11
11
Systems of Nonlinear
Differential Equations
EXERCISES 11.1
Autonomous Systems
3.
The corresponding plane autonomous system is
x
0
=
y,
y
0
=
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.
6.
The corresponding plane autonomous system is
x
0
=
y
0
=
−
x
+
²x

x

.
If (
x, y
) is a critical point,
y
= 0 and
−
x
+
²x

x

=
x
(
−
1+
²

x

) = 0. Hence
x
=0,1
/²
,
−
1
/²
. The critical points
are (0
,
0), (1
/²,
0) and (
−
1
/²,
0).
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.
12.
Adding the two equations we obtain 10
−
15
y/
(
y
+ 5) = 0. It follows that
y
= 10, and from
−
2
x
+
y
+10=0
we can conclude that
x
= 10. Therefore (10
,
10) is the sole critical point of the system.
15.
From
x
(1
−
x
2
−
3
y
2
)=0wehave
x
=0or
x
2
+3
y
2
=1.If
x
= 0, then substituting into
y
(3
−
x
2
−
3
y
2
) gives
y
(3
−
3
y
2
) = 0. Therefore
y
=0,1,
−
1. Likewise
x
2
=1
−
3
y
2
yields 2
y
= 0 so that
y
= 0 and
x
2
−
3(0)
2
=1.
The critical points of the system are therefore (0
,
0), (0
,
1), (0
,
−
1), (1
,
0), and (
−
1
,
0).
18. (a)
From Exercises 10
.
2, Problem 6,
x
=
c
1
+2
c
2
e
−
5
t
and
y
=3
c
1
+
c
2
e
−
5
t
, which is not periodic.
(b)
From
X
(0) = (3
,
4) it follows that
c
1
=
c
2
= 1. Therefore
x
=1+2
e
−
5
t
and
y
=3+
e
−
5
t
gives
y
=
1
2
(
x
−
1)+3.
(c)
21. (a)
From Exercises 10
.
2, Problem 35,
x
=
c
1
(sin
t
−
cos
t
)
e
4
t
+
c
2
(
−
sin
t
−
cos
t
)
e
4
t
and
y
=2
c
1
(cos
t
)
e
4
t
+
2
c
2
(sin
t
)
e
4
t
. Because of the presence of
e
4
t
, there are no periodic solutions.
(b)
From
X
(0) = (
−
1
,
2) it follows that
c
1
= 1 and
c
2
= 0. Therefore
x
= (sin
t
−
cos
t
)
e
4
t
and
y
= 2(cos
t
)
e
4
t
.
194