6. An Introduction to Differential Equations

Motivating Questions

In this section, we strive to understand the ideas generated by the following important questions:

• What is a differential equation and what kinds of information can it tell us?

• How do differential equations arise in the world around us?

• What do we mean by a solution to a differential equation?

• What is a slope field and how can we use a slope field to obtain qualitative information about the solutions of a differential equation?

• What are stable and unstable equilibrium solutions of an autonomous differential equation?

Introduction

In previous chapters, we have seen that a function's derivative tells us the rate at which the function is changing. More recently, the Fundamental Theorem of Calculus helped us to determine the total change of a function over an interval when we know the function's rate of change. For instance, an object's velocity tells us the rate of change of that object's position. By integrating the velocity over a time interval, we may determine by how much the position changes over that time interval. In particular, if we know where the object is at the beginning of that interval, then we have enough information to accurately predict where it will be at the end of the interval.

In this section, we will introduce the concept of differential equations and explore this idea in more depth. Simply said, a differential equation is an equation that provides a description of a function's derivative, which means that it tells us the function's rate of change. Using this information, we would like to learn as much as possible about the function itself. For instance, we would ideally like to have an algebraic description of the function.

Preview Activity 6

The position of a moving object is given by the function s( t), where s is measured in feet and t in seconds. We determine that the velocity is v(t) = 4t + 1 feet per second.

(a) How much does the position change over the time interval [0, 4]?

(b) Does this give you enough information to determine s(4), the position at time t = 4? If so, what is s(4)? If not, what additional information would you need to know to determine s( 4) ?

(c) Suppose you are told that the object's initial position s(0) = 7. Determine -s(2), the object's position 2 seconds later.

(d) If you are told instead that the object's initial position is s(0) = 3,

what is s(2)?

(e) If we only know the velocity v(t) = 4t + 1, is it possible that the object's position at all times is s(t) = 2t² + t — 4? Explain how you know.

(f) Are there other possibilities for s(t)? If so, what are they?

(g) If, in addition to knowing the velocity function is v(t) = 4t + 1, we know the initial position s(0), how many possibilities are there for

s(t)?

What is a differential equation?

A differential equation is an equation that describes the derivative, or derivatives, of a function that is unknown to us. For instance, the equation

= x sin( x)

is a differential equation since it describes the derivative of a function y( x) that is unknown to us.

As many important examples of differential equations involve quantities that change in time, the independent variable in our discussion will frequently be time t. For instance, in the preview activity, we considered the differential equation

= 4t +1.

Knowing the velocity and the starting position of the object, we were able to find the position at any later time.

Because differential equations describe the derivative of a function, they give us information about how that function changes. Our goal will be to take this information and use it to predict the value of the function in the future; in this way, differential equations provide us with something like a crystal ball.

Differential equations arise frequently in our every day world. For instance, you may hear a bank advertising:

Your money will grow at a 3% annual interest rate with us.

This innocuous statement is really a differential equation. Let's translate: A(t) will be amount of money you have in your account at time t. On one hand, the rate at which your money grows is the derivative dA/dt. On the other hand, we are told that this rate is 0.03A. This leads to the differential equation

= 0.03A.

This differential equation has a slightly different feel than the

previous equation

= 4t + 1. In the earlier example, the rate

of change depends only on the independent variable t, and we may find s(t) by integrating the velocity 4t + 1. In the banking example, however, the rate of change depends on the dependent variable A, so we'll need some new techniques in order to find

A(t).

Activity 6-1

Express the following statements as differential equations. In each case, you will need to introduce notation to describe the important quantities in the statement so be sure to clearly state what your notation means.

(a) The population of a town grows at an annual rate of 1.25%.

(b) A radioactive sample loses 5.6% of its mass every day.

(c) You have a bank account that earns 4% interest every year. At the same time, you withdraw money continually from the account at the rate of $1000 per year.

(d) A cup of hot chocolate is sitting in a 70° room. The temperature of the hot chocolate cools by 10% of the difference between the hot chocolate's temperature and the room temperature every minute.

(e) A can of cold soda is sitting in a 70^° room. The temperature of the soda warms at the rate of 10% of the difference between the soda's temperature and the room's temperature every minute.

