Direction fields and Euler's method — visual - numerical intuition first: Worked examples

This page is the drill sheet for the parent topic. We are going to hit every kind of situation Euler's method and direction fields can throw at you — positive slopes, negative slopes, a slope of exactly zero, a curve that bends up, a curve that bends down, a step so big it lies to you, a real word problem, and an exam-style trap.

Before any symbol appears, one reminder in plain words:

The scenario matrix

Every Euler / direction-field problem is some combination of the cells below. The worked examples that follow are each tagged with the cell(s) they cover, so by the end no cell is left blank.

#	Cell (case class)	What is special about it	Covered by
A	Slope $f > 0$ (rising)	height goes up each step	Ex 1, Ex 4
B	Slope $f < 0$ (falling)	height goes down each step	Ex 2, Ex 3
C	Slope $f = 0$ (flat point)	a step that moves only sideways	Ex 3
D	Convex solution $y^{''} > 0$ (bowl)	Euler under-shoots (tangents below curve)	Ex 1, Ex 3, Ex 6
E	Concave solution $y^{''} < 0$ (dome)	Euler over-shoots (tangents above curve)	Ex 2
F	Equilibrium / degenerate $f \equiv 0$ line	solution is a flat line, Euler stays put forever	Ex 3b
G	Step-size effect (halve $h$ )	error roughly halves	Ex 4
H	Big $h$ overshoot / instability	Euler can blow past and oscillate	Ex 5
I	Real-world word problem	naming $f$ , units, interpreting the answer	Ex 6
J	Exam twist (backward vs forward, sign trap)	the classic marks-losing mistakes	Ex 7

Example 1 — Cell A + D: rising slope, convex (bowl) curve

Worked example

$\frac{d y}{d x} = x + y, y (0) = 1, h = 0.5$ , find $y (1)$ . Forecast: The slope $f = x + y$ starts at $0 + 1 = 1$ (positive) and only grows as both $x$ and $y$ grow — the slope is increasing, so $y^{''} > 0$ : a bowl (convex). Before computing: will Euler land above or below the true curve? Guess now.

Step 1 ( $n = 0$ ): read the slope machine at the start. $f (0, 1) = 0 + 1 = 1.$ Why this step? Euler always evaluates the slope at the point it is standing on — the left end of the step. Take the step: $y_{1} = 1 + 0.5 \cdot 1 = 1.5, x_{1} = 0.5.$ Why this step? rise = run × slope = $0.5 \times 1 = 0.5$ , so height climbs from $1$ to $1.5$ .

Step 2 ( $n = 1$ ): re-read the machine at the new point. $f (0.5, 1.5) = 0.5 + 1.5 = 2.$ $y_{2} = 1.5 + 0.5 \cdot 2 = 2.5, x_{2} = 1.0.$ Why this step? The slope must be recomputed — it changed from $1$ to $2$ because we moved.

Answer: $y (1) \approx 2.5$ .

Verify: The exact solution is $y = 2 e^{x} - x - 1$ (this is a separable-adjacent linear ODE — compare against Separable First-Order ODEs techniques). Then $y (1) = 2 e - 2 \approx 3.44$ . Euler gave $2.5$ , an under-shoot — matching the forecast, because for a bowl ( $y^{''} > 0$ ) each straight tangent step lies below the true curve. Look at the figure: every red step ends beneath the black curve.

Figure 1 — Euler's straight steps (red) lie below the true convex "bowl" curve: an undershoot.

Example 2 — Cell E: a concave (dome) curve, and how to spot one

Worked example

$\frac{d y}{d x} = \frac{1}{2 x + 1}, y (0) = 1, h = 1$ , find $y (3)$ . (A genuine dome, Cell E.) Forecast: The slope starts at $\frac{1}{2 1} = 0.5$ and shrinks as $x$ grows — a decreasing slope, so $y^{''} < 0$ : a dome (concave). Before computing: will Euler land above or below the true curve? Guess now.

Step 1 ( $n = 0$ , at $x = 0$ ): the slope depends only on $x$ . $f (0) = \frac{1}{2 1} = 0.5.$ $y_{1} = 1 + 1 \cdot 0.5 = 1.5, x_{1} = 1.$ Why this step? rise = run × slope = $1 \times 0.5 = 0.5$ .

Step 2 ( $n = 1$ , at $x = 1$ ): $f (1) = \frac{1}{2 2} \approx 0.3536.$ $y_{2} = 1.5 + 1 \cdot 0.3536 = 1.8536, x_{2} = 2.$ Why this step? The slope is shallower than before ( $0.354 < 0.5$ ) — the slope is decreasing, the signature of a dome ( $y^{''} < 0$ ).

