Euler's Method

Quick answer

Euler's method approximates the solution of dy/dx = f(x, y) through a point without solving the equation. From the current point, follow the tangent line for a step h: the new point is x + h and y + h·f(x, y). Repeating the step traces a broken line that follows the slope field. Smaller steps give better approximations, with the error shrinking roughly in proportion to h. When the solution curve is concave up the method underestimates, and when it is concave down it overestimates.

What you'll learn

Carry out Euler's method for a given step size
Explain Euler's method as repeated linear approximation
Decide from concavity whether an estimate is too high or too low
Describe how the error changes with the step size

Following the slope field

A slope field shows the direction a solution travels at every point, but most differential equations have no formula for their solutions. Euler’s method walks along the field instead. From the starting point, move a short distance in the direction the slope gives, look up the slope at the new point, and move again.

Definition

To approximate the solution of $\tfrac{dy}{dx} = f(x, y)$ through $(x_0, y_0)$ with step size $h$ , repeat

x_{n+1} = x_n + h \qquad\qquad y_{n+1} = y_n + h \cdot f(x_n, y_n)

exact solution y = eˣ

Euler's method for dy/dx = y with step 0.5

Why each step is a tangent line

A single step is the linear approximation from the lesson on derivatives: $y(x + h) \approx y(x) + h\,y'(x)$ . The differential equation supplies the derivative, $y' = f(x, y)$ , so no formula for $y$ is needed. Euler’s method is linear approximation, repeated, with the slope field supplying each slope.

Each step errs a little, because the solution curves away from its tangent line, and the errors add up. Halving $h$ makes each step’s error about four times smaller but doubles the number of steps, so the total error at a fixed point is roughly halved.

Example 1 — Four steps

Use Euler’s method with $h = 0.5$ to estimate $y(2)$ for $\tfrac{dy}{dx} = y$ , $y(0) = 1$ .

$x_n$	$y_n$	slope $f(x_n, y_n) = y_n$	$y_{n+1} = y_n + 0.5 \cdot \text{slope}$
$0$	$1$	$1$	$1.5$
$0.5$	$1.5$	$1.5$	$2.25$
$1$	$2.25$	$2.25$	$3.375$
$1.5$	$3.375$	$3.375$	$5.0625$

So $y(2) \approx 5.06$ . The exact solution is $y = e^x$ , and $e^2 \approx 7.39$ : four large steps fall well short.

Over or under?

The concavity of the solution decides the direction of the error. Where the solution is concave up, its tangent lines lie below it, so each step lands low and the estimate is too small, as in the graph. Where it is concave down, the estimate is too large. Find the concavity by differentiating the equation itself: for $\tfrac{dy}{dx} = y$ , $\tfrac{d^2y}{dx^2} = \tfrac{dy}{dx} = y > 0$ .

Worked examples

Common mistakes

Practice problems

Use two steps of size $0.1$ to estimate $y(0.2)$ for $\tfrac{dy}{dx} = y$ , $y(0) = 2$ .

Answer

$2.42$

Full solution

$y(0.1) \approx 2 + 0.1(2) = 2.2$ , then $y(0.2) \approx 2.2 + 0.1(2.2) = 2.42$ .
Use two steps of size $0.5$ to estimate $y(1)$ for $\tfrac{dy}{dx} = x$ , $y(0) = 0$ . Compare with the exact solution.

Answer

$0.25$ ; the exact value is $0.5$

Full solution

The slope at $(0, 0)$ is $0$ , so $y(0.5) \approx 0$ . The slope at $(0.5, 0)$ is $0.5$ , so $y(1) \approx 0.25$ . The exact solution is $y = \tfrac{x^2}{2}$ , with $y(1) = 0.5$ .
Use two steps of size $0.5$ to estimate $y(1)$ for $\tfrac{dy}{dx} = x - y$ , $y(0) = 1$ .

Answer

$0.5$

Full solution

The slope at $(0, 1)$ is $-1$ : $y(0.5) \approx 0.5$ . The slope at $(0.5, 0.5)$ is $0$ : $y(1) \approx 0.5$ .
Use two steps of size $0.5$ to estimate $y(2)$ for $\tfrac{dy}{dx} = \tfrac{1}{x}$ , $y(1) = 0$ . Is the estimate too high or too low?

