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Coordinated Calculus

Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

Section 8.1 An Introduction to Differential Equations

In Section 5.3, we learned that 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. In this chapter we will learn several methods for solving differential equations. We start by revisiting what a differential equation is.
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 \(t\text{.}\) The rate at which your money grows is the derivative \(dA/dt\text{,}\) and we are told that this rate is \(0.03 A\text{.}\) This leads to the differential equation
\begin{equation*} \frac{dA}{dt} = 0.03 A\text{.} \end{equation*}
This differential equation has a slightly different feel than the previous equation \(\frac{ds}{dt} = 4t+1\text{.}\) In the earlier example, the rate of change depends only on the independent variable \(t\text{,}\) and we may find \(s(t)\) by integrating the velocity \(4t+1\text{.}\) In the banking example, however, the rate of change depends on the dependent variable \(A\text{,}\) so we’ll need some new techniques in order to find \(A(t)\text{.}\)

Example 8.1.

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.
  1. The population of a town grows continuously at an annual rate of 1.25%.
  2. A radioactive sample loses 5.6% of its mass every day.
  3. You have a bank account that continuously earns 4% interest every year. At the same time, you withdraw money continually from the account at the rate of $1000 per year.
  4. A cup of hot chocolate is sitting in a 70\(^\circ\) room. The temperature of the hot chocolate cools continuously by 10% of the difference between the hot chocolate’s temperature and the room temperature every minute.
  5. A can of cold soda is sitting in a 70\(^\circ\) room. The temperature of the soda warms continuously at the rate of 10% of the difference between the soda’s temperature and the room’s temperature every minute.
Hint.
  1. Small hints for each of the prompts above.
Answer.
  1. Let \(P\) be the population \(t\) the time in years; \(\frac{dP}{dt} = 0.0125P\text{.}\)
  2. Let \(m\) be the mass \(t\) the time in days; \(\frac{dm}{dt} = -0.056m\text{.}\)
  3. Let \(B\) be the balance \(t\) be time in years; \(\frac{dB}{dt} = 0.04B - 1000\text{.}\)
  4. Let \(t\) be time in minutes \(H\) the temperature of the hot chocolate; \(\frac{dH}{dt} = -0.1(H - 70)\text{.}\)
  5. Let \(t\) be time in minutes and \(H\) the temperature of the soda;
    \begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}
Solution.
  1. Let \(P\) be the population of the town and let \(t\) be the time in years. The population of the town grows continuously at an annual rate of \(1.25\%\) means that
    \begin{equation*} \frac{dP}{dt} = 0.0125P\text{.} \end{equation*}
  2. Let \(m\) be the mass of a radioactive sample and let \(t\) be the time in days. The sample loses 5.6% f its mass every day means that
    \begin{equation*} \frac{dm}{dt} = -0.056m\text{.} \end{equation*}
  3. Let \(B\) be the balance (in dollars) in the account and let \(t\) be the time in years. The money in the account continuously earns 4% interest every year and $1000 is withdrawn every year means that
    \begin{equation*} \frac{dB}{dt} = 0.04B - 1000\text{.} \end{equation*}
  4. The temperature of a room is \(70^\circ\text{.}\) Let \(t\) be time in minutes. The temperature \(H\) of a cup of hot chocolate cools continuously by 10% of the difference between the hot chocolate’s temperature and the room temperature every minute means that
    \begin{equation*} \frac{dH}{dt} = -0.1(H - 70)\text{.} \end{equation*}
  5. The temperature of a room is \(70^\circ\text{.}\) Let \(t\) be time in minutes. The temperature \(H\) of a can of soda warms continuously by 10% of the difference between the soda’s temperature and the room temperature every minute means that
    \begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}

Subsection 8.1.1 Differential Equations in the World Around Us

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, which 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 \(mg\text{,}\) the mass of the object times the gravitational constant \(g\text{.}\) We therefore have
\begin{align*} ma =\mathstrut \amp mg, \ \text{or}\\ \frac{dv}{dt} =\mathstrut \amp g\text{,} \end{align*}
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.

Example 8.2.