Step 3 ( $n = 2$ , at $x = 2$ ): $f (2) = \frac{1}{2 3} \approx 0.2887.$ $y_{3} = 1.8536 + 1 \cdot 0.2887 = 2.1423, x_{3} = 3.$

Answer: $y (3) \approx 2.142$ .

Verify: Exact $y = x + 1$ , so $y (3) = 4 = 2$ . Euler gave $2.142$ — an over-shoot, exactly as a dome ( $y^{''} < 0$ ) predicts: each tangent line sits above the true curve, so every step lands too high. This is Cell E done right. Sanity: the slopes fell $0.5 \to 0.354 \to 0.289$ , confirming the curve was bending down the whole way.

Figure 2 — For a concave "dome" curve the red Euler steps ride above the true curve: an overshoot.

Example 3 — Cell B + D and Cell C + F: a decay, and its equilibrium

Worked example Part (a) — a

falling, convex decay: $\frac{d y}{d x} = - y, y (0) = 2, h = 0.5$ , find $y (1)$ . Forecast: $f = - y < 0$ for $y > 0$ (Cell B, falling). Is it bowl or dome? Slope-of-slope: $y^{''} = \frac{d}{d x} (- y) = - y^{'} = - (- y) = + y > 0$ — a bowl (convex). So predict an under-shoot even though the curve is falling: "falling" does not decide over/undershoot; $y^{''}$ does.

Step 1 ( $n = 0$ ): $f (0, 2) = - 2. y_{1} = 2 + 0.5 (- 2) = 1, x_{1} = 0.5.$ Why this step? Negative slope ⇒ the step goes down: rise $= 0.5 \times (- 2) = - 1$ .

Step 2 ( $n = 1$ ): $f (0.5, 1) = - 1. y_{2} = 1 + 0.5 (- 1) = 0.5, x_{2} = 1.0.$ Why this step? Slope recomputed at the lower point; it is shallower ( $- 1$ vs $- 2$ ) — the slope is rising toward zero, i.e. increasing, confirming $y^{''} > 0$ .

Answer: $y (1) \approx 0.5$ .

Verify: Exact $y = 2 e^{- x}$ , $y (1) = 2 e^{- 1} \approx 0.736$ . Euler gave $0.5$ — an under-shoot, matching the bowl prediction ( $y^{''} = 2 e^{- x} > 0$ ). Units/sanity: $y$ stayed positive and shrank, as any decay must. (This is the case that trips people: falling but convex.)

Worked example Part (b) — Cell C + F:

zero slope on an equilibrium line. $\frac{d y}{d x} = y (1 - y), y (0) = 1, h = 0.5$ , find $y (1.5)$ . Forecast: $f = y (1 - y) = 0$ exactly at $y = 1$ . At $y = 1$ the machine returns zero — a horizontal dash. What happens if Euler starts on that flat line?

Step 1 ( $n = 0$ ): $f (x, 1) = 1 (1 - 1) = 0. y_{1} = 1 + 0.5 \cdot 0 = 1, x_{1} = 0.5.$ Why this step? Slope exactly $0$ ⇒ the step is purely sideways — height unchanged.

Steps 2 & 3: still $y = 1$ , so $f = 0$ again: $y_{2} = 1, x_{2} = 1.0; y_{3} = 1, x_{3} = 1.5.$ Why this step? $y = 1$ is an equilibrium (a degenerate case: $f \equiv 0$ along a whole horizontal line). Euler is stuck on it forever — which is exactly correct.

Answer: $y (1.5) = 1$ (flat).

Verify: The exact solution with $y (0) = 1$ is the constant $y \equiv 1$ (an equilibrium studied in stability analysis). Euler reproduces it perfectly: zero slope ⇒ zero rise. In the direction field this is the horizontal row of dashes at height $y = 1$ ; nearby curves are drawn toward it. See figure: the red equilibrium line is dead flat.

Figure 3 — Logistic direction field: every dash on the red line $y = 1$ is horizontal, so Euler never leaves it.

Example 4 — Cell A + G: halve the step, watch the error shrink

Worked example Same as Example 1 (

$y^{'} = x + y, y (0) = 1$ ) but $h = 0.25$ , four steps to $y (1)$ . Forecast: Example 1 with $h = 0.5$ gave $2.5$ ; the truth is $3.44$ . Euler is a first-order method — in plain words, that means the total error shrinks in proportion to the step $h$ : halve $h$ , and you roughly halve the error (not quarter it — proportion, not square). So with half the step we expect an answer meaningfully closer to $3.44$ (around $2.9$ ). Guess before computing.

Step 1 ( $n = 0$ ): at $(0, 1.0000)$ , slope $f = 0 + 1 = 1.0000$ ; step $y_{1} = 1.0000 + 0.25 (1.0000) = 1.2500$ , $x_{1} = 0.25$ . Why this step? Read the slope at the start point and rise by run×slope $= 0.25 \times 1 = 0.25$ .