Answer

About $0.833$ , too high

Full solution

$y(1.5) \approx 0 + 0.5(1) = 0.5$ and $y(2) \approx 0.5 + 0.5 \cdot \tfrac{1}{1.5} \approx 0.833$ . The solution is $\ln x$ , with $\ln 2 \approx 0.693$ . It is concave down, since $y'' = -\tfrac{1}{x^2} < 0$ , so the estimate is too high.
Use two steps of size $0.5$ to estimate $y(1)$ for $\tfrac{dy}{dx} = 2xy$ , $y(0) = 1$ .

Answer

$1.5$

Full solution

The slope at $(0, 1)$ is $0$ : $y(0.5) \approx 1$ . The slope at $(0.5, 1)$ is $1$ : $y(1) \approx 1.5$ .
For $\tfrac{dy}{dx} = -y$ , $y(0) = 1$ , use two steps of size $0.5$ to estimate $y(1)$ , and decide from concavity whether the estimate is too high or too low.

Answer

$0.25$ , too low

Full solution

$y(0.5) \approx 1 - 0.5 = 0.5$ and $y(1) \approx 0.5 - 0.25 = 0.25$ . Differentiating the equation, $y'' = -y' = y > 0$ , so the solution is concave up and the estimate is low. Indeed $e^{-1} \approx 0.368$ .
Use one step of size $-0.5$ to estimate $y(0.5)$ for $\tfrac{dy}{dx} = x + y$ , $y(1) = 2$ .

Answer

$0.5$

Full solution

The slope at $(1, 2)$ is $3$ , so $y(0.5) \approx 2 + (-0.5)(3) = 0.5$ .
A function has $f(2) = 5$ and $f'(x) = x^2 - 1$ . Use two steps of size $0.5$ to estimate $f(3)$ , and compare with the exact value.

Answer

$9.125$ ; the exact value is $\tfrac{31}{3} \approx 10.33$

Full solution

The slope at $x = 2$ is $3$ : $f(2.5) \approx 6.5$ . The slope at $x = 2.5$ is $5.25$ : $f(3) \approx 6.5 + 2.625 = 9.125$ . Exactly, $f(3) = 5 + \int_2^3 (x^2 - 1)\,dx = 5 + \tfrac{16}{3}$ .
For $\tfrac{dy}{dx} = y$ , $y(0) = 1$ , Euler’s method with step $h$ gives $y(1) \approx (1 + h)^{1/h}$ . What happens as $h \to 0$ ?

Answer

The estimates approach $e$ .

Full solution

Each of the $\tfrac{1}{h}$ steps multiplies $y$ by $1 + h$ . The limit of $(1 + h)^{1/h}$ as $h \to 0$ is $e$ , the exact value of $y(1)$ .
For $\tfrac{dy}{dx} = y$ , $y(0) = 1$ , a student estimates $y(0.1)$ with $h = 0.1$ as $1 + 1 = 2$ . What went wrong?

Hint

How far does $y$ change over a run of $0.1$ at slope $1$ ?

Answer

The student added the slope instead of $h$ times the slope. The estimate is $1.1$ .

Full solution

$y(0.1) \approx y(0) + h \cdot f(0, 1) = 1 + 0.1 \cdot 1 = 1.1$ , close to $e^{0.1} \approx 1.105$ .

Frequently asked questions

What is the formula for Euler's method?

From (xₙ, yₙ), take x₍ₙ₊₁₎ = xₙ + h and y₍ₙ₊₁₎ = yₙ + h·f(xₙ, yₙ), where dy/dx = f(x, y) and h is the step size.

Why does Euler's method work?

Each step is a linear approximation: it follows the tangent line, whose slope the differential equation supplies, for a short distance.

How do I know if Euler's method overestimates or underestimates?

Check the concavity of the solution. Concave up means the tangent lines lie below the curve, so the estimate is too low; concave down means it is too high.

How does the step size affect the error?

Smaller steps give smaller errors. Halving h roughly halves the error at a fixed point, at the cost of twice as many steps.

Can Euler's method step backward?

Yes. Use a negative step h to estimate values to the left of the starting point.

What to learn next

Logistic GrowthThe logistic equation dP/dt = kP(1 − P/L): growth that levels off at a carrying capacity, its S-shaped solution, and where growth is fastest.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Following the slope field

Why each step is a tangent line

Over or under?

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What is the formula for Euler's method?

Why does Euler's method work?

How do I know if Euler's method overestimates or underestimates?

How does the step size affect the error?

Can Euler's method step backward?

What to learn next

More Calculus lessons