Differential-Equations · Unit 2 · Video 5 · Interactive Practice

Ask the Direction Field Twice: Improved Euler and Why Halving $h$ Quarters the Error

IKey Formulas

Formula	Name	Role in one step
$A_n = f(x_n, y_n)$	First slope	The field where you stand
$\tilde y_{n+1} = y_n + h\,A_n$	Temporary point	Predict: take the old Euler step
$B_n = f\!\left(x_{n+1},\, \tilde y_{n+1}\right)$	Second slope	Poll the field at that temporary point
$y_{n+1} = y_n + h\,\dfrac{A_n + B_n}{2}$	Improved Euler step	Correct: step with the average

Key Insight: The step in $x$ is untouched, $x_{n+1} = x_n + h$ ; only the slope changes. Euler's error runs like $C_1 h$ , this one like $C_2 h^2$ — so halving the step quarters the error.

IIOne Step, Five Lines

Two readings of the field, one temporary point, and an average that lands near the curve.

Line 1

Line 1 — the slope where you stand

A_n = f(x_n, y_n) \qquad\Longrightarrow\qquad A_0 = f(1,1) = 1

The field here is

y' = y

, so the line element at

(1,1)

has slope

1

— exactly the slope Euler's method uses.

Line 2 — predict: the temporary Euler point

\tilde y_{n+1} = y_n + h\,A_n \qquad\Longrightarrow\qquad \tilde y_1 = 1 + (1)(1) = 2

The tilde is the whole point: this is not

y_1

. It is a scratch value, and on a convex curve it always lands too low.

Line 3 — poll the field again

B_n = f\!\left(x_{n+1},\, \tilde y_{n+1}\right) \qquad\Longrightarrow\qquad B_0 = f(2,2) = 2

This single extra evaluation of

f

is the entire cost of the improvement: two evaluations per step instead of one.

Line 4 — the step in x

x_{n+1} = x_n + h \qquad\Longrightarrow\qquad x_1 = 1 + 1 = 2

Unchanged from Euler's method — the horizontal march never depends on which slope you chose.

Line 5 — correct with the average

y_{n+1} = y_n + h\,\frac{A_n + B_n}{2} \qquad\Longrightarrow\qquad y_1 = 1 + (1)\,\frac{1 + 2}{2} = 2.5

The shallow slope

A_0

undershoots and the steep

B_0

overshoots; the exact value is

e = 2.71828

, so Euler is low by

0.718

and the average by only

0.218

💡 One method, four names: Heun's method, improved Euler, modified Euler, and RK2 — Runge–Kutta with two slope readings per step.

IIIWhat the Second Reading Buys

Start anywhere in the field of $y' = x^2 - y^2$ : how far apart do the two slopes land?

IVThe Order of the Method

Errors against step size, on a logarithmic scale: the steeper line is the second-order method.

Halvings 1

💡 "Second order" describes the method, not the equation: the exponent belongs to $h$ , and the order of the differential equation itself plays no part in it.

VQuiz Questions

Problem 1 · One Improved Euler Step

Given: $y' = x + y$ with $y(0) = 1$ and $h = 0.2$ — find $y_1$ by the improved Euler method.

✅ Correct! The average slope is

\tfrac{1 + 1.4}{2} = 1.2

, so

y_1 = 1 + (0.2)(1.2) = 1.24

— against the exact

1.2428

, while plain Euler's

1.20

is off by more than ten times as much.

❌ That is $\tilde y_1$ , not $y_1$ .

1.20

is the temporary Euler point — the scratch value you feed back into

f

to get

B_0

. The step still has to be retaken with the average.

❌ You stepped with $B_0$ alone.

1 + (0.2)(1.4) = 1.28

uses only the second slope. The method uses the average of

A_0

and

B_0

❌ The division by 2 went missing.

1 + (0.2)(1 + 1.4) = 1.48

adds both slopes instead of averaging them.

