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Chapter 5
dy
dt
=
d
dt
(
e
t
)
=
e
t
=
y.
Thus, given pair of functions is a solution. To sketch the trajectory of this solution, we express
x
as a function of
y
.
x
=
e
3
t
=
(
e
t
)
3
=
y
3
for
y
=
e
t
>
0
.
Since
y
=
e
t
is an increasing function, the ﬂow arrows are directed away from the origin. See
Figure B.29 in the answers of the text.
3.
In this problem,
f
(
x, y
)=
x
−
y
,
g
(
x, y
x
2
+
y
2
−
1. To ±nd the critical point set, we solve
the system
x
−
y
=0
,
x
2
+
y
2
−
1=0
⇒
x
=
y,
x
2
+
y
2
=1
.
Eliminating
y
yields
2
x
2
⇒
x
=
±
1
√
2
.
Substituting
x
into the ±rst equation, we ±nd the corresponding value for
y
. Thus the critical
points of the given system are (1
/
√
2
,
1
/
√
2) and (
−
1
/
√
2
,
−
1
/
√
2).
5.
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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