It is easy to see that the population doubles when 120 = 60
e
0
.
14384
t
, so 0
.
14384
t
d
= ln(2) or the
doubling time is
t
d
=
ln(2)
r
= 4
.
819 weeks
.
b. We begin by separating variables, so the general solution satisfies:
Z
dP
P
=
Z
(
a

b t
)
dt
or
ln(
P
(
t
)) =
a t

bt
2
2
+
C
or
P
(
t
) =
e
C
e
a t

bt
2
2
.
Since the initial value is
P
(0) = 60, it follows that
e
C
= 60. Thus,
P
(
t
) = 60
e
a t

bt
2
2
.
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We now use the data at
t
= 2 and 4 weeks. It follows from the solution above that
80
=
60
e
2
a

2
b
90
=
60
e
4
a

8
b
.
We rearrange the terms and take logarithms of both sides to get
2
a

2
b
=
ln
4
3
4
a

8
b
=
ln
3
2
.
We solve these equations simultaneously to obtain
2
b
= ln
4
3

1
2
ln
3
2
,
so
b
= 0
.
042475. But
a
=
b
+
1
2
ln(4
/
3) or
a
= 0
.
1863. It follows that the solution is
P
(
t
) = 60
e
0
.
1863
t

0
.
021237
t
2
.
The population reaches a maximum when the derivative is zero, which occurs when
t
max
=
a
b
=
4
.
3865, so the maximum population is
P
(
t
max
) = 90
.
286.
15. a. For the differential equation
dy
dt
=
t
(2

y
), the Euler formula is given by
y
n
+1
=
y
n
+
h
(
t
n
(2

y
n
)) =
y
n
+ 0
.
25 (
t
n
(2

y
n
))
.
For this problem,
y
0
= 4, we can use the Euler’s formula to create the following table:
t
0
= 0
y
0
= 4
t
1
= 0
.
25
y
1
=
y
0
+ 0
.
25 (
t
0
(2

y
0
)) = 4 + 0
.
25(0)(2

4) = 4
t
2
= 0
.
5
y
2
=
y
1
+ 0
.
25 (
t
1
(2

y
1
)) = 4 + 0
.
25(0
.
25)(2

4) = 3
.
875
t
3
= 0
.
75
y
3
=
y
2
+ 0
.
25 (
t
2
(2

y
2
)) = 3
.
875 + 0
.
25(0
.
5)(2

3
.
875) = 3
.
6406
t
4
= 1
.
0
y
4
=
y
3
+ 0
.
25 (
t
3
(2

y
3
)) = 3
.
6406 + 0
.
25(0
.
75)(2

3
.
6406) = 3
.
3330
Thus, the approximate the solution at
t
= 1 is
y
4
’
y
(1) = 3
.
3330
.
b. The differential equation is a separable differential equation.
If we write the differential
equation
dy
dt
=

t
(
y

2), then we have the following integrals:
Z
dy
y

2
=

Z
t dt
ln

y

2

=

t
2
2
+
C
y
(
t
)

2
=
e

t
2
/
2+
C
=
Ae

t
2
/
2
y
(
t
)
=
2 +
Ae

t
2
/
2
With the initial condition, we find that
A
= 2. Thus, the solution to the initial value problem is
y
(
t
) = 2 + 2
e

t
2
/
2
.
It follows that
y
(1) = 2 + 2
e

0
.
5
≈
3
.
21306. The error between the actual and Euler’s solution is
100
(
y
4

y
(1))
y
(1)
= 100
(3
.
3330

3
.
21306)
3
.
21306
= 3
.
73%
.
16. a. For the differential equation,
dR
dt
=

0
.
05
R
+0
.
2
e

0
.
01
t
with
R
(0) = 10 and
h
= 1, the Euler’s
formula is
R
n
+1
=
R
n
+
h
(

0
.
05
R
n
+ 0
.
2
e

0
.
01
t
n
) =
R
n

0
.
05
R
n
+ 0
.
2
e

0
.
01
t
n
.
Iterating this, we create a table
t
0
= 0
R
0
= 10
t
1
= 1
R
1
=
R
0

0
.
05
R
0
+ 0
.
2
e

0
.
01
t
0
= 10

0
.
5 + 0
.
2 = 9
.
7
t
2
= 2
R
2
=
R
1

0
.
05
R
1
+ 0
.
2
e

0
.
01
t
1
= 9
.
7

0
.
485 + 0
.
198 = 9
.
413
t
3
= 3
R
3
=
R
2

0
.
05
R
2
+ 0
.
2
e

0
.
01
t
2
= 9
.
413

0
.
471 + 0
.
096 = 9
.
138
Thus, the approximate the solution at
t
= 3 is
R
3
≈
R
(3) = 9
.
138
.
b. The DE is linear and can be written
dR
dt
+ 0
.
05
R
= 0
.
2
e

0
.
01
t
with
μ
(
t
) =
e
0
.
05
t
.
It follows that it can be written
d
dt
e
0
.
05
t
R
(
t
)
= 0
.
2
e
0
.
04
t
or
e
0
.
05
t
R
(
t
) = 0
.
2
Z
e
0
.
04
t
dt
= 5
e
0
.
04
t
+
C.
Thus,
R
(
t
) = 5
e

0
.
01
t
+
Ce

0
.
05
t
,
or
R
(0) = 10 = 5 +
C.
The solution is
R
(
t
) = 5
e

0
.
05
t
+ 5
e

0
.
01
t
. The correct solution at
t
= 3 is
R
(3) = 9
.
15576. The
percent error between the correct solution and the Euler solution
100
R
3

R
(3)
R
(3)
= 100
9
.
138

9
.
15576
9
.
15576
=

0
.
190%
.
17. (Allee effect) Consider the DE given by the model:
dP
dt
=
P
9

0
.
01(
P

70)
2
=
A
(
P
)
.
The equilibria of this population model satisfy
P
(
9

0
.
01(
P

70)
2
)
= 0
.
Thus,
P
e
= 0, 40, and
100. From the phase portrait below, it is easy to see that the equilibria
P
e
= 0 and 100 are stable,
while
P
e
= 40 is unstable. The carrying capacity for this population is
P
e
= 100, and the critical
threshold number of animals required to avoid extinction is
P
e
= 40.
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0
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200
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<
>
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>
>
P
P
(
9

0
.
 Fall '08
 staff