Euler's Method — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

\frac{dy}{dx} = x + y

, cannot be solved by the methods of this unit. You can still find out roughly where a solution goes: stand at a known point, walk a short way along the slope the equation gives, read the new slope, and repeat. That walk is Euler's method.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $f(2) = 5$ and $f'(2) = 3$ . The tangent line at $x = 2$ estimates $f(2.1)$ as

5.3
5.03
8
3.5

Show the answer

$5 + 3(0.1) = 5.3$ .

Correct: 5.3:
5.03:
8:
3.5:

2. For $\frac{dy}{dx} = xy$ , the slope field segment at $(1, 2)$ has slope

2
3
$\frac{1}{2}$
1

Show the answer

$\frac{dy}{dx} = 1 \cdot 2 = 2$ .

Correct: 2:
3:
$\frac{1}{2}$ :
1:

Part 4 · See it

See it first

The slope field of dy/dx = y for x from −0.25 to 1.25, with the solution through (0, 1), y = e to the x, drawn in blue. Two violet Euler steps of size 0.5 start at (0, 1): slope 1 to (0.5, 1.5), then slope 1.5 to (1, 2.25). The true value at x = 1 is e, about 2.718, above the last violet point. The solution is concave up, so the steps fall below it and 2.25 is an underestimate. — Two Euler steps of size 0.5 on dy/dx = y from (0, 1): slope 1 to (0.5, 1.5), then slope 1.5 to (1, 2.25). The solution through (0, 1) is concave up, so both steps fall below it, and 2.25 underestimates its value e ≈ 2.718 at x = 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The differential equation gives the slope at every pointat a known point you know the direction of the solution curve
Follow that slope for a short run $\Delta x$ the rise is $\Delta y = \frac{dy}{dx}\,\Delta x$ , and the new point is $(x + \Delta x, y + \Delta y)$
At the new point, read a new slope from the equationtake the next step; repeat until you reach the $x$ you want
The solution curves are concave up (or down) along the wayeach step falls below (or above) them, so the estimate is too small (or too large)

Part 6 · Key ideas

Key ideas

One step: $y_{\text{new}}$ $= y + \frac{dy}{dx}\Big|_{(x, y)} \cdot \Delta x$ , using the slope at the point you are at.
Lay the work out in a table with columns $x$ , $y$ , $\frac{dy}{dx}$ , $\Delta x$ , $\Delta y$ ; take exactly the steps asked for.
Smaller steps give better estimates. A negative $\Delta x$ steps to the left.
Over or under: find $\frac{d^2y}{dx^2}$ by differentiating the equation. Concave up curves lie above their tangent steps (underestimate); concave down, below (overestimate).

Part 7 · Misconception

A common mistake

The wrong idea: For $\frac{dy}{dx} = x + y$ from $(0, 1)$ with $\Delta x = 0.5$ : the slope is 1, so $y(0.5) \approx 1.5$ and $y(1) \approx 2$ .

What actually happens: Each step needs a new slope at its own point. At $(0.5, 1.5)$ the slope is $0.5 + 1.5 = 2$ , so $y(1)$ $\approx 1.5 + 2(0.5)$ $= 2.5$ . Reusing the first slope is just one long tangent line.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Two Euler steps on the slope field of $\frac{dy}{dx} = x + y$

The slope field of

\frac{dy}{dx} = x + y

is shown with two Euler steps of size

\Delta x = 0.5

, starting at the point

(0, 1)

on a solution curve

y = f(x)

.

1. What slope does the first Euler step use?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $(0, 1)$ : $\frac{dy}{dx} = 0 + 1 = 1$ .

Answer: 1

2. What is the Euler approximation of $f(1)$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Step 1: $1 + 1(0.5) = 1.5$ at $x = 0.5$ . Step 2: slope $0.5 + 1.5 = 2$ , so $1.5 + 2(0.5) = 2.5$ at $x = 1$ .

Answer: 2.5

3. For $x \ge 0$ and $y > 0$ , $\frac{d^2y}{dx^2} = 1 + x + y$ . Is the estimate of $f(1)$ too large or too small?

Too large: the slopes increase from one step to the next
Too small: $\frac{dy}{dx} = x + y > 0$ , so the solution is increasing
No conclusion is possible without solving the equation
Too small: $\frac{d^2y}{dx^2} > 0$ , so the solution curves are concave up and the steps fall below them

Show the answer

$\frac{d^2y}{dx^2}$ $= \frac{d}{dx}(x + y)$ $= 1$ $+ \frac{dy}{dx}$ $= 1$ $+ x$ $+ y$ , positive in this region, where every Euler point lies. So every solution curve there is concave up and lies above its tangent segments: the first step ends below $f(0.5)$ , and the second step ends below the solution curve through $(0.5, 1.5)$ , which stays below $f$ (solution curves do not cross). So 2.5 underestimates $f(1)$ .

