4.2 Direction Fields and Numerical Methods

1. Direction Fields and Numerical Methods

Title

Calculus II · Section 4.2

Seeing, and computing, solutions you cannot write down

2. What this lesson gives you

Objectives

Section 4.1 told you what a solution of a differential equation is and how to check one. It gave no way of finding one, and most differential equations have no formula for their solutions at all. This lesson gives you two ways around that: a picture and an algorithm.

• Draw the direction field of a first-order equation, by hand and by isoclines
• Sketch a solution curve by following the field from a starting point
• Find equilibrium solutions and classify them as stable, unstable or semi-stable
• Approximate a solution numerically with Euler's method
• Explain how Euler's error behaves as the step size shrinks, and when a step is too big

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 365-376 — learning objectives 4.2.1 to 4.2.3

In Section 4.1 you learned to check a proposed solution by substituting it into the equation. That is useful, but it assumes somebody hands you the solution. For most differential equations nobody can, because the solutions cannot be written with the functions you know.

This lesson gives you two tools that work anyway. The first is a picture, the direction field, which shows the shape of every solution at once and lets you read off long-term behaviour, including which constant solutions attract their neighbours and which repel them. The second is an algorithm, Euler's method, which turns the equation into a table of numbers you can compute by hand, in a spreadsheet, or in a few lines of code.

Both rest on one idea you already own: the derivative at a point is the slope of the tangent line there. Keep that in mind and nothing in this lesson will feel arbitrary.

3. Before anything new: the tangent line

Warm-up

Write your tangent line before revealing the answer. The recipe is point plus slope: the curve passes through (4, 2), and the derivative of the square root, one over twice the root, is one quarter there.

Evaluating the line at 4.1 costs one multiplication, and the answer 2.025 is already right to three decimal places. The line and the curve agree to first order near the point of tangency, and drift apart only slowly.

Hold on to this. A differential equation of the form y prime equals f of x and y is a machine that hands you the slope at any point you ask about. Give a tangent line a point and a slope and it will predict the next height. Euler's method, at the end of this lesson, is nothing but this warm-up repeated over and over.

4. The field

Section

Part 1

5. An equation that hands out slopes

Concept

A first-order equation of the form below does not tell you y. It tells you the slope of y at any point you name.

\[ y' = f(x, y) \]

Name the point (1, 2) for the equation y prime equals 3x plus 2y minus 4, and it answers with a number:

\[ y' = 3(1) + 2(2) - 4 = 3 \]

So any solution curve that passes through (1, 2) must pass through it with slope 3. Draw that as a short segment, and repeat at every point you like.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 366 — the slope at (1, 2)

Read the equation as a rule, not as a formula to solve. You name a point, and it tells you how steeply any solution through that point must be climbing or falling. At (1, 2) the answer is 3, so a solution curve through (1, 2) leaves it rising three units for every one unit to the right.

The figure shows eight such answers, each drawn as a short segment through its point. Look at how the slopes change: moving right adds 3 to the slope, moving up adds 2. That is just the coefficients of x and y in the equation, now visible as a pattern.

Nothing has been solved yet. You do not know any solution curve. But you already know the direction every solution curve must take at eight different places, and that turns out to be a remarkable amount of information.

6. Building the field of 3x + 2y - 4

Worked example

Compute enough slopes of the equation below to see the pattern of its direction field.

\[ y' = 3x + 2y - 4 \]

\[ f(0, 0) = 0 + 0 - 4 = -4 \]

\[ f(1, 0) = -1, \quad f(2, 0) = 2 \]

\[ f(0, 1) = -2, \quad f(0, 2) = 0 \]

\[ 3x + 2y - 4 = 0 \iff y = 2 - \tfrac32 x \]

\[ 3x + 2y - 4 = m \iff y = 2 + \tfrac{m}{2} - \tfrac32 x \]

\[ f(0, 2) = 0 + 4 - 4 = 0, \qquad f(2, -1) = 6 - 2 - 4 = 0 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 366 — Figure 4.7

The first three steps are plain substitution. Do a handful by hand once, because it makes the pattern obvious: the slope grows by 3 per unit right and by 2 per unit up.

The fourth and fifth steps are the efficient way to work. Instead of computing slopes point by point, ask where the slope takes a particular value. Setting the right-hand side equal to zero gives a whole line of flat segments; setting it equal to any other constant m gives a parallel line of segments with slope m. These lines are isoclines, and with three or four of them you can draw the whole field in a minute.

The check substitutes two points lying on the flat line and gets zero both times, which confirms the algebra. Look at the figure: the segments on the yellow dashed line really are horizontal, and those on the other two dashed lines really do share one slope each.

7. What a direction field is

Concept

direction field (slope field) — A picture of a first-order differential equation: at each point of a grid, a short segment whose slope equals the slope that the equation assigns to a solution passing through that point.

\[ \text{segment at } (x_0, y_0) \text{ has slope } f(x_0, y_0) \]

The field contains exactly the information the equation does, arranged for the eye. Nothing has been solved, and no constant of integration has been chosen: every solution curve of the equation is in the picture at once, waiting to be traced.

Segments are drawn at a fixed length rather than a fixed horizontal run, so a very steep slope still gives a short, readable mark.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 366 — Definition of a direction field

Here is the book's definition in one sentence: at each point, a segment with the slope that a solution through that point would have. The field is the differential equation, drawn.

A useful way to think about it: every solution of the equation is already in the picture. The general solution has an arbitrary constant, and each value of the constant gives one curve through the field. The initial condition is what picks out one of them.

The detail about fixed length matters when you draw by hand. If you drew each segment with the same horizontal run, a slope of 50 would give a segment fifty times taller than it is wide, spearing through the neighbouring points. Fixing the length keeps every mark inside its own cell.

8. Isoclines: where the slopes agree

Concept

An isocline is a curve along which the field has one constant slope. Setting the right-hand side equal to a constant gives it.

\[ f(x, y) = m \]

For the equation y prime equals x squared minus y squared, the isoclines are hyperbolas, and the zero-slope isocline is the pair of diagonals:

\[ x^2 - y^2 = 0 \iff y = \pm x, \qquad x^2 - y^2 = m \text{ a hyperbola for } m \ne 0 \]

Draw a few isoclines, put the matching segments along each, and the field is done. The zero isocline is the most valuable one: it is the only place a solution can have a horizontal tangent.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 366 — the field of x squared minus y squared

The word isocline means equal slope. Setting the right-hand side equal to a constant gives a curve, and all along that curve the segments are parallel. For the previous equation these curves were straight lines; for this one they are hyperbolas.

Look at the figure. On the two yellow diagonals the segments are flat. Above the upper diagonal and below the lower one, y squared beats x squared, so the slope is negative. In the left and right wedges x squared wins and the slope is positive. You now know the sign of the slope everywhere without computing a single point.

The zero isocline deserves special attention in every problem. It is the only place where a solution curve can have a horizontal tangent, so it is the only place a rising curve can turn into a falling one or back. You will use exactly this in Checkpoint 4.7.

9. Reading y' = f(x, y) piece by piece

Notation

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 365 — first-order equations of the form y′ = f(x, y)

Step through the notes one at a time. The first two are the reading you have been using: the left side is a slope, the right side is a rule that produces slopes from points.

The third note is the reason these equations are hard. If the right-hand side involved only x, you could integrate it, as you did for years. Here it involves y, the unknown function itself. To know the slope you need the height, and to know the height you need the slope. The direction field and Euler's method both break that circle by working locally: at one point at a time, the height is known and so the slope is known.

