Exercises 5.4
17.
For critical points, we solve the system
f
(
x, y
)=0
,
g
(
x, y
⇒
2
x
+13
y
=0
,
−
x
−
2
y
⇒
2(
−
2
y
)+13
y
,
x
=2
y
⇒
y
,
x
.
Therefore, the system has just one critical point, (0
,
0).
The direction ±eld is shown in
Figure B.33 in the text. From this picture we conclude that (0
,
0) is a center (stable).
19.
We set
v
=
y
0
.Then
y
0
=(
y
0
)
0
=
v
0
and so given equation is equivalent to the system
y
0
=
v,
v
0
−
y
⇒
y
0
=
v
0
=
y.
In this system,
f
(
y,v
)=
v
and
g
(
y
. For critical points we solve
f
(
v
,
g
(
y
⇒
y
,
v
and conclude that, in
yv
plane, the system has only one critical point, (0
,
0). In the upper
halfplane,
y
0
=
v>
0 and, therefore,
y
increases and solutions ﬂow to the right; similarly,
solutions ﬂow to the left in the lower halfplane. See Figure B.34 in the answers of the text.
The phase plane equation for the system is
dv
dy
=
dv/dx
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Critical Point, saddle point, Stationary point, hessian matrix, yv plane

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