Too large: the slopes increase from one step to the next: Increasing slopes mean concave up, and a tangent segment lies below a concave up curve, so the estimate is too small.
Too small: $\frac{dy}{dx} = x + y > 0$ , so the solution is increasing: The conclusion is right but the reason is not: increasing says nothing about the gap. Concavity, from $\frac{d^2y}{dx^2}$ , decides it.
No conclusion is possible without solving the equation: $\frac{d^2y}{dx^2}$ comes from the differential equation itself, so its sign can be read without solving.
Correct: Too small: $\frac{d^2y}{dx^2} > 0$ , so the solution curves are concave up and the steps fall below them: Right: each tangent step lies below the concave up solution curve it starts on, and a curve that starts lower stays below the graph of $f$ , since solution curves do not cross.

4. The function $f$ satisfies $\frac{dy}{dx} = x^2 - y$ and $f(1) = 2$ . Use one Euler step of size $0.1$ to approximate $f(1.1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f(1.1)$ $\approx 2 + (1 - 2)(0.1)$ $= 1.9$ .

Answer: 1.9

5. For $\frac{dy}{dx} = F(x, y)$ and step size $\Delta x$ , which formula is one Euler step?

$y_{\text{new}} = y \cdot F(x, y)\,\Delta x$
$y_{\text{new}}$ $= y + F(x + \Delta x, y)\,\Delta x$
$y_{\text{new}} = y + F(x, y)\,\Delta x$
$y_{\text{new}} = y + \Delta x$

Show the answer

Follow the tangent line: new $y$ = old $y$ + slope × $\Delta x$ , with the slope from the differential equation at the current point.

$y_{\text{new}} = y \cdot F(x, y)\,\Delta x$ : The change is added to $y$ , not multiplied by it.
$y_{\text{new}}$ $= y + F(x + \Delta x, y)\,\Delta x$ : Euler's method uses the slope at the point you are at, $(x, y)$ , not ahead of it.
Correct: $y_{\text{new}} = y + F(x, y)\,\Delta x$ : Right: the slope at the current point times the run gives the rise.
$y_{\text{new}} = y + \Delta x$ : That adds the run, not the rise. Multiply the run by the slope first.

6. A solution $y = f(x)$ has $f(2) = 4$ and $f'(2) = 0.6$ , and its graph is concave down for $2 \le x \le 2.5$ . One Euler step gives $f(2.5) \approx 4.3$ . Which is true?

$4.3$ is an overestimate: the tangent step lies above the graph
$4.3$ is an underestimate, because $f'(2) > 0$ means $f$ is increasing
$4.3$ is exact, because one step of Euler's method is exactly the tangent line
$4.3$ is an underestimate, because the graph is concave down

Show the answer

One Euler step from $(2, 4)$ is the tangent line: $4 + 0.6(0.5) = 4.3$ . A concave down graph lies below its tangent lines, so $f(2.5) < 4.3$ .

Correct: $4.3$ is an overestimate: the tangent step lies above the graph: Right: a concave down graph lies below its tangent lines, so the estimate is too large.
$4.3$ is an underestimate, because $f'(2) > 0$ means $f$ is increasing: Increasing is not the reason. Concave down puts the graph below its tangent line, so the estimate is too large.
$4.3$ is exact, because one step of Euler's method is exactly the tangent line: It is the tangent line value, which differs from $f(2.5)$ whenever the graph bends.
$4.3$ is an underestimate, because the graph is concave down: Concave down graphs lie below their tangent lines, which makes a tangent estimate too large.

Part 9 · Summary

Summary

Euler's method approximates a solution of

\frac{dy}{dx} = F(x, y)

by short tangent steps: from a known point,

y_{\text{new}} = y + F(x, y)\,\Delta x

, then repeat from the new point. Smaller steps follow the curve more closely. The sign of

\frac{d^2y}{dx^2}

tells whether the steps fall below the solution (concave up, an underestimate) or above it (concave down, an overestimate).

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Two Euler steps on the slope field of dydx=x+y\frac{dy}{dx} = x + y

Summary

What comes next

Connections

Two Euler steps on the slope field of $\frac{dy}{dx} = x + y$