Differential-Equations · Unit 2 · Video 3 · Interactive Practice

Convex Means Too Low: How the Equation Itself Grades Euler's Answer

IKey Formulas

Formula	Name	What it gives
$y' = x^2 - y^2, \quad y(0) = 1$	The initial value problem	No elementary solution
$y'' = 2x - 2y\,y'$	Differentiate the equation	$y''$ from $x$ , $y$ , $y'$ alone
$y''(0) = 2(0) - 2(1)(-1) = 2$	Evaluate at the start	A positive number, so convex
$y'' > 0 \Rightarrow \text{convex} \Rightarrow \text{Euler too low}$	The corner-cutting rule	$y'' < 0$ reverses both arrows

Key Insight: The solution is never found — only the sign of $y''$ at the starting point, which the differential equation hands over for free. That sign decides which side of the true curve the broken line falls on.

IIWhich Side Does the Broken Line Miss On?

Euler's broken line always cuts the corner; the sign of $y''$ decides which corner.

Curvature y″ = k 1/2

The field here depends on $x$ alone, so each strut is exactly parallel to the curve's tangent at that $x$ ; for a general $f(x,y)$ that parallelism is only approximate, and the corner-cutting is the same.

IIIReading the Sign Off the Equation

For $y' = x^2 - y^2$ , the starting point alone fixes $y'' = 2x - 2y\,y'$ and the verdict.

Start x₀ 0 Start y₀ 1

Crossing the dashed curve flips the verdict — which is why a sign computed at one point is trustworthy only near that point.

IVTwo Steps, Then the Verdict

The equation grades its own approximation: $y''(0) = 2 > 0$ , so $0.82$ is too small.

Step 1

Step 1 — The slope at the start

y' = x^2 - y^2, \qquad y(0) = 1

y'(0) = 0^2 - 1^2 = -1

The first strut leaves

(0, 1)

along the line element of slope

-1

Step 2 — Two Euler steps, h = 0.1

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

y'(0.1,\, 0.9) = 0.01 - 0.81 = -0.8

y_2 = 0.9 + (0.1)(-0.8) = 0.82

Step 3 — Differentiate the equation

y'' = \frac{d}{dx}\left(x^2 - y^2\right) = 2x - 2y\,y'

Since

y

is itself a function of

x

, the term

y^2

needs the chain rule. The right-hand side asks only for

x

y

and

y'

— never for a solution formula.

Step 4 — Evaluate at the start point

y_1''(0) = 2(0) - 2(1)(-1) = 2 > 0

A positive second derivative bends the solution up off its tangent: locally

y \approx 1 - x + x^2

, the dashed curve.

Step 5 — The verdict

\text{convex} \Rightarrow \text{broken line below} \Rightarrow y(0.2) > 0.82

A numerical reference gives

y(0.2) = 0.8358

, low by

0.0158

— but the inequality came from the equation, before any curve was drawn.

VQuiz Questions

Problem 1 · Differentiate the Equation

Given: $y' = x^2 - y^2$ , where $y$ is a solution — find $y''$ by differentiating both sides with respect to $x$ .

✅ Correct!

y

is a function of

x

, so

\left(y^2\right)' = 2y\,y'

. The right-hand side asks only for

x

y

and

y'

— all available at the starting point.

❌ The chain rule is missing.

y

depends on

x

, so

\left(y^2\right)' = 2y\,y'

, not

2y

❌ Check the sign. The term being differentiated is

-y^2

, so it contributes

-2y\,y'

❌ Not quite. Differentiate term by term:

\left(x^2\right)' = 2x

and

\left(y^2\right)' = 2y\,y'

Show solution

Differentiate both sides of $y' = x^2 - y^2$ with respect to $x$ :

y'' = \frac{d}{dx}\left(x^2\right) - \frac{d}{dx}\left(y^2\right) = 2x - 2y\,y'

The second term needs the chain rule because $y = y(x)$ : $\dfrac{d}{dx}\left(y^2\right) = 2y \cdot \dfrac{dy}{dx} = 2y\,y'$ .

