11
Systems of Nonlinear Differential Equations
Exercises 11.1
1.
The corresponding plane autonomous system is
x
=
y,
y
=
−
9sin
x.
If (
x, y
) is a critical point,
y
= 0 and
−
9sin
x
= 0. Therefore
x
=
±
nπ
and so the critical points are (
±
nπ,
0)
for
n
= 0, 1, 2,
. . .
.
2.
The corresponding plane autonomous system is
x
=
y,
y
=
−
2
x
−
y
2
.
If (
x, y
) is a critical point, then
y
= 0 and so
−
2
x
−
y
2
=
−
2
x
= 0. Therefore (0
,
0) is the sole critical point.
3.
The corresponding plane autonomous system is
x
=
y,
y
=
x
2
−
y
(1
−
x
3
)
.
If (
x, y
) is a critical point,
y
= 0 and so
x
2
−
y
(1
−
x
3
) =
x
2
= 0. Therefore (0
,
0) is the sole critical point.
4.
The corresponding plane autonomous system is
x
=
y,
y
=
−
4
x
1 +
x
2
−
2
y.
If (
x, y
) is a critical point,
y
= 0 and so
−
4
x
1 +
x
2
−
2(0) = 0. Therefore
x
= 0 and so (0
,
0) is the sole critical
point.
5.
The corresponding plane autonomous system is
x
=
y,
y
=
−
x
+
x
3
.
If (
x, y
) is a critical point,
y
= 0 and
−
x
+
x
3
= 0. Hence
x
(
−
1 +
x
2
) = 0 and so
x
= 0,
1
/
,
−
1
/
. The
critical points are (0
,
0), (
1
/ ,
0) and (
−
1
/ ,
0).
6.
The corresponding plane autonomous system is
x
=
y,
y
=
−
x
+
x

x

.
If (
x, y
) is a critical point,
y
= 0 and
−
x
+
x

x

=
x
(
−
1+

x

) = 0. Hence
x
= 0, 1
/
,
−
1
/
. The critical points
are (0
,
0), (1
/ ,
0) and (
−
1
/ ,
0).
7.
From
x
+
xy
= 0 we have
x
(1+
y
) = 0. Therefore
x
= 0 or
y
=
−
1. If
x
= 0, then, substituting into
−
y
−
xy
= 0,
we obtain
y
= 0, Likewise, if
y
=
−
1, 1 +
x
= 0 or
x
=
−
1. We may conclude that (0
,
0) and (
−
1
,
−
1) are
critical points of the system.
8.
From
y
2
−
x
= 0 we have
x
=
y
2
. Substituting into
x
2
−
y
= 0, we obtain
y
4
−
y
= 0 or
y
(
y
3
−
1) = 0. It follows
that
y
= 0, 1 and so (0
,
0) and (1
,
1) are the critical points of the system.
9.
From
x
−
y
= 0 we have
y
=
x
. Substituting into 3
x
2
−
4
y
= 0 we obtain 3
x
2
−
4
x
=
x
(3
x
−
4) = 0. It follows
that (0
,
0) and (4
/
3
,
4
/
3) are the critical points of the system.
10.
From
x
3
−
y
= 0 we have
y
=
x
3
. Substituting into
x
−
y
3
= 0 we obtain
x
−
x
9
= 0 or
x
(1
−
x
8
). Therefore
x
= 0, 1,
−
1 and so the critical points of the system are (0
,
0), (1
,
1), and (
−
1
,
−
1).
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