Math 2015
Lesson 10
A Generalization of Exponential Growth
We will look at a new differential equation, with a solution closely related to the
exponential equation.
Growth Proportional to a Difference
We will consider the differential equation
where k and A are both constants. Note that where before we considered growth which is
proportional to y, we are now considering growth which is proportional to the difference
between y and some constant A. In other words, growth (or decay) is faster the further
we are away from A.
Note first that the function y = A is a solution to the differential equation:
Solving the Equation
Let’s solve the equation, using our solution for y ′ = ky :
dy
= k( y − A)
dt
If we make the substitution u = y – A, then we see that u′ =
So our equation becomes
We know the solution to this equation is ______________________
so we get
Thus, the solution to y ′ = k(y − A) is y = A + Ce kt , where C can be any constant. (Note
that A however is the constant given in the equation!)
Example:
Solve the initial value problem y ′ = 2(y −10) , y(0) = 8.
We know the general solution is ___________. Then since y(0) = _____,
we must have C = ___. Thus the solution is _______________.
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Math 2015
Lesson 10
We can solve a variety of differential equations with this new formula:
Example:
Solve the initial value problem y ′ = −2y − 6 , y(0) = 1. (HINT: You must
first put this into the appropriate form.)
What if we instead had y ′ = −2y + 6 and y(0) = 3? What solution would
we get then?
Newton’s Law of Cooling
Newton’s law of cooling states that the rate at which an object cools down (or warms up)
is proportional to the difference between the temperature of the object and the
temperature of its surroundings. (The surrounding temperature is called the ambient
temperature.) Thus, if we let y be the temperature of some object which is heating up or
cooling down, and we let A represent the ambient temperature, Newton’s law of cooling
states that
y ′ = k(y − A)
Therefore, the temperature of the object after time t has passed is
y = A + Ce kt
where A + C is the temperature at time t = 0. In practice, we must determine the constant
k based on the temperature of the object at a later time.
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Math 2015
Example:
Lesson 10
A cup of very hot coffee is at 120º (Fahrenheit) when it is served. If left on
the table in a room with ambient temperature of 70º, it is found to have
cooled to 115º in one minute. How many minutes will the coffee have to
sit before the temperature drops to 100º?
Here, our differential equation is
(The ambient temperature is 70º, and we have not yet determined k.) Thus
the temperature after t minutes is given by
Next, let us calculate C. We will use the fact that at t = 0, y = 120:
Finally, we must use the additional information to calculate k. We know
that y(1) = 115, and so we set up and solve the appropriate equation:
Thus we have k ≈ _______, and a solution of
Setting the solution equal to 100º and solving yields
Thus the coffee has cooled to 100º after about ____ minutes, or __minutes
___ seconds.
Stable and Unstable Equilibrium Solutions
An equilibrium solution to a family of exponential functions is a horizontal line that that
the exponential functions either tend toward or go away from. It has the form
y = a constant.
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Math 2015
Lesson 10
Since an equilibrium solution is a constant, we need to find where
Since we know y ′ = k(y − A) , we must determine where
This is obviously just where
So
If
.
.
, if k is not itself zero. If it is, we really just have y ′ = 0 .
, we just have our exponential equation
, so
If other solutions tend toward the equilibrium solution as t goes to infinity, we say that
the equilibrium solution is
If other solutions move away from the equilibrium solution, we say it is
For example, if we solve y ′ = ky , we get solutions of the form y = Cekt . If k is negative,
then we have a negative exponential, which goes towards 0 as t goes to infinity.
If k is positive, then we get positive exponentials which go to infinity or minus infinity
as t goes to infinity. All the solutions move away from y = 0!
Below are several solutions to y ′ = ky
where k = –1:
where k = 1:
2
40
1
20
0.5
-1
-2
1
1.5
2
2.5
0.5
3
1
1.5
2
2.5
3
-20
-40
Notice in the graphs above on the left that all the solutions tend toward y = 0, regardless
of the initial condition. This means y = 0 is a stable equilibrium solution.
Notice in the graphs above on the right that all the solutions move away from y = 0,
regardless of the initial condition. This means y = 0 is an unstable equilibrium solution.
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Math 2015
Lesson 10
Now let’s look at our new equation: y ′ = k(y − A)
We know the solutions are all of the form
is
. Is this solution stable or unstable?
, and the equilibrium solution
What happens to y = A + Ce kt as t goes to infinity if k is negative?
What happens to y = A + Ce kt as t goes to infinity if k is positive?
Which diagram below corresponds to k > 0 and which to k < 0? Which is stable and
which is unstable?
A
A
1
2
3
1
2
3
Summary
Today, we have
•
Interpreted the differential equation y ′ = k(y − A) as meaning the growth or decay
of something is proportional how far it is away from a fixed quantity.
•
Learned that the solution to y ′ = k(y − A) is y = A + Ce kt , where C is a constant.
•
Solved problems involving Newton’s law of cooling (or warming).
•
Determined equilibrium solutions to a differential equation, and identified
whether they were stable or unstable.
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