Module 2 · Lesson 1

Differential Equations: Laws of change: differential equations from planetary orbits to population growth

You'll see how science states its laws as rules about how fast things change, and how to get from such a rule to a prediction. You'll step a growing population forward by hand and see why so many quantities in nature grow or shrink exponentially.

What you will be able to do

Describe how a differential equation relates a quantity to its rate of change and read the behavior of a simple solution.

0% of this lesson · 9 items · 1h 3m total · 48m without the optional problem set

Contents of this lesson9 items

ReadingWhy Hot Coffee Cools Fast at First3 min
ReadingLaws of Change from Newton to Poincaré4 min
ReadingDifferential Equations: Rules That Link a Quantity to Its Rate of Change4 min
ReadingStepping Forward: Numerical Solutions and Equilibrium4 min
ReadingExponential Growth and Decay: The Landmark Solution4 min
Guided ReadingWorked Example: Stepping a Growing Population Forward by Hand7 min
Guided ConversationWrite the Law of Change for Something You Know12 min
Problem Set · optionalStepping Models Forward15 min
Knowledge CheckLaws of change: differential equations from planetary orbits to population growth10 min

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Reading 3 min Done

Why Hot Coffee Cools Fast at First

Pour a cup of coffee at about 90 °C and leave it on a table in a room at 20 °C. Check it with a kitchen thermometer every few minutes and you'll see a pattern you probably already sense. In the first few minutes it loses heat fast. Later, once it's only warm, it creeps down a degree at a time. It never drops below room temperature, and it takes a long while to get all the way there.

Why should the same cup cool at such different speeds? Before you read on, try to name what changes between the first minute and the twentieth. The coffee, the cup, and the room all stay the same.

What changes is the gap. At first the coffee is 70 degrees hotter than the room. Later it might be only 10 degrees hotter. Newton's law of cooling, a good approximation for a cup on a table, says that the rate at which an object cools is proportional to how much hotter it is than its surroundings (Strang and Herman 2016b, sec. 4.2). Proportional means that doubling the gap doubles the rate, and halving the gap halves it.

Put numbers on it. Suppose the coffee loses one-tenth of the gap each minute. When the gap is 70 degrees, it drops about 7 degrees in a minute. When the gap is 10 degrees, it drops about 1 degree. When the gap is zero, it stops changing. That one rule accounts for the fast start, the slow finish, and the floor at room temperature.

Notice what kind of rule this is. It doesn't tell you the temperature at a quarter past eight. It tells you how fast the temperature is changing, and it states that speed in terms of the temperature itself. A rule that links a quantity to its own rate of change is called a differential equation (Strang and Herman 2016b, sec. 4.1).

That's an odd kind of knowledge to hold. You know the rate right now. But the rate depends on the temperature, the temperature keeps falling, and so the rate keeps falling too. Getting from the rule to a forecast, such as "the coffee will be down to 60 °C in about six minutes," takes work. You can step forward a minute at a time, recomputing the rate after each step, or you can find a formula that gives the temperature at every moment.

That work is the business of the branch of mathematics called differential equations: turning a rule about rates into a prediction about the future. The same move describes the orbit of a planet, the size of a population, the amount of a medicine left in the body, and the weather. Some of those predictions hold for centuries. Others fail after about a week, and the reasons they fail belong to the subject too.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016b. Calculus Volume 2. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 2

Reading 4 min Done

Laws of Change from Newton to Poincaré

Introduction

Physics states its laws as rules about rates: how fast velocity changes, how fast heat flows, how fast a fluid speeds up. That habit is a little over three centuries old, and it came with a surprise about its own limits. This reading follows how laws of change took that form, from Newton through the Bernoullis and Euler, to Poincaré's discovery that some of them can't be followed far into the future.

Newton States Motion as a Rate

In 1687 Isaac Newton published the Principia, which set out his laws of motion and his law of universal gravitation (Katz 2018, chap. 16). A law of motion is a rule that says how a body's motion changes under the forces acting on it. Newton's second law, in its modern form, says that a body's acceleration equals the force on it divided by its mass. Acceleration is the rate at which velocity changes, and velocity is the rate at which position changes.

So the law doesn't hand you a planet's position on a given night. It tells you how the planet's velocity is changing right now, given where the planet is relative to the Sun. That makes it a differential equation: a rule that links a quantity to its own rate of change. To get an orbit, you have to work forward from the rule. Working this way, the Principia showed that an attraction toward the Sun that weakens with the square of the distance fits the planets' motions.