The last note is the initial condition. It is the one piece of information that chooses a single curve out of the infinitely many in the field.

10. Newton's law of cooling as a field

Picture it

\[ T'(t) = -0.4\left(T - 72\right) \]

The slope depends only on how far the temperature is from the room's 72 degrees. Above 72 the segments point down, below 72 they point up, and on 72 itself they are flat.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 365 — Figure 4.6

This is the book's opening example, Figure 4.6. T is the temperature of an object and 72 degrees is the temperature of the room. The equation says the object's temperature changes at a rate proportional to its distance from room temperature, and the minus sign says the change is always toward the room.

Look at how the field shows that. High up, far above 72, the segments point steeply down. Close to 72 they flatten out. Below 72 they tilt upward. The line at 72 is entirely flat segments.

The two curves are a hot object cooling and a cold one warming. Both obey the same law, both approach 72, and both approach it more and more slowly. Notice that the field also tells you something the curves only hint at: the object never overshoots room temperature, because the segments on 72 are flat. You will prove this carefully in Part 3.

11. Where is the field of (2x + 4)/(y - 2) vertical?

Prediction

\[ y' = \frac{2x + 4}{y - 2} \]

\[ 2x + 4 = 0 \iff x = -2 \;(\text{flat}), \qquad y - 2 = 0 \iff y = 2 \;(\text{vertical}) \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 366 — the third example equation

Commit to one option before you reveal. The equation is a quotient, and a quotient is zero where its top is zero and blows up where its bottom is zero.

The top, 2x plus 4, is zero when x is minus 2. So all along the vertical line x equal to minus 2 the segments are flat. The bottom, y minus 2, is zero when y is 2. Approaching that horizontal line the slope becomes enormous, so the segments stand up vertically, and on the line itself the equation is not defined.

People swap these because the lines point the opposite way to their segments: a vertical line of flat segments and a horizontal line of vertical segments. Say it slowly: flat segments where the slope is zero; vertical segments where the slope is infinite.

12. Which variable does the field depend on?

Sorting

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 377 — Exercises 74-83

Sort by looking only at which letters appear on the right-hand side. You do not need to compute anything.

When only y appears, two points at the same height get the same slope, wherever they are left or right. So each horizontal row of the field is a row of parallel segments, and the whole field looks the same when you slide it sideways. These equations are called autonomous, and they are the ones whose equilibria you will classify in Part 3.

When only x or t appears, points in the same vertical column get the same slope, and the equation is really just an antiderivative problem: every solution is one antiderivative shifted up or down. When both appear, as in 3y plus xy, neither pattern holds. Notice that 3y plus xy factors as (3 plus x) times y, which is still a mixture of both.

13. Match each equation to its field

Matching

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 378 — Exercises 84-88

Before matching, spend ten seconds on each small field and ask three questions. Where is it flat? Do the rows or the columns repeat? Where is it steepest?

Field B is flat along the horizontal axis and every column looks the same, so its slope depends only on y and vanishes at y equal to 0: that is minus 3y. Field D is flat along the vertical axis and every row is the same: slope depends only on t and vanishes at t equal to 0, which is minus 3t. Field C rises everywhere, gently on the left and steeply on the right, which is the exponential.

Field A is the only one flat on both axes, with signs that change from quadrant to quadrant: minus t times y is zero when either factor is, and negative when t and y have the same sign. Matching by features like these is a skill you will use on exam questions that show a field and ask for its equation.

14. Exercise 74: the field of y' = t cubed and its solutions

Worked example

Draw the field, solve the equation, and check that the solutions follow the field.

\[ y' = t^3 \]

\[ f(t, y) = t^3 \quad \text{(independent of } y\text{)} \]

\[ t < 0: y' < 0, \qquad t = 0: y' = 0, \qquad t > 0: y' > 0 \]

\[ y = \int t^3\,dt = \frac{t^4}{4} + C \]

\[ \frac{d}{dt}\left(\frac{t^4}{4} + C\right) = t^3 \;\checkmark \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 377 — Exercise 74

This exercise asks you to do something the section encourages throughout: draw the field, then solve the equation when you can, and see whether the solutions really follow the arrows.

Because the right-hand side involves only t, every column of the field is a column of parallel segments, flat at t equal to 0, falling to the left and rising to the right. You can read the shape of every solution from that: a curve that falls, flattens at t equal to 0, and rises. That is the shape of a quartic.

Integrating confirms it. The general solution is t to the fourth over 4 plus a constant, and in the figure you can see that the constant just slides the same curve up and down. The check differentiates the family and gets t cubed back, so every member fits the field exactly.

15. Following the field

Section

Part 2

16. Stepping off the point (0, 1)

Worked example

A solution of y prime equals 3x plus 2y minus 4 passes through (0, 1). Estimate its height at x equal to 0.1.

\[ f(0, 1) = 3(0) + 2(1) - 4 = -2 \]

\[ L(x) = 1 - 2(x - 0) = 1 - 2x \]

\[ L(0.1) = 1 - 0.2 = 0.8 \]

\[ y(0.1) = -0.15 + 1.25 - 0.25e^{0.2} \approx 0.79465 \quad (\text{error } 0.0054) \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 367 — the solution through (0, 1)

This is the book's first use of a field to predict a solution, and it is also the first step of Euler's method, although the book does not name it yet.

You are told only that a solution passes through (0, 1). The equation gives its slope there, minus 2, and a point and a slope give a tangent line. Walk along that line for a run of 0.1 and you land at height 0.8. That is your estimate of the solution at x equal to 0.1.

The check uses the exact solution, which the book supplies, and finds 0.7946. So one short step along a tangent line was off by about five thousandths. In the figure the tangent line and the true curve are almost on top of each other for such a short run; they separate only as you walk further. That separation is the error of the method, and you will measure it carefully in Part 5.

17. The solution curve through (0, 1)

Picture it

\[ y = -\tfrac32 x + \tfrac54 + Ce^{2x}, \qquad C = -\tfrac14 \text{ through } (0, 1) \]

Repeat the tangent step with ever smaller runs and the steps fuse into a smooth curve that is tangent to the field everywhere: the solution curve. Every other starting point gives another member of the same family.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 367 — Figure 4.8

Imagine repeating the last slide's step with a run of a thousandth, then a millionth, re-reading the slope after every step. The broken line you trace fuses into a smooth curve that follows the field everywhere. That limiting curve is the solution curve through (0, 1), the highlighted curve in the figure.

The formula underneath is a family with one constant C. The value minus one quarter is the member through (0, 1). The value zero gives a straight line, and it is worth checking that a straight line can be a solution: its slope is minus three halves, and substituting it into the right-hand side gives minus three halves too.

Look at what that line does to its neighbours. Curves just above it bend upward and escape; curves just below it bend down and plunge. The field tells you this at a glance, which is exactly the kind of long-term information a picture can give you without any algebra.

18. Checking the solution through (0, 1)

Worked example

Confirm that the book's formula really is the solution of the initial-value problem.

\[ y' = 3x + 2y - 4, \quad y(0) = 1, \qquad y = -\tfrac32 x + \tfrac54 - \tfrac14 e^{2x} \]

\[ y' = -\tfrac32 - \tfrac12 e^{2x} \]

\[ 3x + 2\left(-\tfrac32 x + \tfrac54 - \tfrac14 e^{2x}\right) - 4 \]

\[ = 3x - 3x + \tfrac52 - \tfrac12 e^{2x} - 4 \]

\[ = -\tfrac32 - \tfrac12 e^{2x} \]

\[ y' = 3x + 2y - 4 \;\checkmark, \qquad y(0) = \tfrac54 - \tfrac14 = 1 \;\checkmark \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 367 — the exact solution after Figure 4.8

This is the Section 4.1 skill applied to the book's formula, and it is worth doing once because it builds trust in the picture. Differentiate the proposed solution, substitute it into the right-hand side, and see whether the two sides agree.

