ODE solvers — Euler's method (derivation, global error): Worked examples

Before any numbers, one reminder of what every symbol means, so nothing is used unexplained:

The scenario matrix

Every problem this topic can throw is one (or a blend) of these cells. The table lists the cell, what makes it distinct, and which example below covers it.

Cell	What makes it special	Covered by
A. Growing solution, positive slope	$f > 0$ , $y$ climbs, errors are one-sided (undershoot)	Example 1
B. Decaying solution, negative slope	$f < 0$ , $y$ falls toward a floor	Example 2
C. Step-halving / order check	verify global error $\propto h$ empirically	Example 3
D. Degenerate: slope $= 0$ at a point	$f = 0$ ⇒ a flat step, $y$ unchanged for that step	Example 4
E. Limiting / unstable: $h$ too large	$	1+h\lambda
F. Autonomous non-linear	$f$ depends on $y$ only, but non-linearly	Example 6
G. Real-world word problem	translate physics → IVP → Euler	Example 7
H. Exam twist: back-solve for $h$ or $y_{0}$	given the answer, find an input	Example 8

We now walk every cell. Throughout, the derivation and error law come from the parent topic and rest on the Taylor Series Expansion.

Example 1 — Cell A: growing solution, positive slope

Worked example

$y^{'} = x + y, y (0) = 1$ , find $y (0.3)$ with $h = 0.1$ Forecast: the true solution $2 e^{x} - x - 1$ curves upward (it is convex). Euler follows straight tangents, so it should land below the true curve every step. Guess: our $y (0.3)$ will be a touch under the true $\approx 1.4$ .

Step 1 — start. $x_{0} = 0, y_{0} = 1$ . Slope $f (0, 1) = 0 + 1 = 1$ . Why? The ODE is the slope-machine; feed it the current point. $y_{1} = 1 + 0.1 \cdot 1 = 1.1, x_{1} = 0.1$

Step 2. Slope $f (0.1, 1.1) = 0.1 + 1.1 = 1.2$ . Why? Re-measure the slope at the new point — the road bent, so the old slope is stale. $y_{2} = 1.1 + 0.1 \cdot 1.2 = 1.22, x_{2} = 0.2$

Step 3. Slope $f (0.2, 1.22) = 0.2 + 1.22 = 1.42$ . Why? Same rule a third time — always re-evaluate $f$ at the freshly landed point $(x_{2}, y_{2})$ before stepping again. $y_{3} = 1.22 + 0.1 \cdot 1.42 = 1.362, x_{3} = 0.3$

Verify: true value $2 e^{0.3} - 0.3 - 1 = 1.39972$ . Our $1.362$ is below it (error $\approx 0.038$ ), exactly as forecast for a convex curve. ✓

The figure below plots our three orange Euler points against the teal true curve. Notice that the orange polyline sits under the teal curve at every $x$ , and the plum arrow at $x = 0.3$ measures the gap — this is the accumulated undershoot the forecast predicted. Because the true curve bends upward, straight tangent steps can only cut the corner and fall short.

Figure — ODE solvers — Euler's method (derivation, global error)

Example 2 — Cell B: decaying solution, negative slope

Worked example

$y^{'} = - y, y (0) = 2$ , find $y (0.4)$ with $h = 0.2$ Here $f = - y$ , i.e. the linear test equation with growth rate $λ = - 1$ . Forecast: true solution $2 e^{- x}$ falls toward $0$ and is convex-from-below. A negative slope means each step subtracts height. Guess: a value near $2 e^{- 0.4} \approx 1.34$ , and (since the curve is convex) Euler will sit slightly below it.

Step 1. $f (0, 2) = - 2$ . $y_{1} = 2 + 0.2 (- 2) = 1.6$ at $x_{1} = 0.2$ . Why? Slope is negative, so the rise $h f = - 0.4$ is a fall.

Step 2. $f (0.2, 1.6) = - 1.6$ . $y_{2} = 1.6 + 0.2 (- 1.6) = 1.28$ at $x_{2} = 0.4$ . Why? Re-measure the slope at the new, lower height $y_{1} = 1.6$ ; a smaller height gives a shallower fall ( $- 1.6$ instead of $- 2$ ), so the curve is levelling off.

Verify: true $2 e^{- 0.4} = 1.34064$ . Euler gives $1.28$ , below it (error $\approx 0.061$ ). The method is stable here because the stability factor $∣1 + hλ ∣ = ∣1 + 0.2 (- 1) ∣ = 0.8 < 1$ — the estimates shrink toward $0$ just like the truth. ✓ (Contrast this with Example 5, where the same idea fails.)

Example 3 — Cell C: the order check (halve $h$ , halve the error)

Worked example

$y^{'} = x + y, y (0) = 1$ , compare final error at $x = 0.2$ for $h = 0.1$ vs $h = 0.05$ Forecast: the parent proved global error $= O (h)$ . So halving $h$ should roughly halve the error. Predict the ratio $\approx 0.5$ .

With $h = 0.1$ (2 steps): from Example 1's first two lines, $y (0.2) = 1.22$ . True $2 e^{0.2} - 1.2 = 1.24281$ . Error $E_{0.1} = 0.02281$ .

