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An introduction to differential equations

Introduction

In previous chapters, we have seen that a function's derivative tells us the rate at which the function is changing. The Fundamental Theorem of Calculus helped us determine the total change of a function over an interval from 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 can determine how much the position changes over that time interval. If we know where the object is at the beginning of that interval, we have enough information to predict where it will be at the end of the interval.

In this chapter, we introduce the concept of differential equations. 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, ideally including 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.

Exploration

What is a differential equation?

A differential equation is an equation that describes one or more of the derivatives of a function that is unknown to us. For instance, the equation \[\begin{aligned}\end{aligned}\] describes the derivative of an unknown function \(y(x)\).

As many important examples of differential equations involve quantities that change over time, the independent variable in our discussion will frequently be time \(t\). In the preview activity, we considered the differential equation \[\begin{aligned}\end{aligned}\].

Knowing the velocity and the starting position of a moving object, we were able to find its 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 use this information 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 everyday 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 \(t\). The rate at which your money grows is the derivative \(\frac{dA}{dt}\), and we are told that this rate is \(0.03 A\). This leads to the differential equation \[\begin{aligned}\end{aligned}\].

This differential equation has a slightly different feel than the previous equation \(\frac{ds}{dt} = 4t+1\). 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 \(4t+1\) (with respect to \(t\)). 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 \(A(t)\).

Differential equations in the world around us

Differential equations give a natural way to describe phenomena we see in physical reality. 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 \(mg\), the mass of the object times the gravitational constant \(g\). We therefore have \[\begin{aligned}ma \amp= mg, \text{ or} \\ \frac{dv}{dt} \amp= g\end{aligned}\], 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.

The point of this activity is to demonstrate how differential equations model processes in physical reality. 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.

Solving a differential equation

A differential equation describes one or more 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 \[\begin{aligned}\end{aligned}\], 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 to a differential equation, it is straightforward to check whether the function is a solution or not. Before we demonstrate how, let's consider the same issue in a simpler context. Suppose we are given the equation \(2x^2 - 2x = 2x+6\) and asked whether \(x=3\) is a solution. To answer this question, we could rewrite the variable \(x\) in the equation with the symbol \(\Box\): \[\begin{aligned}\end{aligned}\].

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 \(\Box\) and see if indeed the two resulting values are equal. Doing so, we observe that \[\begin{aligned}\end{aligned}\], and \[\begin{aligned}\end{aligned}\].

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

We will do the same thing with differential equations. Consider the differential equation \[\begin{aligned}\frac{dv}{dt} \amp= 1.5 - 0.5v, \ \text{or} \\ \frac{d\Box}{dt} \amp= 1.5 - 0.5\Box\end{aligned}\].

Let's ask whether \(v(t) = 3 - 2e^{-0.5t}\) is a solution (for now, don't worry about why we chose this function; we will learn some techniques for finding solutions to differential equations soon). Using this formula for \(v\), observe first that \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\].

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.

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

Condensed — the full section is in Boelkins, Active Calculus.

Summary

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 derivative(s) satisfy the equation's description. Differential equations typically have infinitely many solutions, parametrized by the initial values.

