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

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
.

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 predator-prey 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).