Fluids, Strings, and Mechanics in the 1700s

No single person founded the subject. It grew alongside physics, as each new physical question produced new equations. Daniel Bernoulli's Hydrodynamica, published in 1738, treated the flow of fluids. He also showed that the motion of a vibrating string, such as one on a musical instrument, is made up of an infinite number of simple harmonic vibrations (O'Connor and Robertson 1998c). His father, Johann Bernoulli, published a rival book on fluids, Hydraulica, in 1739, and backdated it to suggest that Daniel had borrowed from him (O'Connor and Robertson 1998c).

Leonhard Euler's Mechanica of 1736 presented Newton's dynamics in the language of calculus. Through the 1750s Euler set up the equations for the motion of an idealized fluid with no internal friction, still called the Euler equations. Problems in physics led him into a wide study of differential equations, including a method for approximating their solutions (O'Connor and Robertson 1998d).

Who	When	What they treated as a law of change
Newton	1687	Motion of bodies and planets under force and gravity
Daniel Bernoulli	1738	Fluid flow; vibrating strings
Euler	1736; 1750s	Mechanics in the language of calculus; ideal fluid flow
Poincaré	1889–1890	Three bodies attracting each other by gravity

Poincaré and the Three-Body Problem

Two bodies under gravity, such as the Sun and a single planet, follow a pattern that repeats. Add a third body and the mathematics becomes far harder. The three-body problem is the problem of predicting the motions of three bodies that attract each other by gravity.

In 1887 Oscar II, King of Sweden and Norway, set up a mathematical competition to mark his sixtieth birthday in 1889. Henri Poincaré won it with a memoir on the three-body problem. As the memoir was being prepared for the journal Acta Mathematica, Phragmén, who was editing it for publication, found an error. Poincaré agreed that he had made one, and a revised version appeared in 1890 (O'Connor and Robertson 2003).

The revised memoir contained the first mathematical description of chaotic motion, and the MacTutor biography says the error "is now regarded as marking the birth of chaos theory" (O'Connor and Robertson 2003). Chaos is behavior of a system that follows exact rules, yet in which tiny differences in the starting state grow until long-range prediction fails. Poincaré had found that a law of change can be exact and still not let you follow its consequences far ahead.

Conclusion

Over the 1700s, differential equations became the usual way to state laws of motion, fluid flow, and vibration, built up by many people rather than founded by one. Poincaré's work on three bodies, completed in 1890, showed that knowing the law exactly doesn't guarantee long-range prediction. That limit later became a field of study in its own right.

Key Terms

Law of motion: A rule that says how a body's motion changes under the forces acting on it.
Differential equation: A rule that links a quantity to its own rate of change.
Three-body problem: The problem of predicting the motions of three bodies that attract each other by gravity.
Chaos: Behavior of a system that follows exact rules, yet in which tiny differences in the starting state grow until long-range prediction fails.

References

Katz, Victor J. 2018. A History of Mathematics: An Introduction. 3rd ed., Classic Version. New York: Pearson.
- Free: None
- Publisher: Pearson
- Free companion: O'Connor and Robertson 1996
O'Connor, J. J., and E. F. Robertson. 1998c. "Daniel Bernoulli." MacTutor History of Mathematics Archive, University of St Andrews. Last updated September 1998.
- Free: MacTutor: Daniel Bernoulli
O'Connor, J. J., and E. F. Robertson. 1998d. "Leonhard Euler." MacTutor History of Mathematics Archive, University of St Andrews. Last updated September 1998.
- Free: MacTutor: Euler
O'Connor, J. J., and E. F. Robertson. 2003. "Jules Henri Poincaré." MacTutor History of Mathematics Archive, University of St Andrews. Last updated October 2003.
- Free: MacTutor: Poincaré

Reading 4 min Done

Differential Equations: Rules That Link a Quantity to Its Rate of Change

Introduction

Many laws in science don't give a formula for a quantity. They give a rule for how fast the quantity is changing. This reading explains what such a rule says, how to read one written in symbols, and what it means to solve one.

Three Laws of Change in Words

A differential equation is a rule that links a quantity to its own rate of change. A rate of change says how much a quantity changes per unit of time, such as degrees per minute or people per year. Here are three differential equations stated in words.

Cooling. A hot drink cools at a rate proportional to how much hotter it is than the room. A gap of 60 degrees means cooling twice as fast as a gap of 30 degrees.
Population. An unchecked population grows at a rate proportional to its size. A population of 2,000 adds twice as many members per year as a population of 1,000 with the same birth and death rates.
Falling. An object dropped near the ground, with air resistance ignored, gains speed at a constant rate. Its velocity goes up by about 9.8 meters per second during each second of the fall.

