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Lesson 98 — Euler's Method (Numerical)

Explicit Euler's method for ODEs: discretization, local error O(h²), global error O(h), implementation, and comparison with Runge-Kutta.

Used in: Cálculo Numérico (UFRGS, USP, UNICAMP) · Spécialité Maths Terminale (França) · Mathematics 4 (IIT-JEE Advanced, Índia)

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

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Rigorous notation, full derivation, hypotheses

Derivation and error analysis

Initial Value Problem

Given the IVP:

$y' = f(x, y), \quad y(x_0) = y_0$

We wish to approximate $y(x)$ on $x \in [x_0, X]$ without a closed-form expression.

Discretization

Divide the interval into $N$ equal subintervals:

$h = \frac{X - x_0}{N}, \qquad x_n = x_0 + n\,h, \quad n = 0, 1, \ldots, N$

"The simplest numerical method for solving $y' = f(x,y)$ , $y(x_0) = y_0$ , is Euler's method. We replace $y'$ with the difference quotient $(y_{n+1} - y_n)/h$ and evaluate $f$ at $x_n$ : this gives $y_{n+1} = y_n + hf(x_n, y_n)$ ." — Lebl, Notes on Diffy Qs §1.7

Error analysis via Taylor series

Comparison of methods

Comparison of single-step methods for ODEs. RK4 is the industry standard for precision; Implicit Euler is for stiff equations.

Solved Examples

Example— 98.1· Euler in 4 steps — exponential ODE

Problem. Use Euler's method with $h = 0.1$ to approximate $y(0.4)$ given $y' = -y$ , $y(0) = 1$ . Compare with the exact value.

Strategy. Apply the recurrence $y_{n+1} = y_n + h(-y_n) = y_n(1-h)$ four times.

Resolution.

$n$	$x_n$	$y_n$	$f = -y_n$	$y_{n+1}$
0	0	1.0000	-1.0000	0.9000
1	0.1	0.9000	-0.9000	0.8100
2	0.2	0.8100	-0.8100	0.7290
3	0.3	0.7290	-0.7290	0.6561

$y_4 = 0.6561$ . Exact: $e^{-0.4} \approx 0.6703$ . Relative error: $\approx 2.12\%$ .

Verification. At each step, we multiply by $(1-h) = 0.9$ , therefore $y_4 = (0.9)^4 = 0.6561$ ✓. The error grows with $h$ — for $h = 0.05$ , the error would drop to $\approx 1.06\%$ .

Source. Lebl, Notes on Diffy Qs §1.7, Example 1.7.1 — CC-BY-SA

Example— 98.2· Euler in non-linear equation

Problem. Estimate $y(1)$ for $y' = y^2$ , $y(0) = 0.5$ , using Euler with $h = 0.25$ .

Strategy. The exact solution is $y = 1/(2-x)$ (blow-up at $x = 2$ ). With 4 steps up to $x = 1$ .

Resolution.

$n=0$ : $f(0, 0.5) = 0.25$ . $y_1 = 0.5 + 0.25 \times 0.25 = 0.5625$ .

$n=1$ : $f(0.25, 0.5625) = 0.31641$ . $y_2 = 0.5625 + 0.25 \times 0.31641 = 0.6416$ .

$n=2$ : $f(0.5, 0.6416) = 0.4117$ . $y_3 = 0.6416 + 0.25 \times 0.4117 = 0.7445$ .

$n=3$ : $f(0.75, 0.7445) = 0.5543$ . $y_4 = 0.7445 + 0.25 \times 0.5543 = 0.8831$ .

Exact: $y(1) = 1/(2-1) = 1$ . Error: $11.7\%$ .

Verification. The error is large because $f = y^2$ grows and the tangents underestimate the curve. Reducing to $h = 0.1$ (10 steps) would reduce it to $\approx 5.8\%$ , confirming order 1.

Source. OpenStax Calculus Volume 2, §4.2, Example 4.2 — CC-BY-NC-SA

Example— 98.3· Comparison Euler versus RK4

Problem. For $y' = x + y$ , $y(0) = 1$ , approximate $y(0.2)$ with $h = 0.1$ using (a) Euler and (b) RK4. Exact solution: $y = 2e^x - x - 1$ .

Strategy. Apply the formulas for each method and compare errors.

Resolution.

Exact: $y(0.2) = 2e^{0.2} - 0.2 - 1 \approx 2 \times 1.2214 - 1.2 = 1.2428$ .

(a) Euler:

$n=0$ : $f(0,1) = 1$ . $y_1 = 1 + 0.1 \times 1 = 1.1$ .

$n=1$ : $f(0.1, 1.1) = 0.1 + 1.1 = 1.2$ . $y_2 = 1.1 + 0.1 \times 1.2 = 1.22$ .

Euler error: $|1.2428 - 1.22| = 0.0228$ .

