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Exercises 3.2
<<<
•
>>>>>
•
<<<<<
0
a/b
Figure 3–A
: The phase line for
p
0
=(
a
−
bp
)
p
.
Putting this value for
k
into the equation for
p
(
t
)g
ives
p
(
t
) = 1000
e
(
t
ln 3)
/
7
= 1000
·
3
t/
7
.
To estimate the population in 2010 we plug
t
= 2010
−
1980 = 30 into this formula to get
p
(30) = 1000
·
3
30
/
7
≈
110
,
868 splakes
.
11.
In this problem, the dependent variable is
p
, the independent variable is
t
, and the function
f
(
t, p
)=(
a
−
bp
)
p
.S
ince
f
(
t, p
)=
f
(
p
), i.e., does not depend on
t
, the equation is autonomous.
To Fnd equilibrium solutions, we solve
f
(
p
)=0
⇒
(
a
−
bp
)
p
=0
⇒
p
1
,p
2
=
a
b
.
Thus,
p
1
(
t
)
≡
0and
p
2
(
t
)
≡
a/b
are equilibrium solutions. ±or
p
1
<p<p
2
,
f
(
p
)
>
0, and
f
(
p
)
<
0when
p>p
2
.(A
l
so
,
f
(
p
)
<
0for
p<p
1
.) Thus the phase line for the given equation
is as it is shown in ±igure 3A. ±rom this picture, we conclude that the equilibrium
p
=
p
1
is
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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