Each of these is a rate equation: an equation that states how fast a quantity is changing, usually in terms of the quantity's current value. In the first two, the rate depends on the quantity itself, which is what makes them interesting. In the third, the rate is the same at every moment.

Reading the Notation Once

Mathematicians write the population rule like this, for a population growing at 10 percent of its size per year:

\frac{dP}{dt} = 0.1P

Here $P$ is the population and $t$ is time, measured in years. The left side, $\frac{dP}{dt}$ , is read "the rate of change of $P$ with time." It's a single symbol, not a fraction to simplify. The whole equation says that the population grows at 10 percent of its size per unit of time, the simplest model of population growth (Strang and Herman 2016b, sec. 4.4).

Reading it at a few moments shows how the rule works. When $P$ is 1,000, the population is growing at 100 per year. When $P$ is 2,000, it's growing at 200 per year. The rule never states the population at any particular time. It states the speed of change at whatever size the population has reached.

A Solution Is a Function

To solve an equation in school algebra, you find a number. To solve a differential equation, you find something larger. A solution of a differential equation is a function, a rule giving the quantity's value at every time, that obeys the equation at every moment (Strang and Herman 2016b, sec. 4.1).

For the population rule, starting from 1,000 people, the solution is

P = 1000e^{0.1t}

where $e$ is a fixed number, about 2.718, that shows up whenever growth is proportional to size. You don't need to compute with $e$ to follow the idea. The solution gives a population of 1,000 at the start, about 1,105.2 after one year, about 1,221.4 after two, and about 1,491.8 after four. It's a whole curve, not a single answer.

Same Rule, Different Starting Points

The rule alone doesn't pin down one curve. A population of 500 obeying the same rule grows along a different curve, $P = 500e^{0.1t}$ , which stays exactly half the size of the first at every moment. The initial value is the quantity's value at the starting time, and it picks out one solution from the many that obey the rule. A differential equation together with an initial value is called an initial-value problem (Strang and Herman 2016b, sec. 4.1).

The cooling rule works the same way. A cup of coffee at 90 °C and a cup of tea at 70 °C, in the same 20 °C room, obey the same rule. They follow different temperature curves, and both curves head toward 20 °C.

Conclusion

A differential equation states how fast a quantity changes, usually in terms of the quantity itself. Its solution is a function of time, not a number, and the initial value decides which of the many possible solutions you get. Reading the rule in words, at a few sample values, shows what it predicts before any formula is found.

Key Terms

Differential equation: A rule that links a quantity to its own rate of change.
Rate equation: An equation that states how fast a quantity is changing, usually in terms of the quantity's current value.
Solution: A function giving a quantity's value at every time that obeys a differential equation at every moment.
Initial value: The value of a quantity at the starting time, which picks out one solution of a differential equation.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016b. Calculus Volume 2. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 2

Reading 4 min Done

Stepping Forward: Numerical Solutions and Equilibrium

Introduction

Many of the differential equations that matter in science have no tidy formula for their solution. They're solved approximately, by computers taking many small steps. This reading explains the simplest stepping method, why smaller steps help, and two kinds of long-run behavior: settling at an equilibrium and growth that levels off.

Use the Rate, Take a Step, Repeat

A differential equation tells you how fast a quantity is changing right now. A numerical method is a procedure that approximates the solution of a differential equation by arithmetic, one step at a time. The simplest is called Euler's method (Strang and Herman 2016b, sec. 4.2). It runs like this:

Use the rule to find the current rate of change.
Multiply the rate by a short stretch of time to get the change over that stretch.
Add the change to the current value to get the next value.
Repeat from the new value.

Take coffee at 90 °C in a 20 °C room, cooling at $0.1 \times (T - 20)$ degrees per minute, where $T$ is the coffee's temperature in degrees Celsius. At 90 °C the rate is $0.1 \times 70 = 7$ degrees per minute, so after one minute the estimate is $90 - 7 = 83$ °C. At 83 °C the rate is $0.1 \times 63 = 6.3$ , so after two minutes the estimate is $83 - 6.3 = 76.7$ °C. Each step uses the newest value to recompute the rate.

Smaller Steps, Better Answers

The step size is the length of time between one computed value and the next. Each step pretends the rate stays fixed for the whole step, but the rate actually changes as the quantity changes. So each step carries a small error, and the errors add up.

Smaller steps keep the pretense closer to the truth. For a population of 1,000 growing at 10 percent of its size per year, the exact four-year figure is about 1,491.8. Here is what Euler's method gives with three step sizes.

