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Mathematics LibreTexts

7.1: An Introduction to Differential Equations

Skills to Develop

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?

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 chapter, 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. As we’ll see, this may be too much to ask in some situations, but we will still be able to make accurate approximations.

Exercise \(\PageIndex{1}\)

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.

  1. How much does the position change over the time interval \([0, 4]\)?
  2. 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 \(s(4)\)?
  3. Suppose you are told that the object’s initial position \(s(0) = 7\). Determine \(s(2)\), the object’s position 2 seconds later.
  4. If you are told instead that the object’s initial position is \(s(0) = 3\), what is \(s(2)\)?
  5. If we only know the velocity \(v(t) = 4t + 1\), is it possible that the object’s position at all times is \(s(t) = 2t 2 + t − 4\)? Explain how you know.
  6. Are there other possibilities for \(s(t)\)? If so, what are they?
  7. 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

\(\dfrac{dy}{dx} = x \sin x\)

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

\(\dfrac{ds}{dt} = 4t + 1. \)

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 \(\dfrac{dA}{dt}\). On the other hand, we are told that this rate is \(0.03A\). This leads to the differential equation

\(\dfrac{dA}{dt} = 0.03A.\)

This differential equation has a slightly different feel than the previous example \(\dfrac{ds}{dt} = 4t + 1. \). 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 \(\PageIndex{1}\):

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° 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° 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.

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

\(ma = mg\)


\(\dfrac{dv}{dt} = g,\)

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 \(\PageIndex{2}\):

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.

  1. Begin with the skydiver’s velocity and use the given graph to measure the rate of change \(dv/dt \) when the velocity is \(v = 0.5, 1.0, 1.5, 2.0, and 2.5\). 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.

  1. 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.
  2. 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.
  3. The relationship you just found is a differential equation. Write a complete sentence that explains its meaning.
  4. By looking at the differential equation, determine the values of the velocity for which the velocity increases.
  5. By looking at the differential equation, determine the values of the velocity for which the velocity decreases.
  6. 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

\(\dfrac{ds}{dt} = 4t + 1,\)

 which describes an unknown function \(s(t)\). We may check that

\(s(t) = 2t^2 + t\)

is a solution because it satisfies this description. Notice that

\(s(t) = 2t^2 + t + 4\)

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

\(2x^2 − 2x = 2 x + 6\)

and asked whether \(x = 3\) is a solution. To answer this question, we could rewrite the variable \(x\) in the equation with the symbol \(\square\):

\(2\square ^2 − 2\square = 2\square + 6.\)

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 \(\square\) in and see if indeed the two resulting values are equal. Doing so, we observe that

\(2 \square ^2 − 2\square = 2 \times 3^2 − 2 \times 3 = 12,\)


\(2 \square + 6 = 2 \times 3 + 6 = 12.\)

Therefore, \(x = 3\) is indeed a solution.

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

\(\dfrac{dv }{dt} = 1.5 − 0.5v\) or, 

\(\dfrac{d \square }{ dt} = 1.5 − 0.5\square .\)

Let’s ask whether \(v(t) = 3 − 2e^{−0.5t}\) is a solution (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

\(\dfrac{dv}{dt} = \dfrac{d\square}{dt} = \dfrac{d}{ dt} [3 − 2e^{−0.5t} ] = −2e^{ −0.5t} (−0.5) = e^{ −0.5t}\)


\(1.5 − 0.5v = 1.5 − 0.5\square = 1.5 − 0.5(3 − 2e^{−0.5t} ) = 1.5 − 1.5 + e^{−0.5t} = e^{−0.5t}.\)

Since \(\frac{dv}{dt}\) and \(1.5 − 0.5v\) agree for all values of \(t\) when \(v = 3 − 2e −0.5t\), we have indeed found a solution to the differential equation.

Activity \(\PageIndex{3}\):

Consider the differential equation

\(\dfrac{dv}{dt} = 1.5 − 0.5v.\)

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

  1. \(v(t) = 1.5t − 0.25t^2 \).
  2. \(v(t) = 3 + 2e −0.5t \).
  3. \(v(t) = 3\).
  4. \(v(t) = 3 + Ce^{−0.5t}\) where \(C\) is any constant.

This activity shows us something interesting. Notice that the differential equation has infinitely many solutions, which are parametrized by the constant \(C\) in \(v(t) = 3 + Ce^{−0.5t} . \)

In Figure 7.1, we see the graphs of these solutions for a few values of \(C\), as labeled.

Figure 7.1: The family of solutions to the differential equation dv dt = 1.5 − 0.5v.

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,

\(\dfrac{dv}{dt} = 1.5 − 0.5v\), with \(v(0) = 0.5\)

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. This demonstrates the following important general property of initial value problems.

Initial value problems

Initial value problems that are “well behaved” have exactly one solution, which exists in some interval around the initial point.

We won’t worry about what “well behaved” means—it is a technical condition that will be satisfied by all the differential equations we consider.

To close this section, we note that 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,

\(\dfrac{dv}{dt} = 1.5 − 0.5v\)

is a first-order equation while

\(\dfrac{d^2 y} {dt^2} = −10y\)

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

\(\dfrac{dv}{dt} = 1.5 − 0.5v\)

is autonomous because the description of the derivative \(\frac{dv}{dt}\) does not depend on time. The equation

\(\dfrac{dy}{dt} = 1.5t − 0.5y,\)

however, is not autonomous.


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, parametrized by the initial values.


Matt Boelkins (Grand Valley State University), David Austin (Grand Valley State University), Steve Schlicker (Grand Valley State University)