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

Make a Table: Euler's Method by Hand, y(0) = 1 to y(0.2) ≈ 0.82

IKey Formulas

Formula	Name	What you need
$x_{n+1} = x_n + h$	Step in $x$	The current $x_n$ and the step size $h$
$A_n = f(x_n,\, y_n)$	Slope at the current point	The right-hand side, evaluated at $(x_n, y_n)$
$y_{n+1} = y_n + hA_n$	Step in $y$	The current $y_n$ and the product $hA_n$
$A_n = x_n^2 - y_n^2$	Specialized to this IVP	$y' = x^2 - y^2$ , $\ y(0) = 1$ , $\ h = 0.1$

Key Insight: $hA_n$ earns a column of its own because the formula adds the product, not the slope: with $h = 0.1$ a slope of $-1$ moves $y$ by only $-0.1$ .

IIVisualization 1 — Building the Table, Row by Row

Two steps of $h = 0.1$ carry $y(0) = 1$ to $y(0.2) \approx 0.82$ , one table entry at a time.

Step 1

$n$	$x_n$	$y_n$	$A_n$	$hA_n$
0
1
2

💡 $y_1 = 0.9$ is Euler's approximation to $y(0.1)$ , not its exact value. As far as anyone knows $y' = x^2 - y^2$ has no elementary solution to check it against — which is why a numerical method is the right tool here.

IIIVisualization 2 — Stepping Along the Direction Field

Each segment leaves its left-hand point along the line element there, wherever the initial condition sits.

IVVisualization 3 — Predict the Next Two Rows

Two more rows carry the table from $x = 0.2$ to $x = 0.4$ — where does the broken line land?

Your guess for y(0.4) 0.78

$n$	$x_n$	$y_n$
2	0.2	0.82
3
4

VQuiz Questions

Problem 1 · The First Slope Entry

Given: $y' = x^2 - y^2$ with $y(0) = 1$ and $h = 0.1$ . Row 0 of the table holds $(x_0, y_0) = (0, 1)$ — find the slope entry $A_0$ .

✅ Correct!

0^2 - 1^2 = -1

: the solution starts out heading downhill at

45\degree

❌ Check the order of the subtraction. The right-hand side is

x^2 - y^2

, not

y^2 - x^2

❌ That is the next column.

-0.1

is the product

hA_0

;

A_0

itself is the slope, before multiplying by

h

❌ Not quite. Evaluate

x^2 - y^2

at the row-0 point

(0, 1)

— both coordinates come from that row.

Show solution

The slope column is the right-hand side of the equation evaluated at the current row's point. With $f(x, y) = x^2 - y^2$ and $(x_0, y_0) = (0, 1)$ :

A_0 = x_0^2 - y_0^2 = 0^2 - 1^2 = 0 - 1 = -1

$A_0$ is a slope, not a change in $y$ . The change in $y$ lives in the next column, $hA_0 = (0.1)(-1) = -0.1$ .

Problem 2 · What the Formula Adds

Given: row 0 is complete — $x_0 = 0$ , $y_0 = 1$ , $A_0 = -1$ , $hA_0 = -0.1$ , with $h = 0.1$ . Find $y_1$ .

✅ Correct!

y_1 = 1 + (-0.1) = 0.9

— the last column is exactly the amount

y

changes.

❌ You added $A_0$ , not $hA_0$ .

1 + (-1) = 0

is a drop ten times too large; the formula adds the product

hA_0 = -0.1

❌ Check the sign.

A_0 = -1

is negative, so

hA_0 = -0.1

and

y

must decrease.

❌ Not quite.

y_1 = y_0 + hA_0

: start from

y_0 = 1

and add the

hA_0

entry.

Show solution

The step in $y$ adds the last column of the row to the current $y$ :

y_1 = y_0 + hA_0 = 1 + (0.1)(-1) = 1 - 0.1 = 0.9

This is exactly why $hA_n$ gets a column of its own — the product, not the slope, is what the formula adds. Adding $A_0 = -1$ directly would give $y_1 = 0$ , a drop ten times too big.