Step size	Number of steps	Estimate after 4 years	Gap from exact
1 year	4	1,464.1	27.7
6 months	8	1,477.5	14.3
3 months	16	1,484.5	7.3

Halving the step roughly halves the gap, but it doubles the number of steps. That trade between accuracy and cost is the everyday business of scientific computing.

Equilibrium: Where the Rate Is Zero

An equilibrium is a value at which the rate of change is zero, so a quantity that starts there stays there. You find one by setting the rate equal to zero and solving (Strang and Herman 2016b, sec. 4.2). For the coffee, $0.1 \times (T - 20) = 0$ when $T = 20$ . Coffee at room temperature stays at room temperature.

This equilibrium is stable: coffee that starts hotter cools toward it, and a drink that starts colder warms toward it. Other equilibria are unstable, so that a small push sends the quantity away. Knowing where the equilibria are, and which kind each is, tells you a lot about long-run behavior before you take a single step.

Growth That Levels Off

Real populations don't grow in proportion to their size forever. Food, space, and predators set a ceiling, which ecologists call the carrying capacity. Logistic growth is growth whose rate is close to proportional to size while the population is small, and falls to zero as the population nears a limit (Strang and Herman 2016b, sec. 4.4).

The logistic rule multiplies the proportional rate by the fraction of room left below the limit. With a growth rate of 10 percent per year and a limit of 10,000, a population of 1,000 grows at $0.1 \times 1000 \times 0.9 = 90$ per year. A population of 9,000 also grows at 90 per year: $0.1 \times 9000 \times 0.1 = 90$ . At 10,000 the rate is zero. The rule has two equilibria, zero and the limit, and any population in between climbs toward the limit along an S-shaped curve.

Conclusion

Many differential equations in practice are solved by stepping: compute the rate, take a short step, repeat. Smaller steps give more accurate results at more cost. Equilibria, where the rate is zero, show where a system will settle, and logistic growth shows how a growing quantity can level off at a limit.

Key Terms

Numerical method: A procedure that approximates the solution of a differential equation by arithmetic, one step at a time.
Step size: The length of time between one computed value and the next in a numerical method.
Equilibrium: A value at which the rate of change is zero, so a quantity that starts there stays there.
Logistic growth: Growth whose rate is close to proportional to size while the population is small, and falls to zero as the population nears a limit.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016b. Calculus Volume 2. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 2

Reading 4 min Done

Exponential Growth and Decay: The Landmark Solution

Introduction

Savings in an account, an unchecked colony of bacteria, and the first weeks of an epidemic all share one pattern: the bigger they get, the faster they grow. This reading explains the solution of that pattern, exponential growth and decay, how to estimate doubling times and half-lives, and where the pattern breaks down.

Growth in Proportion to Size

When a quantity grows at a rate proportional to its size, each unit of time multiplies it by the same factor. Exponential growth is growth in which the rate of change is proportional to the current size, so the quantity multiplies by the same factor in each equal stretch of time.

Money earning interest that's added continuously behaves this way, since interest is paid on interest. So does a population with plenty of food and no predators, and so does an infection in its early days, when almost everyone is still susceptible. The rule is one of the simplest differential equations there is: the rate of change equals a fixed number $r$ times the quantity. Its solution is

P = P_0 e^{rt}

Here $P_0$ is the starting amount, $r$ is the growth rate as a decimal (0.07 for 7 percent per year), $t$ is time, and $e$ is a fixed number, about 2.718 (Strang and Herman 2016b, sec. 4.4). You don't need to compute with $e$ to use the ideas below.

Decay in Proportion to Size

Run the same rule with a negative rate and the quantity shrinks. Exponential decay is decline in which the rate of change is proportional to the current size, so the quantity shrinks by the same factor in each equal stretch of time.

Radioactive material decays this way: a fixed fraction of the remaining atoms decays in each period, however much is left. Many medicines are cleared from the body in roughly the same way, with the body removing a fixed fraction of the remaining amount per hour. Large amounts fall fast, and small amounts fall slowly, much like a hot drink cooling toward room temperature.

Doubling Time and Half-Life

Exponential change has a fixed clock. The doubling time is the time an exponentially growing quantity takes to double, and it's the same whatever the starting size. A handy estimate is

\text{doubling time} \approx 70 \div (\text{percent growth rate})

So 7 percent growth per year doubles a quantity in about $70 \div 7 = 10$ years. The 70 comes from the solution: for continuous growth the exact doubling time is about $69.3 \div$ (percent rate), and 70 is easier to divide. The table compares the estimate with the exact continuous figure.

