Euler Method Visualizer

change step size and watch global error accumulate

Worked example — Euler Method Visualizer

Euler y(xn)	7.000266
Exact y(xn)	7.000671
Absolute global error	0.000405
Error / step size	0.002025
Endpoint error with h/2	0.000234
Error ratio	1.731285
Euler multiplier	0.8

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Halving the step only halves the error 🖖

Euler steps forward along the tangent line: y_n+1 = y_n + h · f(x_n, y_n). One step is wrong by about h²/2 times the second derivative of the solution. That is small. But shrinking h also multiplies the steps — 1/h of them across a fixed interval — so those h² errors accumulate into a total proportional to h, not h². One power of h is what first order means, and it prices accuracy: a further correct decimal digit costs ten times the arithmetic.

Following the slope one step at a time 🖖

Euler's method is the simplest honest guess: from where you are, head in the direction the equation says the curve is going, walk a short distance h, then stop and re-aim. The exact curve keeps bending, but a straight step ignores that bend and drifts off. And it drifts the same way every time: below a curve that bends upward, above one that bends down. All seven presets here obey that. The panel prints the endpoint error twice, at h and at h/2, so the ratio between them is there to read.

Where the number e hides inside Euler 🖖

Apply Euler's method to the equation y' = y starting at y = 1, and each step simply multiplies by (1 + h). Take N steps of size h = 1/N across the interval [0, 1] and the result is exactly (1 + 1/N)^N — the very formula for compound interest. As N grows it converges to e ≈ 2.71828, so Euler's crude straight-line stepping quietly reconstructs continuous compounding and the definition of the exponential.

EULER'S METHOD — THE STEP SIZE DECIDES WHETHER THE ANSWER MEANS ANYTHING

Which Step-Size Case Are You In?

Euler takes the slope where it is standing and walks in a straight line for h before looking again: yₙ₊₁ = yₙ + h·f(xₙ, yₙ). Everything that goes wrong is about h. Small enough and the error shrinks in proportion; too large and the answer oscillates or explodes; and no h at all protects you from a solution that runs to infinity while the numbers on screen stay calm.

A sensible step — the error is there but small y_n+1 = y_n + h f(x_n, y_n)

Step too big — the answer oscillates around the truth |1 + hλ| > 1

A fine step — you pay in work, not in accuracy ε ∝ h

The exact solution escapes to infinity — and Euler does not notice y′ = y² ⇒ y → ∞, x < ∞

A sensible step — the error is there but small

What you know: A smooth, stable equation with a step well inside the stability limit. The numerical curve tracks the exact one and lags slightly behind it.

Step: y_n+1 = y_n + h f(x_n, y_n)

Worked example: y′ = x − y from y(0) = 1 with h = 0.2 → at x = 2 Euler gives 1.2147 against the exact 1.2707, an error of 0.056

Open this case: linear stable

$A sensible step — the error is there but small. The straight-line steps stay close to the true curve and lean consistently to one side of it. A smooth, stable equation with a step well inside the stability limit. The numerical curve tracks the exact one and lags slightly behind it.$

The straight-line steps stay close to the true curve and lean consistently to one side of it.

Step too big — the answer oscillates around the truth

What you know: A decaying equation where h has passed the point at which one step overshoots the equilibrium. The exact solution never changes sign; the numerical one does.

Step: |1 + hλ| > 1

Worked example: y′ = −2y with h = 0.6 → 1, then −0.2, 0.04, −0.008: the first step alone jumps past zero, while the exact value at x = 0.6 is 0.301

Open this case: decay coarse h

Step too big — the answer oscillates around the truth. One step overshoots the axis, so the numbers flip sign while the true solution never does. A decaying equation where h has passed the point at which one step overshoots the equilibrium. The exact solution never changes sign; the numerical one does. — One step overshoots the axis, so the numbers flip sign while the true solution never does.

A fine step — you pay in work, not in accuracy

What you know: The same equation with a much smaller step. The error falls in proportion to h, and the step count rises in inverse proportion.

Step: ε ∝ h

Worked example: y′ = −2y at x = 0.6: with h = 0.1 the error is 0.039, with h = 0.05 it is 0.019 — half the step, half the error, twice the steps

Open this case: decay fine h

A fine step — you pay in work, not in accuracy. Halving the step halves the gap to the exact curve, and doubles the number of steps. The same equation with a much smaller step. The error falls in proportion to h, and the step count rises in inverse proportion. — Halving the step halves the gap to the exact curve, and doubles the number of steps.

The exact solution escapes to infinity — and Euler does not notice

What you know: A nonlinear equation whose true solution has a vertical asymptote at a finite x. The numerical solution sails past it and keeps returning ordinary numbers.

Step: y′ = y² ⇒ y → ∞, x < ∞

Worked example: y′ = y² from y(0) = 0.18 blows up at x = 1/0.18 = 5.56, yet Euler with h = 0.18 reports a comfortable 0.68 at x = 4.32, where the true value is already 0.81

Open this case: quadratic growth

The exact solution escapes to infinity — and Euler does not notice. The true curve turns vertical at finite x; the numerical one strolls straight past. A nonlinear equation whose true solution has a vertical asymptote at a finite x. The numerical solution sails past it and keeps returning ordinary numbers. — The true curve turns vertical at finite x; the numerical one strolls straight past.