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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 \[\begin{aligned}\end{aligned}\].
1. Supposes that \(T(0)=105\). What does the differential equation give us for the value of \(\frac{dT}{dt}\vert_{T=105}\)? Explain in a complete sentence the meaning of these two facts.
2. Is \(T\) increasing or decreasing at \(t=0\)?
3. What is the approximate temperature at \(t=1\)?
4. On the graph below, make a plot of \(dT/dt\) as a function of \(T\).
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 \(T(0) = 105\). What happens to this solution after a long time?
Kusonyeza yankho
1. Since \(\frac{dT}{dt}= -\frac1{15}T+5\) and \(T(0)=105\), \(\frac{dT}{dt}\vert_{T=105} = -\frac{1}{15}(105) + 5 = -7 + 5 = -2\). The fact that \(T(0) = 105\) tells us the coffee's temperature at time \(t = 0\) is \(T = 105\) degrees F, and \(\frac{dT}{dt}\vert_{T=105} = -2\) tells us that at the instant \(T = 105\) (i.e. when \(t = 0\)), the coffee's temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
2. At \(t = 0\), \(T = 105\), and since \(\frac{dT}{dt}\vert_{T=105}\) is negative, \(T\) decreasing at \(t=0\).
3. Using the linear approximation \(T(1) \approx T(0) + \frac{dT}{dt}\vert_{t=0} (1-0)\) along with the given information, it follows that \[\begin{aligned}\end{aligned}\] degrees F.
4. We note that \(\frac{dT}{dt}\) is a linear function of \(T\) with slope \(-\frac{1}{15}\) and vertical intercept \((0,5)\), so its graph is as shown below. Note particularly that \(\frac{dT}{dt}\vert_{T = 75} = 0\).
5. For \(T \lt 75\), \(\frac{dT}{dt} \gt 0\), so for these \(T\)-values, \(T\) increases. For \(T \gt 75\), \(\frac{dT}{dt} \lt 0\), so for these \(T\)-values, \(T\) decreases.
6. In the previous question, we observed that \(75\) degrees F is the critical temperature that changes the behavior of \(T\). There are several reasons to think that \(75\) degrees F is room temperature: if the coffee's temperature is above room temperature (\(75\)), its temperature will decrease to room temperature, while if the coffee is below room temperature, its temperature will increase to room temperature. Moreover, if the coffee's temperature is \(75\) (which would match the room temperature), then its rate of change is \(0\) and the coffee's temperature doesn't change.
7. Verify that \(T(t) = 75 + 30e^{-t/15}\) is the solution to the differential equation with initial value \(T(0) = 105\). What happens to this solution after a long time? Assuming that \(T(t) = 75 + 30e^{-t/15}\), we observe that \[\begin{aligned}\end{aligned}\]. In addition, \[\begin{aligned}\end{aligned}\]. The two preceding equations show that \[\begin{aligned}\end{aligned}\] when we substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\), and thus this function is a solution to the differential equation. Moreover, \(T(0) = 75 + 30e^0 = 75 + 30 = 105\), so the function satisfies the initial condition, too. Finally, because \(e^{-t/15} \to 0\) as \(t \to \infty\), it follows that \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\), which shows that the solution function's value tends to room temperature as time increases without bound, as we expect.
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 \[\begin{aligned}\end{aligned}\]. Determine whether the following functions are solutions to the given differential equation.
  1. \(y(t) = t + 1 + 2e^t\)
  2. \(y(t) = t + 1\)
  3. \(y(t) = t + 2\)
2. When you weigh bananas in a scale at the grocery store, the height \(h\) of the bananas is described by the differential equation \[\begin{aligned}\end{aligned}\] 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 \(h(t) = 4\sin(3t)\). What is the value of the spring constant?
Kusonyeza yankho
1. To see if \(y(t) = t + 1 + 2e^t\) is a solution to \(\frac{dy}{dt} = y - t\), we compute \(\frac{dy}{dt}\) and \(y - t\) separately and see if they are equal. First, using \(y(t) = t + 1 + 2e^t\), \[\begin{aligned}\end{aligned}\]. Next, \[\begin{aligned}\end{aligned}\]. Thus, \(y(t) = t + 1 + 2e^t\) satisfies the DE and is indeed a solution.
  For \(y(t) = t + 1\), observe that \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\] as well, so \(y(t) = t + 1\) is a solution to the DE.
  If \(y(t) = t + 2\), then \[\begin{aligned}\end{aligned}\] but \[\begin{aligned}\end{aligned}\] as well, so \(\frac{dy}{dt} \ne y-t\), and therefore \(y(t) = t + 2\) is a not solution to the DE.
2. In order for \(h(t) = 4\sin(3t)\) to be a solution to \[\begin{aligned}\end{aligned}\], it must make the equation true. To find \(k\), we substitute the formula for \(h\) into the DE and see what is required of \(k\). We observe that for \(h(t) = 4\sin(3t)\), \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. To satisfy the DE \[\begin{aligned}\end{aligned}\], it follows that we must have \(-36\sin(3t) = -kh = -k \cdot 4\sin(3t)\). Thus, we need \(-36 = -4k\), and hence \(k = 9\).