The left side is straightforward: the derivative of the exponential term brings down a factor 2. For the right side, substitute the whole formula for y, multiply out, and watch the 3x terms cancel. What is left, minus three halves minus one half of e to the 2x, is exactly the derivative.

Finally the initial condition: at x equal to 0 the exponential is 1, and five quarters minus one quarter is 1. Both requirements are met, so this is the solution, and by uniqueness it is the only one.

19. Checkpoint 4.7: the curve through (-1, 2)

Worked example

Use the field of the equation below to sketch the solution through (-1, 2).

\[ y' = x^2 - y^2, \qquad y(-1) = 2 \]

\[ f(-1, 2) = 1 - 4 = -3 \]

\[ x^2 - y^2 = 0 \iff y = \pm x \]

\[ y > |x| \;\Longrightarrow\; y' < 0 \]

\[ \text{lowest point on } y = x: \quad (0.578,\, 0.578) \]

\[ |x| > |y| \;\Longrightarrow\; y' > 0 \]

\[ f(0, 0.749) = 0 - 0.561 = -0.561 < 0 \;\checkmark \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 367 — Checkpoint 4.7

This checkpoint asks for a sketch, not a formula, and the way to get a good sketch is to decide the sign of the slope everywhere before drawing anything. The isoclines slide gave you that: negative above and below both diagonals, positive in the left and right wedges, zero on the diagonals.

The starting point (-1, 2) is above both diagonals, and the slope there is minus 3, so the curve heads down. It keeps falling as long as it stays above both diagonals. It can only stop falling where the slope is zero, and the first diagonal it can meet is the line y equal to x, which it reaches at about (0.578, 0.578). After that it is in the right-hand wedge, where slopes are positive, so it rises.

The check picks a point on the way down. An accurate numerical solution puts the curve at 0.749 when x is 0, and the slope there is negative, as it must be if the turning point comes later. You will compute that 0.749 yourself with Euler's method in Part 4.

20. Solution curves never cross

Concept

Suppose two solution curves met at a point. Both would pass through it, so both would solve the same initial-value problem there.

\[ y_1(x_0) = y_2(x_0) = y_0, \qquad y_1' = y_2' = f(x_0, y_0) \]

For the equations of this chapter, with f and its partial derivative in y continuous, an initial-value problem has exactly one solution. So two curves through one point are the same curve.

\[ y = x - 1 + Ce^{-x}: \quad \text{different } C \text{ never meet, since } Ce^{-x} \ne C'e^{-x} \]

Non-crossing is what makes sketching reliable: once you have drawn one solution, every other solution must stay on its own side of it.

This fact is used in every sketch, so it is worth knowing why it holds. If two solution curves met at a point, each would solve the initial-value problem that starts at that point. For the equations in this chapter, where the right-hand side and its partial derivative with respect to y are continuous, that initial-value problem has exactly one solution. So the two curves are the same curve.

The figure shows the family for y prime equals x minus y. Different constants give curves that crowd closer and closer to the line y equal to x minus 1 but never touch it or each other, because the exponential term never vanishes.

In practice, non-crossing is what lets you fence a sketch in. Once you have drawn one solution, every other solution stays on its own side of it forever. The most useful fences of all are the flat, constant solutions of Part 3.

21. Find the error: a sketch that cuts through y = 2

Error analysis

Try to find the flaw before revealing the notes. The sketch describes a curve starting at height 3 that falls, passes through 2 and settles at 0.

The first note is the key fact: y equal to 2 makes the right-hand side zero for every x, so the horizontal line y equal to 2 is itself a solution. A curve that crossed it would share a point with another solution, which the previous slide ruled out.

The third note shows the sketch is wrong twice over. Even if the curve somehow got below 2, the field between 0 and 2 points upward, so it could not drift down to 0. The correct sketch falls from 3, flattens, and approaches 2 from above without ever reaching it. Whenever a sketch passes through a flat line of segments, stop and check whether that line is a solution.

22. Before sketching: find where the field is flat

Step zero

\[ y' = y^2 - 2y, \qquad y(0) = 3, \; 1, \; -1 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 377 — Exercises 70-73

Write your answer before revealing. The instinct is to start drawing arrows from each starting point, but three seconds of algebra first saves a lot of guesswork.

Factor the right-hand side and set it to zero: y equal to 0 and y equal to 2 are constant solutions, fences that no other curve can cross. Then test the sign of the right-hand side once in each region. At 3 it is positive, at 1 it is negative, at minus 1 it is positive.

Now every sketch is decided. From 3, above both fences, the curve rises and pulls away from 2. From 1, between the fences, it falls toward 0. From minus 1, below both, it rises toward 0. Notice that 0 attracts from both sides and 2 repels on both sides; that observation is exactly what the next part of the lesson names.

23. Put the sketching steps in order

Ranking

Arrange the steps before checking. Most of the order is natural: you cannot read a slope before you have a point, and you cannot repeat before you have moved.

The one step people put late is drawing the flat lines. Putting it first is the professional habit. The flat lines tell you where no curve may cross and where a curve can turn around, so they constrain every sketch you draw afterwards, and they cost almost nothing to find.

The last three steps, read the slope, move a little, read again, are also a description of Euler's method. You are doing numerical analysis with a pencil.

24. Equilibria and stability

Section

Part 3

25. Constant solutions: equilibria

Concept

Try a constant function y equal to k in the equation y prime equals (x minus 3)(y squared minus 4). Its derivative is zero, so it is a solution exactly when the right-hand side is zero for every x.

\[ 0 = (x - 3)(k^2 - 4) \quad \text{for all } x \]

\[ k^2 - 4 = 0 \iff k = -2 \text{ or } k = 2 \]

equilibrium solution — A constant solution y equal to k of y prime equals f(x, y). To find them, solve f(x, k) equal to zero for every x in the domain.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 368 — Figure 4.9 and the definition of equilibrium solution

A constant function has derivative zero, so it solves the equation exactly when the right-hand side is zero along it, for every x. That turns a differential equation problem into an ordinary algebra problem.

Here the right-hand side is (x minus 3) times (y squared minus 4). For a constant k to work, the product must vanish for every x, not just at x equal to 3. The only way is for the second factor to be zero, which gives k equal to 2 or minus 2. The book's argument shows these are the only constant solutions.

In the figure, the two equilibrium lines are rows of flat segments running all the way across. A solution starting between them, at 0.5, rises and hugs the line at 2. Remember this curve: two slides from now you will follow it further to the right and see something the book's window does not show.

26. Trap: a flat diagonal is not an equilibrium

Trap

27. Stable, unstable and semi-stable

Concept

An equilibrium is a solution you could sit on forever. The question is what happens if you start slightly off it.

asymptotically stable — Every solution starting close enough to k, above or below, approaches k as x goes to infinity.

asymptotically unstable — No solution starting close to k (other than k itself) approaches k as x goes to infinity.

semi-stable — Neither: solutions approach from one side and leave on the other.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 369 — Definition of stability

Equilibria come in kinds, and the kind is decided by what happens to solutions that start close by. Think of a ball resting at the bottom of a bowl, on top of a hill, or on a ledge partway down a slope.