(b) RK4 (step $n=0 \to n=1$ , $h=0.1$ ):

$k_1 = 0 + 1 = 1$ ; $k_2 = (0.05) + (1 + 0.05) = 1.1$ ; $k_3 = 0.05 + (1 + 0.05) = 1.1$ ; $k_4 = 0.1 + (1 + 0.1) = 1.2$ .

$y_1 = 1 + (0.1/6)(1 + 2.2 + 2.2 + 1.2) = 1 + 0.1 \times 1.1\overline{0}\approx 1.1103$ .

RK4 error (two steps): $< 10^{-6}$ — orders of magnitude better than Euler.

Verification. Ratio of errors for $h$ halved: Euler 2:1 (1st order); RK4 16:1 (4th order) ✓.

Source. REAMAT UFRGS, Cálculo Numérico §8.3 — CC-BY-SA

Example— 98.4· Implicit Euler — stability in stiff equation

Problem. $y' = -100y$ , $y(0) = 1$ . Compare explicit Euler with $h = 0.03$ (unstable) and $h = 0.01$ (stable).

Strategy. Analyze $|1 + h\lambda|$ with $\lambda = -100$ .

Resolution.

Amplification factor explicit Euler: $r = 1 + h\lambda = 1 - 100h$ .

$h = 0.03$ : $r = 1 - 3 = -2$ . $|r| = 2 > 1$ — unstable: $y_n = (-2)^n \to \pm\infty$ .
$h = 0.01$ : $r = 1 - 1 = 0$ . $|r| = 0 < 1$ — stable (at the boundary), $y_n = 0$ in 1 step (over-damped).
$h = 0.02$ : $r = 1 - 2 = -1$ . $|r| = 1$ — oscillates without damping.

Euler stability region: $h < 2/|\lambda| = 0.02$ for this equation. The exact solution $e^{-100x}$ decays at $x \sim 0.01$ — steps much smaller than the scale of interest.

Verification. Implicit Euler: $r_{imp} = 1/(1-h\lambda) = 1/(1+100h)$ . For any $h > 0$ : $|r_{imp}| < 1$ ✓ — always stable.

Source. REAMAT UFRGS, Cálculo Numérico §8.4 — CC-BY-SA

Example— 98.5· Euler for system of ODEs — oscillator

Problem. System $x' = v$ , $v' = -x$ (harmonic oscillator), IC: $x(0) = 1$ , $v(0) = 0$ , $h = 0.1$ , simulate 3 steps.

Strategy. Apply Euler to each component simultaneously.

Resolution.

State: $x_n$ , $v_n$ . Functions $f_1 = v$ and $f_2 = -x$ .

Step 0: $x_0 = 1$ , $v_0 = 0$ . Slopes: $\dot x = 0$ , $\dot v = -1$ .

$x_1 = 1 + 0.1 \times 0 = 1$ ; $v_1 = 0 + 0.1 \times (-1) = -0.1$ .

Step 1: $x_1 = 1$ , $v_1 = -0.1$ . Slopes: $\dot x = -0.1$ , $\dot v = -1$ .

$x_2 = 1 + 0.1 \times (-0.1) = 0.99$ ; $v_2 = -0.1 + 0.1 \times (-1) = -0.2$ .

Step 2: $x_2 = 0.99$ , $v_2 = -0.2$ . Slopes: $\dot x = -0.2$ , $\dot v = -0.99$ .

$x_3 = 0.99 + 0.1 \times (-0.2) = 0.97$ ; $v_3 = -0.2 + 0.1 \times (-0.99) = -0.299$ .

Exact at $t = 0.3$ : $x = \cos(0.3) \approx 0.9553$ , $v = -\sin(0.3) \approx -0.2955$ . Euler overestimates $x$ — does not conserve energy. Symplectic Euler solves this: calculates $v_{n+1}$ first, then uses it for $x_{n+1}$ .

Source. Lebl, Notes on Diffy Qs §1.7, Exercise 1.7.4 — CC-BY-SA

Exercise list

28 exercises · 7 with worked solution (25%)

