Differential-Equations · Unit 2 · Video 1 · Interactive Practice

One Line Element at a Time: Euler's Method and Its Three Equations

IKey Formulas

Formula	Name	What you need
$y' = f(x, y), \qquad y(x_0) = y_0$	Initial value problem (IVP)	An equation and a starting point
$A_n = f(x_n, y_n)$	Slope of the line element where you stand	The right-hand side, evaluated at $(x_n, y_n)$
$x_{n+1} = x_n + h$	Advance $x$	One step size $h$ , the same at every step
$y_{n+1} = y_n + h A_n$	Advance $y$ along the line element	The current $y_n$ and the current slope $A_n$

Key Insight: The three equations are one right triangle. Its run is $h$ , its hypotenuse lies along the line element and so has slope $A_n$ , and therefore its rise must be $h A_n$ . Read backwards, that is the difference quotient $\dfrac{y_{n+1} - y_n}{h} = A_n$ : a slope computed from two points standing in for the derivative.

IIVisualization 1 — One Step Is a Triangle

Run $h$ , hypotenuse of slope $A_n$ — so the rise can only be $h A_n$ .

IIIVisualization 2 — Five Steps, and the Gap They Open

Every slope after the first is measured at a point the true solution never visits.

Steps taken 0.0

Step 1 — from $(0,\ 1)$

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

y_1 = 1 + 0.5(-1) = 0.5

x_1 = 0.5

Step 2 — from $(0.5,\ 0.5)$

A_1 = f(0.5, 0.5) = 0.5 - 0.5 = 0

y_2 = 0.5 + 0.5(0) = 0.5

x_2 = 1

Step 3 — from $(1,\ 0.5)$

A_2 = f(1, 0.5) = 1 - 0.5 = 0.5

y_3 = 0.5 + 0.5(0.5) = 0.75

x_3 = 1.5

Step 4 — from $(1.5,\ 0.75)$

A_3 = f(1.5, 0.75) = 1.5 - 0.75 = 0.75

y_4 = 0.75 + 0.5(0.75) = 1.125

x_4 = 2

Step 5 — from $(2,\ 1.125)$

A_4 = f(2, 1.125) = 2 - 1.125 = 0.875

y_5 = 1.125 + 0.5(0.875) = 1.5625

x_5 = 2.5

💡 A program stores only these six points; the smooth curve a screen shows you is drawn through them afterwards, for display alone.

IVVisualization 3 — What the Step Size Costs

The step size $h$ is yours to choose, and the gap at $x = 2$ is what the choice buys.

Step size h 0.5

💡 Each step starts where the previous one ended, so its error is inherited: the gap at $x = 2$ is the accumulation of all $N$ steps, not the error of the last one.

VQuiz Questions

Problem 1 · One Step of the Recursion

Given: the IVP $y' = x - y$ with $y(0) = 1$ , and step size $h = 0.5$ — find the point $(x_1, y_1)$ produced by the first Euler step.

✅ Correct!

A_0 = f(0,1) = -1

, so the step falls:

y_1 = 1 + 0.5(-1) = 0.5

x_1 = 0.5

❌ Check the sign. The right-hand side is

x - y

, so

f(0,1) = 0 - 1 = -1

. The line element at

(0,1)

points downhill, and the step must follow it down.

❌ Close, but the slope is not the rise. The vertical side of the step triangle is

h A_n

, not

A_n

0.5 \times (-1) = -0.5

, so

y_1 = 1 - 0.5 = 0.5

❌ Check $x_1$ . The first equation is

x_{n+1} = x_n + h

, and

h = 0.5

, not

1

— you moved a full unit along the

x

-axis.

❌ Not quite. Compute

A_0 = f(x_0, y_0)

first, then apply

x_1 = x_0 + h

and

y_1 = y_0 + h A_0

Show solution

Take the three equations in the order the computer takes them — slope first, because the line for $y$ uses it:

A_0 = f(x_0, y_0) = 0 - 1 = -1

x_1 = x_0 + h = 0 + 0.5 = 0.5

y_1 = y_0 + h A_0 = 1 + 0.5(-1) = 0.5

The new point is $(0.5,\ 0.5)$ . Geometrically this is the step triangle: run $0.5$ , hypotenuse of slope $-1$ , so the rise is $0.5 \times (-1) = -0.5$ .

Problem 2 · Where the Next Slope Is Measured

Given: the same walk — $y' = x - y$ , $h = 0.5$ — has reached $(x_1, y_1) = (0.5,\ 0.5)$ . Find the next point $(x_2, y_2)$ .

✅ Correct!

A_1 = f(0.5, 0.5) = 0

, a horizontal line element, so the second step runs flat:

y_2 = y_1 = 0.5

❌ That is the old slope.

