Download Section 2.1

Section 2.1
MODELING VIA SYSTEMS
A tale of rabbits and foxes
Suppose you have two populations: rabbits and foxes.
R(t) represents the population of rabbits at time t.
F(t) represents the population of foxes at time t.
• What happens to the rabbits if there are no foxes?
Try to write a DE.
• What happens to the foxes if there are no rabbits?
Try to write a DE.
• What happens when a rabbit meets a fox?
• If R is the number of rabbits and F is the number of foxes, the
number of “rabbit-fox interactions” should be proportional to
what quantity?
The predator-prey system
A system of DEs that might describe the behavior of the
populations of predators and prey is
dR
 2R 1.2RF
dt
dF
 F  0.9RF
dt
1.
2.
3.
4.
What happens if there are no predators? No prey?
Explain the coefficients of the RF terms in both equations.
What happens when both R = 0 and F = 0?

Are there other situations in which both populations are
constant?
5. Modify the system so that the prey grows logistically if there are
no predators.
Exercises
Page 164, 1-6. I will assign either system (i) or (ii).
Graphing solutions
Here are some solutions to
P(0) = 0
dR
 2R 1.2RF
dt
dF
 F  0.9RF
dt

prey
predators
R(0) = 0
predators
prey
A startling picture!
Here’s what happens if we start with R(0) = 4 and
F(0) = 1.
predators
prey
The phase plane
Look at PredatorPrey demo.
This is the graph of the
parametric equation
(x,y) = (R(t), F(t)) for the IVP.
R(0) = 4
F(0) = 1
Exercises
• p. 165 #7a, 8ab
• Look at GraphingSolutionsQuiz in the Differential
Equations software (hard!)
Spring break!
Now for something completely different…
Suppose a mass is suspended on a spring.
• Assume the only force acting on the mass is the force
of the spring.
• Suppose you stretch the spring and release it. How
does the mass move?
Quantities:
y(t) = the position of the mass at time t.
– y(0) = resting
– y(t) > 0 when the spring is stretched
– y(t) < 0 when the spring is compressed
Newton’s Second Law: force = mass  acceleration
d2y
Fm 2
dt
Hooke’s law of springs: the force exerted by a spring is
proportional to the spring’s displacement from rest.

Fs  ky
k is called the spring constant and depends on how
powerful the spring is.
DE for a simple
harmonic oscillator
Combine Newton and Hooke:
d2y
Fs  ky  m 2
dt
Sooo….

d2y k
 y0
2
m
dt
which is the equation for a simple (or undamped)
 oscillator. It is a second-order DE
harmonic
because it contains a second derivative (duh).
How to solve it!
Now we do something really clever. We don’t have any
methods to solve second-order DEs.
Let v(t) = velocity of the mass at time t.
Then v(t) = dy/dt and dv/dt = d2y/dt2. Now our DE
becomes a system:
dy
v
dt
dv
k
 y
dt
m
Comes from our assumption
Comes from the original DE
Exercises
p. 167 #19
• Rewrite the DE as a system of first-order DEs.
• Do (a) and (b).
• Check (b) using the MassSpring tool.
• Do (c) and (d).
Homework
(due 5pm Thursday)
• Read 2.1
• Practice: p. 164-7, #7, 9, 11, 15, 17, 19
• Core: p. 164-7, #10, 16, 20, 21
Some of the problems in this section are really wordy.
You don’t have to copy them into your HW.