MATH 2280-002
Lecture Notes: 01/07/2013
Math 2280 Section 002 [SPRING 2013]
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Introduction
The goal of this course is to introduce you to the tools you will need for mathematical modeling.
Real-world
situation ffLL
LLInterpret
LLL
LLL
r
r
Formulate
rrr
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xx rr
r
Mathematical
model
Analyze
// Mathematical
results
We will learn how to formulate a real-world question as a mathematical problem and then analyze
this problem to get mathemtical results. Interpreting these results give us insight into the original
question.
To see what I mean, let’s consider the following example.
Example. Suppose a colony of chupacabras is growing at a rate of 1.5% each month. If there are
10 chupacabras now, how many will there be in 2 years?
FORMULATION:
How do we express this situation as a mathematical equations? Let P (t) be
the population at time t (in months). The growth rate is 1.5% of the current
population. That is,
P ′ (t) = .015P (t).
When t = 0, the population is 10.
P (0) = 10.
Remark. P (0) = 10 is called the initial condition for the differential equation P ′ (t) = .015P (t).
An initial condition is just the value of the population at a specified time t. For example, P (3)
(if it was provided) would also be an initial condition. A differential equation together with some
initial condition(s) is called an initial value problem.
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MATH 2280-002
ANALYSIS:
Lecture Notes: 01/07/2013
Solve the differential equation accounting for the initial condition. Our differential
equation and initial condition has the solution
P (t) = 10e.015t .
(We can use the separation of variables technique that we’ll review later this week.)
That means that when t = 24 the population is
P (24) = 10e.015(24) ≈ 14.
Question. Why is our solution for P (t) unique?
In this case, because we used separation of variables, we know we get a solution that is unique up
to a constant, and we can uniquely determine this constant using the initial condition. Sometimes
we will not know a technique to solve an initial value problem, but we can still determine whether
any solution exists and is unique. More on that later.
INTERPRETATION:
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What does our math model tell us about our original scenario? We estimate that
in 2 years there will be 14 chupacabras.
Math Models
Over the course of the semester, we’ll see many examples of differential equations or systems of
differential equations used to model physical phenomenon. We’ve seen one example already. One
simple model for population change is
dP
= kP,
dt
where P is the population at time t. We’ll see that k > 0 (as in our chupacabra example) indicates
growth and k < 0 indicates decay.
Newton’s law of cooling can be represented by the differential equation
dT
= −k(T − A),
dt
where T is the temperature of an object and A is the ambient temperature. We can use this to
approximate how long a turkey removed from the oven will take to cool off or estimate time of death
for a recent corpse.
A third math model we’ll look at soon is Torricelli’s law,
√
dV
= −k y,
dt
which descibes the rate at which the volume V of a liquid in a leaking tank changes over time as a
function of the height h of the liquid.
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Verify a Solution
Obtaining the solution to a differential equation can be difficult, but checking a solution is easy. All
you have to do is plug in your guess!
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MATH 2280-002
Lecture Notes: 01/07/2013
Exercise: Verify that y(x) = xex − ex is a solution to the differential equation
y + y ′′ = 2y ′ .
Remark. There are actually infinitely many solutions to the above DEQ. Can you describe these
solutions?
Sometimes the easiest way to solve a differential equation is by clever guessing. For example, an
easy (and correct!) guess for the differential equation
y + y ′′ = 2y ′ .
is ex . We’ll see later in the semester that the solution space for this differential equation is two
dimensional, and since xex − ex and ex are linearly independent (also not obvious), xex − ex and ex
span the solution space for y + y ′′ = 2y ′ . In other words, every solution is a linear combination of
xex − ex and ex .
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