Section 9.1: Stability and the Phase Plane 515
In Problems 13–16, the given x– and y-equations are independent exponential differential equa-
tions that we can solve immediately by inspection.
13. Solution:
2
0
t
x
txe
,
2
0
t
yt ye
.
14. Solution:
2
0
t
x
txe,
2
0
t
yt ye
.
Then 00
x
yxy k, so the trajectories are rectangular hyperbolas. Thus the origin is an
unstable saddle point like the one shown.
5
x
Problem 14
5
x
Problem 15
15. Solution:
2
0
t
x
txe
,
0
t
yt ye
.
x
x
516 Chapter 9: Nonlinear Systems and Phenomena
16. Solution:
0
t
x
txe,
3
0
t
yt ye.
The origin is an unstable improper node. The trajectories consist of the y-axis and curves
of the form 3
y
kx, departing from the origin as shown.
5
Problem 16
5
Problem 17
17. Differentiation of the first equation and substitution using the second one gives
x
yx
, so 0xx
. Solving this differential equation for
x
t and using the fact
that yx
lead to the general solution
cos si cos s , ninxt A t B t yt B t A t .
18. Elimination of y as in Problem 17 gives 40xx
. Solving this differential equation
for x and using the fact that
y
x
lead to the general solution
Section 9.1: Stability and the Phase Plane 517
5
x
Problem 18
5
x
Problem 19
19. Elimination of y as in Problem 17 gives 40xx
, and then using 1
2
y
x
we get the
general solution
20. Substitution of yx
from the first equation into the second one gives
554
x
xy x x
,
so 450xxx
. The characteristic roots of this equation are 2ri , so we get
the general solution
y
518 Chapter 9: Nonlinear Systems and Phenomena
5
Problem 20
21. We want to solve the system
22
10,
ky x x y
22. After separation of variables, a partial-fraction decomposition gives
2
11 1 1 1 1
ln ln 1 ln 1
2121 222
1
dr
tdrrrr
rr r
rr
,
23. The equation dy x
dx y
separates to 0xdx ydy, so 22
x
yC. Thus the trajecto-
Section 9.1: Stability and the Phase Plane 519
5
x
Problem 23
5
x
Problem 24
25. The equation 4
dy x
dx y
separates to 40xdx ydy
, so 22
4
x
yC. Thus the trajec-
tories consist of the origin
0, 0 and the ellipses 22
40xyC, as shown.
5
Problem 25
4
Problem 26
520 Chapter 9: Nonlinear Systems and Phenomena
26. The equation
3
3
dy x
dx y
separates to 33
0xdx ydy, so 44
x
yC. Thus the trajec-
tories consist of the origin
0, 0 and the ovals of the form 44
x
yC, as illustrated.
27. If
txt
and
tyt
, then
28. If
txt
and
tyt
, then
SECTION 9.2
LINEAR AND ALMOST LINEAR SYSTEMS
In Problems 1–10 we first find the roots 1
and 2
of the characteristic equation of the coeffi-
cient matrix of the given linear system. We can then read the type and stability of the critical
point
0, 0 from the table of Figure 9.2.4 in the text.
Section 9.2: Linear and Almost Linear Systems 521
5
x
Problem 1
5
x
Problem 2
2. The roots 12
and 23
of the characteristic equation 2560
are both posi-
tive, so
0, 0 is an unstable improper node.
5
Problem 3
5
Problem 4
4. The roots 12
and 24
of the characteristic equation 2230
have differ-
ent signs, so
0, 0 is an unstable saddle point.
522 Chapter 9: Nonlinear Systems and Phenomena
5
x
Problem 5
5
x
Problem 6
6. The roots 12
2
of the characteristic equation 2440
are positive and
equal, so
0, 0 is an unstable node.
7. The roots 12
,12i
of the characteristic equation 2250
are complex con-
5
Problem 7
5
Problem 8
Section 9.2: Linear and Almost Linear Systems 523
9. The roots 12
,2i
of the characteristic equation 240
are pure imaginary, so
0, 0 is a stable (but not asymptotically stable) center.
5
x
Problem 9
5
x
Problem 10
10. The roots 12
,3i
of the characteristic equation 290
are pure imaginary, so
0, 0 is a stable (but not asymptotically stable) center.
524 Chapter 9: Nonlinear Systems and Phenomena
5
Problem 11
5
Problem 12
12. The Jacobian matrix 12
J has characteristic equation 2560
and eigen-
13. The Jacobian matrix 21
32
J has characteristic equation 210
and eigenvalues
11
, 21
having different signs. Hence the critical point
2,2 is an unstable
saddle point.