Step 2 ( $n = 1$ ): at $(0.25, 1.2500)$ , slope $f = 0.25 + 1.25 = 1.5000$ ; step $y_{2} = 1.2500 + 0.25 (1.5000) = 1.6250$ , $x_{2} = 0.50$ . Why this step? Slope recomputed at the new point — it grew from $1$ to $1.5$ because both $x$ and $y$ rose.

Step 3 ( $n = 2$ ): at $(0.50, 1.6250)$ , slope $f = 0.50 + 1.625 = 2.1250$ ; step $y_{3} = 1.6250 + 0.25 (2.1250) = 2.1563$ , $x_{3} = 0.75$ . Why this step? Again a fresh slope at the current spot; steeper still, so the rise per step keeps growing.

Step 4 ( $n = 3$ ): at $(0.75, 2.1563)$ , slope $f = 0.75 + 2.1563 = 2.9063$ ; step $y_{4} = 2.1563 + 0.25 (2.9063) = 2.8828$ , $x_{4} = 1.00$ . Why this step? Final slope read where we now stand; one more run×slope lands us at $x = 1$ .

Collected as a table:

$n$	$x_{n}$	$y_{n}$	$f = x_{n} + y_{n}$	$y_{n + 1} = y_{n} + 0.25 f$
0	0.00	1.0000	1.0000	1.2500
1	0.25	1.2500	1.5000	1.6250
2	0.50	1.6250	2.1250	2.1563
3	0.75	2.1563	2.9063	2.8828

Answer: $y (1) \approx 2.883$ .

Verify: Truth is $3.44$ . Errors: $h = 0.5$ gave error $3.44 - 2.50 = 0.94$ ; $h = 0.25$ gives $3.44 - 2.88 = 0.56$ . The ratio $0.56/0.94 \approx 0.60$ — not exactly $\frac{1}{2}$ because "error proportional to $h$ " is only the leading behaviour that becomes exact as $h$ gets tiny; here $h$ is still fairly big. But the error clearly shrank with smaller $h$ , as first-order accuracy promises. To close the gap much faster you'd switch to RK4, whose error shrinks like $h^{4}$ . See figure comparing the two staircases.

Figure 4 — Two Euler staircases: the finer step (red) hugs the true curve more closely than the coarse one.

Example 5 — Cell H: a step so big Euler goes unstable

Worked example

$\frac{d y}{d x} = - 5 y, y (0) = 1$ . Try $h = 0.5$ for four steps, then explain. Forecast: The true solution $y = e^{- 5 x}$ decays smoothly to $0$ . But the slope is steep (factor $5$ ). What could a big step do to a fast-decaying curve? Guess: smooth decay, or wild oscillation?

Steps using $y_{n + 1} = y_{n} + h (- 5 y_{n}) = y_{n} (1 - 5 h) = y_{n} (1 - 2.5) = - 1.5 y_{n}$ : $y_{1} = - 1.5, y_{2} = 2.25, y_{3} = - 3.375, y_{4} = 5.0625.$ Why this step? The amplification factor is $1 - 5 h = 1 - 2.5 = - 1.5$ . Its size $∣ - 1.5∣ = 1.5 > 1$ , so each step magnifies and flips sign — the numbers explode and oscillate.

Answer: Euler gives ${- 1.5, 2.25, - 3.375, 5.0625}$ — total nonsense versus the true $y (0.5) = e^{- 2.5} \approx 0.082$ .

Verify: The stability rule for explicit Euler on $y^{'} = λ y$ is $∣1 + hλ ∣ \leq 1$ . Here $λ = - 5$ , so we need $∣1 - 5 h ∣ \leq 1$ , i.e. $0 \leq h \leq 0.4$ . Our $h = 0.5$ violates it, hence the blow-up. Redo with $h = 0.1$ : factor $= 1 - 0.5 = 0.5$ , giving $y_{1} = 0.5, y_{2} = 0.25, \dots \to 0$ — well-behaved. This is exactly the "stiff equation" warning of Stability and Stiff Equations: small $h$ is not just for accuracy, sometimes it's for survival. See figure: the exploding red zig-zag vs the tame decay.

Figure 5 — With an oversized step the red Euler iterates zig-zag and blow up, while the true solution quietly decays.

Example 6 — Cell I + D: real-world word problem

Worked example A cup of coffee cools by Newton's law: the temperature

$T$ (°C) obeys $\frac{d T}{d t} = - k (T - 20)$ with $k = 0.2 min^{- 1}$ , room at $20°$ C, starting at $T (0) = 90°$ C. Estimate $T$ after $2$ minutes using $h = 1$ min. Forecast: Coffee is far above room temperature, so it should cool fast at first, then slow down (slope $\propto$ gap $T - 20$ ). Will $2$ minutes of cooling drop it a lot or a little? Guess a range.