The three panels are drawn from real equations, each with the equilibrium y equal to 0. For y prime equals minus y, solutions are multiples of e to the minus x, and every one of them flows into 0: stable. For y prime equals y, they are multiples of e to the x, and every one except 0 itself runs away: unstable. For y prime equals y squared, the slope is never negative, so everything moves up. Solutions below 0 rise into it; solutions above rise away from it: semi-stable.

When you classify an equilibrium, you are always asking the same two questions: what happens just above it, and what happens just below it.

28. Reading the definition of stability

Notation

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 369 — Definition, parts 1-3

The book's definition is precise, and each piece of it matters. Step through the annotations and connect each to the pictures on the previous slide.

The small number epsilon says the definition only cares about starting values near k. A stable equilibrium may still lose solutions that start far away; the stable equilibrium of y prime equals y squared minus 1 at minus 1 is an example, since solutions starting above 1 run off to infinity.

The phrase as x approaches infinity matters too, and the next few slides show why. A solution may sit near an equilibrium for a long stretch and still leave it later. Stability is about where solutions end up, not where they spend their time. Semi-stable is defined as neither, and in the examples of this section it always means in from one side, out on the other.

29. Autonomous equations: one column says it all

Concept

When the right-hand side depends on y alone, the equation is called autonomous, and the field looks the same in every column.

\[ y' = f(y) \]

Then the sign of f between two equilibria is the whole story: positive means every solution there rises, negative means it falls, and it can never cross the equilibria that fence it in.

\[ f(y) > 0 \;\Rightarrow\; y \nearrow, \qquad f(y) < 0 \;\Rightarrow\; y \searrow \]

So to classify an equilibrium, test the sign of f once just below it and once just above it.

Most of the equilibria you will classify belong to autonomous equations, where the right-hand side depends on y alone. For these, the field is the same in every column, so you can study one column and know the whole picture.

Between two neighbouring equilibria the right-hand side has no zeros, and a continuous function with no zeros keeps one sign. So in each band between equilibria every solution is either rising the whole time or falling the whole time.

That gives a two-test method for classifying any equilibrium: pick a value just below it and a value just above it, and check the sign of f at each. Arrows pointing toward it from both sides mean stable, away on both sides unstable, one of each semi-stable.

30. Trapped between two fences

Intuition

Take an autonomous equation and a solution that starts strictly between two neighbouring equilibria. It cannot cross either of them, and the sign of f does not change in between, so the solution moves in one direction only.

\[ k_1 < y(x_0) < k_2 \;\Longrightarrow\; k_1 < y(x) < k_2 \text{ for all } x \]

A quantity that only increases and never passes a ceiling must level off, and the only place it can level off is where the slope is zero. So it approaches the equilibrium it is heading toward.

That is why the sign pattern is enough: every solution of an autonomous equation either runs to an equilibrium or runs off to infinity.

This slide explains why the sign test is enough. A solution that starts between two neighbouring equilibria can never cross either one, so it stays in that band forever. Inside the band the slope has one sign, so the solution only ever moves one way.

Now use a fact about increasing functions: a function that keeps increasing but never passes a ceiling must level off at some height. A solution can only level off where its slope becomes zero, and in that band the slope is zero only at the fence it is heading toward.

Put together: for an autonomous equation there are only two possible fates for a solution. It approaches an equilibrium, or there is no fence in its way and it runs off to infinity. The phase line on the next slides is simply a map of which fate each starting value gets.

31. Example 4.8: three equilibria, three verdicts

Worked example

Find and classify the equilibrium solutions.

\[ y' = (y - 3)^2(y^2 + y - 2) \]

\[ y^2 + y - 2 = (y + 2)(y - 1) \]

\[ (y - 3)^2(y + 2)(y - 1) = 0 \iff y = -2, \; 1, \; 3 \]

\[ f(-3) = 144, \quad f(0) = -18, \quad f(2) = 4, \quad f(4) = 18 \]

\[ y = -2: \; + \text{ below}, \; - \text{ above} \;\Rightarrow\; \text{stable} \]

\[ y = 1: \; - \text{ below}, \; + \text{ above} \;\Rightarrow\; \text{unstable} \]

\[ y = 3: \; + \text{ below}, \; + \text{ above} \;\Rightarrow\; \text{semi-stable} \]

\[ y(0) = 0.92 \to y(3) \approx -2.000, \qquad y(0) = 2.2 \to y(3) \approx 2.965 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 370-371 — Example 4.8 and Figure 4.12

The equation is autonomous, so the two-test method applies. First factor completely: the quadratic splits as (y plus 2)(y minus 1), and together with the squared factor that gives three equilibria, at minus 2, 1 and 3.

Next, one test value in each of the four intervals they create. The signs come out plus, minus, plus, plus. Read each equilibrium from the signs on either side. At minus 2, solutions below rise and solutions above fall: both come in, so it is stable. At 1 it is the reverse, so it is unstable. At 3 the squared factor keeps the sign positive on both sides, so solutions below rise into it and solutions above rise away from it: semi-stable.

The figure confirms all three, and the check puts numbers on it. A solution starting at 0.92, just under the unstable equilibrium at 1, has fallen all the way to minus 2 by x equal to 3. One starting at 2.2 has crept up to 2.965, approaching 3 from below, slowly, because the squared factor makes the slope tiny near 3.

32. The phase line: the whole field in one column

Picture it

Because the field of an autonomous equation repeats in every column, one column carries all of it. Plot f against y, read off where it is positive and negative, and draw the arrows on a single line.

Because an autonomous field repeats in every column, all its information fits on one line. The top of the figure is the graph of the right-hand side against y. Where the graph is above the axis, solutions increase; where it is below, they decrease.

The arrows on the line underneath record exactly that: right for increasing, left for decreasing. Now each equilibrium is classified by looking at the two arrows beside it. Both pointing in, as at minus 2, means stable. Both pointing out, as at 1, means unstable. Both pointing the same way, as at 3, means semi-stable.

Notice the graph at 3: it touches the axis and turns back instead of crossing. That touch is the picture of a squared factor, and it is always the signature of a semi-stable equilibrium.

33. Complete the sign table for y' = y squared minus 1

Comparison

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 377 — Exercise 79

Fill in each blank before checking. This is Exercise 79, and the table is the whole of the two-test method written down.

The equilibria are at minus 1 and 1, where y squared equals 1. Below minus 1, y squared is bigger than 1, so the slope is positive and solutions rise toward minus 1. Between the two, y squared is less than 1, so solutions fall, toward minus 1 again. Above 1 the slope is positive and solutions rise away.

So minus 1 has arrows pointing in from both sides and is stable; 1 has arrows pointing out on both sides and is unstable. If your field sketch disagrees with your table, trust the table and redraw the sketch.

34. Checkpoint 4.8: a factor that depends on x

Worked example

Find and classify the equilibria.

\[ y' = (x + 5)(y + 2)(y^2 - 4y + 4) \]

\[ y^2 - 4y + 4 = (y - 2)^2 \]

\[ (x + 5)(k + 2)(k - 2)^2 = 0 \; \forall x \iff k = -2 \text{ or } 2 \]

\[ x > -5: \quad \operatorname{sign} f = \operatorname{sign}\big[(y + 2)(y - 2)^2\big] \]

\[ f(0, -3) = -125, \quad f(0, 0) = 40, \quad f(0, 3) = 25 \]

\[ y = -2: \; - \text{ below}, \; + \text{ above} \;\Rightarrow\; \text{unstable} \]

\[ y = 2: \; + \text{ below}, \; + \text{ above} \;\Rightarrow\; \text{semi-stable} \]

\[ y(-4.6) = 0.8 \;\longrightarrow\; y(3) \approx 1.992 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 371 — Checkpoint 4.8

This equation is not autonomous: the factor x plus 5 changes sign at x equal to minus 5, so the whole field reverses direction there. You can see it in the figure, where every segment tilts one way to the left of the dashed line and the other way to the right.