5
Problem 13
5
Problem 14
Section 9.2: Linear and Almost Linear Systems 525
14. The Jacobian matrix 11
31
J has characteristic equation 240
and eigenvalues
15. The Jacobian matrix 11
53
J has characteristic equation 2220
and eigen-
values 12
,1i
that are complex conjugates with negative real part. Hence the crit-
ical point
1,1 is an asymptotically stable spiral point.
5
Problem 15
5
Problem 16
16. The Jacobian matrix 12
13
J has characteristic equation 2450
and eigen-
17. The Jacobian matrix 15
11
J has characteristic equation 240
and pure imagi-
526 Chapter 9: Nonlinear Systems and Phenomena
5
Problem 17
5
Problem 18
18. The Jacobian matrix 45
54
J has characteristic equation 290
and pure imagi-
nary eigenvalues 12
,3i
. Hence
2, 1 is a stable (but not asymptotically stable)
center.
19. 12 32
46
y
x
y
x
J
At
0, 0 : The Jacobian matrix 13
46
J has characteristic equation 2560
and eigenvalues 13
, 22
that are both negative. Hence
0, 0 is an asymptoti-
Section 9.2: Linear and Almost Linear Systems 527
2
Problem 19
3
Problem 20
20. 62 5
212
x
y
J
At
0, 0 : The Jacobian matrix 65
21
J has characteristic equation 2540
and eigenvalues 11
, 24
that are both positive. Hence
0, 0 is an unstable node
21. 12 22
23 23
x
y
y
x
J
528 Chapter 9: Nonlinear Systems and Phenomena
At
0.51, 2.12 : The Jacobian matrix 0.014 2.236
8.354 0.479
J has complex conjugate
eigenvalues 12
, 0.25 4.32i
with negative real parts. Hence
0.51, 2.12 is a
spiral sink.
5
Problem 21
2
4
6
Problem 22
22.
2
2
142
22 1
y
xy
x
yx
J
At
0, 0 : The Jacobian matrix 14
21
J has characteristic equation 290
and
seigenvalues 12
,3
that have different signs. Hence
0, 0 is a saddle point of the
given almost linear system.
Section 9.2: Linear and Almost Linear Systems 529
23.
2
3
23 5
464
x
y
J
At
0, 0 : The Jacobian matrix 25
J has characteristic equation 2480
1
2
3
Problem 23
5
Problem 24
24.
22
22
52 3 3
52 3
x
yxy
xy x y
J
530 Chapter 9: Nonlinear Systems and Phenomena
25. 13 23
22 32
y
x
x
y
J
At
0, 0 : The Jacobian matrix 12
23
J has characteristic equation 2210
and equal negative eigenvalues 11
, 11
. Hence
0, 0 is either a nodal sink or a
spiral sink of the given almost linear system.
At
0.121,0.074 : The Jacobian matrix 1.222 1.636
1.758 3.148
J has real eigenvalues
12.34
, 20.42
with different signs. Hence
0.121,0.074 is a saddle point.
5
Problem 25a
0.2
Problem 25b
Section 9.2: Linear and Almost Linear Systems 531
26. 32 22
23 13
x
y
y
x
J
At
0, 0 : The Jacobian matrix 32
21
J has characteristic equation 2210
and equal positive eigenvalues 12
1
. Hence
0, 0 is either a nodal source or a spi-
ral source of the given almost linear system.
3
Problem 26
2
Problem 27a
532 Chapter 9: Nonlinear Systems and Phenomena
27.
3
3
14 12
22 14
x
y
x
y
J
At
0, 0 : The Jacobian matrix 11
21
J has characteristic equation 2
10
and
equal positive eigenvalues 12
,i
. Hence
0, 0 is either a center or a spiral point,
but its stability is not determined by Theorem 2.
At
0.254, 0.507 : The Jacobian matrix 0.934 0.014
2.508 1.521
J has real eigenvalues
Figure a shows these four critical points. Figure b, a close-up, suggests that the origin
may be a stable center; note both the trajectory in grey that seems to spiral inward, and
the oval-shaped trajectories that lie closer to the origin.
0.6
Problem 27b
1
Problem 28
Section 9.2: Linear and Almost Linear Systems 533
28.
22
33 13
13 3 3 3
x
y
y
x
J
At
0, 0 : The Jacobian matrix 31
13 3
J has characteristic equation 240
and
29. 11
21x
J
At
0, 0 : The Jacobian matrix 11
01
J has characteristic equation 2
10
and
534 Chapter 9: Nonlinear Systems and Phenomena
2
Problem 29
3
Problem 30
30. 01
21x
J
At
1,1 : The Jacobian matrix 01
21
J has characteristic equation 220
31. 2
02
31
y
x
J