Differential equations may be classified based on certain characteristics they may possess. Indeed, you may see many different types of differential equations in a later course in differential equations. For now, we would like to introduce a few terms that are used to describe differential equations.

A first-order differential equation is one in which only the first derivative of the function occurs. For this reason,

= 1.5 — 0.5v

is a first-order equation while

= — 10y

is a second-order equation.

A differential equation is autonomous if the independent variable does not appear in the description of the derivative. For instance,

= 1.5 — 0.5v

is autonomous because the description of the derivative dv/dt does not depend on time. The equation

= 1.5t — 0.5y,

however, is not autonomous.

Differential equations in the world around us

As we have noted, differential equations give a natural way to describe phenomena we see in the real world. For instance, physical principles are frequently expressed as a description of how a quantity changes. A good example is Newton's Second Law, an important physcial principle that says:

The product of an object's mass and acceleration equals the force applied to it.

For instance, when gravity acts on an object near the earth's surface, it exerts a force equal to mg, the mass of the object times the gravitational constant g. We therefore have

ma = mg, or

= ^g

where v is the velocity of the object, and g = 9.8 meters per second squared. Notice that this physical principle does not tell us what the object's velocity is, but rather how the object's velocity changes.

Activity 6-2

Shown are two graphs depicting the velocity of falling objects. One is the velocity of a skydiver, while the other is the velocity of a meteorite entering the Earth's atmosphere.

(a) Begin with the skydiver's velocity and use the given graph to measure the rate of change dv/dt when the velocity is v = 0.5,1.0,1.5,2.0, and 2.5. Plot your values on the graph below. You will want to think carefully about this: you are plotting the derivative dv/ dt as a function of velocity.

Figure 6.43: Graphs of velocities used in Activity6-2.

(b) Now do the same thing with the meteorite's velocity: use the given graph to measure the rate of change dv/dt when the velocity is v = 3.5,4.0,4.5, and 5.0. Plot your values on the graph above.

(c) You should find that all your points lie on a line. Write the equation of this line being careful to use proper notation for the quantities on the horizontal and vertical axes.

(d) The relationship you just found is a differential equation. Write a complete sentence that explains its meaning.

(e) By looking at the differential equation, determine the values of the velocity for which the velocity increases.

(f) By looking at the differential equation, determine the values of the velocity for which the velocity decreases.

(g) By looking at the differential equation, determine the values of the velocity for which the velocity remains constant.

The point of this activity is to demonstrate how differential equations model processes in the real world. In this example, two factors are influencing the velocities: gravity and wind resistance. The differential equation describes how these factors influence the rate of change of the objects' velocities.

Solving a differential equation

We have said that a differential equation is an equation that describes the derivative, or derivatives, of a function that is unknown to us. By a solution to a differential equation, we mean simply a function that satisfies this description.

For instance, the first differential equation we looked at is

= 4t +1,

which describes an unknown function s(t). We may check that s(t) = 2t² +1 is a solution because it satisfies this description. Notice that s(t) = 2t² + t + 4 is also a solution.

If we have a candidate for a solution, it is straightforward to check whether it is a solution or not. Before we demonstrate, however, let's consider the same issue in a simpler context. Suppose we are given the equation 2x² — 2x = 2x + 6 and asked whether x = 3 is a solution. To answer this question, we could rewrite the variable x in the equation with the symbol

To determine whether x = 3 is a solution, we can investigate the value of each side of the equation separately when the value 3 is

placed in □

and see if indeed the two resulting values are equal.

Doing so, we observe that

and

Therefore, x = 3 is indeed a solution.

We will do the same thing with differential equations. Consider the differential equation

= 1.5 — 0.5v, or

Let's ask whether v(t) = 3 — 2e ^0.5t is a solution⁴.

⁴ At this time, don't worry about why we chose this function; we will learn techniques for finding solutions to differential equations soon enough.

Using this

formula for v, observe first that

and

Since

and 1.5 — 0.5v agree for all values of t when v = 3 —

2e , we have indeed found a solution to the differential equation.

Activity 6-3

Consider the differential equation

= 1.5 — 0.5v.