Application 17Understanding 3Modeling 5Challenge 1Proof 2

Ex. 98.1Application
Use Euler with $h = 0.5$ to approximate $y(1)$ given $y' = y^2$ , $y(0) = 0$ .
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Ex. 98.2Application
Use Euler with $h = 0.1$ to approximate $y(0.2)$ , given $y' = x + y$ , $y(0) = 1$ . Compare with the exact value $y = 2e^x - x - 1$ .
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Ex. 98.3ApplicationAnswer key
Use Euler with $h = 0.25$ to approximate $y(1)$ , given $y' = -y$ , $y(0) = 1$ . Exact: $e^{-1}$ .
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Ex. 98.4Application
Repeat exercise 98.3 with $h = 0.1$ . Compare the errors and verify the 1st order of the method.
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Ex. 98.5Application
Use Euler with $h = 0.5$ for $y' = 2x$ , $y(0) = 0$ , and estimate $y(2)$ . Compare with the exact value.
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Ex. 98.6Application
Use Euler with $h = 0.2$ for $y' = y - x^2 + 1$ , $y(0) = 0.5$ , and estimate $y(0.4)$ .
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Ex. 98.7ApplicationAnswer key
For $y' = y$ , $y(0) = 1$ , estimate the local error of Euler's method with $h = 0.1$ on $[0, 1]$ .
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Ex. 98.8Application
Determine the maximum step size $h_{\max}$ for stability of explicit Euler in $y' = -2y$ .
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Ex. 98.9Application
Apply implicit Euler with $h = 0.5$ for $y' = -y$ , $y(0) = 1$ , and estimate $y(1)$ .
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Ex. 98.10ApplicationAnswer key
Apply Heun's method (RK2) with $h = 0.5$ for $y' = -y$ , $y(0) = 1$ , and estimate $y(0.5)$ .
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Ex. 98.11Application
For $y' = x + y$ , $y(0) = 1$ : calculate the errors in $y(0.2)$ with Euler for $h = 0.1$ and $h = 0.05$ . Verify the 1st order.
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Ex. 98.12Application
How many Euler steps are needed for $y' = y$ , $y(0) = 1$ , with global error less than $10^{-4}$ on $[0, 1]$ ?
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Ex. 98.13ApplicationAnswer key
Simulate the oscillator $x'' + x = 0$ , $x(0) = 1$ , $x'(0) = 0$ with Euler and $h = 0.1$ . Calculate $(x_1, v_1)$ , $(x_2, v_2)$ , $(x_3, v_3)$ .
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Ex. 98.14Application
Verify that Euler's method does not conserve the energy of the oscillator $x'' + x = 0$ . Compare with symplectic Euler.
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Ex. 98.15Modeling
$P' = 0.3P(1 - P/1000)$ , $P(0) = 100$ . Use Euler with $h = 1$ to estimate $P(12)$ (12 months). Sketch the graph of the calculated points.
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Ex. 98.16ModelingAnswer key
RLC circuit: $L = 1$ H, $R = 0.5$ Ω, $C = 1$ F, $Q(0) = 1$ , $I(0) = 0$ . Use Euler with $h = 0.1$ to simulate $Q(t)$ for 3 steps.
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Ex. 98.17Modeling
$T' = -0.1(T - 20)$ , $T(0) = 90$ °C. Use Euler with $h = 5$ min to estimate $T(10)$ .
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Ex. 98.18Modeling
Carbon-14 has a half-life of 5730 years. Use Euler with $h = 500$ years to estimate the fraction remaining after 5000 years.
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Ex. 98.19Understanding
Why does Euler's method have a global error of $O(h)$ if each step has a local error of $O(h^2)$ ?
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Ex. 98.20Understanding
In which situation does explicit Euler's method become impractical due to numerical instability?
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Ex. 98.21Understanding
What is the main advantage of RK4 over Euler's method?
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Ex. 98.22ApplicationAnswer key
Use Euler with $h = \pi/4$ to approximate $y(\pi/2)$ given $y' = \cos x$ , $y(0) = 0$ . Compare with $\sin(\pi/2) = 1$ .
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Ex. 98.23Application
Use Euler with $h = 0.5$ for $y' = \sqrt{y}$ , $y(0) = 1$ . Estimate $y(1)$ and compare with the exact value $(1.5)^2 = 2.25$ .
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Ex. 98.24Application
For $y' = \sqrt{y}$ , $y(0) = 1$ , compare Euler and Heun (RK2) with $h = 0.5$ to estimate $y(0.5)$ . Exact: $y(0.5) = (1.25)^2 = 1.5625$ .
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Ex. 98.25Modeling
Describe how to experimentally verify the order of a numerical method by comparing errors for $h$ and $h/2$ .
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Ex. 98.26Proof
Derive the local error of Euler's method using the Taylor series of $y(x_{n+1})$ around $x_n$ .
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Ex. 98.27Proof
Derive the stability region of explicit Euler's method in the $h\lambda$ plane and show it is the disk $|1 + h\lambda| < 1$ .
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Ex. 98.28ChallengeAnswer key
Apply RK4 with $h = 0.1$ to $y' = y$ , $y(0) = 1$ . Compare the error with Euler's and confirm that RK4 is 4th order.
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Sources

Lebl, Jiří. Notes on Diffy Qs: Differential Equations for Engineers. Version 6.4. CC-BY-SA. jirka.org/diffyqs — §1.7 covers Euler's method with Taylor error analysis.
UFRGS Reamat. Cálculo Numérico (Python version). CC-BY-SA. ufrgs.br/reamat/CalculoNumerico — Ch. 8: Euler, Heun, RK4, stability, and error analysis with Python code.
OpenStax. Calculus Volume 2. CC-BY-NC-SA. openstax.org/details/books/calculus-volume-2 — §4.2: direction fields and Euler's method with graphical interpretation.