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

• 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?• Modify the system so that the prey grows logistically if there are

no predators.

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dR

dt= 2R−1.2RF

dF

dt= −F + 0.9RF

Exercises

Page 164, 1-6. I will assign either system (i) or (ii).

Graphing solutions

Here are some solutions to

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dR

dt= 2R−1.2RF

dF

dt= −F + 0.9RF

prey

predators

P(0) = 0

predators

prey

R(0) = 0

A startling picture!

Here’s what happens if we start with R(0) = 4 and F(0) = 1.

prey

predators

The phase plane

Look at PredatorPrey demo.

R(0) = 4F(0) = 1

This is the graph of the parametric equation (x,y) = (R(t), F(t)) for the IVP.

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

Hooke’s law of springs: the force exerted by a spring is proportional to the spring’s displacement from rest.

k is called the spring constant and depends on how powerful the spring is.

€

F = md2y

dt2

€

Fs = −ky

DE for a simple harmonic oscillator

Combine Newton and Hooke:

Sooo….

which is the equation for a simple (or undamped) harmonic oscillator. It is a second-order DE because it contains a second derivative (duh).

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Fs = −ky = md2y

dt2

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d2y

dt2+k

my = 0

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

dt= v

dv

dt= −

k

my

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.