Naming $f$ : here $f (t, T) = - 0.2 (T - 20)$ . Units check: $[k] = 1/ min$ , $[T - 20] = °$ C, so $[f] = ° C / min$ — a genuine rate of temperature change. Good.

Step 1 ( $n = 0$ , at $t_{0} = 0, T_{0} = 90$ ): $f = - 0.2 (90 - 20) = - 0.2 \cdot 70 = - 14 ° C/min .$ $T_{1} = 90 + 1 \cdot (- 14) = 76° C, t_{1} = t_{0} + h = 1.$ Why this step? rise = run × rate = $1 min \times (- 14°/ min) = - 14°$ .

Step 2 ( $n = 1$ , at $t_{1} = 1, T_{1} = 76$ ): $f = - 0.2 (76 - 20) = - 0.2 \cdot 56 = - 11.2° C/min .$ $T_{2} = 76 + 1 \cdot (- 11.2) = 64.8° C, t_{2} = t_{1} + h = 2.$ Why this step? The gap shrank, so the cooling rate shrank too ( $- 11.2$ vs $- 14$ ) — the "slows down" behaviour we forecast.

Answer: $T (2) \approx 64.8°$ C.

Verify: Exact solution $T = 20 + 70 e^{- 0.2 t}$ , so $T (2) = 20 + 70 e^{- 0.4} \approx 20 + 70 (0.6703) = 66.9°$ C. Euler's $64.8$ under-shoots the true $66.9$ : the cooling curve is a bowl in $T$ ( $T^{''} > 0$ ), tangents lie below, consistent with Cell D. Physically the estimate is sensible (still hot coffee), and the direction of error is exactly what convexity predicts.

Example 7 — Cell J: the exam twist (forward vs backward Euler, sign trap)

Worked example

$\frac{d y}{d x} = 2 x - y, y (1) = 3$ . An exam asks for one forward-Euler step with $h = - 0.5$ to estimate $y (0.5)$ . (Note the negative step — we walk left.) Forecast: A negative $h$ means moving to a smaller $x$ . The formula is unchanged; the trap is (a) using the wrong point's slope, and (b) mishandling the sign of $h$ . Watch both.

Step 1 ( $n = 0$ , at $x_{0} = 1, y_{0} = 3$ ): $f (1, 3) = 2 \cdot 1 - 3 = - 1.$ Why this step? Explicit (forward) Euler uses the slope at the point you are leaving, $(x_{0}, y_{0})$ — not at the destination. Using the destination point would be backward/implicit Euler, a different method (and unsolvable here without algebra). $y_{1} = y_{0} + h f = 3 + (- 0.5) (- 1) = 3 + 0.5 = 3.5, x_{1} = 1 + (- 0.5) = 0.5.$ Why this step? Two negatives multiply to a plus: stepping left along a downward slope makes $y$ go up. Both sign flips must be respected.

Answer: $y (0.5) \approx 3.5$ .

Verify: Exact solution of $y^{'} = 2 x - y, y (1) = 3$ is $y = 2 x - 2 + 3 e^{1 - x}$ . Check: at $x = 1$ , $y = 2 - 2 + 3 = 3$ ✓. At $x = 0.5$ : $y = 1 - 2 + 3 e^{0.5} = - 1 + 3 (1.6487) = 3.946$ . Euler's single big backward step gave $3.5$ — the right ballpark, and it correctly moved upward despite a negative slope, because $h < 0$ . The two classic mark-losers (destination-slope trap, sign of $h$ ) are both handled.

Recall

Recall Which cell caused the under/overshoot?

Sign of $f$ (rising/falling) decides direction; sign of $y^{''}$ (bowl/dome) decides over- or under-shoot. ::: Convex $y^{''} > 0$ (bowl) ⇒ tangents below ⇒ Euler undershoots; concave $y^{''} < 0$ (dome) ⇒ tangents above ⇒ Euler overshoots.

Recall Stability threshold for

$y^{'} = λ y$ ? What must $h$ satisfy for explicit Euler not to blow up? ::: $∣1 + hλ ∣ \leq 1$ ; for $λ = - 5$ this needs $h \leq 0.4$ .

Recall Negative step size — what changes?

Does the update formula change when $h < 0$ ? ::: No — same formula $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ ; the negative $h$ automatically walks left.

Connections

Separable First-Order ODEs — where the exact comparison curves come from.
Taylor's Theorem — the origin of the "error proportional to $h$ " (first-order) accuracy.
Runge-Kutta Methods (RK4) — how to beat Euler's slow first-order accuracy.
Existence and Uniqueness (Picard–Lindelöf) — guarantees the single curve Euler is chasing exists.
Stability and Stiff Equations — Example 5's blow-up, in full.