Math 224 Exam 2 Solutions
September 28, 2015
1. (20 points; 10 points each) Consider the two following models of interacting populations:
A:
dx
= 2x − xy,
dt
dy
= −2y + xy,
dt
B:
dx
= 2x − xy,
dt
dy
= 2y − xy.
dt
(a) Identify which vector field is for which system, and explain your answer.
Solution: The first vector field is for system A and the second vector field is
for system B. There are many ways to tell the difference. One is to look at
the vector at the point (1, 1). For system A the vector at that point is (1, −1),
pointing southeast, which matches the first vector field but not the second one.
For system B the vector at that point is (1, 1), pointing northwest, which matches
the second vector field but not the first one.
Remark: Lots of people tried to answer this part by observing that the first
vector field looks like the one for the predator-prey example from class, and the
second vector field looks like the one for the competing species example from class.
But we don’t know that all systems of those types have vector fields that look
like this.
This part is much simpler because it doesn’t involve (hard) mathematical modeling issues; it just needs some (easy) math.
(b) Identify which system is a predator/prey model and which is a competing species
model, and explain your answer.
Solution: In system A, interactions between species decrease the growth of
species x and increase the growth of species y; moreover species y has a negative growth rate in the absence of species x. Thus this is a predator/prey model
in which species x is the prey and species y is the predator.
In system B each species has a positive growth rate on its own, but interactions
between species decrease the growth of both species. Thus this is a competing
species model.
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2. (20 points) Solve the system
dx
= 2x − 8y 2 ,
dt
dy
= −3y.
dt
Solution: This is a partially decoupled system. We start with the second equation,
which has general solution y = Ae−3t . Substituting into the first equation, we get the
nonhomogeneous linear equation
dx
= 2x − 8A2 e−6t .
dt
dx
= 2x has general solution xh = Be2t . To find a particdt
ular solution to the nonhomogeneous equation, we try xp = Ce−6t , so that
The homogeneous equation
dxp
= −6Ce−6t
dt
and
2xp − 8A2 e−6t = (2C − 8A2 )e−6t .
Equating these, we get
−6C = 2C − 8A2 ,
so we see that our guess for xp works if C = A2 .
Putting everything together, the general solution to the system is
x = A2 e−6t + Be2t ,
y = Ae−3t .
2
3. (20 points) Solve the initial value problem
d2 y
dy
+ 5 + 6y = 0,
2
dt
dt
y(0) = 1,
y 0 (0) = 1.
Solution: This second-order equation with constant coefficients has the characteristic
equation
s2 + 5s + 6 = 0,
which has solutions s = −2, −3. So the general solution to the differential equation is
y = Ae−2t + Be−3t .
From this we find
y(0) = A + B,
y 0 = −2Ae−2t − 3Be−3t ,
and so
y 0 (0) = −2A − 3B.
So to fit the initial conditions we need to solve
A + B = 1,
−2A − 3B = 1.
This has solution A = 4, B = −3, so the solution to our IVP is
y = 4e−2t − 3e−3t .
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4. (20 points; 10 points each) In the movie I Am Legend, zombie-like Infecteds work
together to increase the number
of Infecteds. We can modify√the SIR model to include
√
this by replacing I with I in the interaction term (since I ≥ I when 0 ≤ I ≤ 1).
We obtain the system
√
dS
= −αS I,
dt
√
dI
= αS I − βI.
dt
(a) Find all the equilibrium points of this model.
√
I = 0 we √
need either S = 0 or I = 0. If I = 0 then
Solution
1:
For
−αS
√
αS I − βI = 0. If S = 0, then αS I − βI = −βI, which is only 0 if I is zero.
Therefore the equilibrium points for this model are exactly those points with
I = 0.
Solution 2: This isn’t really a different
an even easier way to see
√ approach, just
dI
dS
how it works out. If dt = 0 then αS I = 0, so dt = −βI. This will happen
= 0 automatically. So the equilibrium
exactly when I = 0, which then makes dS
dt
points are those with I = 0.
(b) Find the region of the phase plane where
dI
> 0.
dt
√
Solution: These are exactly
the
points
where
αS
I − βI > 0, or more simply
√
the points where αS > β I. Even easier to understand on the phase plane, it’s
where
2
α
I<
S 2,
β
so it’s the region under a particular parabola.
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5. (20 points; 10 points each) Consider the following model for Paul and Bob’s cafés, as
discussed in the textbook:
dx
= x − y,
dt
dy
= x + y,
dt
where x(t) is Paul’s daily profit and y(t) is Bob’s daily profit.
(a) Describe briefly the interaction between the cafés. (For example, if Paul is making
a profit, how does that affect each of them?)
Solution: In this model, if Paul is making a profit, that tends to increase both
cafés’ profits. If Bob is making a profit, that tends to increase Bob’s profits but
decrease Paul’s profit.
(b) Suppose that both stores are making a profit on day 0 (i.e., x(0) > 0 and y(0) > 0).
According to this model, will there ever be a day on which both stores make
exactly 0 profit? Explain your answer.
Solution: The point (0, 0) is an equilibrium point of this autonomous system.
Therefore, according to the Uniqueness Theorem, that equilibrium solution is the
only solution which ever passes through that point in the phase plane. So if either
of x(0) and y(0) is nonzero, then x(t) and y(t) will never both be 0 for the same
t.
Remark: A common mistake on both parts of this problem was to mix up the profit
or dy
.
x(t) or y(t) with the rate of change of profit dx
dt
dt
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