The general solution for this problem is
x
(
t
) =
c
1
e

t
4 cos(2
t
)
2 sin(2
t
)
+
c
2
e

t
4 sin(2
t
)

2 cos(2
t
)
+

2
2
.
We solve the IVP, so
x
(0) =
c
1
4
0
+
c
2
0

2
+

2
2
=
5
2
or
4
c
1

2
c
2
=
7
0
It follows that
c
1
=
7
4
and
c
2
= 0, so
x
(
t
) =
7
e

t
4
4 cos(2
t
)
2 sin(2
t
)
+

2
2
.
4. From the information, we write the rate
of change in
amounts
,
A
1
(
t
) and
A
2
(
t
).
The rate of change in amount is concentra
tion times flow rate.
dA
i
dt
=
amount enter

amount leave.
For Tank 1, the amount entering is
f
1
q
1
and
f
3
c
2
, while the amount leaving is
f
4
c
1
. Simi
lar expressions give the equation for Tank 2.
The concentration equations follow by simply dividing the amount equations by the appropriate
volumes.
With the data the amount equations are
dA
1
dt
=
0
.
3
·
8 + 0
.
4
·
c
2

0
.
7
·
c
1
,
dA
2
dt
=
0
.
2
·
15 + 0
.
2
·
c
1

0
.
4
·
c
2
.
Subscribe to view the full document.
Dividing by the volumes gives the concentration equations
dc
1
dt
=

7
2000
c
1
+
1
500
c
2
+
3
250
,
dc
2
dt
=
1
500
c
1

1
250
c
2
+
3
100
.
The equilibria are found first by solving:
7
2000
c
1
e

1
500
c
2
e
=
3
250
,

1
500
c
1
e
+
1
250
c
2
e
=
3
100
.
It follows that
c
1
e
= 10
.
8 g/l, while
c
2
e
= 12
.
9 g/l. This gives the equilibrium solution, which will
be the asymptotic limit for the concentrations.
We make a change of variables
z
1
(
t
) =
c
1
(
t
)

10
.
8 and
z
2
(
t
) =
c
2
(
t
)

12
.
9. The homogeneous
equation, ˙
z
=
Az
, is
˙
z
1
˙
z
2
=

0
.
0035
0
.
002
0
.
002

0
.
004
z
1
z
2
.
The characteristic equation is
det

0
.
0035

λ
0
.
002
0
.
002

0
.
004

λ
=
λ
2
+ 0
.
0075
λ
+ 0
.
00001 = 0
,
which gives
λ
1
=

0
.
0057656 with its associated eigenvector
v
1
=
1

1
.
13278
and
λ
2
=

0
.
0017344 with its associated eigenvector
v
2
=
1
.
13278
1
. This results in the general solution
x
1
(
t
)
x
2
(
t
)
=
c
1
1

1
.
13278
e

0
.
0057656
t
+
c
2
1
.
13278
1
e

0
.
0017344
t
+
10
.
8
12
.
9
,
which is a
stable node
. The solution to the IVP satisfies
1
1
.
13278

1
.
13278
1
c
1
c
2
=
2

10
.
8
3

12
.
9
=

8
.
8

9
.
9
,
which gives
c
1
= 1
.
05752 and
c
2
=

8
.
70206. It follows that the unique solution to the IVP is
x
1
(
t
)
x
2
(
t
)
=
1
.
05752

1
.
19794
e

0
.
0057656
t
+

9
.
85754

8
.
70206
e

0
.
0017344
t
+
10
.
8
12
.
9
.
This solution clearly shows that the trajectory converges asymptotically to the equilibrium solution
as expected.
5. The predatorprey model is given by:
dH
dt
=
0
.
1
H

0
.
0005
H
2

0
.
016
HP
=
f
1
(
H, P
)
,
dP
dt
=
0
.
005
HP

0
.
2
P
=
f
2
(
H, P
)
.
The equilibria satisfy:
H
e
(0
.
1

0
.
0005
H
e

0
.
016
P
e
)
=
0
,
P
e
(0
.
005
H
e

0
.
2)
=
0
.
One equilibrium is (
H
e
, P
e
) = (0
,
0), the extinction equilibrium.
When
P
e
= 0, then there is
another equilibrium at
H
e
= 200, the carrying capacity. Finally, there is a third equilibrium with
0
.
005
H
e

0
.
2 = 0 and 0
.
1

0
.
0005
H
e

0
.
016
P
e
= 0. The first equation gives
H
e
= 40, which gives
P
e
= 5 in the second equation. Thus, there is a coexistence equilibrium, (
H
e
, P
e
) = (40
,
5).
As we did in class, we use Taylor’s theorem to linearize this system about the equilibria (finding
the
Jacobian matrix
). If
h
(
t
) =
H
(
t
)

H
e
and
p
(
t
) =
P
(
t
)

P
e
, then the linearized system can
be written:
˙
h
˙
p
=
∂f
1
(
H
e
,P
e
)
∂h
∂f
1
(
H
e
,P
e
)
∂p
∂f
2
(
H
e
,P
e
)
∂h
∂f
2
(
H
e
,P
e
)
∂p
h
p
=
0
.
1

0
.
001
H
e

0
.
016
P
e

0
.
016
H
e
0
.
005
P
e
0
.
005
H
e

0
.
2
!
h
p
.
The linear system about (
H
e
, P
e
) = (0
,
0) is
˙
h
˙
p
=
0
.
1
0
0

0
.
2
h
p
,
which has eigenvalues
λ
1
=

0
.
2 with associated eigenvector
v
1
=
0
1
and
λ
2
= 0
.
1 with
associated eigenvector
v
2
=
1
0
.
This is a
saddle node
at the extinction equilibrium.
The
general linear solution is given by
h
(
t
)
p
(
t
)
=
c
1
0
1
e

0
.
2
t
+
c
2
1
0
e
0
.
1
t
.
Thus, if there are both predators and prey, then the solution moves away from extinction (unstable).
Subscribe to view the full document.
 Fall '08
 staff