Which of the following functions are solutions of this differential equation?

(a) v(t) = 1.5t 0.25t².

(b) v(t) = 3 + 2e^—0.5t.

(c) v(t) = 3.

(d) v(t) = 3 + Ce^—0.5t where C is any constant.

Figure 6.44: The family of solutions to the differential equation ^ = 1.5 — 0.5v.

This activity shows us something interesting. Notice that the differential equation has infinitely many solutions, which are parameterized by the constant C in v(t) = 3 + Ce^—0.5t. In Figure 6.44, we see the graphs of these solutions for a few values of C, as labeled.

Notice that the value of C is connected to the initial value of the velocity v(0), since v(0) = 3 + C. In other words, while the differential equation describes how the velocity changes as a function of the velocity itself, this is not enough information to determine the velocity uniquely: we also need to know the initial velocity. For this reason, differential equations will typically have infinitely many solutions, one corresponding to each initial value. We have seen this phenomenon before, such as when given the velocity of a moving object v(t), we were not able to uniquely determine the object's position unless we also know its initial position.

If we are given a differential equation and an initial value for the unknown function, we say that we have an initial value problem. For instance,

= 1.5 — 0.5v, v(0) = 0.5

is an initial value problem. In this situation, we know the value of v at one time and we know how v is changing. Consequently, there should be exactly one function v that satisfies the initial value problem.

Figure 6.45: Beginnings of the slope field for

t — 2..

Slope Fields

We may sketch the solution to an initial value problem if we know an appropriate collection of tangent lines. Because we may use a given differential equation to determine the slope of the tangent line at any point of interest, by plotting a useful collection of these, we can get an accurate sense of how certain solution curves must behave.

Let's investigate the differential equation

= t — 2. If t = 0,

this equation says that dy/dt = 0 — 2 = —2. Note that this value holds regardless of the value of y. We will therefore sketch tangent lines for several values of y and t = 0 with a slope of — 2; see Figure 6.45.

Let's continue in the same way: if t = 1, the differential equation tells us that dy/ dt = 1 — 2 = — 1, and this holds regardless of the value of y. We now sketch tangent lines for several values of y and t = 1 with a slope of —1; see Figure 6.45-(a).

Similarly, we see that when t = 2, dy/dt = 0 and when t = 3, dy/dt = 1. We may therefore add to our growing collection of tangent line plots; see Figure 6.45-(b). In this figure, you may see the solutions to the differential equation emerge. However, for the sake of clarity, we will add more tangent lines to provide the more complete picture shown in Figure 6.46-(c).

Figure 6.46: Generating the slope field for

t — 2..

Figure 6.46-(c) is called a slope field for the differential equation, allows us to sketch solutions of the differential equation. Here, we will begin with the initial value y(0) = 1 and start sketching the solution by following the tangent line, as shown in Figure 6.47.

We then continue using this principle: whenever the solution passes through a point at which a tangent line is drawn, that line is tangent to the solution. Doing so leads us to the following sequence of images.

Figure 6.47: Sketching a solution curve for

t — 2..

In fact, we may draw solutions for any possible initial value, and doing this for several different initial values for y(0) results in the graphs shown in Figure 6.48. Just as we have done for the most recent example with

t — 2, we can construct a slope field for any differential equation of interest. The slope field provides us with visual information about how we expect solutions to the differential equation to behave.

Figure 6.48: Several solution curves for

= t — 2..

Activity 6-4

Consider the autonomous differential equation

(a) Make a plot of

versus y on the axes provided. Looking at the

graph, for what values of y does y increase and for what values of y does y decrease?

(b) Next, sketch the slope field for this differential equation on the axes provided.

(c) Use your work in (b) to sketch the solutions that satisfy y(0) = 0, y(0)= 2, y(0)= 4 and y(0) = 6.

(d) Verify that y(t) = 4 + 2e^—1/2 is a solution to the given differential equation with the initial value y( 0) = 6. Compare its graph to the one you sketched in (c).

(e) What is special about the solution where y(0) = 4?