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 \[\begin{aligned}\end{aligned}\] 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 \(P(0) = 3\), how will the population change in time?
4. If the initial population satisfies \(0\lt P(0)\lt 1\), what will happen to the population after a very long time?
5. If the initial population satisfies \(1\lt P(0)\lt 3\), what will happen to the population after a very long time?
6. If the initial population satisfies \(3\lt P(0)\), 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.
Kusonyeza yankho
1. \(P\) increases for \(1 \lt P \lt 3\) since \(\frac{dP}{dt} \gt 0\) for these \(P\)-values.
2. \(P\) decreases for \(P \lt 1\) and \(3 \lt P \lt 4\) since \(\frac{dP}{dt} \lt 0\) for these \(P\)-values.
3. If \(P(0) = 3\), \(\frac{dP}{dt}\vert_{P=3} = 0\), so \(P\) will not change at all with time since its derivative is \(0\).
4. If the initial population satisfies \(0\lt P(0)\lt 1\), we expect that the population will decrease toward \(P = 0\) with \(P\) always being positive. Since \(\frac{dP}{dt}\) is always negative for \(0\lt P \lt 1\), \(P\) is always decreasing on this interval. The population can't drop below \(P = 0\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
5. If the initial population satisfies \(1\lt P(0)\lt 3\), we expect that the population will increase toward \(P = 3\) with \(P\) always being between \(1\) and \(3\). Since \(\frac{dP}{dt}\) is always positive for \(1 \lt P \lt 3\), \(P\) is always increasing on this interval. The population can't increase above \(P = 3\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
6. If the initial population satisfies \(3\lt P(0)\), we expect that the population will decrease toward \(P = 3\) with \(P\) always being above \(3\). Since \(\frac{dP}{dt}\) is always negative for \(P \gt 3\), \(P\) is always decreasing on this interval. The population can't drop below \(P = 3\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
7. It looks like there are two thresholds for the population: there's a minimum threshold of \(P = 1\), since below that population, the population decreases to \(0\), and there's a maximum threshold of \(P = 3\), where if we are above the minimum but below the maximum, the population works to increase toward the maximum threshold, while if the population is above the maximum, its value will fall back toward it.
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 \[\begin{aligned}\end{aligned}\].
1. Supposes that \(T(0)=105\). What does the differential equation give us for the value of \(\frac{dT}{dt}\vert_{T=105}\)? Explain in a complete sentence the meaning of these two facts.
2. Is \(T\) increasing or decreasing at \(t=0\)?
3. What is the approximate temperature at \(t=1\)?
4. On the graph below, make a plot of \(dT/dt\) as a function of \(T\).
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 \(T(0) = 105\). What happens to this solution after a long time?
Kusonyeza yankho
1. Since \(\frac{dT}{dt}= -\frac1{15}T+5\) and \(T(0)=105\), \(\frac{dT}{dt}\vert_{T=105} = -\frac{1}{15}(105) + 5 = -7 + 5 = -2\). The fact that \(T(0) = 105\) tells us the coffee's temperature at time \(t = 0\) is \(T = 105\) degrees F, and \(\frac{dT}{dt}\vert_{T=105} = -2\) tells us that at the instant \(T = 105\) (i.e. when \(t = 0\)), the coffee's temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
2. At \(t = 0\), \(T = 105\), and since \(\frac{dT}{dt}\vert_{T=105}\) is negative, \(T\) decreasing at \(t=0\).
3. Using the linear approximation \(T(1) \approx T(0) + \frac{dT}{dt}\vert_{t=0} (1-0)\) along with the given information, it follows that \[\begin{aligned}\end{aligned}\] degrees F.
4. We note that \(\frac{dT}{dt}\) is a linear function of \(T\) with slope \(-\frac{1}{15}\) and vertical intercept \((0,5)\), so its graph is as shown below. Note particularly that \(\frac{dT}{dt}\vert_{T = 75} = 0\).
5. For \(T \lt 75\), \(\frac{dT}{dt} \gt 0\), so for these \(T\)-values, \(T\) increases. For \(T \gt 75\), \(\frac{dT}{dt} \lt 0\), so for these \(T\)-values, \(T\) decreases.
6. In the previous question, we observed that \(75\) degrees F is the critical temperature that changes the behavior of \(T\). There are several reasons to think that \(75\) degrees F is room temperature: if the coffee's temperature is above room temperature (\(75\)), its temperature will decrease to room temperature, while if the coffee is below room temperature, its temperature will increase to room temperature. Moreover, if the coffee's temperature is \(75\) (which would match the room temperature), then its rate of change is \(0\) and the coffee's temperature doesn't change.
7. Verify that \(T(t) = 75 + 30e^{-t/15}\) is the solution to the differential equation with initial value \(T(0) = 105\). What happens to this solution after a long time? Assuming that \(T(t) = 75 + 30e^{-t/15}\), we observe that \[\begin{aligned}\end{aligned}\]. In addition, \[\begin{aligned}\end{aligned}\]. The two preceding equations show that \[\begin{aligned}\end{aligned}\] when we substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\), and thus this function is a solution to the differential equation. Moreover, \(T(0) = 75 + 30e^0 = 75 + 30 = 105\), so the function satisfies the initial condition, too. Finally, because \(e^{-t/15} \to 0\) as \(t \to \infty\), it follows that \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\), which shows that the solution function's value tends to room temperature as time increases without bound, as we expect.