With $h = 0.05$ (4 steps):

$y_{1} = 1 + 0.05 (0 + 1) = 1.05$ at $0.05$
$f = 0.05 + 1.05 = 1.1$ , $y_{2} = 1.05 + 0.05 (1.1) = 1.105$ at $0.10$
$f = 0.10 + 1.105 = 1.205$ , $y_{3} = 1.105 + 0.05 (1.205) = 1.16525$ at $0.15$
$f = 0.15 + 1.16525 = 1.31525$ , $y_{4} = 1.16525 + 0.05 (1.31525) = 1.2310125$ at $0.20$

Error $E_{0.05} = 1.24281 - 1.23101 = 0.01180$ .

Verify: ratio $E_{0.05} / E_{0.1} = 0.01180/0.02281 = 0.517 \approx \frac{1}{2}$ . First order confirmed. ✓ Why not exactly $0.5$ ? The $O (h)$ law is a leading-order statement; tiny higher-order terms nudge it off exact half. As $h \to 0$ the ratio tightens to $0.5$ .

Example 4 — Cell D: degenerate slope $f = 0$ (a flat step)

Worked example

$y^{'} = x^{2} - 1, y (0) = 0.5$ , find $y (2)$ with $h = 1$ This is a case people fumble: at some step the slope is exactly zero, so $y$ does not move at all.

Forecast: at $x = 1$ the slope $1^{2} - 1 = 0$ . So the step starting at $x = 1$ will be perfectly flat — $y$ stays put across that step. Watch for it.

Step 1. $f (0, 0.5) = 0^{2} - 1 = - 1$ . $y_{1} = 0.5 + 1 \cdot (- 1) = - 0.5$ at $x_{1} = 1$ . Why the drop? The slope was negative, so height fell.

Step 2. $f (1, - 0.5) = 1^{2} - 1 = 0$ . $y_{2} = - 0.5 + 1 \cdot 0 = - 0.5$ at $x_{2} = 2$ . Why unchanged? ==Zero slope ⇒ zero rise ⇒ $y_{n + 1} = y_{n}$ .== The step is horizontal.

Verify: solve exactly: $y = \frac{x ^{3}}{3} - x + 0.5$ , so $y (2) = \frac{8}{3} - 2 + 0.5 = 1.16 \overline{6}$ . Euler's crude $h = 1$ gives $- 0.5$ — wildly off because $h$ is huge, but the mechanics of the flat step are correct: $y$ genuinely did not change when $f = 0$ . ✓ (Degenerate inputs don't break Euler; they just produce a flat segment.)

The figure traces the two orange steps. The first segment slopes down (slope $- 1$ ); the second, drawn thick in plum, is perfectly horizontal because $f = 0$ there — the visual proof that a zero slope produces a flat step with $y_{2} = y_{1}$ . Reading the picture left to right is the calculation.

Example 5 — Cell E: $h$ too large ⇒ instability

Worked example

$y^{'} = - 2 y, y (0) = 1$ , take $h = 1.5$ , find $y (3)$ This is the linear test equation with growth rate $λ = - 2$ . Forecast: true solution $e^{- 2 x}$ decays smoothly to $\approx 0$ . But the stability factor is $1 + hλ = 1 + 1.5 (- 2) = - 2$ , and $∣ - 2∣ = 2 > 1$ . Numbers multiplied by something bigger than $1$ in size will grow and flip sign. Forecast: oscillating, exploding values — the opposite of the true decay.

Step 1. $y_{1} = 1 + 1.5 (- 2) (1) = 1 - 3 = - 2$ at $x = 1.5$ . Step 2. $y_{2} = - 2 + 1.5 (- 2) (- 2) = - 2 + 6 = 4$ at $x = 3$ .

Verify: each step multiplies by the stability factor $1 + hλ = - 2$ : $1 \to - 2 \to 4$ , i.e. $y_{n} = (- 2)^{n}$ . True $e^{- 6} = 0.00248$ . Euler screams to $4$ — a stability failure, not a mere accuracy error. ✓

The cure: need $∣1 + hλ ∣ < 1$ , i.e. $∣1 - 2 h ∣ < 1 \Rightarrow 0 < h < 1$ . Any $h < 1$ decays correctly; $h = 1.5$ violates it. This is precisely the gateway to Numerical Stability and to Backward Euler & Implicit Methods, which stay stable for any $h$ here.

The figure contrasts the smooth teal decay curve $e^{- 2 x}$ (heading gently to $0$ ) with the orange Euler points that leap $1 \to - 2 \to 4$ . The plum annotation flags the culprit — a stability factor of size $2 > 1$ — and the orange zig-zag straddling the axis shows the tell-tale sign-flipping of an unstable run. When your numbers alternate sign and grow, this picture is what is happening.

Example 6 — Cell F: autonomous non-linear $f$

Worked example

$y^{'} = y^{2}, y (0) = 1$ , find $y (0.2)$ with $h = 0.1$ Here $f$ depends on $y$ squared — a non-linear slope machine. (The true solution $y = \frac{1}{1 - x}$ blows up at $x = 1$ ; we stay safely left of it.)

