Direction fields and Euler's method — visual - numerical intuition first: Question bank

True or false — justify

Every answer below demands a reason, never a bare yes/no.

The length of a direction-field segment tells you how fast the solution changes.

False. Only the tilt (slope) of a segment carries meaning; length is purely cosmetic. A steep short dash and a steep long dash mean the identical slope

f (x, y)

For

y^{'} = f (x, y)

, two different solution curves can cross each other at a point.

False in general. Where

f

is nice (continuous with bounded

\partial f / \partial y

), Existence and Uniqueness (Picard–Lindelöf) guarantees exactly one curve through each point, so they cannot cross — a crossing would force two slopes at one point.

f (x, y)

depends only on

x

, every dash in the same vertical column is parallel.

True. The slope

f (x)

is fixed once

x

is fixed, so it does not change as you move up/down that column — all those dashes share one tilt.

f (x, y)

depends only on

y

, every dash in the same horizontal row is parallel.

True. By the same logic: fixing

y

fixes the slope

f (y)

, so along a horizontal line (constant

y

) all dashes match.

Halving the step size

h

roughly halves Euler's total error.

True. Global error is

O (h)

(first order), so error scales linearly with

h

— halve

h

, halve the error, approximately.

Halving

h

halves the error per step.

False. The per-step (local truncation) error is

O (h^{2})

, so halving

h

makes each step's error about four times smaller — but you take twice as many steps, netting the

O (h)

global result.

Euler's method is exact whenever the true solution is a straight line.

True. If

y^{''} = 0

everywhere the discarded term

\frac{h ^{2}}{2} y^{''} (ξ)

vanishes, so the tangent step lands exactly on the curve — Euler has zero error.

Euler always undershoots the true solution.

False. It undershoots only when the curve is convex (

y^{''} > 0

), because tangents lie below it. For a concave curve (

y^{''} < 0

) tangents lie above, so Euler overshoots.

Isoclines are themselves solution curves of the ODE.

False. An isocline

f (x, y) = m

is just the set of points sharing slope

m

; a solution is tangent to the field, not required to follow an isocline. They only coincide by accident.

Taking more and more Euler steps guarantees you approach the true answer.

False in practice. Truncation error

\to 0

h \to 0

, but rounding error accumulates with more steps, and stiff equations can blow up — so smaller

h

is not unconditionally safe.

Spot the error

Each statement contains one flaw. Name it and fix it.

"Euler step:

y_{n + 1} = y_{n} + f (x_{n}, y_{n})

Missing the step size.

f

is a slope, not a rise; you must multiply by the run

h

y_{n + 1} = y_{n} + h f (x_{n}, y_{n})

. Check units:

Δ y = (Δ y /Δ x) \cdot Δ x

"To be accurate, evaluate the slope at the point you're heading to:

f (x_{n + 1}, y_{n + 1})

That is a different method — implicit (backward) Euler. Explicit Euler must use the slope at the point it is leaving,

f (x_{n}, y_{n})

, because

y_{n + 1}

is not yet known.

"The direction field arrows show the direction the particle moves through time."

They are not velocity vectors. Each segment only marks the tangent direction of the solution at that point; it has no speed and no time-orientation.

"Since

y^{'} = x + y

(0, 1)

gives slope

1

, after a step of

h = 0.5

the new

y

1 + 1 = 2

Forgot to scale by

h

. Rise

= h \cdot slope = 0.5 \cdot 1 = 0.5

, so

y_{1} = 1.5

, not

2

"An isocline is the curve where the solution has slope zero."

That is only the special isocline

f = 0

. An isocline is

f (x, y) = m

for any fixed

m

; the

m = 0

one (horizontal dashes) is just one member of the family.

"Because Euler is convergent, its answer for finite

h

equals the exact solution."

Convergence means error

\to 0

h \to 0

, a limiting statement. For any finite

h

there is a genuine

O (h)

error; convergent

\neq =

exact.

y^{''} > 0

means the curve slopes upward, so Euler overshoots."

y^{''} > 0

means the curve is convex (bends upward), not that it rises. Tangents to a convex curve lie below it, so Euler undershoots, not overshoots.