❌ Not quite. Run the five lines in order:

A_0

, then

\tilde y_1

, then

B_0

, then

x_1

, then

y_1

Show solution

Line 1 — the slope where you stand:

A_0 = f(0, 1) = 0 + 1 = 1

Line 2 — the temporary Euler point:

\tilde y_1 = 1 + (0.2)(1) = 1.2

Line 3 — poll the field there (at $x_1 = 0.2$ , not at $x_0$ ):

B_0 = f(0.2,\ 1.2) = 0.2 + 1.2 = 1.4

Lines 4 and 5 — advance and correct:

x_1 = 0 + 0.2 = 0.2, \qquad y_1 = 1 + (0.2)\,\frac{1 + 1.4}{2} = 1 + (0.2)(1.2) = \mathbf{1.24}

The exact solution is $y = 2e^x - x - 1$ , giving $y(0.2) = 1.2428$ . Improved Euler misses by $0.0028$ ; Euler's $1.20$ misses by $0.0428$ .

Problem 2 · Where Is the Second Slope Measured?

Given: the improved Euler step for $y' = f(x,y)$ — identify the correct definition of the second slope $B_n$ .

✅ Correct!

B_n

is the field's slope at the far end of the Euler strut — new

x

, temporary

y

. That endpoint is the only place the strut actually reaches.

❌ Circular.

y_{n+1}

is the answer you are still computing; it is produced from

B_n

. The tilde exists precisely to keep these two names apart.

❌ The $x$ never moved. The strut runs from

x_n

x_{n+1}

, so its endpoint is

\left(x_{n+1}, \tilde y_{n+1}\right)

— you kept the old

x

with the new

y

❌ That is a different method. Sampling the field at the midpoint of the strut is the midpoint method (also second order), not the averaging of two endpoint slopes.

❌ Not quite.

B_n

is read where the Euler strut lands: the new

x

, paired with the temporary

y

Show solution

The strut is the plain Euler step. It starts at $(x_n, y_n)$ and ends at

\left(x_{n+1},\, \tilde y_{n+1}\right) = \left(x_n + h,\ y_n + h\,A_n\right).

The second slope is simply the direction field evaluated there:

B_n = f\!\left(x_{n+1},\, \tilde y_{n+1}\right)

The tilde matters. $\tilde y_{n+1}$ is a temporary value used to make one more evaluation of $f$ ; the real new point is the one the averaged slope produces,

y_{n+1} = y_n + h\,\frac{A_n + B_n}{2},

and it is generally different from $\tilde y_{n+1}$ . Writing $f(x_{n+1}, y_{n+1})$ would define $B_n$ in terms of a number that does not exist yet.

Problem 3 · Second Order in Practice

Given: on $y' = x^2 - y^2$ , $y(0) = 1$ , improved Euler's error at $x = 0.2$ is about $1.0 \times 10^{-3}$ when $h = 0.1$ — estimate its error when $h = 0.025$ .

✅ Correct! Two halvings quarter the error twice:

10^{-3} \div 16 = 6.3 \times 10^{-5}

. The measured value is

6.1 \times 10^{-5}

❌ Only one halving, and only a halving. Going from

h = 0.1

h = 0.025

is two halvings, and each one divides a second-order error by

4

, not by

2

❌ That is first-order behaviour. Dividing by

2

per halving is Euler's rule. Here

e \approx C_2 h^2

, so each halving divides by

4

❌ One halving too few.

1.3 \times 10^{-4}

is roughly the error at

h = 0.05

; halving once more quarters it again.

❌ Not quite. Count the halvings from

0.1

0.05

0.025

, then apply the factor a second-order method promises for each one.

Show solution

Improved Euler is second order, so its error behaves like

e \approx C_2 h^2 .

Replacing $h$ by $h/2$ replaces $h^2$ by $h^2/4$ :

C_2\left(\frac{h}{2}\right)^2 = \frac{1}{4}\,C_2 h^2 .

