Introduction to Differential Equations

Many  applications in both the physical and social sciences involve solving equations that involve the derivative of a function. Such equations are called differential equataions . In general, a differential equation may also involve higher order derivatives of a function. The topic of differential equations is a vast area of study in mathematics that continues to generate research today. Two simple examles of differential equations that occur in population models are

and

In the first example,sometimes referred to as the simple population equation, k  is a constant and y  is assumed to be a function of the variable t , which is interpreted as time  measured in some appropriate unit such as seconds, hours, months, etc. The second equation is known as the logistic  equation. In the logistic equation, k  and M  are constants and again, y  is assumed to be a function of t . We will study these equations in some detail.

Since the above equations involve both a function y ( t ) and the derivative , in order to solve for the function y ( t ), it is reasonable to expect that integration should be involved.

The Simple Population Equation

In the simple population equation y ( t ) represents the size of a certain population at a given time t. The population may be the size of a bacteria culture, the number of a particular species in a particular reserve or the number of individuals in a particular city or country. Since  represents the rate of change of y , the population, with respect to t , the differential equation

states that the population function changes at a rate proportional (equal to a constant times) to the size of the current population. In other words, the larger the size of the population, the greater the increase of the population per unit of time. If we think of the differential equation mathematically, we are trying to find a function y ( t ) having the property that its derivative is equal to the constant k  times the same function. If we think about all the differentiation rules, it is not difficult to determine such a function. Recall that the natural exponential function  has the property that the derivative is the same as the original function. In other words this function satisfies the differential equation

This is the simple population equation with k  = 1.

Project - Part 1

Determine how to modify the exponential function to become a solution of the simple population equation .

Systematic Solution of

The simple population equation is simple enough to guess a solution based on our knowledge of basic differentiation formulas. For more complicated differential equations, this is not a very useful approach. Another question that arises is whether there are solutions other than the one we determined. We now solve the same differential equation in a more systematic manner by using integration. We first rewrite the equation in the form

and then we integrate both sides of this equation with respect to the variable t .

By the chain rule the left-hand side of this equation is  which evaluates to . Since k  is a constant, the right-hand side of the equation evaluates to . We could accomplish this by using the Maple command

>	int(1/y,y)=int(k,t);

Recall that when we take the antiderivative of an expression we should add an arbitrary constant. In fact, we should do this on both sides of the equation after performing the integration. However, the two arbitrary constants can be moved to one side  the equation and replaced by a single arbitrary constant C . Thus after performing the integration on both sides, we obtain

To solve this equation for y , we raise e  to bothe sides of this equation:

This gives

Since c  is an arbitrary constant,  is an arbitrary positive constant, which we relabel as A . Thus we write our solution in the form

Recall that y ( t ) represents the population at time t. More precisely, t  represents the time since the population was first introduced. In other words, the population was initially introduced at time t  = 0. Since , it follows that the constant A  represets the initial population . The fact that y (0) = A  is often referred to as the initial condition .

Project - Part 2

1) A population of bacteria is modeled by the simple population equation. The initial population was 4000 and after 1 hour the population was 6500. Use the solution derived above for the population function and determine the numerical values of A  and k . Sketch the population function for t  ranging between 0 and 3 hours.

2) Determine the population of the bacteria 3 and a half hours after the population was first introduced.

3) Determine the time required for the population of bacteria to double.

4) Determine the time required for the population to reach 10,000.

Solving the differential equation with Maple

Maple has a package for solving differential equations called DEtools. In order to use this package, it must first be loaded. The command in this package that we want to use is dsolve . This command takes a set of arguments, the first being the differential equation and the second being the initial condition. The diff  notation is used for the derivative and y  must be represented as y ( t )

>	with(DEtools):

>	dsolve({diff(y(t),t)=k*y(t),y(0)=4000},y(t));

Note that once the differential equation has been solved, one must still define the solution as a function in Maple in order to manipulate the solution. It is advisable to use a letter different than y for the population function since once y  is a specific function, the dsolve  command will not operate properly. We will use the letter P  for the population function.

>	P:=t->4000*exp(k*t);

>

One can now use the solve  command to solve for the constant k .

>	solve(P(1)=6500,k);

We can convert this to a decimal and assign the value to the variable k .

>	k:=evalf(ln(13/8));

We can now work with the population function just like any other function that has been defined in Maple.

>

P(t);

Project - Part 3

Use the population function defined above to answer the following questions:

1) Determine the population of the bacteria after 6 hours.

2) Determine the time required for the population to triple.

3) Determine the time required for the population to reach 20,000.

The Logistic Population Model

The solutions of the simple population equation grow exponentially as time increases and become unbounded as t  approaches . For many models this is unrealistic. The logistic population model has a maximum carrying capacity  or maximum sustainable population M which takes into account limitations imposed by space or other resources. The logistic differential equation is

Notice that as y  gets close to M , the factor  gets smaller and smaller. This tends to make the rate of growth of the population smaller. In the logistic model there are two competing forces that control the rate of growth - one is the size of the population, which tends to increase the rate of growth and the other is how close the population is to the maximum carrying capacity, which tends to decrease the rate of growth. In most applications of the logistic population model, the initial population is (substantially) smaller than the maximum carrying capacity.

Note that it is not easy to guess a solution of this equation based on our knowledge of the basic differentiation formulas. Even if we try to solve the equation by integration, the procedure is non-trivial and requires the method of partial fractions

In this case it is more convenient to use the dsolve  command. In order to avoid typing more complicated expressions, we first assign the differential equation to a variable eq . We will use a more general initial condition .

>	restart:with(DEtools):

>	eq:=diff(y(t),t) = k*y(t)*(1-y(t)/M);

>	dsolve({eq,y(0)=A},y(t));

Again we define the population function P ( t ) as the right-hand side of the output.

>	P:=t->M/(1+exp(-k*t)*(M-A)/A);

Project - Part 4

Verify that the population function returned by the dsolve  command does indeed satisfy the logistic differential equation.

Project - Part 5

A population of 200 red foxes is introduced into a natural preserve that can sustain a maximum population of 600. After two years the population is estimated to be 235. Assume the population is modeled by the logistic differential equation.

1) Determine the population function for the red fox population.

2) Plot the graph of the red fox population over the interval from 0 to 40 years. Discuss the general properties of the graph including concavity.

3) Determine the time required for the red fox population in the preserve to reach 450.

4) Based on the graph of the population function, what appears to be happening to the population as t  approaches ? Can you verify your answer by using the formula for the population function.

5) Determine the time required for the red fox population to reach 90% of the carrying capacity. At what rate is the population increasing at this time?

Extra Credit

Determine the time at which the red fox population changes concavity. What is the population at this time? At what rate is the population changing at this time? What is the maximum rate of increase of the red fox population?