Why questions

Why does the ODE give a field of slopes rather than a single curve?

Because

f (x, y)

assigns a slope to every point

(x, y)

in the plane; you only pin down one curve after you also fix an initial condition to say where to start.

Why do we drop the

\frac{h ^{2}}{2} y^{''} (ξ)

term to get Euler?

We usually don't know

y^{''}

, and for small

h

that term is tiny compared to

h y^{'}

. Dropping it turns the exact Taylor expansion into a computable one-step rule — at the cost of a known

O (h^{2})

per-step error.

Why is the global error one order worse (

O (h)

) than the local error (

O (h^{2})

To cross a fixed interval of length

L

you take

N = L / h

steps, each contributing

\sim h^{2}

: total

\sim (L / h) \cdot h^{2} = L h = O (h)

. More steps dilute the extra accuracy.

Why does explicit Euler use the left endpoint's slope?

Because at the moment of stepping you only know

y_{n}

; the slope

f (x_{n}, y_{n})

is computable, while

f (x_{n + 1}, y_{n + 1})

needs the unknown

y_{n + 1}

you're trying to find.

Why do isoclines make hand-sketching so fast?

Along one isocline every dash has the same tilt, so you draw a whole family of parallel dashes at once instead of computing

f

point by point — the field appears almost for free.

Why do methods like RK4 beat Euler?

They sample the slope at several points within each step and blend them, cancelling more Taylor terms, so their error is

O (h^{4})

— dramatically smaller for the same

h

Why can't Euler follow a solution through a point where

f

blows up?

An infinite slope has no finite tangent step; the local linear approximation breaks down, and near such singularities uniqueness may fail too — the "follow your nose" picture stops making sense.

Edge cases

h = 0

, what does one Euler step give?

y_{n + 1} = y_{n} + 0 = y_{n}

— you never move. A zero step is degenerate: no progress, no error either, and infinitely many steps to go anywhere.

What does the direction field look like where

f (x, y) = 0

A horizontal dash (slope

0

). Solutions passing through are momentarily flat there; if

f = 0

on a whole curve it may be an equilibrium the solution levels off toward.

What happens to Euler on the exact straight-line solution of

y^{'} =

constant?

It is exact at every step, since

y^{''} = 0

kills the truncation term. Euler and the true solution coincide regardless of

h

For a separable ODE you can solve exactly, why still bother with Euler?

To verify the numerical method against a known answer and to build intuition; but if a closed form exists, the exact solution is preferred — Euler shines when no formula is available.

What can happen on a stiff equation even with a "small"

h

Explicit Euler can become unstable and oscillate or blow up unless

h

is extremely tiny; the step size must respect a stability limit, not just accuracy — this is the domain of Stability and Stiff Equations.

At a point where two solution curves would have to touch but not cross, is the field ambiguous?

If uniqueness holds, only one slope exists there so no ambiguity; ambiguity signals

f

\partial f / \partial y

misbehaving, where the single-arrow picture cannot be trusted.

If the true solution is concave (

y^{''} < 0

) everywhere on the interval, which way does Euler err?

It overshoots: tangents to a concave curve lie above it, so each step lands too high, the mirror image of the convex undershoot.

Active recall

Recall One-line traps to self-quiz

Segment length meaning? → None; only tilt matters.
Curves crossing under uniqueness? → Impossible.
Missing factor in wrong Euler formula? → The step $h$ .
Euler on a straight-line solution? → Exact.
Convex curve → Euler? → Undershoots.
Concave curve → Euler? → Overshoots.
Global error order? → $O (h)$ .

Connections

Direction fields and Euler's method — visual - numerical intuition first — the parent this bank drills.
Taylor's Theorem — why the dropped term is $O (h^{2})$ .
Runge-Kutta Methods (RK4) — the accurate successor.
Existence and Uniqueness (Picard–Lindelöf) — why curves don't cross.
Stability and Stiff Equations — why "more steps" can still explode.
Separable First-Order ODEs — exact answers to test against.