Shown below are two graphs depicting the velocity of falling objects. On the left is the velocity of a skydiver, while on the right is the velocity of a meteorite entering the Earth’s atmosphere.
Figure 8.3. A skydiver’s velocity.
Figure 8.4. A meteorite’s velocity.
  1. Begin with the skydiver’s velocity and use the given graph to measure the rate of change \(dv/dt\) when the velocity is \(v=0.5, 1.0, 1.5, 2.0\text{,}\) and \(2.5\text{.}\) 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.
  2. 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 \(v=3.5,4.0,4.5\text{,}\) and \(5.0\text{.}\) Plot your values on the graph above.
  3. 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.
  4. The relationship you just found is a differential equation. Write a complete sentence that explains its meaning.
  5. By looking at the differential equation, determine the values of the velocity for which the velocity increases.
  6. By looking at the differential equation, determine the values of the velocity for which the velocity decreases.
  7. By looking at the differential equation, determine the values of the velocity for which the velocity remains constant.
Hint.
  1. Hints for each of the prompts above.
Answer.
  1. For the skydiver:
    \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*}
  2. For the meteorite:
    \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp \approx -0.3 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp \approx -0.6\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp \approx -0.9 \amp \left. \frac{dv}{dt}\right|_{(v = 5)} \amp \approx -1.2 \end{align*}
    A graph of the points from parts (a) and (b) is shown in the following diagram:
  3. \(\frac{dv}{dt} = -0.6v + 1.8\text{.}\)
  4. The rate of change of velocity with respect to time is a linear function of velocity.
  5. \(0 \lt v \lt 3\text{.}\)
  6. \(3 \lt v \lt 5\text{.}\)
  7. \(v = 3\text{.}\)
Solution.
Each graph is a graph of velocity \(v\) as a function of time \(t\text{.}\) To estimate \(\frac{dv}{dt}\) at a particular time, we must estimate the slope of the tangent line for the graph of \(v\) at that point. For parts (a) and (b), we draw tangent lines and measure the slopes of the tangent lines. These values are approximations.
  1. For the skydiver:
    \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*}
  2. For the meteorite:
    \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp \approx -0.3 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp \approx -0.6\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp \approx -0.9 \amp \left. \frac{dv}{dt}\right|_{(v = 5)} \amp \approx -1.2 \end{align*}
    A graph of the points from parts (a) and (b) is shown in the following diagram:
  3. Using the values in parts (a) and (b), we obtain \(\frac{dv}{dt} = -0.6v + 1.8\text{.}\)
  4. This differential equation means that the rate of change of velocity with respect to time is a linear function of velocity.
  5. The velocity increases when \(\frac{dv}{dt} \gt 0\text{.}\) So the velocity increases when \(0 \lt v \lt 3\text{.}\)
  6. The velocity decreases when \(\frac{dv}{dt} \lt 0\text{.}\) So the velocity decreases when \(3 \lt v \lt 5\text{.}\)
  7. The velocity will remain constant when \(\frac{dv}{dt} = 0\text{,}\) which is when \(v = 3\text{.}\)
The point of this example is to demonstrate how differential equations model processes in the real world. In this example, two factors influence the velocities: gravity and wind resistance. The differential equation describes how these factors influence the rate of change of the velocities.

Subsection 8.1.2 Types of Differential Equations

To close this section, we note that differential equations may be classified based on certain characteristics they may possess. 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.

Order of a differential equation.

The order of a differential equation is the order of the highest derivative that appears in the equation.
For example, a first-order differential equation is one in which only the first derivative of the function occurs. For this reason,
\begin{equation*} \frac{dv}{dt} = 1.5-0.5v \end{equation*}
is a first-order equation while
\begin{equation*} \frac{d^2 y}{dt^2} = -10y \end{equation*}
is a second-order equation.

Autonomous differential equation.

A differential equation is autonomous if the independent variable does not appear in the description of the derivative.
For instance,
\begin{equation*} \frac{dv}{dt} = 1.5-0.5v \end{equation*}
is autonomous because the description of the derivative \(dv/dt\) does not depend on time. The equation
\begin{equation*} \frac{dy}{dt} = 1.5t - 0.5y\text{,} \end{equation*}
however, is not autonomous.

