CHAPTER
17

Differential Equations

In This Chapter

• What are differential equations?
• Separation of variables
• Initial conditions and differential equations
• Modeling exponential growth and decay

Most calculus courses contain some discussion of differential equations, but that discussion is extremely limited to the basics. Most math majors will tell you that they had to suffer through an entire course on solving differential equations at some point in their math career. This is because differential equations are extremely useful in modeling real-life scenarios, and are used extensively by scientists.

A differential equation is nothing more than an equation containing a derivative. In fact, you have created more than your fair share of differential equations simply by finding derivatives of functions in the first half of the book. In this chapter, you’ll begin with the differential equation (i.e., the derivative) and work your way backward to the original equation. Sound familiar? Basically you’re just going to be applying integration methods, as you have for numerous chapters now.

However, solving differential equations is not the same thing as integrating. There are lots of complicated differential equations (that we won’t be exploring). Luckily, the most popular differential equation application in beginning calculus (exponential growth and decay) requires you to use a very simple solution technique called separation of variables. Let’s start there.

Separation of Variables

If a differential equation is nothing more than an equation containing a derivative, and solving a differential equation basically means finding the antiderivative, then what’s so hard about solving differential equations, and why does it get treated as a separate topic? The reason is that differential equations are usually not as straightforward as this one:

Clearly, the solution to this differential equation is y = 3sin x + x + C. All you have to do is integrate both sides of the equation. Most differential equations are all twisted up and knotted together with variables all over the place, like this:

It looks like someone chewed up a whole bunch of equations and spat them out in random order (which is both puzzling and unappetizing). In order to solve this differential equation, you’ll have to separate the variables. In other words, move all the y’s to the left side of the equation and all the x’s to the right side. Once that is done, you’ll be able to integrate both sides of the equation separately. This process, appropriately called separation of variables, solves any basic differential equations you’ll encounter.

Example 1: Solve the differential equation , where k is a constant.

Solution: You need to move y to the left side of the equation and move dx to the right side. Because k is a constant, it’s not clear whether or not you should move it. As a rule of thumb, move all constants to the right side of the equation. Your goal is to solve for y, so you don’t want any non-y things on the left side of the equation. Start by moving the y, so divide both sides of the equation by y to get:

Now shoot that dx to the right side of the equation by multiplying both sides by dx:

At this point, you can integrate both sides of the equation. Because k is a constant, its antiderivative is kx, just as the antiderivative of 5 would be 5x:

You’re not quite done yet. Your final answer to a differential equation should be solved for y. To cancel out the natural log function and accomplish this, you have to use its inverse function, e^x, like this: . (Drop the absolute value signs around the y now—they were only needed because the domain of the natural log function is only positive numbers. As the natural log disappears, let the absolute value bars go with it.)

In other words, rewrite the equation so that both sides are the powers of the natural exponential function. This gives you y = e^kx + ^C, because e^x and ln x are inverse functions, and as such, . You could stop here, but go just one step further. Remember the basic exponential rule that said x^a ∙ x^b = x^a+^b? The preceding equation looks like x^a+^b, so you can break it up into x^a ∙ x^b:

y = e^kx ∙ e^C

Almost done. I promise. Because you have no idea what value C has, you have no idea what e^C will be. You know it’ll be some number, but you have no idea what number that is. As you’ve done in the past, rewrite e^C as C, signifying that even though it’s not the same value as the original C, it’s still some number you don’t know: y = Ce^kx.

That’s the solution to the differential equation. It took you a while to get here, but this is a very important equation, and you’ll need it in a few pages.

Types of Solutions

Just like integrals, solutions to differential equations come in two forms: with and without a “+ C” term. Definite integrals had no such term, because their final answers were numbers rather than equations. Whereas the solution to a differential equation will always come in delicious, equation form (with candy-shaped marshmallows), in some cases, you’ll be able to determine exactly what the value of C should be, so you can provide a more specific answer.

Family of Solutions

If you are only given a differential equation, you can only get a general solution. Example 1 and Problem 1 are two such instances. Remember, integration cannot usually tell you exactly what a function’s antiderivative is, because any functions differing only by a constant will have the same derivative.

The solution to a differential equation containing a “+ C” term is actually a family of solutions, because it technically represents an infinite number of possible solutions to the differential equation. Think about the differential equation . If you use the separation-of-variables technique, you get a solution of y = x² + 7x + C. You can plug in any real-number value for C, and the result is a solution to the original differential equation.

For example, all have a derivative of 2x + 7. These three (plus an infinite number of other equations) make up the family of solutions. The members of a family of solutions have nearly identical graphs, differing only in their vertical position along the y-axis.

Knowing a family of solutions is sometimes not enough. Differential equations are often used as mathematical models to illustrate real-life examples and situations. In such cases, you’ll need to be able to find specific solutions to differential equations, but to do so you’ll need a little more information up front.

Example 2: Graph the family of solutions for the differential equation when C = –4, –2, 0, 2, and 4.