Finding the equilibria still works the same way: the right-hand side must be zero for every x, so one of the y factors must be zero, which gives minus 2 and 2. For stability, remember that the definition is about x going to infinity. Once x is past minus 5, the factor x plus 5 is positive and the signs are decided by the y factors alone.

The squared factor at 2 should now make you suspect semi-stable, and the test values confirm it: positive below 2 and positive above. At minus 2 the signs are negative below and positive above, so solutions leave on both sides. The numerical check follows a solution from 0.8 and finds it at 1.992 by x equal to 3, creeping up to 2 from below exactly as predicted.

35. Follow the arrows past x = 3

Anomaly

\[ \int\frac{dy}{y^2 - 4} = \int(x - 3)\,dx \;\Longrightarrow\; \tfrac14\ln\left|\frac{y - 2}{y + 2}\right| = \tfrac12(x - 3)^2 + C \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 368-370 — Figures 4.9 to 4.11

Commit to an answer before revealing. This slide revisits the equation of the book's Figures 4.9 to 4.11, whose field depends on x through the factor x minus 3.

The book's figures show the solution from 0.5 over a short window, where it rises to 2 and sits there, and the book concludes that 2 is stable. But to the right of x equal to 3 the factor x minus 3 turns positive and every arrow between minus 2 and 2 flips to point down. The solution sits within a ten-thousandth of 2 until about x equal to 5, then peels away, passes back through 0.5 at x equal to 6, and settles onto minus 2.

The formula at the bottom explains the symmetry. Separating variables, a method from Section 4.3, gives a solution that depends on x only through (x minus 3) squared, so every solution is a mirror image of itself about x equal to 3. For large x, minus 2 attracts from both sides and 2 repels, the opposite of the book's reading. The lesson: stability is about x going to infinity, so never judge it from a window that stops early, especially when the right-hand side involves x.

36. Trap: calling a semi-stable equilibrium stable

Trap

37. An equation with no equilibrium

Counterexample

Write your counterexample before revealing. The claim sounds plausible because every equation you have met so far had some flat line in its field.

An equilibrium needs the right-hand side to be zero for some constant y, for every x. For y squared plus 1 that never happens, because a square plus one is at least one. Every segment in the field slopes upward, and every solution climbs forever; in fact the solutions are tangent functions, which climb to infinity in finite time.

The equation y prime equals e to the t is another counterexample, and so is y prime equals x minus y from the earlier trap. Equilibria are special features that some equations have, not something every equation is owed.

38. Euler's method

Section

Part 4

39. Why compute instead of solve

Intuition

The equation y prime equals x squared minus y squared looks innocent, but none of the solution methods of this chapter can solve it, and its solutions cannot be written with the functions you know.

The field still shows you their shape. What it cannot give you is a number: the height of the solution through (-1, 2) at x equal to 0, to four decimal places. For that you follow the field with arithmetic instead of with a pencil.

That arithmetic is simple enough to run in a spreadsheet or a ten-line program, which is how almost every differential equation in science and engineering actually gets solved.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 365 — numerical methods

The next section teaches a real solution method, separation of variables, and later sections add more. But each method works only for equations of a particular shape, and the equation y prime equals x squared minus y squared is not of any of those shapes. Its solutions exist, and the field shows them clearly, but there is no formula in elementary functions.

A picture answers qualitative questions: does the solution rise or fall, where does it turn, what does it approach. It cannot tell you that the solution through (-1, 2) is at 0.749 when x is 0. Engineers, physicists and economists need that kind of number.

Euler's method gets it by following the field numerically. It is the simplest numerical method there is, and the ideas you learn with it, step size, local error, accumulated error and instability, carry over to the more accurate methods used in real software.

40. From a tangent line to a step

Concept

At a point on the solution, the equation gives the slope, and a slope and a point give a tangent line.

\[ L(x) = y_0 + f(x_0, y_0)(x - x_0) \]

Step right by a small amount h, called the step size, and read the new height off the tangent line:

\[ x_1 = x_0 + h \]

\[ y_1 = L(x_1) = y_0 + h\, f(x_0, y_0) \]

Now stand at the new point, ask the equation for a new slope, and step again. That is the whole method.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 372 — Figure 4.13

This is the warm-up, rewritten for a differential equation. At the starting point you know the height and, from the equation, the slope. That gives the tangent line, the first line on the slide.

Choose a step size h, move right by h, and read the height off the tangent line. Because the run is h and the slope is the right-hand side at the starting point, the rise is h times that slope. That gives the formula for the new height.

The figure shows one step on the parabola from the book's Figure 4.13. From (0, 3) the slope is minus 3, so a run of 0.5 gives a fall of 1.5 and lands at 1.5. The true curve is at 1.75 there, because it bends upward while the tangent line goes straight. Then the method starts again from the new point, with a new slope.

41. Reading Theorem 4.1 piece by piece

Notation

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 373 — Theorem 4.1 and equation 4.2

Theorem 4.1 is the step of the previous slide, written once for every n. Step through the notes in order.

The x-values march forward in equal steps of h. The new height is the old height plus a rise, and the rise is h times a slope. The most important detail is which slope: the one at the start of the step, evaluated at the previous point. That is the only point where both coordinates are known, since the true solution is unknown and the next point has not been computed yet.

The second note is the source of accumulated error. Each step starts from the previous estimate, not from the true solution, so any error made earlier is carried into every later step. Keep that in mind when you reach the error slides in Part 5.

42. Euler on y' = 2x - 3 with h = 0.5

Worked example

Run four steps of Euler's method from (0, 3), then compare with the exact solution.

\[ y' = 2x - 3, \quad y(0) = 3, \quad h = 0.5 \]

\[ y_1 = 3 + 0.5(2 \cdot 0 - 3) = 3 - 1.5 = 1.5 \]

\[ y_2 = 1.5 + 0.5(2 \cdot 0.5 - 3) = 1.5 - 1 = 0.5 \]

\[ y_3 = 0.5 + 0.5(2 \cdot 1 - 3) = 0.5 - 0.5 = 0 \]

\[ y_4 = 0 + 0.5(2 \cdot 1.5 - 3) = 0 \]

\[ y = x^2 - 3x + C, \quad C = 3 \]

\[ y(2) = 4 - 6 + 3 = 1, \qquad y_4 = 0, \qquad \text{error} = 1 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 371-372 — Figure 4.13

This is the book's first Euler example, chosen because the exact solution is known, so you can see the error directly. The right-hand side depends only on x, which makes every slope easy to compute.

Each step has the same shape: evaluate the slope at the current point, multiply by 0.5, add. The slopes are minus 3, minus 2, minus 1 and 0, and the heights go 3, 1.5, 0.5, 0, 0. Writing every product in full is slow, but it is the habit that prevents the traps on the next few slides.

The check compares with the exact solution, a parabola, at x equal to 2: the true value is 1 and Euler reached 0. The figure shows why the error keeps growing. The parabola curves upward, every tangent line runs below it, and each new step starts from a point that is already too low. The red bars grow by exactly 0.25 each step, a pattern you will explain in Part 5.