Growth rate per year	Estimate (70 ÷ rate)	Exact, continuous growth
1 percent	70 years	69.3 years
2 percent	35 years	34.7 years
5 percent	14 years	13.9 years
7 percent	10 years	9.9 years
10 percent	7 years	6.9 years

For interest added once a year instead of continuously, the true figure runs a little longer: about 10.2 years at 7 percent.

Decay has the matching idea. The half-life is the time an exponentially decaying quantity takes to fall to half its size. Take a hypothetical medicine with a half-life of 6 hours. A 100 mg dose falls to 50 mg after 6 hours, 25 mg after 12, and 12.5 mg after 18. Each half-life halves whatever is left.

Why It's the Landmark, and Where It Fails

Most differential equations have no exact formula for their solution. This one does, and the same solution recurs across finance, biology, medicine, and physics. That's why it's the first result anyone learns in the subject, and why "exponential" has entered everyday speech.

It also fails in a predictable way. Unchecked growth goes on without bound, which no real population can do. Strang and Herman call the pure exponential model unrealistic for exactly this reason, since food, space, and predators eventually slow growth (2016b, sec. 4.4). Exponential growth describes the early stage of many processes well and their later stages badly.

Conclusion

When a quantity changes at a rate proportional to its size, it grows or decays exponentially, with a fixed doubling time or half-life. The rule of 70 gives a quick estimate of either. The model fits the early stages of growth well and breaks down once limits begin to bite.

Key Terms

Exponential growth: Growth in which the rate of change is proportional to the current size, so the quantity multiplies by the same factor in each equal stretch of time.
Exponential decay: Decline in which the rate of change is proportional to the current size, so the quantity shrinks by the same factor in each equal stretch of time.
Doubling time: The time an exponentially growing quantity takes to double, the same whatever the starting size.
Half-life: The time an exponentially decaying quantity takes to fall to half its size.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016b. Calculus Volume 2. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 2

Guided Reading 7 min Done

Worked Example: Stepping a Growing Population Forward by Hand

Introduction

A differential equation tells you how fast something is changing right now, not where it will be next year. To get a forecast, you have to work forward from the rule. This worked example does that by hand for a growing population, using only multiplication and addition, and then checks the result against the exact answer to see how far off the hand method is and why.

The method used here is the simplest numerical method, called Euler's method: find the current rate of change, multiply it by a short stretch of time, add the result to the current value, and repeat (Strang and Herman 2016b, sec. 4.2). A numerical method is a procedure that approximates the solution of a differential equation by arithmetic, one step at a time.

The Problem

A population of 1,000 grows at 10 percent of its size per year. Written as a differential equation, a rule that links a quantity to its own rate of change, the law is

\frac{dP}{dt} = 0.1P

Read it in words: "the rate of change of the population $P$ over time $t$ , in years, equals 0.1 times the population." When the population is 1,000, it's growing at 100 per year. When it's 1,500, it's growing at 150 per year.

There are three questions. First, what does stepping forward one year at a time predict for the population after four years? Second, how does that compare with the exact solution? Third, what happens to the estimate if you step forward six months at a time instead?

Working It Through

Step 1: Use the current rate to step forward one year

At the start the population is 1,000. The rule says the current rate of growth is $0.1 \times 1000 = 100$ people per year. The step size, the length of time between one computed value and the next, is one year. Multiply the rate by the step to get the change: $100 \times 1 = 100$ . Add it to the current value:

1000 + 100 = 1100

So the estimate after one year is 1,100.

Step 2: Repeat for four years

Now start again from 1,100. The rate is $0.1 \times 1100 = 110$ per year, so the next value is $1100 + 110 = 1210$ . From 1,210 the rate is 121, giving 1,331. From 1,331 the rate is 133.1, giving 1,464.1.

Year	Population at start of year	Rate (0.1 × population)	Population at end of year
1	1,000	100	1,100
2	1,100	110	1,210
3	1,210	121	1,331
4	1,331	133.1	1,464.1

Notice that each year's increase is bigger than the last. The rate depends on the population, so as the population grows, so does the rate. After four years the step-by-step estimate is 1,464.1.

Step 3: Compare with the exact solution

This equation is one of the few with an exact formula for its solution. For a starting population of 1,000, the solution is $P = 1000e^{0.1t}$ , where $e$ is a fixed number, about 2.718 (Strang and Herman 2016b, sec. 4.4). After four years, $t = 4$ , and

1000e^{0.4} \approx 1491.8

The hand method gave 1,464.1, so it falls short by about 27.7. The same shortfall shows up every year, and it grows: the exact values after one, two, and three years are about 1,105.2, 1,221.4, and 1,349.9, against stepped values of 1,100, 1,210, and 1,331.

