Chapter 2
Even-Numbered Homework Solutions
2.1
2. Find all equilibrium points for the two systems. Explain the significance of these points in terms of the predator and
prey populations.
System (i):
dx
dy
Set
= 0 and
= 0. We have:
dt
dt
x
0 = 10x 1 −
− 20xy ⇒ 0 = x (10 − x − 20y) ⇒ 0 = x and 0 = 10 − x − 20y
10
xy
0 = −5y +
⇒ 0 = 20y(−100 + x) ⇒ 0 = y(−100 + x) ⇒ 0 = y and 100 = x
20
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One equilibrium point is given by (0, 0). Solving for the other gives: 0 = 10 − 100 − 20y ⇒ −20y = 90 ⇒ y =
.
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We can also find 100, −
as an equilibrium point, but note that a negative population is impossible. The
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third equilibrium point is (10, 0).
System (ii):
Using the same approach
that we used for system (i), verify that the equilibrium points for this system are
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(0, 0), (0, 15), and
, 30 .
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4. For each system, describe the behavior of the prey population if the predators are extinct. (Sketch the phase line for the
prey population assuming that the predators are extinct, and sketch the graphs of the prey population as a function of time
for several solutions. Then interpret these graphs for the prey population.)
dx
x
= 10x 1 −
= 10x − x2 . Thus, the critical points are x = 0 and x = 10. For x < 0,
dt
10
dx
dx
dx
is negative, for 0 < x < 10,
is positive, and for x > 10,
is negative. The population of the prey will attempt to
dt
dt
dt
reach 10, which is its equilibrium.
For the first system,
dx
dx
= 0.3x, and the phase line should include x = 0 as a critical point, and
should be
dt
dt
0.3t+C
positive for x > 0 and negative for x < 0. Also, x(t) = e
. The (approximate) graph of x(t) below shows that the
population of the prey in system (ii) will grow exponentially following the extinction of their predators.
For the second system,
system ii.jpg
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6. For each system, describe the behavior of the predator population if the prey are extinct. (Sketch the phase line for the
predator population assuming that the prey are extinct, and sketch the graphs of the predator population as a function of
time for several solutions. Then interpret these graphs for the predator population.)
dy
For system (i),
= −5y, so y(t) = e−5t+C , where C is a constant. The (approximate) graph of y(t) below shows
dt
that the predators in this system will die off following the extinction of their prey. The phase line should have y = 0 as a
dy
critical point with
negative for x > 0.
dt
system i.jpg
For system (ii),
15e15t+C
dy
= 15y − y 2 , and y(t) = 15t+C
. The phase line should have y = 0 and y = 15 as critical
dt
e
+1
dy
points, and
increasing for 0 < x < 15 and decreasing for x > 15. Thus, the population of predators tries to reach
dt
equilibrium at y = 15.
2.2
8. Convert the second-order differential equation
d2 y
+ 2y = 0
dt2
into a first-order system in terms of y and v, where v =
dy
.
dt
(a) Determine the vector field associated with the first-order system.
The system can be written as:
dy
=v
dt
dv
= −2y.
dt
The vector field is given by V(y, v) = (v, −2y).
(b) Sketch enough vectors in the vector field to get a sense of its geometric structure.
The vector field has an ovular structure.
10. Consider the system
dx
= −2x + y
dt
dy
= −2y
dt
and its corresponding direction field (page 179 of the textbook).
(a) Sketch a number of different solution curves on the phase plane.
The following picture was generated by HPGSystemSolver:
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solution curves.png
(b) Describe the behavior of the solution that satisfies the initial condition (x0 , y0 ) = (0, 2).
From the solution curves above, it is obvious that as t increases, x and y tend to the origin. Thus,
y → 0 and x → 0 as t → ∞.
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