Solution: To solve the differential equation, begin by separating the variables. In other words, multiply both sides of the equation by dx to isolate the y-variables on the left side of the equal sign and the x-variables on the right side:

dy = (3x² + 2x – 6) dx

Now integrate both sides of the equation separately.

This is a family of solutions, because substituting any real number C into the equation creates a new solution whose derivative (with respect to x) is the original differential equation. In this problem, you are asked to graph specific members of the family of solutions. Plug each given value of C into the solution to create five unique solutions:

The only difference in each equation is the constant, the value at which the graph crosses the y-axis when x = 0. Therefore, the graphs are all vertical translations (or shifts) of each other, as illustrated in Figure 17.1.

Figure 17.1

The family of solutions to the differential equation when C = –4, –2, 0, 2, and 4.

Specific Solutions

To determine exactly what C equals for any differential equation solution, you’ll need to know at least one coordinate pair of the differential equation’s antiderivative. With that information, you can plug in the (x,y) pair and solve for C. To explain what I mean, I have thrown together a little example for those game show fans out there.

I don’t know what it is that makes people (by which I mean my wife and me) so excited to watch other people agonize about winning dishwashers by throwing comically oversized dice, but it doesn’t stop us from watching game shows. However, the sudden trend in these programs isn’t playing silly games for prizes anymore; instead, they subject the contestants to peril in order to win vast sums of money.

Example 3: A new television game show in the works already is making quite a stir. Terminal Velocity will suspend contestants by their ankles on a bungee cord. Producers are still working out the details, but one of the show’s features will be dropping the participants from the ceiling and allowing them to repeatedly lurch their way toward the studio floor as the length of the bungee slowly increases. Also, the audience will throw things at them (like small rocks or maybe piranhas if ratings begin to sag).

Suppose that the velocity of a contestant (in ft/sec), for the first 10 seconds of her fall, is given by . If the initial position of the doomed individual is 115 feet off the ground, find her position equation.

Solution: You are given a differential equation representing velocity. The solution to the differential equation will then be the antiderivative of velocity, position. The problem also tells you that the initial position is 115 feet high. This means that the contestant’s position at time equals 0 is 115, so s(0) = 115. You’ll use that in a second to find C, but first things first—you need to apply separation of variables to solve the differential equation:

There you have it—the position equation. Remember that you should get 115 if you plug in 0 for t; make that substitution, and you can find C easily:

Therefore, the exact position equation is s(t) = 40cos(2t) – 4t + 75.

Exponential Growth and Decay

Most people have an intuitive understanding of what it means to exhibit exponential growth. Basically, it means that things are increasing in an out-of-control way, like a virus in a horror movie. One infected person spreads the illness to another person, then those two spread it to other people. Two infected people become four, four become eight, eight become sixteen, until it’s an epidemic and Jackie Chan has to come in to save the day, possibly with karate kicks.

Truth be told, actual exponential growth doesn’t happen a lot. An exponential growth model assumes that there is an infinite amount of resources from which to draw. In our epidemic example, the rate of increase of the illness cannot go on uninhibited, because eventually, everyone will already be sick. To get around such restrictions, many problems involving exponential growth and decay deal with exciting things like bacterial growth. Bacteria are small, so it takes them much longer to conquer the world (thanks in no small part, once again, to Jackie Chan).

Notice that in Figure 17.2, the logistic growth curve changes concavity (from concave up to concave down) about midway through the interval. This change in concavity indicates the point at which growth begins to slow.

Figure 17.2

Two kinds of growth. Neither explains that weird mole on your neck.

A more realistic example of growth and decay is logistic growth. In this model, growth begins quickly (it basically looks exponential at first) and then slows as it reaches some limiting factor (as our virus could only spread to so many people before everyone was already dead—isn’t that a pleasant thought?). Although it is not beyond our abilities to examine logistic growth, it is by far more complicated to understand and model, so we’ll stick with exponential growth.

Mathematically, exponential growth is pretty neat. We say that a population exhibits exponential growth if its rate of change is directly proportional to the population itself. Thus, a population P grows (or decays) exponentially if and P are in proportion to each other.

So how fast something grows or decays is based on how much of it there is. Without getting into a lot of detail (too late for that, isn’t it?), we can say that these two things are in proportion when they’re equal to each other, if one of the terms is multiplied by a constant (for instance, one thing is two times or five times as big as the other).

Recognize that? It’s Example 1! Because you already solved this differential equation earlier in the chapter, you know that a population showing exponential growth has this equation:

y = Ne^kt

(I know I used C as the constant before, but I like having the N there better, because when you read the formula, it looks like the word naked, and I am immature enough to think that’s pretty funny.) In this formula, N represents the beginning or initial population, k is a constant of proportionality, and t stands for time. (The e is a constant—Euler’s number, which you’ve undoubtedly seen lurking about in your precalculus work. Most calculators, even scientific ones, have a button for Euler’s number, so you don’t have to memorize it.) The y represents the total population after time t has passed.

Your first step in the majority of exponential growth and decay problems is to find k, because you will almost never be able to determine what k is based on the problem. Don’t even try to guess k—it’s rarely, if ever, obvious. For example, if your population increased in the familiar sequence 2, 4, 8, 16, 32, etc., you might be tempted to think that k = 2 because the population constantly doubles, but instead, k ≈ 0.693147, which is actually ln 2.