Step 4: Explain the gap

Each step used the rate at the start of the year, as if the population stayed at its starting size for the whole twelve months. It doesn't. In the first year the population climbs from 1,000 toward 1,100, and as it climbs, the real rate climbs from 100 toward 110. By using 100 for the whole year, the step undercounts the growth.

So the hand method is always a little behind for a growing population. The error isn't a slip in the arithmetic. It's built into the method, which pretends the rate stays fixed during each step.

Step 5: Halve the step to six months

If the problem is that the rate changes during a step, the cure is shorter steps. With a step of half a year, the change in each step is the rate times 0.5. From 1,000, the rate is 100 per year, so six months adds $100 \times 0.5 = 50$ , giving 1,050. That's the same as adding 5 percent per half-year.

From 1,050, the next half-year adds $105 \times 0.5 = 52.5$ , giving 1,102.5 at the end of year one. Carrying on for eight half-year steps gives these values at each full year (rounded to one decimal place):

Year	Six-month steps	Exact solution
1	1,102.5	1,105.2
2	1,215.5	1,221.4
3	1,340.1	1,349.9
4	1,477.5	1,491.8

The four-year estimate is now 1,477.5, short by about 14.3 instead of 27.7. Halving the step roughly halved the gap. Halving it again, to three-month steps, gives about 1,484.5, short by about 7.3. The estimates close in on the exact value as the steps shrink, at the cost of more arithmetic each time.

Key Considerations

Yearly stepping is compound interest. Adding 10 percent of the current value once a year is exactly how yearly compound interest works. Stepping every six months matches interest compounded twice a year. The exact solution matches growth that's added continuously, at every instant, which is why it always comes out a little ahead.

Fractions of people are fine in a model. A population of 1,464.1 can't exist, but the model treats the population as a smoothly varying quantity. For large populations the fractions don't matter. You'd round only when reporting a result.

The common mistake: adding 100 every year forever. A 10 percent rate on a starting population of 1,000 gives 100 new members in the first year, and it's tempting to think it gives 100 every year after. That treats the rate as fixed, when the rule says the rate depends on the current population. Adding 100 a year gives 1,400 after four years, which is short by about 92. Over twenty years the difference is dramatic: steady additions give 3,000, while the exact solution gives about 7,389. Proportional growth is multiplication repeated, not addition repeated.

Smaller steps are more accurate, not exact. However small the step, Euler's method stays slightly behind for a growing population. In practice, computers take many small steps, often with more refined methods that sample the rate more than once per step, until the error is small enough for the purpose at hand.

Summary

Stepping forward means using the current rate to estimate the next value, then repeating. For a population of 1,000 growing at 10 percent of its size per year, yearly steps undercount, and halving the step roughly halves the error:

After year Yearly steps Six-month steps Three-month steps Exact
1 1,100.0 1,102.5 1,103.8 1,105.2
2 1,210.0 1,215.5 1,218.4 1,221.4
3 1,331.0 1,340.1 1,344.9 1,349.9
4 1,464.1 1,477.5 1,484.5 1,491.8
Gap at year 4 27.7 14.3 7.3 0

After year	Yearly steps	Six-month steps	Three-month steps	Exact
1	1,100.0	1,102.5	1,103.8	1,105.2
2	1,210.0	1,215.5	1,218.4	1,221.4
3	1,331.0	1,340.1	1,344.9	1,349.9
4	1,464.1	1,477.5	1,484.5	1,491.8
Gap at year 4	27.7	14.3	7.3	0

Each column uses the same rule; only the step size changes.
Every stepped value sits below the exact value, because each step uses the rate at the start of the step.
The check that the method is working: as the step shrinks, the estimates move steadily toward the exact solution.

References

Strang, Gilbert, and Edwin "Jed" Herman. 2016b. Calculus Volume 2. Houston, TX: OpenStax. CC BY-NC-SA 4.0.
- Free: OpenStax: Calculus Volume 2

Guided Conversation 12 min Done

Write the Law of Change for Something You Know

In this conversation you'll pick a process from your own life, such as a cooling drink or a growing following online, and work out what its rate of change depends on. With hints when you want them, you'll write the rule in words and then in plain notation, and predict where it leads in the long run. You'll leave with a law of change of your own and its equilibrium, if it has one.

You'll have this conversation with an AI assistant, using your own account. Choose a button to open a new chat with the prompt already filled in, then press send to start. If the chat opens empty, copy the prompt and paste it in.

Run this conversation in whichever assistant you already use:

Open in Claude Open in ChatGPT

Claude desktop app

To use another LLM, simply copy and paste the prompt into its chat window.