43. The book's first two steps for y' = x squared minus y squared

Worked example

Take two steps of size 0.1 from (-1, 2), writing each tangent line in full, the way the book does.

\[ y' = x^2 - y^2, \quad y(-1) = 2, \quad h = 0.1 \]

\[ f(-1, 2) = 1 - 4 = -3 \]

\[ L(x) = 2 - 3(x + 1) = -3x - 1 \]

\[ y_1 = L(-0.9) = 2.7 - 1 = 1.7 \]

\[ f(-0.9, 1.7) = 0.81 - 2.89 = -2.08 \]

\[ L(x) = 1.7 - 2.08(x + 0.9) = -2.08x - 0.172 \]

\[ y_2 = L(-0.8) = 1.664 - 0.172 = 1.492 \]

\[ y_2 = y_1 + h\,f(x_1, y_1) = 1.7 + 0.1(-2.08) = 1.492 \;\checkmark \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 372-373 — the worked steps before Table 4.2

The book introduces Euler's method on this equation by writing out each tangent line in full. It is slower than the formula, but it shows clearly that every step really is a linear approximation.

From (-1, 2) the slope is minus 3 and the tangent line simplifies to minus 3x minus 1. Evaluating at minus 0.9 gives 1.7. From (-0.9, 1.7) the new slope is 0.81 minus 2.89, which is minus 2.08, and the new tangent line gives 1.492 at minus 0.8.

The check shows that you never need to write the line out. Old height plus step times slope gives 1.492 in one line. From now on use the formula, but remember what it means: stand at a point, look along the field, walk a short way.

44. Table 4.2: ten steps to x = 0

Worked example

Continue to x equal to 0 with the formula, keeping every digit and rounding only for display.

\[ y_3 = 1.492 + 0.1(0.64 - 2.226064) = 1.492 - 0.158606 = 1.333394 \]

\[ y_4 = 1.333394 + 0.1(0.49 - 1.777938) = 1.204600 \]

\[ y(0) \approx 0.7490, \quad y_{10} = 0.7156, \quad \text{error} \approx 0.033 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 373 — Table 4.2

The remaining steps are pure bookkeeping. Keep six or more decimal places while computing and round only when you write the table, or rounding errors will creep into the last digits.

A warning if you compare with the printed Table 4.2: the book's row for n equal to 6 shows 1.0004, which is 1.0005 when correctly rounded, and the entries for n equal to 7 to 10 are misprinted with a leading 1, as 1.9164 and so on. The correct values fall steadily, as the negative slopes demand: 0.9164, 0.8414, 0.7746, 0.7156.

The check compares with an accurate numerical solution, 0.7490 at x equal to 0. Euler is low by about 0.033. The figure shows the reason: the true curve is bending upward as it approaches its turning point, so every tangent-line step undershoots it.

45. Trap: taking the slope at the new point

Trap

46. Find the error: the step size went missing

Error analysis

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 375 — Example 4.9, first row

Read the line and find the mistake before revealing the notes. Every piece of arithmetic in it is correct.

The error is conceptual: the slope was added to the height directly. A slope is a rate, rise per unit of run. To turn it into a rise you multiply by the run, which is the step size 0.1. Without it, the computation takes a step of length 1, ten times too long.

The last note is a habit worth keeping. With a small step the height should change only a little. A jump from 2 to minus 1 in a step of 0.1 would need a slope of minus 30, and the equation gives minus 3. Whenever a single step moves further than you expect, check that h is in the formula.

47. Example 4.9: a table from x = 0 to 1

Worked example

Use Euler's method with step 0.1 to tabulate the solution for x between 0 and 1.

\[ y' = 3x^2 - y^2 + 1, \quad y(0) = 2, \quad h = 0.1 \]

\[ y_1 = 2 + 0.1(-3) = 1.7 \]

\[ y_2 = 1.7 + 0.1(-1.86) = 1.514 \]

\[ y_3 = 1.514 + 0.1(-1.172196) = 1.396780 \]

\[ y(1) \approx 1.6698, \quad y_{10} = 1.6202, \quad \text{error} \approx 0.050 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, pp. 374-375 — Example 4.9 and Table 4.3

This is the book's full worked example, and the first three rows are shown with every slope computed so you can follow the pattern. Each row needs the right-hand side at the previous point: three times x squared, minus y squared, plus 1.

Rows 4 to 10 continue the same way. Notice the slope changes sign between rows 4 and 5, so the heights stop falling and start rising. The accurate solution turns around near x equal to 0.54, and the Euler table turns at about the same place.

The check compares with an accurate numerical value at x equal to 1, which is 1.6698. Euler's 1.6202 is low by about 0.05. The figure shows where it went wrong: the first step, aimed with slope minus 3, drops too far while the true curve is already flattening, and the table never makes up the difference.

48. Checkpoint 4.9: when the solution steepens

Worked example

Use step 0.1 to tabulate the solution for x between 1 and 2.

\[ y' = x^3 + y^2, \quad y(1) = -2, \quad h = 0.1 \]

\[ y_1 = -2 + 0.1(5) = -1.5 \]

\[ y_2 = -1.5 + 0.1(3.581) = -1.1419 \]

\[ y_3 = -1.1419 + 0.1(3.031936) = -0.838706 \]

\[ y(2) \approx 3.5628, \quad y_{10} = 2.6962, \quad \text{error} \approx 0.87 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 376 — Checkpoint 4.9

The procedure is identical; only the equation changes. Keep careful track of signs early on: y starts negative, but y squared is positive, so every slope in this problem is positive and the heights rise at every step.

Once y turns positive around x equal to 1.6, both terms on the right-hand side are growing and the slope takes off: 5 at the start, over 15 by the end.

The check shows why this matters. An accurate solution reaches 3.563 at x equal to 2, while Euler reaches only 2.696, an error of 0.87. When the slope increases quickly along a step, the slope at the start of the step is badly out of date by the end, and each step's shortfall feeds the next. The same step size that gave an error of 0.05 in Example 4.9 gives an error almost twenty times as large here.

49. Two Euler steps, by hand

Fill the middle

\[ y' = y + t^2, \quad y(0) = 3, \quad h = 0.2 \]

\[ y_1 = y_0 + 0.2\left(y_0 + t_0^2\right), \qquad y_2 = y_1 + 0.2\left(y_1 + t_1^2\right) \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 379 — Exercise 97

Fill in both blanks before checking. This is Exercise 97, and the only way to get the second blank is to get the first one right, which is exactly how Euler's method works.

First step: at (0, 3) the slope is 3 plus 0, so the rise is 0.2 times 3, which is 0.6, and the new height is 3.6. Second step: now at (0.2, 3.6) the slope is 3.6 plus 0.04, which is 3.64, and the rise is 0.728. The new height is 4.328.

The exact solution, which the book gives, is 4.499 at t equal to 0.4, so two steps are already low by 0.17. The solution grows like an exponential, curving up away from every tangent line, which is the same reason the parabola example came out low.

50. Slide h and watch Euler's polygon

Tweak it

For y prime equals y, each Euler step multiplies the height by 1 plus h, so the polygon's corners are at the powers of 1 plus h. Start with h at 1: two steps take you to 4, while the true solution is at 7.389.

Now slide h down slowly. At 0.5 the polygon reaches about 5.06, at 0.25 about 5.96, and at 0.05 about 7.04. Every polygon lies below the true exponential, because the exponential curves upward away from each tangent line, and the gap shrinks steadily as the steps get shorter.

You might recognise the limit: 1 plus h, raised to the power 2 over h, approaches e squared as h goes to zero. Euler's method for this equation is the compound-interest formula, and shrinking the step is compounding more often.

