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Sec 3.2
12.
First, we equate the expression (21) for
0
p
at time
a
t
and
b
t
, we get
1
1
1
0
1
(1
)
a
a
Ap t
a
Ap t
a
p p e
p
p
p
e


=


and
1
1
1
0
1
(1
)
b
b
Ap t
b
Ap t
b
p p e
p
p
p
e


=


.
Since
2
b
a
t
t
=
, we set
1
a
Ap t
e
χ

=
and
1
2
b
Ap t
e

=
, hence
2
1
1
2
1
1
(1
)
(1
)
a
b
a
b
p p
p p
p
p
p
p
=




1
1
(
)
(
)
a
b
b
a
p
p
p
p p
p

=

.
Then substituting
1
1
(
)
(
)
a
b
b
a
p
p
p
p p
p

=

into
1
0
1
(1
)
a
a
p p
p
p
p
=


and solving for
1
p
,
1
1
1
0
1
1
1
(
)
(
)
(
)
(1
)
(
)
a
b
a
b
a
a
b
a
b
a
p
p
p
p p
p p
p
p
p
p
p
p
p
p p
p


=




1
1
0
1
1
1
1
(
)
(
)
(
(
)
(
))
a
a
b
b
a
a
b
a
a
b
p p p
p
p
p
p p p
p
p
p p
p
p
p
p

=





2
1
0
2
1
(
)
2
a
b
b
a
b
a
p
p
p
p
p p
p p
p

=

+
0
0
1
2
0
2
(
)
a
b
b
a
a
a
b
p p
p p
p p
p
p
p
p p

+
=

.
Now we find
A
from
1
a
Ap t
e

=
, i.e., from
1
1
1
(
)
(
)
a
Ap t
a
b
b
a
p
p
p
e
p p
p


=

, which yields:
1
1
1
(
)
1
ln[
]
(
)
b
a
a
a
b
p p
p
A
p t
p
p
p

=

0
0
2
0
0
0
1
2
0
2
((
)
)
1
ln[
]
2
((
)
)
a
b
b
a
b
a
a
a
b
a
b
b
a
a
a
a
b
a
b
p p
p p
p p
p
p
p
p
p p
p p
p p
p p
p t
p
p
p
p
p p

+


=

+


=
2
0
0
0
2
1
0
0
0
((
2
)
)
1
ln[
]
(
2
)
(
)
b
a
b
b
a
a
b
a
a
b
b
a
a
b
a
b
p
p p
p p
p p
p
p p
p t
p p
p p
p p
p
p p
p p

+



+


0
1
0
(
)
1
ln[
]
(
)
b
a
a
b
a
p p
p
p t
p p
p

=

.
14.
If we set
0
t
=
to be the year 1970, then substituting
0
300
p
=
into formula
0
( )
kt
p t
p e
=
we have:
( )
300
kt
p t
e
=
.
(*)
We also know that by 1980 (
10
t
=
years) the population had grown to 1500, thus
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(10)
300
1500
k
p
e
=
=
ln 5
10
k
=
Substituting this value into equation (*), we have:
ln5
10
( )
300
t
p t
e
=
.
Hence, the population in 2010 (
40
t
=
years) is
ln5
40
4
10
(40)
300
300 5
187500
p
e
=
=
=
.
15.
Similarly, we set
0
t
=
to be the year 1970, then
0
300
p
=
. We also know that
(5)
1200
p
=
(population in 1975) and
(10)
1500
p
=
(population in 1980). Then we
substitute
(5)
1200
p
=
and
(10)
1500
p
=
into the formula
(
29
1
0
1
0
1
0
( )
Ap t
p p
p t
p
p
p e

=
+

,
and obtain two nonlinear equations for the two unknowns
1
p
and
A
:
(
29
1
1
5
1
300
1200
300
300
Ap
p
p
e

=
+

,
(
29
1
1
10
1
300
1500
300
300
Ap
p
p
e

=
+

.
To solve
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This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.
 Spring '08
 Schecter
 Differential Equations, Equations

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