A_n

is recomputed at every step:

A_1 = f(0.5, 0.5) = 0

, not the

A_0 = -1

used before.

❌ The slope is measured before the step, not after.

A_1 = f(x_1, y_1) = f(0.5, 0.5)

, not

f(x_2, y_1) = f(1, 0.5)

— the line element belongs to the point you are standing on.

❌ That is the exact solution.

y = x - 1 + 2e^{-x}

gives

y(1) = 2e^{-1} \approx 0.736

, but the broken line left the true curve at the end of step 1 and never returns to it.

❌ Not quite. Evaluate

f

at the point you have reached,

(0.5,\ 0.5)

, then step.

Show solution

Recompute the slope where the last step landed:

A_1 = f(x_1, y_1) = 0.5 - 0.5 = 0

x_2 = 0.5 + 0.5 = 1, \qquad y_2 = 0.5 + 0.5(0) = 0.5

So $(x_2, y_2) = (1,\ 0.5)$ : the second step is horizontal, because the line element at $(0.5,\ 0.5)$ is horizontal.

Notice what has happened. The exact solution through $(0,1)$ is $y = x - 1 + 2e^{-x}$ , which passes through $(0.5,\ 0.713)$ — so $(0.5,\ 0.5)$ is not on it. The slope $A_1 = 0$ is the field's honest slope, but at a point the true solution never visits.

Problem 3 · Reading the Difference Quotient Backwards

Given: one step of Euler's method carried a computation from $(1.2,\ 3)$ to $(1.5,\ 3.6)$ — find the step size $h$ and the slope $A_n$ that step used.

✅ Correct!

h

is the run,

1.5 - 1.2 = 0.3

, and

A_n

is the difference quotient

0.6 / 0.3 = 2

❌ That is the rise, not the slope. The vertical side is

h A_n = 0.6

; divide by

h = 0.3

to recover

A_n

❌ Check which coordinate gives $h$ . The step size is the horizontal advance

x_{n+1} - x_n

;

0.6

is the vertical advance.

❌ The quotient is upside down. Slope is rise over run:

\dfrac{y_{n+1} - y_n}{h}

, not

\dfrac{h}{y_{n+1} - y_n}

❌ Not quite. Use

x_{n+1} = x_n + h

for

h

, then

\dfrac{y_{n+1} - y_n}{h} = A_n

Show solution

The first equation gives the step size directly:

h = x_{n+1} - x_n = 1.5 - 1.2 = 0.3

The second equation, written as a difference quotient, gives the slope:

\frac{y_{n+1} - y_n}{h} = A_n \quad \Longrightarrow \quad A_n = \frac{3.6 - 3}{0.3} = \frac{0.6}{0.3} = 2

Check it forwards: $y_{n+1} = y_n + h A_n = 3 + 0.3(2) = 3.6\ \checkmark$ . The vertical side $0.6$ is $h A_n$ , never $A_n$ itself.

Problem 4 · Whose Slope Is the Second Slope?

Given: for $y' = x - y$ with $h = 0.5$ , Euler's method steps from $(0,\ 1)$ to $(0.5,\ 0.5)$ , while the exact solution through $(0,1)$ has $y(0.5) \approx 0.713$ . The second step therefore uses $A_1 = f(0.5,\ 0.5) = 0$ . Which statement about that slope is correct?

✅ Correct!

A_1

is computed honestly from

f

— but at

(0.5,\ 0.5)

, a point off the true curve, so it is the correct direction for a different integral curve.

❌ $f$ does depend on $y$ here. On the exact solution,

f(0.5,\ 0.713) = 0.5 - 0.713 = -0.213

, not

0

— two different

y

values at the same

x

give two different line elements.

❌ The slope belongs to the point you stand on.

A_n = f(x_n, y_n)

is evaluated before the step; using

f(1,\ 0.5)

would need a point not yet computed.

❌ Only the first corner sits on the true solution. The broken line is tangent to the direction field at every corner, but after step 1 those corners are no longer on the original integral curve.

❌ Not quite. Ask where

A_1

was measured, and whether the exact solution passes through that point.

Show solution

The first step is the only one that begins on the true solution. It ends at $(0.5,\ 0.5)$ , while the exact solution $y = x - 1 + 2e^{-x}$ is at

y(0.5) = 0.5 - 1 + 2e^{-0.5} \approx 0.713

so the broken line is already $0.213$ below the curve. The method then does the only thing it can: it reads the line element where it actually is,

A_1 = f(0.5,\ 0.5) = 0.5 - 0.5 = 0

That number is not an error in arithmetic — it is the exact slope of the integral curve through $(0.5,\ 0.5)$ , which is a different solution of the same equation. Euler's method is exact in direction at every corner and correct as a solution only at the very first one.

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