Equilibrium solutions and stability

As our work in Activity 6-4 demonstrates, first-order autonomous solutions may have solutions that are constant. In fact, these are quite easy to detect by inspecting the differential equation dy/ dt = f( y) : constant solutions necessarily have a zero derivative so dy/dt = 0 = f (y).

For example, in Activity6-4, we considered the equation

Constant solutions are found by setting f(y) =

(y 4) = 0,

which we immediately see implies that y = 4.

Values of y for which f (y) = 0 in an autonomous differential

equation

are usually called or equilibrium solutions of

the differential equation.

Activity 6-5

Consider the autonomous differential equation

(a) Make a plot of

versus y. Looking at the graph, for what values

of y does y increase and for what values of y does y decrease?

(b) Identify any equilibrium solutions of the given differential equation.

(c) Now sketch the slope field for the given differential equation.

(d) Sketch the solutions to the given differential equation that correspond to initial values y (0) = —1,0,1,..., 5.

(e) An equilibrium solution

is called stable if nearby solutions con-

verge to

This means that if the initial condition varies slightly

from

then

Conversely, an equilibrium solution

is called unstable if nearby

solutions are pushed away from

Using your work above, classify the equilibrium solutions you found in (b) as either stable or unstable.

(f) Suppose that y(t) describes the population of a species of living organisms and that the initial value y( 0) is positive. What can you say about the eventual fate of this population?

(g) Remember that an equilibrium solution

satisfies

If we

graph dy/ dt = f( y) as a function of y, for which of the following

differential equations is

a stable equilibrium and for which is

unstable? Why?

Summary

In this section, we encountered the following important ideas:

• A differential equation is simply an equation that describes the derivative(s) of an unknown function.

• Physical principles, as well as some everyday situations, often describe how a quantity changes, which lead to differential equations.

• A solution to a differential equation is a function whose derivatives satisfy the equation's description. Differential equations typically have infinitely many solutions, parameterized by the initial values.

• A slope field is a plot created by graphing the tangent lines of many different solutions to a differential equation.

• Once we have a slope field, we may sketch the graph of solutions by drawing a curve that is always tangent to the lines in the slope field.

• Autonomous differential equations sometimes have constant solutions that we call equilibrium solutions. These may be classified as stable or unstable, depending on the behavior of nearby solutions.

Exercises

Problems

1) Suppose that T(t) represents the temperature of a cup of coffee set out in a room, where T is expressed in degrees Fahrenheit and t in minutes. A physical principle known as Newton's Law of Cooling tells us that

(a) Supposes that T(0) = 105. What does the differ-

ential equation give us for the value of

Explain in a complete sentence the meaning of these two facts.

(b) Is T increasing or decreasing at t = 0?

(c) What is the approximate temperature at t = 1?

(d) On the graph below, make a plot of dT/dt as a function of T.

(e) For which values of T does T increase? For which values of T does T decrease?

(f) What do you think is the temperature of the room? Explain your thinking.

(g) Verify that T(t) = 75 + 30e^—1/15 is the solution to the differential equation with initial value T( 0) = 105. What happens to this solution after a long time?

2) Suppose that the population of a particular species is described by the function P( t), where P is expressed in millions. Suppose further that the population's rate of change is governed by the differential equation

where f( P) is the function graphed below.

(a) For which values of the population P does the population increase?

(b) For which values of the population P does the population decrease?

(c) If P(0) = 3, how will the population change in time?

(d) If the initial population satisfies

what will happen to the population after a very long time?

(e) If the initial population satisfies

what will happen to the population after a very long time?

(f) If the initial population satisfies

what

will happen to the population after a very long time?

(g) This model for a population's growth is sometimes called "growth with a threshold." Explain why this is an appropriate name.

3) In this problem, we test further what it means for a function to be a solution to a given differential equation.

(a) Consider the differential equation

Determine whether the following functions are solutions to the given differential equation.

(i) y(t) = t + 1 + 2e^t

(ii) y(t) = t +1

(iii) y(t) = t +2

(b) When you weigh bananas in a scale at the grocery store, the height h of the bananas is described by the differential equation

where k is the spring constant, a constant that depends on the properties of the spring in the scale. After you put the bananas in the scale,

observe that the height of the baby h(t) = 4sin(3t). What is the ring constant?