMOQPLNMOQPw w
6 00 1
15
1 11 4
50t
or
′ = +′ = −
w ww w
1 1
2 2
6 tt .
Solving these, yields
w
w
1 16
2 2
6136
1
t c e t
t c e t
t
t
( ) = − −
( ) = + + .
504 CHAPTER 6 Linear Systems of Differential Equations
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
SECTION 6.4 Uncoupling a Linear DE System 505
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
, , ,, , ,
, , ,, , , .
506 CHAPTER 6 Linear Systems of Differential Equations
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
( ) = −
( ) =
( ) = +
( ) = − +
−
− .
SECTION 6.4 Uncoupling a Linear DE System 507
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
508 CHAPTER 6 Linear Systems of Differential Equations
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.
!!!! Suggested Journal Entry
17. Student Project
SECTION 6.5 Stability and Linear Classification 509
6.5 Stability and Linear Classification
!!!! Classification Verification
1. ′ =−LNMOQPx x
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.
2. ′ =−LNMOQPx x
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.
6. ′ =− −LNMOQPx x
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.
510 CHAPTER 6 Linear Systems of Differential Equations
!!!! 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
SECTION 6.5 Stability and Linear Classification 511
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
512 CHAPTER 6 Linear Systems of Differential Equations
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
SECTION 6.5 Stability and Linear Classification 513
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.
514 CHAPTER 6 Linear Systems of Differential Equations
!!!! 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.
SECTION 6.5 Stability and Linear Classification 515
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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