Exercises 8.1.3 Exercises

1. Matching solutions with equations.

Match the solutions to the differential equations. If there is more than one solution to an equation, select the answer that includes all solutions.
  1. \(\displaystyle \frac{d^2y}{dx^2}=-16 y\)
  2. \(\displaystyle \frac{dy}{dx}=-4 y\)
  3. \(\displaystyle \frac{d^2y}{dx^2}=16 y\)
  4. \(\displaystyle \frac{dy}{dx}=4 y\)
  1. \(\displaystyle y = \sin(4 x)\)
  2. \(\displaystyle y = 4\sin(x)\)
  3. \(y = e^{-4 x}\) or \(y = e^{4 x}\)
  4. \(\displaystyle y = e^{-4 x}\)
  5. \(\displaystyle y = e^{4 x}\)
  6. \(y = \sin(4 x)\) or \(y = 4\sin(x)\)

2. Finding constant to complete solution.

Find a positive value of \(k\) for which \(y = \cos(k t)\) satisfies
\begin{equation*} \frac{d^2y}{dt^2} + 9 y = 0. \end{equation*}
\(k =\)

3. Choosing solution of \(dy/dt=k(1-Ay)\).

Let \(A\) and \(k\) be positive constants.
Which of the given functions is a solution to \(\frac{dy}{dt}=-k(y+A)\text{?}\)
  • \(\displaystyle y = A + C e^{kt}\)
  • \(\displaystyle y = -A + C e^{-kt}\)
  • \(\displaystyle y = -A + C e^{kt}\)
  • \(\displaystyle y = A^{-1} + C e^{Akt}\)
  • \(\displaystyle y = A + C e^{-kt}\)
  • \(\displaystyle y = A^{-1} + C e^{-Akt}\)

4. Analyzing Newton’s Law of Cooling.

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
\begin{equation*} \frac{dT}{dt}= -\frac1{15}T+5\text{.} \end{equation*}
  1. Supposes that \(T(0)=105\text{.}\) What does the differential equation give us for the value of \(\frac{dT}{dt}\vert_{T=105}\text{?}\) Explain in a complete sentence the meaning of these two facts.
  2. Is \(T\) increasing or decreasing at \(t=0\text{?}\)
  3. What is the approximate temperature at \(t=1\text{?}\)
  4. On the graph below, make a plot of \(dT/dt\) as a function of \(T\text{.}\)
  5. For which values of \(T\) does \(T\) increase? For which values of \(T\) does \(T\) decrease?
  6. What do you think is the temperature of the room? Explain your thinking.
  7. Verify that \(T(t) = 75 + 30e^{-t/15}\) is the solution to the differential equation with initial value \(T(0) = 105\text{.}\) What happens to this solution after a long time?

5. Population growth.

Suppose that the population of a particular species is described by the function \(P(t)\text{,}\) where \(P\) is expressed in millions. Suppose further that the population’s rate of change is governed by the differential equation
\begin{equation*} \frac{dP}{dt} = f(P) \end{equation*}
where \(f(P)\) is the function graphed below.
  1. For which values of the population \(P\) does the population increase?
  2. For which values of the population \(P\) does the population decrease?
  3. If \(P(0) = 3\text{,}\) how will the population change in time?
  4. If the initial population satisfies \(0\lt P(0)\lt 1\text{,}\) what will happen to the population after a very long time?
  5. If the initial population satisfies \(1\lt P(0)\lt 3\text{,}\) what will happen to the population after a very long time?
  6. If the initial population satisfies \(3\lt P(0)\text{,}\) what will happen to the population after a very long time?
  7. This model for a population’s growth is sometimes called “growth with a threshold.” Explain why this is an appropriate name.

6. A look at solutions to differential equations.

In this problem, we test further what it means for a function to be a solution to a given differential equation.
  1. Consider the differential equation
    \begin{equation*} \frac{dy}{dt} = y - t\text{.} \end{equation*}
    Determine whether the following functions are solutions to the given differential equation.
    1. \(\displaystyle y(t) = t + 1 + 2e^t\)
    2. \(\displaystyle y(t) = t + 1\)
    3. \(\displaystyle y(t) = t + 2\)
  2. When you weigh bananas in a scale at the grocery store, the height \(h\) of the bananas is described by the differential equation
    \begin{equation*} \frac{d^2h}{dt^2} = -kh \end{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, you (cleverly) observe that the height of the bananas is given by \(h(t) = 4\sin(3t)\text{.}\) What is the value of the spring constant?
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