References (1)

The method, as Euler set it out: L. Euler, Institutiones calculi integralis, Volume I. Academia Imperialis Scientiarum, Saint Petersburg, 1768.

Common wrong intuition

More steps do not guarantee accuracy by themselves. If h is still too large for curvature, Euler remains biased.

Problem solved in full

Euler's answer at x = 8 for y′ = x − y 5 steps
Write down Euler's answer at x = 8 in closed form, without taking a single step, and then explain why halving h does not halve the error. This is y′ = x − y started at (0, 1) with h = 0.2 and 40 steps.
1. Start with the answer Euler is aiming at. The equation is linear, x − 1 is a particular solution, and the initial condition fixes the coefficient of the decaying part at 2.
  
  $y^{'} = x - y, y (0) = 1 ⟹ y = x - 1 + 2 e^{- x}, y (8) = 7 + 2 e^{- 8} = 7.000671$
2. Euler's rule on this equation is not a general nonlinear mess — it is an affine recurrence. And the same x − 1 that solved the differential equation solves the recurrence exactly, step for step. Euler gets the straight part of the answer perfectly right.
  
  $y_{n + 1} = y_{n} + h (x_{n} - y_{n}) = (1 - h) y_{n} + h x_{n}, y_{n} = x_{n} - 1 ⟹ y_{n + 1} = x_{n} - 1 + h = x_{n + 1} - 1$
3. So the entire error lives in what is left, and what is left is a geometric sequence with ratio 1 − h. Forty factors of 0.8 land on the panel's endpoint with no stepping done at all.
  
  $y_{n} = x_{n} - 1 + 2 (1 - h)^{n} ⟹ y_{40} = 7 + 2 (0.8)^{40} = 7 + 2.658456 \times 1 0^{- 4} = 7.000266$
4. Now put the two decaying parts side by side: the true one is e^−x, Euler's is (1 − h)^x/h. Both are exponentials in x, and δ is the whole difference between their rates. Its series starts at xh/2 — that leading h is the entire content of the phrase “first order”.
  
  $(1 - h)^{x / h} = e^{- x - δ}, δ = x (\frac{- ln ( 1 - h )}{h} - 1) = \frac{x h}{2} + \frac{x h ^{2}}{3} + \dots = 0.925742$
5. The error is what survives when the two exponentials are subtracted. First order predicts that halving h halves it, so the ratio should be 2.
  
  $E_{h} = 2 e^{- x} (1 - e^{- δ}) : E_{0.2} = 0.000405, E_{0.1} = 0.000234, \frac{E _{0.2}}{E _{0.1}} = \frac{1 - e ^{- 0.925742}}{1 - e ^{- 0.428841}} = \frac{0.603763}{0.348737} = 1.731285$
Answer

Euler ends at 7.000266 against an exact 7.000671, a global error of 0.000405; halve the step and it falls only to 0.000234, a ratio of 1.731285 where first order promised 2. The promise is not broken — it has not started. It needs δ ≈ xh/2 to be small, and at x = 8 with h = 0.2 that number is 0.8, so the linear term in the error is still competing with every term behind it; dragging the observed ratio up to 1.9933 takes h = 0.005, which is 1600 steps across the same eight units. The closed form gives away the sharper reading: Euler is not approximating this solution, it is solving a different equation exactly, with geometric decay 0.8ⁿ standing in for e^−x — and 0.8 is the multiplier 1 + hλ the panel already prints. Measured on that decaying part alone, Euler is wrong by 60%. The global error looks tiny only because the part it gets wrong has almost died out by the time x reaches 8.

$\frac{E _{h}}{E _{h /2}} = \frac{1 - e ^{- δ_{h}}}{1 - e ^{- δ_{h /2}}} ⟶ 2 (δ ≪ 1), δ_{h} \approx \frac{x h}{2} = 0.8$

Learning path

Three numerical methods, and where each one gives out

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

linear stable - The one equation here whose Euler answer has a closed form: 7 + 2(0.8)⁴⁰, no stepping needed, so the error is checkable to the last digit.
decay coarse h - At h = 0.6 the Euler multiplier is -0.2, so the approximation flips sign at every step while the true solution never leaves the positive side.
logistic medium h - Logistic growth from 0.2, where the slope rises, peaks, and dies away. Euler overshoots to 0.9993 against 0.9987, because the curve bends down and the tangent does not.
decay fine h - The same decay equation at h = 0.05: the multiplier is 0.9 and the sign-flipping stops. Twelve times the steps buys stability, not only accuracy.
cooling curve - Coffee at 90 degrees cooling toward a 20 degree room. At the end of the run Euler reads 20.06 against an exact 20.12 -- six hundredths, after a fall of seventy.
sine forcing - y' = cos(x) has no y on the right, so Euler here is exactly the left Riemann sum of the cosine: 1.079 where the true integral gives 0.970.
quadratic growth - y' = y² runs to infinity at x = 5.56 and the panel stops at 4.32, reading 0.68 where the truth is 0.81. It looks calmest where it is least trustworthy.