From $h = 0.1$ to $h = 0.025$ is two halvings, so the error is divided by $4 \times 4 = 16$ :

e \approx \frac{1.0 \times 10^{-3}}{16} = 6.3 \times 10^{-5}.

Measured values on this equation: $1.0 \times 10^{-3}$ at $h = 0.1$ , then $2.5 \times 10^{-4}$ , then $6.1 \times 10^{-5}$ — ratios of about $4$ each time.

Euler's method on the same problem goes $1.58 \times 10^{-2}$ , $7.5 \times 10^{-3}$ , $3.7 \times 10^{-3}$ : ratios of about $2$ , because $e \approx C_1 h$ .

Problem 4 · The Whole Step, Start to Finish

Given: $y' = y - x$ with $y(0) = 2$ and $h = 0.5$ , so that $A_0 = 2$ and $\tilde y_1 = 3$ — find the second slope and then the improved value of $y_1$ .

What is the second slope $B_0$ ?

What is $y_1$ ?

✅ Correct!

B_0 = 3 - 0.5 = 2.5

, the average is

2.25

, and

y_1 = 2 + (0.5)(2.25) = 3.125

— against the exact

3.1487

, where Euler's

3.000

is more than five times further off.

❌ You reported $\tilde y_1$ , not $f$ of it.

B_0

is the field's slope at

(0.5,\, 3)

f = y - x = 3 - 0.5 = 2.5

❌ That is $A_0$ again.

2

is the slope back at

(0, 2)

. The second reading is taken at the far end of the strut, where both coordinates have moved.

❌ Only $x$ advanced.

f(0.5,\, 2) = 1.5

keeps the old

y

. The strut carried

y

up to

\tilde y_1 = 3

as well.

❌ Not quite. Evaluate

f(x,y) = y - x

at the strut's endpoint

\left(x_1, \tilde y_1\right) = (0.5,\, 3)

❌ That is plain Euler.

2 + (0.5)(2) = 3

uses

A_0

alone and ignores the second reading entirely.

❌ You stepped with $B_0$ alone.

2 + (0.5)(2.5) = 3.25

swaps one single slope for another instead of averaging the two.

❌ The average was never taken.

2 + (0.5)(2 + 2.5) = 4.25

adds the slopes; the formula divides that sum by

2

❌ Not quite. Average the two slopes first, then take one step of width

h

with that average.

Show solution

Here $f(x,y) = y - x$ , $x_0 = 0$ , $y_0 = 2$ , $h = 0.5$ .

Lines 1 and 2 (given): $A_0 = f(0,2) = 2 - 0 = 2$ , so $\tilde y_1 = 2 + (0.5)(2) = 3$ .

Line 3 — the second slope, at the strut's endpoint $(x_1, \tilde y_1) = (0.5,\, 3)$ :

B_0 = f(0.5,\ 3) = 3 - 0.5 = \mathbf{2.5}

Lines 4 and 5 — advance and correct:

x_1 = 0.5, \qquad \frac{A_0 + B_0}{2} = \frac{2 + 2.5}{2} = 2.25

y_1 = 2 + (0.5)(2.25) = \mathbf{3.125}

Check against the exact solution. $y = x + 1 + e^{x}$ satisfies $y' = 1 + e^x = y - x$ and $y(0) = 2$ , so

y(0.5) = 0.5 + 1 + e^{0.5} = 3.1487.

Euler's $3.000$ is low by $0.1487$ ; improved Euler's $3.125$ is low by $0.0237$ — about six times closer for one extra evaluation of $f$ . Both fall short because $y'' = e^{x} > 0$ makes the solution convex.

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Ask the Direction Field Twice: Improved Euler and Why Halving hhh Quarters the Error

IKey Formulas

IIOne Step, Five Lines

IIIWhat the Second Reading Buys

IVThe Order of the Method

VQuiz Questions

Ask the Direction Field Twice: Improved Euler and Why Halving $h$ Quarters the Error