Forecast: $y^{2}$ grows fast as $y$ grows, so slopes accelerate. Euler, using the earlier smaller slope, will undershoot the true steep climb.

Step 1. $f (0, 1) = 1^{2} = 1$ . $y_{1} = 1 + 0.1 (1) = 1.1$ at $0.1$ . Why square? Just evaluate $f$ : $(y_{0})^{2} = 1$ .

Step 2. $f (0.1, 1.1) = 1. 1^{2} = 1.21$ . $y_{2} = 1.1 + 0.1 (1.21) = 1.221$ at $0.2$ . Why? Re-evaluate $f$ at the new height $y_{1} = 1.1$ ; squaring a larger height gives a steeper slope ( $1.21 > 1$ ), so the climb is accelerating.

Verify: true $\frac{1}{1 - 0.2} = 1.25$ . Euler gives $1.221$ , below it (error $\approx 0.029$ ), as forecast. ✓ Non-linearity changes nothing in the recipe — you still just plug the current point into $f$ .

Example 7 — Cell G: real-world word problem

Worked example A cup of coffee at

$9 0^{\circ} C$ sits in a $2 0^{\circ} C$ room. Newton's law of cooling gives $\frac{d T}{d t} = - 0.1 (T - 20)$ , with $t$ in minutes. Estimate the temperature at $t = 20$ min using $h = 10$ min. Translate: $f (t, T) = - 0.1 (T - 20)$ , $T_{0} = 90$ . The negative sign means it cools toward room temperature.

Forecast: true cooling is $20 + 70 e^{- 0.1 t}$ ; at $t = 20$ that is $20 + 70 e^{- 2} \approx 29. 5^{\circ}$ . With a coarse $h = 10$ , Euler over-cools (slope biggest at the start, applied over a long step) — expect a value below $29.5$ .

Step 1. $f (0, 90) = - 0.1 (90 - 20) = - 7$ . $T_{1} = 90 + 10 (- 7) = 2 0^{\circ}$ at $t = 10$ . Why so drastic? The start slope is steep and we held it for a whole $10$ minutes.

Step 2. $f (10, 20) = - 0.1 (20 - 20) = 0$ . $T_{2} = 20 + 10 (0) = 2 0^{\circ}$ at $t = 20$ . Why flat? At $2 0^{\circ}$ we hit room temperature exactly, so the slope is $0$ — a degenerate step (Cell D again, arising naturally!).

Verify: Euler says $2 0^{\circ}$ ; truth $29. 5^{\circ}$ . Error $\approx 9. 5^{\circ}$ — large because $h = 10$ is coarse and the stability margin $∣1 + hλ ∣ = ∣1 + 10 (- 0.1) ∣ = ∣0∣ = 0$ collapses the estimate straight to the floor in one jump. Units check: temperature stays in $^{\circ}$ C throughout. ✓ Smaller $h$ (say $h = 2$ ) would track the smooth $29. 5^{\circ}$ far better.

Example 8 — Cell H: exam twist (given the answer, find the input)

Recall Which cell was hardest — and why?

Cell E (instability). Every other cell only produces accuracy error, controlled by $h$ via the $O (h)$ law. Cell E produces a qualitative failure: the numbers diverge from a truth that decays. The lesson: $h$ must satisfy both accuracy and stability constraints.

Quick self-test

Positive slope Euler estimate vs a convex true curve — above or below?

Below (tangents undercut a convex curve).

For

y^{'} = - y

, which

h

keeps Euler stable?

Any

h

with

∣1 - h ∣ < 1

, i.e.

0 < h < 2

What does a step with

f = 0

do to

y

Nothing —

y_{n + 1} = y_{n}

, a flat horizontal step.

Halving

h

changes the global error by roughly what factor?

About

\frac{1}{2}

(first-order,

O (h)

Given

y_{1} = y_{0} + h f

and a known

y_{1}

, how do you find

h

Solve the linear equation

h = (y_{1} - y_{0}) / f (x_{0}, y_{0})

Connections

Parent topic — derivation & error law these examples exercise
Taylor Series Expansion — source of the local $O (h^{2})$ term each example inherits
Runge-Kutta Methods — would cut the errors seen above dramatically ( $O (h^{4})$ )
Backward Euler & Implicit Methods — stays stable in Example 5 for any $h$
Numerical Stability — the $∣1 + hλ ∣ < 1$ test used in Examples 2, 5, 7
Lipschitz Continuity — guarantees the error bound behind Example 3
Finite Difference Approximations — the forward-difference view of every step here

Worked examples — ODE solvers — Euler's method (derivation, global error)

The scenario matrix

Example 1 — Cell A: growing solution, positive slope

Example 2 — Cell B: decaying solution, negative slope

Example 3 — Cell C: the order check (halve $h$ , halve the error)

Example 4 — Cell D: degenerate slope $f = 0$ (a flat step)

Example 5 — Cell E: $h$ too large ⇒ instability

Example 6 — Cell F: autonomous non-linear $f$

Example 7 — Cell G: real-world word problem

Example 8 — Cell H: exam twist (given the answer, find the input)

Quick self-test

Connections

Case Map