Section 9.1
453
CHAPTER 9
NONLINEAR SYSTEMS AND PHENOMENA
SECTION 9.1
STABILITY AND THE PHASE PLANE
1.
The only solution of the homogeneous system
2
0,
3
0
x
y
x
y
−
=
−
=
is the origin
(0, 0).
The only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point
at the origin is Fig. 9.1.13.
Thus the only critical point of the given autonomous system
is the saddle point (0, 0) shown in Figure 9.1.13 in the text.
2.
The only solution of the system
0,
3
4
0
x
y
x
y
−
=
+
+
=
is the point
(1, 1).
The only
figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (1, 1) is Fig.
9.1.15.
Thus the only critical point of the given autonomous system is the node (1, 1)
shown in Figure 9.1.15 in the text.
3.
The only solution of the system
2
3
0,
2
0
x
y
x
y
−
+
=
−
+
=
is the point
(–1, 1).
The
only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (–1, 1) is
Fig. 9.1.18.
Thus the only critical point of the given autonomous system is the stable
center (

1, 1) shown in Figure 9.1.18 in the text.
4.
The only solution of the system
2
2
4
0,
4
3
0
x
y
x
y
−
−
=
+
+
=
is the point
(1, –1).
The
only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (1, –1) is
Fig. 9.1.12.
Thus the only critical point of the given autonomous system is the spiral
point (1,

1) shown in Figure 9.1.12 in the text.
5.
The first equation
2
1
0
y
−
=
gives
y
=
1
or
y
=

1
at a critical point.
Then the
second equation
2
0
x
y
+
=
gives
x
=

2
or
x
=
2,
respectively.
The only figure
among Figs. 9.1.11 through 9.1.18 showing two critical points at (–2, 1) and (2, –1) is
Fig. 9.1.11.
Thus the critical points of the given autonomous system are the spiral point
(

2, 1) and the saddle point (2, 1) shown in Figure 9.1.11 in the text.
6.
The second equation
2
4
0
x
−
=
gives
x
=
2
or
x
=

2
at a critical point.
Then the
first equation
2
4
15
0
x
y
−
−
=
gives
y
=

2/5
or
x
=
2/3,
respectively.
The only
figure among Figs. 9.1.11 through 9.1.18 showing two critical points at (–2, 2/3) and
(2, –2/5) is Fig. 9.1.17.
Thus the critical points of the given autonomous system are the
spiral point (

2, 2/3) and the saddle point (2,

2/5) shown in Figure 9.1.17 in the text.
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454
Chapter 9
7.
The first equation
3
4
0
x
x
−
=
gives
x
=

2,
x
=
0,
or
x
=
2
at a critical point.
Then the second equation
2
0
x
y
−
=
gives
y
=

1,
y
=
0,
or
y
=
1,
respectively.
The only figure among Figs. 9.1.11 through 9.1.18 showing three critical points at
(–2, –1),
(0, 0), and (2, 1) is Fig. 9.1.14.
Thus the critical points of the given
autonomous system are the spiral point (0, 0) and the saddle points (

2, 1) and (2, 1)
shown in Figure 9.1.14 in the text.
8.
The second
2
0
y
x
−
−
=
equation gives
y
=

x
2
at a critical point.
Substitution of this
in the first equation
2
0
x
y
x
xy
−
−
+
=
then gives
x

x
3
=
0,
so
x
=

1,
x
=
0,
or
x
=
1.
The only figure among Figs. 9.1.11 through 9.1.18 showing three critical points
at
(–1, –1),
(0, 0), and (1, –1) is Fig. 9.1.16.
Thus the critical points of the given
autonomous system are the spiral point (

1,

1), the saddle point (0, 0), and the node
(1,

1) shown in Figure 9.1.16 in the text.
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 Spring '08
 Zhu
 Differential Equations, Equations, Critical Point, Multivariable Calculus, Linear Systems, Complex number, saddle point, Stability theory

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