The new point is $(x_1, y_1) = (0.1, 0.9)$ .

Problem 3 · Row One, Both Working Columns

Given: $y' = x^2 - y^2$ , $h = 0.1$ , and row 1 holding $(x_1, y_1) = (0.1, 0.9)$ — find the slope $A_1$ (to two decimal places) and then $y_2$ .

What is $A_1$ ?

What is $y_2$ ?

✅ Correct!

A_1 = 0.01 - 0.81 = -0.80

and

y_2 = 0.9 - 0.08 = 0.82

, so

y(0.2) \approx 0.82

❌ You kept the old $y$ .

0.01 - 1 = -0.99

uses

y_0 = 1

; the slope must be evaluated at the new point, with

y_1 = 0.9

❌ That is $hA_1$ , not $A_1$ .

-0.08

is the slope already multiplied by

h = 0.1

❌ Check $A_1$ . Square both new coordinates:

(0.1)^2 = 0.01

and

(0.9)^2 = 0.81

, then subtract in that order.

❌ You reused the first slope.

0.9 + (0.1)(-1) = 0.80

recycles

A_0

; every row gets its own slope, and here

A_1 = -0.80

❌ You added $A_1$ instead of $hA_1$ .

0.9 + (-0.80) = 0.10

; multiply by

h

first:

hA_1 = -0.08

❌ Check $y_2$ .

y_2 = y_1 + hA_1

, with

y_1 = 0.9

and

hA_1 = (0.1)(-0.80)

Show solution

Step 1 — the slope, re-evaluated at the new point. Both coordinates are new:

A_1 = x_1^2 - y_1^2 = (0.1)^2 - (0.9)^2 = 0.01 - 0.81 = -0.80

Step 2 — the working column.

hA_1 = (0.1)(-0.80) = -0.08

Step 3 — the step in $y$ .

y_2 = y_1 + hA_1 = 0.9 + (-0.08) = 0.82

Two steps of size $0.1$ have carried the table from $x = 0$ to $x = 0.2$ , so $y(0.2) \approx 0.82$ .

Common mistakes:

Keeping $y_0 = 1$ in the slope: $0.01 - 1 = -0.99$ .
Reusing $A_0 = -1$ for the second step: $y_2 = 0.9 - 0.1 = 0.80$ .

Problem 4 · A New Equation, Two Steps

Given: $y' = x + y$ with $y(0) = 1$ and $h = 0.2$ — find the Euler approximation to $y(0.4)$ after two steps.

✅ Correct!

A_0 = 1

A_1 = 1.4

, and

y_2 = 1.2 + 0.28 = 1.48

. The same five columns work for any right-hand side.

❌ You froze the slope.

1 + 0.2 + 0.2 = 1.40

reuses

A_0 = 1

twice; row 1 needs its own slope,

A_1 = 1.4

❌ You evaluated $A_1$ at the old $y$ .

A_1 = x_1 + y_1 = 0.2 + 1.2 = 1.4

, not

0.2 + 1 = 1.2

❌ That is $y_1$ . With

h = 0.2

one step reaches only

x = 0.2

;

x = 0.4

takes a second row.

❌ Not quite. Build the table:

A_n = x_n + y_n

, then

hA_n = 0.2A_n

, then

y_{n+1} = y_n + hA_n

Show solution

Only the slope column changes: here $A_n = x_n + y_n$ , with $h = 0.2$ .

Row 0 — $(x_0, y_0) = (0, 1)$ :

A_0 = 0 + 1 = 1, \qquad hA_0 = (0.2)(1) = 0.2

Row 1 — $x_1 = 0 + 0.2 = 0.2$ and $y_1 = 1 + 0.2 = 1.2$ :

A_1 = 0.2 + 1.2 = 1.4, \qquad hA_1 = (0.2)(1.4) = 0.28

Row 2 — $x_2 = 0.4$ :

y_2 = y_1 + hA_1 = 1.2 + 0.28 = 1.48

So $y(0.4) \approx 1.48$ . Every entry to the right of $y_n$ was built from entries already in its own row, which is what makes the table checkable.

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