51. Step size and error

Section

Part 5

52. One step's miss is of size h squared

Concept

Start a step exactly on the true solution. Taylor's theorem compares where the curve goes with where the tangent line goes:

\[ y(x_0 + h) = y_0 + h\,y'(x_0) + \tfrac12 h^2 y''(c) \]

\[ \text{Euler: } y_1 = y_0 + h\,y'(x_0) \]

\[ y(x_0 + h) - y_1 = \tfrac12 h^2 y''(c) \]

The miss depends on the curvature and on h squared. For the parabola of y prime equals 2x minus 3, the second derivative is 2, so every step from a point on the curve misses by exactly h squared.

\[ y'' = 2 \;\Rightarrow\; \text{miss} = h^2: \quad h = 0.5 \text{ gives } 0.25 \]

To understand Euler's error, first look at a single step that starts exactly on the true solution. Taylor's theorem with remainder, which you will meet properly in Chapter 6, says the true curve equals its tangent line plus a correction involving the second derivative and h squared.

Euler keeps the tangent line and throws the correction away, so the miss on one step is one half of h squared times the second derivative somewhere in the step. Two things control it: how sharply the solution bends, and the square of the step size.

For the parabola of y prime equals 2x minus 3, the second derivative is exactly 2 everywhere, so every step that starts on the curve misses by exactly h squared. With h equal to 0.5 that is 0.25, which is exactly the first red bar in the parabola figure. Halve the step and each miss becomes a quarter as big.

53. Halving h on y' = 2x - 3

Worked example

Find Euler's value exactly for any step size, and compare two step sizes at x equal to 4.

\[ y_n = 3 + h\sum_{k=0}^{n-1}(2kh - 3) \]

\[ y_n = 3 + h^2 n(n - 1) - 3nh \]

\[ y_n = 3 + x(x - h) - 3x \]

\[ \text{error} = (x^2 - 3x + 3) - y_n = xh \]

\[ h = 0.5: \; y = 5, \; \text{error } 2; \qquad h = 0.25: \; y = 6, \; \text{error } 1 \]

\[ y(4) = 7: \quad 7 - 5 = 2 = 4(0.5), \quad 7 - 6 = 1 = 4(0.25) \;\checkmark \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 374 — Figure 4.14

Here you can find Euler's answer exactly for every step size, which is rare and very instructive. Because the right-hand side ignores y, the heights are just the starting value plus a sum of rises, and the rises form an arithmetic series.

Summing the series and writing everything in terms of x, which is n times h, gives Euler's height as a simple quadratic. Subtract it from the exact parabola and almost everything cancels, leaving an error of exactly x times h. So the error grows in proportion to how far you have travelled, and in proportion to the step size.

The figure is the book's Figure 4.14 with the error marked. At x equal to 4 the true value is 7; with h equal to 0.5 Euler reaches 5, and with 0.25 it reaches 6. The check compares the formula with the step-by-step values and they agree exactly: halving the step halved the error from 2 to 1.

54. Adding up the misses: error proportional to h

Concept

Reaching a fixed x takes x over h steps. Each step misses by roughly a constant times h squared, and the misses accumulate:

\[ \frac{x}{h} \text{ steps} \times Kh^2 \text{ each} = Kx\,h \]

So halving the step roughly halves the error at a given x. That makes Euler a first-order method.

Errors made early are also carried forward: each step starts from the previous step's wrong height, so errors pile up instead of cancelling.

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 373 — step size and accuracy

Put the last two slides together. To reach a fixed x you take x over h steps. Each misses by something like a constant times h squared. The total is x over h times h squared, which is a constant times x times h. Double the number of steps and each miss drops by a factor of four, so the total halves.

The table tests this on a different equation, y prime equals y, where the misses also get carried forward and grow. The errors at x equal to 1 still shrink in proportion to h, and in the figure the dots for small h fall onto a straight line through the origin with slope about e over 2; for large h the carried-forward errors bend them below it.

Methods whose error is proportional to h are called first order. That is honest but slow: every extra correct decimal place costs ten times as many steps. It is the same trade-off you saw with the numerical integration rules of Section 3.6, where Simpson's rule improved on the midpoint rule by a higher power of the step.

55. Guess the error at h = 0.001

Estimation

\[ e - (1.001)^{1000} = 2.718282 - 2.716924 = 0.001358 \]

Commit to a guess before revealing. The previous slide is all you need: the error is roughly proportional to the step size.

Going from a step of 0.01 to a step of 0.001 divides h by ten, so the error should drop by a factor of about ten, from 0.0135 to about 0.0014. The actual computation, a thousand multiplications by 1.001, gives 0.001358.

Now look at the cost. Ten times as many steps bought one more correct decimal place. To get six correct decimals you would need about a million steps. That is why real software uses higher-order methods, such as the Runge-Kutta methods, whose error shrinks like the fourth power of h, and why the accurate values in this lesson were computed with one of them.

56. Exercise 94: five steps of y' = -3y

Worked example

Estimate the solution at t equal to 1 with 5 steps, and compare with the exact solution.

\[ y' = -3y, \quad y(0) = 1, \quad h = 0.2 \]

\[ y_{n+1} = y_n + 0.2(-3y_n) = 0.4\,y_n \]

\[ y_n = 0.4^n, \qquad y_5 = 0.4^5 = 0.01024 \]

\[ y = e^{-3t}, \qquad y(1) = e^{-3} \approx 0.049787 \]

\[ y' = -3e^{-3t} = -3y \;\checkmark, \qquad \frac{0.049787 - 0.01024}{0.049787} \approx 0.794 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 379 — Exercise 94

When the right-hand side is a constant times y, every Euler step multiplies the height by the same factor, here 1 minus 0.6, which is 0.4. So five steps give 0.4 to the fifth power, about one hundredth.

The exact solution is the exponential e to the minus 3t, and at t equal to 1 it is about 0.0498. The check confirms the exponential satisfies the equation, then measures the error: Euler's value is only about a fifth of the true one, a relative error of 79 percent.

The figure shows why. Over each step the true solution shrinks by a factor of e to the minus 0.6, about 0.549, but Euler shrinks it by 0.4. The tangent line of a decaying exponential is steeper than the curve's average slope over the step, so Euler decays too fast. Five steps is simply too few for a solution that changes this quickly.

57. Exercise 96: Euler against an exact solution

Worked example

Five steps over the interval from 0 to 1, compared with the exact solution the book supplies.

\[ y' = 3t - y, \quad y(0) = 1, \quad h = 0.2 \]

\[ y_1 = 1 + 0.2(0 - 1) = 0.8, \quad y_2 = 0.8 + 0.2(0.6 - 0.8) = 0.76 \]

\[ y_3 = 0.76 + 0.2(0.44) = 0.848, \quad y_4 = 0.848 + 0.2(0.952) = 1.0384 \]

\[ y_5 = 1.0384 + 0.2(1.3616) = 1.31072 \]

\[ y' = 3 - 4e^{-t} = 3t - y \;\checkmark, \qquad 1.4715 - 1.3107 = 0.1608 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 379 — Exercise 96

Five steps, each with the slope 3t minus y at the start of the step. Keep the t-values straight: they are 0, 0.2, 0.4, 0.6 and 0.8 for the five slopes.

The table puts the Euler values beside the exact ones. The exact solution dips to a minimum near t equal to 0.29 and then rises; Euler dips too, but lower, and turns later. From then on it runs parallel but below, ending about 0.16 short at t equal to 1.