Show the full prompt (it lists misreadings to watch for, so skip it if you would rather come to the conversation fresh)

Guided Conversation: Write the Law of Change for Something You Know (about 12 minutes)

Note to the learner: press send to start. Everything below is facilitator guidance for the AI. It lists misconceptions to watch for, so skip it if you'd rather come to the conversation fresh.

Please facilitate a coached problem session with me. I'm an adult who is comfortable with basic high-school algebra but may be rusty or uneasy with math, exploring what differential equations are and whether to study them further. I'm studying laws of change: differential equations from planetary orbits to population growth. Follow this guidance for the whole conversation.

GOAL
Describe how a differential equation relates a quantity to its rate of change. The problem: I pick a process I know, say what its rate of change depends on, write the rule in words and then in plain notation, and predict its long-run behavior.

HOW TO RUN THE CONVERSATION
- Ask one question at a time, then wait for my reply. Keep each of your turns under about 120 words.
- Don't lecture. Explain a point only when I need it to continue, then return to my process.
- Be curious, collegial, and patient. Give hints and ask questions; don't solve the problem for me. Define terms briefly on first use (rate of change, differential equation, proportional, equilibrium). Welcome disagreement when I give a reason.
- Always ask for my approach before giving any hint. Give one hint at a time. Never state the rule or the long-run behavior for my process before I've attempted it.
- Write math in plain notation, such as: rate of change of T = -0.1 x (T - 20).
- Plain conversation only: don't search the web or create files or documents.
- Aim for about 12 minutes, mostly on topics 2 and 3. If my replies are brief, offer one concrete prompt, such as "does it change faster when there's more of it, or when it's further from some target?", and move on. If I seem uncertain, shorten the conversation to 5-7 minutes. Always reach the final topic.
- Start now. Open with one or two warm sentences: this is a conversation, not a quiz; a rough rule I can explain matters more than a polished formula; I can ask you to clarify anything. Then ask me to pick a process, offering three examples: a cooling drink, a bathtub draining, or a growing following online.

TOPICS, IN ORDER
1. My process. Ask what quantity is changing, in what units, and what its rate of change depends on. Hint if needed: does it change faster when there's more of it, or when it's further from something? Follow-up: when does it change fastest?
2. The rule. Ask me to say the rule as a sentence, then write it with a letter for the quantity. Hint if needed: "rate of change of Q = (some number) x (something about Q)". Follow-up: what would doubling the quantity, or the gap, do to the rate?
3. Long-run behavior. Ask me to predict: growth without end, decay toward zero, or settling at some value. Hint if needed: look for a value where the rate would be zero. Follow-up: could growth here really go on forever?
4. Closing. Ask me to state my rule's equilibrium, if it has one: the value where the rate of change is zero. Then ask what one short time step from a starting value of my choice gives. This is something I can take into practice stepping models forward by hand.

CORRECT METHOD AND ANSWER
- Method: name the quantity; decide what its rate depends on; write rate of change = (constant) x (that dependence); set the rate to zero to find any equilibrium; read the long run from the sign of the rate on each side.
- Cooling drink: rate of change of T = -k x (T - room temperature), k positive, such as 0.1 per minute. Equilibrium: room temperature. It settles there, fast at first, then slowly.
- Draining bathtub, no tap: water leaves faster when deeper, so the level falls ever more slowly toward empty. With the tap running, it settles where inflow equals outflow.
- Growing following: early on, rate of change of F = r x F, exponential growth. Realistically it slows near a limit (logistic growth), with equilibria at 0 and at the limit.
- One step: new value = old value + rate x step. Drink at 90, room 20, k = 0.1, one minute: rate = -7, new value 83.
- Any reasonable rule tying the rate to the current value, with a sensible equilibrium, is a good answer. There's no single correct rule.

MISCONCEPTIONS TO CORRECT GENTLY
When one appears, name the accurate version briefly, then return to my process.
- "A rate tells you the future directly": it tells you only how fast things change now; you have to step it forward or solve it.
- "Exponential growth goes on forever": real systems hit limits, such as food, space, or audience size.
- "A differential equation is a single answer": its solution is a whole function, and different starting values give different solutions.

LIMITS
- No solution techniques beyond stepping forward; no calculus rules or the formula with e unless I ask.
- Plain notation only. Work with the process I choose.

TO FINISH
After I state my equilibrium and my one-step result, close in one short turn:
- Affirm one specific thing I worked out, in my own words where possible.
- Suggest one or two next steps that fit how the conversation went. Possible steps: step my rule forward three more times; halve my step and compare; write a rule for a process that grows instead of settling; review what "proportional" means with a small example.
- Restate my rule and its equilibrium on their own line, labeled "My law of change", so I can copy it.