Example 4: Even after the movie Pay It Forward came out, the movement promoted by the film never really caught on. Its premise was that you should do a big favor for three different people, something they couldn’t accomplish on their own. In turn, they would provide favors for three other people, and so on. Unfortunately, a new movement called Punch It Forward is catching on instead. It’s the same premise, but with punching instead of favors. On the first day of Punch It Forward, 19 people are involved in the movement. After 10 days, 193 people are involved. How many people will be involved 30 days after Punch It Forward begins, assuming that exponential growth is exhibited during that time?

Solution: Use the exponential growth and decay formula y = Ne^kt. N represents the initial population (19). You know that after 10 days have elapsed, the new population is 193. Therefore, when t = 10, y = 193. Plug all these values in and solve for k:

Therefore, the exponential growth model is y = 19e^0.231825t. To determine the population after the first 30 days, plug in 30 for t:

Approximately 19,914 people have been inducted (and possibly indicted) into the Punch It Forward society after only one month. It’s a brave new world, my friend.

Example 5: According to Sir Isaac Newton (the very same cofounder of calculus discussed in Chapter 1), the rate at which an object cools is directly proportional to the temperature of its environment. This property is known as Newton’s Law of Cooling. Now that I have shared this tasty nugget of information with you, how about a couple of questions to help you digest it?

(a) Assume that an object cools at a rate of , where T is the temperature of the object after time t has elapsed. Furthermore, the object’s environment has temperature T_E and the constant of proportionality described by Newton is equal to k. Create a differential equation based on Newton’s Law of Cooling and solve it for T. Then, identify the specific solution stating that the original temperature of the object (when t = 0 and no time has passed) equals T₀.

(b) According to scientists, the optimum temperature for tea is 140°F. You leave a cup of tea at this temperature in a 72°F room, and exactly 17.5 minutes later it reaches 113°F, the minimum temperature at which scientists deem the tea still drinkable. At approximately what time was the tea’s temperature 125°F?

Solution:

(a) The first part of this problem requires you to generate Newton’s Law of Cooling. It states that the rate at which an object’s temperature T changes is proportional to the difference between the object’s temperature T and the temperature of the environment T_E.

Recall that if two values are proportional, it means you can multiply one of those values by a constant (in this case k) to get the other value. Separate the variables, dividing both sides of the equation by T – T_E and multiplying both sides by dt:

Integrate both sides of the equation:

Note that T_E and k are constants, so treat them as you would any other number. That means ∫ k dt = k · t + C. Use u-substitution to integrate the left side of the equation. If u = T – T_E, then du = dt and . Set these two new integrals equal to each other as described in the above differential equation:

The absolute value bars are unnecessary. The object will eventually cool to room temperature (T_E), but it will not get colder than that. Because T ≥ T_E, you can conclude that T – T_E ≥ 0.

ln (T – T_E) = kt + C

Don’t lose sight of your goal, to solve this equation for T. Exponentiate both sides of the equation, making them powers of e, to eliminate the natural logarithm.

Recall that e^C is another unknown constant, which can be represented once again by C.

Whew! That’s a lot of work but finally you have an equation solved for T. This is a family of solutions, and to complete this problem, you need to identify the specific solution for which T = T⁰ when t = 0. Your goal is to figure out what C should equal:

Substitute this value of C into your solution:

The final equation is the solution you’re looking for. It translates Newton’s Law of Cooling into a handy formula that you can use any time you want. For example, you could use it in part (b) of the problem!

(b)  Begin by taking inventory of the information you are given. You know that the original temperature of your tea was T₀ = 140 in a room with temperature T_E = 72. After t = 17.5 minutes, the tea has a new temperature of t = 113. Plug all of this information into the equation you created in part (a).

The only information the problem does not give you is the unique value of k for this situation. Luckily, you can solve this equation to determine the value of k:

Now that you know the value of k, you can finally take the t out of “mystery.” Come to think of it, that would leave you with “mysery,” which sounds like “misery,” which is an apt description for a problem this long.

You want to know what time t elapses between your drink’s original temperature of T₀ = 140 and the new temperature T = 125. You already know that T_E = 72 and you just discovered a shiny new value of k to use. Let’s break out Newton’s Law of Cooling one last time:

Your tea cools to 125°F after approximately 8.62 minutes. There are 60 seconds in a minute, so 0.62 minutes is equal to (0.62)(60) = 37.2 seconds. Therefore, your tea reaches 125°F after approximately 8 minutes, 37.2 seconds.

The Least You Need to Know

• Differential equations contain derivatives; solutions to basic differential equations are simply the antiderivatives solved for y.
• If a problem contains sufficient information, you can find a specific solution for a differential equation; it won’t contain a “+ C” term.
• If a population’s rate of growth or rate of decay is proportional to the size of the population, the growth or decay is exponential in nature.
• Exponential growth and decay are modeled with the equation y = Ne^kt.