The check does two things. It confirms the book's exact solution by differentiating it and substituting, which is always worth doing before using a formula as a benchmark. Then it measures the error at the end of the interval. Compare with the previous exercise: here the error is about 11 percent, not 79, because this solution changes more gently over each step.

58. Exercise 116: a step that is far too big

Worked example

Approximate the solution at t equal to 10 using a step of 5.

\[ y' = -2y, \quad y(0) = 2, \quad h = 5 \]

\[ y_{n+1} = y_n + 5(-2y_n) = -9\,y_n \]

\[ y_1 = -18, \qquad y_2 = 162 \]

\[ y = 2e^{-2t}, \qquad y(10) = 2e^{-20} \approx 4.1 \times 10^{-9} \]

\[ |1 - 2h| < 1 \iff 0 < h < 1, \qquad h = 5: \; 1 - 10 = -9 \]

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 380 — Exercises 114-118

This exercise shows the worst thing a step size can do. With h equal to 5, each step multiplies the height by 1 minus 10, which is minus 9. The true solution decays smoothly toward zero; Euler jumps to minus 18 and then to 162.

The figure makes the absurdity plain. The true solution is indistinguishable from zero by t equal to 2, while the Euler polygon swings wildly and grows. This is not a small inaccuracy but a qualitatively wrong answer: a decaying solution computed as an exploding one.

The check finds the rule. Each step multiplies by 1 minus 2h, and for the computed values to decay that factor must lie strictly between minus 1 and 1, which needs h between 0 and 1. The general lesson: for a rapidly decaying solution, a step that is too big does not just lose accuracy, it makes the method unstable. Exercises 117 and 118 push the same equation further.

59. Putting it together

Section

Part 6

60. Pattern: studying y' = f(x, y) without solving it

Pattern

1. Draw the field: substitute points, or draw isoclines f(x, y) = m, starting with m = 0.
2. Find the equilibria: solve f(x, k) = 0 for every x. They are horizontal lines no solution crosses.
3. Classify each: test the sign of f just below and just above (for x large, if f involves x).
4. Sketch: start at the initial point, stay tangent to the segments, never cross another solution.
5. For numbers, run Euler: yₙ = yₙ₋₁ + h·f(xₙ₋₁, yₙ₋₁); halve h to roughly halve the error; keep h small enough not to overshoot.

This is the order to work in when you meet a first-order equation you cannot solve, and it runs from cheap to expensive. Drawing the field and finding the equilibria cost only algebra. Classifying the equilibria costs a few sign tests. Together they tell you the long-term fate of every solution.

The diagram is Euler's method as a loop: at a point, ask the equation for the slope, take a step of rise h times the slope and run h, and repeat from the new point. The equation enters at exactly one place, the slope box, which is why the same program works for any equation.

The last item is the numerical rule of thumb from Part 5. Halving h roughly halves the error, so you can judge an answer's accuracy by running it twice and comparing, and a step that is too large can make the method blow up.

61. A cooling cup of coffee, step by step

Real world

OpenStax Calculus Volume 2, §4.2 Direction Fields and Numerical Methods §4.2, p. 365 — Newton's law of cooling, Figure 4.6

Try the three steps yourself before revealing. The equation is Newton's law of cooling from the start of the lesson, with the coffee starting at 180 degrees.

Each step subtracts 0.4 times the current distance from 72, which means the distance to room temperature shrinks by a factor of 0.6 per step: 108, then 64.8, then 38.88, then 23.328. Adding back the 72 gives 136.8, 110.88 and 95.328.

The exact solution at 3 minutes is about 104.5 degrees, so Euler is about 9 degrees too cold. The figure shows why the error is always in that direction: the coffee cools fastest at the start of each minute, Euler uses that fastest rate for the whole minute, and so it overshoots the cooling every time. If you needed the answer to a degree, the halving rule says to shrink the step several times.

62. Check: reading a slope off the equation

Check

This is the basic move of every direction field: substitute the point into the right-hand side. The only care needed is with signs.

With x equal to 2 and y equal to minus 1, x squared is 4, and minus 3 times minus 1 is plus 3, so the slope is 7. The wrong answers come from dropping the sign on y, swapping the coordinates, or reading a coordinate as the slope.

63. Check: classifying equilibria

Check

Use the sign test: the equilibria are minus 1, 0 and 2, and the right-hand side alternates sign between them, starting negative on the far left.

That makes 0 the only equilibrium with arrows pointing in from both sides. At minus 1 and at 2 the arrows point away on both sides, so they are unstable. With three simple zeros the pattern must alternate, so stable and unstable equilibria take turns.

64. Check: running Euler's method

Check

Two steps are needed to get from 0 to 1 with a step of 0.5. The first uses the slope at (0, 1), which is 1; the second uses the slope at (0.5, 1.5), which is 2.

The wrong answers show three different slips: mixing the new x with the old y, stopping after one step, and reporting the exact solution instead of Euler's estimate. The exact value, 2e minus 2, is higher than Euler's, because the solution curves upward away from each tangent line.

65. Explain why Euler's errors pile up

Explain it to yourself

Write your explanation before revealing. The strongest answers mention both halves of the story: how big each step's miss is, and why the misses add up instead of cancelling.

Each step misses by h squared because the tangent line cannot follow the parabola's bend. Each step starts from the previous step's answer, so the misses are carried forward and accumulate. The number of steps needed to reach x is x over h, so the total miss is x over h times h squared, which is x times h.

That one line explains both observations: the error grows as x grows because more steps have been taken, and halving h halves the error because you take twice as many steps, each missing by a quarter as much.

66. Exit ticket

Exit ticket

\[ y' = y(3 - y), \qquad y(0) = 1 \]

This combines the two halves of the lesson. The equilibria come from factoring: y times (3 minus y) is zero at 0 and at 3. The sign test shows the right-hand side is negative below 0, positive between 0 and 3, and negative above 3, so 0 is unstable and 3 is stable.

The Euler steps are quick: slope 2 at height 1, rise 1, then slope 2 at height 2, rise 1. After two steps the estimate is exactly 3, and since the slope at 3 is zero, Euler would stay there forever.

That should surprise you. The true solution approaches 3 but never reaches it, since a solution cannot meet the equilibrium line; at x equal to 1 it is about 2.728. A large step overshot the curve's gentle approach and landed on the fence. Smaller steps would approach 3 gradually, the way the true solution does.

67. Recap

Recap

\[ y_n = y_{n-1} + h\,f(x_{n-1}, y_{n-1}), \qquad \text{error} \approx Kh \]

Next, Section 4.3 solves a whole class of equations exactly, the separable ones, and you will be able to compare those formulas with the fields and Euler tables of this lesson.

Stewart, Calculus: Early Transcendentals 8e, §9.2 Direction Fields and Euler's Method §9.2, pp. 591-598 — the same material in Stewart

The direction field is the equation drawn: one segment of the prescribed slope at each point. Isoclines, especially the zero isocline, let you draw it quickly and show where solutions can turn. Solution curves follow the segments and never cross.

Equilibria are the horizontal lines along which the right-hand side vanishes for every x. For an autonomous equation, two sign tests beside each equilibrium classify it as stable, unstable or semi-stable, and the phase line records the result. When the right-hand side involves x, judge stability by what happens as x goes to infinity, not by a short window.

Euler's method follows the field numerically, one tangent-line step at a time. Its error is roughly proportional to the step size, and a step that is too large can make it blow up. In Section 4.3 you will solve separable equations exactly and be able to test both the pictures and the tables of this lesson against formulas.