Problem Set 15 minOptional Done

Stepping Models Forward

Overview

This optional problem set gives you five short problems on stepping simple models of change forward by hand. Each model is a differential equation, a rule that links a quantity to its own rate of change. For each step, find the current rate from the rule, multiply it by the step, and add the result to the current value. A calculator helps but isn't required. Try each problem, then compare your work with the answers and methods given.

Problems

Problem 1

A tank holds 10 liters of water and fills at a constant 3 liters per minute. How much water is in the tank after 5 minutes?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 1Hide the solution

Answer: 25 liters.

Method:

The rate is constant: 3 liters per minute, whatever the amount in the tank.
Over 5 minutes, the change is $3 \times 5 = 15$ liters.
Add it to the starting amount: $10 + 15 = 25$ liters.

Most common error: Writing $3 \times 5 = 15$ and stopping. The 15 liters is the change; the tank started with 10 liters already in it.

Problem 2

A population of 100 grows at 20 percent of its size per year. Step it forward one year at a time for two years. What are the populations after one year and after two years?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 2Hide the solution

Answer: 120 after one year, then 144 after two years.

Method:

The rate at the start is 20 percent of 100, which is $0.2 \times 100 = 20$ per year. One year's step adds 20, giving 120.
Now recompute the rate from the new value: $0.2 \times 120 = 24$ per year. The second step adds 24, giving 144.

Most common error: Adding 20 again in the second year and getting 140. The rule says the rate is 20 percent of the current size, so once the population is 120, the rate is 24.

Problem 3

A cup of coffee at 90 °C sits in a room at 20 °C. It cools at $0.1 \times (T - 20)$ degrees per minute, where $T$ is its current temperature in degrees Celsius. Step forward one minute at a time for two minutes. What are the temperatures after one minute and after two minutes?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 3Hide the solution

Answer: 83 °C after one minute, then 76.7 °C after two minutes.

Method:

At 90 °C, the gap above room temperature is $90 - 20 = 70$ . The rate of cooling is $0.1 \times 70 = 7$ degrees per minute. After one minute: $90 - 7 = 83$ °C.
At 83 °C, the gap is $83 - 20 = 63$ . The rate is $0.1 \times 63 = 6.3$ degrees per minute. After two minutes: $83 - 6.3 = 76.7$ °C.

Most common error: Taking one-tenth of the temperature itself, 9 degrees, instead of one-tenth of the gap above room temperature. The rule is about how much hotter the coffee is than the room.

Problem 4

For the coffee in Problem 3, at what temperature does the coffee stop cooling?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 4Hide the solution

Answer: 20 °C, room temperature.

Method:

The coffee stops cooling when its rate of change is zero.
Set the rate equal to zero: $0.1 \times (T - 20) = 0$ . That happens only when $T - 20 = 0$ , so $T = 20$ .
This value, where the rate is zero, is called the equilibrium. Coffee hotter than 20 °C cools toward it.

Most common error: Answering 0 °C. The coffee doesn't cool toward zero; it cools toward the temperature of the room around it.

Problem 5

A quantity grows at 7 percent of its size per year. About how many years does it take to double?

Once you've tried the problem, check your work against the solution.Show the solution to Problem 5Hide the solution

Answer: About 10 years.

Method:

Use the rule of 70: the doubling time is about 70 divided by the percent growth rate.
$70 \div 7 = 10$ years.
For growth added continuously, the exact figure is about 9.9 years, so the estimate is close. For growth added once a year, it's about 10.2 years.

Most common error: Reasoning that 7 percent a year needs about $100 \div 7 \approx 14$ years to add another 100 percent. That treats the growth as a fixed amount each year. Because each year's growth builds on the last, doubling comes sooner.

Self-Check

When you're done, check that:

Each new value you computed equals the old value plus the rate times the step.
You recomputed the rate from the newest value before each step.
Your cooling values move toward room temperature, not toward zero.
Your doubling-time estimate for 7 percent growth is close to 10 years.

Knowledge Check 10 min Done

Laws of change: differential equations from planetary orbits to population growth

This ungraded knowledge check assesses your understanding of how differential equations state laws of change. You'll be asked about the history of differential equations from Newton to Poincaré, what a differential equation and its solution are, numerical stepping and equilibrium, and exponential growth and decay.

Note: Use this to test yourself, review the feedback on any questions you miss, and retry until you feel confident before moving forward.

5 questions · ungraded · retry as often as you like