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 \[\begin{aligned}\end{aligned}\]. Determine whether the following functions are solutions to the given differential equation.
  1. \(y(t) = t + 1 + 2e^t\)
  2. \(y(t) = t + 1\)
  3. \(y(t) = t + 2\)
2. When you weigh bananas in a scale at the grocery store, the height \(h\) of the bananas is described by the differential equation \[\begin{aligned}\end{aligned}\] 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 \(h(t) = 4\sin(3t)\). What is the value of the spring constant?
Kusonyeza yankho
1. To see if \(y(t) = t + 1 + 2e^t\) is a solution to \(\frac{dy}{dt} = y - t\), we compute \(\frac{dy}{dt}\) and \(y - t\) separately and see if they are equal. First, using \(y(t) = t + 1 + 2e^t\), \[\begin{aligned}\end{aligned}\]. Next, \[\begin{aligned}\end{aligned}\]. Thus, \(y(t) = t + 1 + 2e^t\) satisfies the DE and is indeed a solution.
  For \(y(t) = t + 1\), observe that \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\] as well, so \(y(t) = t + 1\) is a solution to the DE.
  If \(y(t) = t + 2\), then \[\begin{aligned}\end{aligned}\] but \[\begin{aligned}\end{aligned}\] as well, so \(\frac{dy}{dt} \ne y-t\), and therefore \(y(t) = t + 2\) is a not solution to the DE.
2. In order for \(h(t) = 4\sin(3t)\) to be a solution to \[\begin{aligned}\end{aligned}\], it must make the equation true. To find \(k\), we substitute the formula for \(h\) into the DE and see what is required of \(k\). We observe that for \(h(t) = 4\sin(3t)\), \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. To satisfy the DE \[\begin{aligned}\end{aligned}\], it follows that we must have \(-36\sin(3t) = -kh = -k \cdot 4\sin(3t)\). Thus, we need \(-36 = -4k\), and hence \(k = 9\).
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 \[\begin{aligned}\end{aligned}\] 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 \(P(0) = 3\), how will the population change in time?
4. If the initial population satisfies \(0\lt P(0)\lt 1\), what will happen to the population after a very long time?
5. If the initial population satisfies \(1\lt P(0)\lt 3\), what will happen to the population after a very long time?
6. If the initial population satisfies \(3\lt P(0)\), 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.
Kusonyeza yankho
1. \(P\) increases for \(1 \lt P \lt 3\) since \(\frac{dP}{dt} \gt 0\) for these \(P\)-values.
2. \(P\) decreases for \(P \lt 1\) and \(3 \lt P \lt 4\) since \(\frac{dP}{dt} \lt 0\) for these \(P\)-values.
3. If \(P(0) = 3\), \(\frac{dP}{dt}\vert_{P=3} = 0\), so \(P\) will not change at all with time since its derivative is \(0\).
4. If the initial population satisfies \(0\lt P(0)\lt 1\), we expect that the population will decrease toward \(P = 0\) with \(P\) always being positive. Since \(\frac{dP}{dt}\) is always negative for \(0\lt P \lt 1\), \(P\) is always decreasing on this interval. The population can't drop below \(P = 0\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
5. If the initial population satisfies \(1\lt P(0)\lt 3\), we expect that the population will increase toward \(P = 3\) with \(P\) always being between \(1\) and \(3\). Since \(\frac{dP}{dt}\) is always positive for \(1 \lt P \lt 3\), \(P\) is always increasing on this interval. The population can't increase above \(P = 3\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
6. If the initial population satisfies \(3\lt P(0)\), we expect that the population will decrease toward \(P = 3\) with \(P\) always being above \(3\). Since \(\frac{dP}{dt}\) is always negative for \(P \gt 3\), \(P\) is always decreasing on this interval. The population can't drop below \(P = 3\) because at that value of \(P\), \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
7. It looks like there are two thresholds for the population: there's a minimum threshold of \(P = 1\), since below that population, the population decreases to \(0\), and there's a maximum threshold of \(P = 3\), where if we are above the minimum but below the maximum, the population works to increase toward the maximum threshold, while if the population is above the maximum, its value will fall back toward it.

Symbols used here

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

\frac{\partial u}{\partial t},\ \nabla^2 u

partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: An introduction to differential equations

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?

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

Sankhani wanu

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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