Nothing here mentions a formula for the solution. Given a point $(x, y)$ , the original equation supplies $y'$ , and this identity then supplies $y''$ .

Problem 2 · Same Equation, New Starting Point

Given: $y' = x^2 - y^2$ with $y(0) = -1$ . Two Euler steps with $h = 0.1$ give $y(0.2) \approx -1.22$ . Which statement about the true value is justified?

✅ Correct!

y'(0) = -1

and

y''(0) = 2(0) - 2(-1)(-1) = -2 < 0

: the solution is concave, so the broken line runs above it. The true value is

\approx -1.2470

❌ That is the convex verdict. Here

y''(0) = -2 < 0

— concave — so Euler runs too high, which puts the true value below the estimate.

❌ The equation does decide it.

y'' = 2x - 2y\,y'

is computable at

(0, -1)

without any solution formula.

❌ Not quite. Evaluate

y'' = 2x - 2y\,y'

(0, -1)

, using

y'(0) = 0^2 - (-1)^2 = -1

Show solution

Step 1 — the slope at the start:

y'(0) = 0^2 - (-1)^2 = -1

Step 2 — the second derivative at the start:

y''(0) = 2(0) - 2(-1)(-1) = 0 - 2 = -2 < 0

Step 3 — apply the rule. A negative second derivative means the solution is concave there: it bends down and away from every tangent, so each strut lands above the curve and the misses accumulate upward.

\text{concave} \Rightarrow \text{Euler too high} \Rightarrow y(0.2) < -1.22

The same equation, a different starting point, and the verdict reverses — the sign of $y''$ is what does the deciding, not the equation alone. (A numerical solution gives $-1.2470$ , indeed below $-1.22$ .)

Problem 3 · A Different Equation

Given: $y' = x - y^2$ with $y(0) = 1$ — find the second derivative at the start, then say which side an Euler estimate of $y(0.2)$ falls on.

What is the second derivative at the start?

Which side does the Euler estimate fall on?

✅ Correct!

y'(0) = -1

and

y'' = 1 - 2y\,y'

, so

y''(0) = 1 - 2(1)(-1) = 3 > 0

: convex, and the broken line cuts the corner from below.

❌ That is the chain-rule slip.

\left(y^2\right)' = 2y\,y'

, not

2y

— using

2y

gives

1 - 2 = -1

instead of

1 - 2(1)(-1) = 3

❌ Check the differentiation. Differentiating

y' = x - y^2

gives

y'' = 1 - 2y\,y'

; now substitute

y(0) = 1

and

y'(0) = 0 - 1^2 = -1

❌ Check the direction of the rule.

y''(0) = 3 > 0

means convex, and a convex solution bends up away from every strut.

Show solution

Step 1 — the slope at the start:

y'(0) = 0 - 1^2 = -1

Step 2 — differentiate the equation:

y'' = \frac{d}{dx}\left(x - y^2\right) = 1 - 2y\,y'

Step 3 — evaluate at $(0, 1)$ :

y''(0) = 1 - 2(1)(-1) = 1 + 2 = 3 > 0

Step 4 — read off the verdict. Positive second derivative means convex, and a convex solution bends up away from each line element, so the struts stay underneath:

\text{convex} \Rightarrow \text{Euler too low}

Two traps live in this problem: dropping the chain rule gives $1 - 2 = -1$ and flips the verdict, and forgetting that $x$ differentiates to $1$ gives $2$ instead of $3$ .

Problem 4 · How Far Does the Verdict Reach?

Given: for $y' = x^2 - y^2$ , $y(0) = 1$ we found $y''(0) = 2 > 0$ and concluded $y(0.2) > 0.82$ . A numerical solution shows $y''$ stays positive until $x \approx 2.05$ and is negative after that. What is the honest scope of the argument?

✅ Correct! The sign was computed at one point, so convexity is certified only near it. Past an inflection the tangents fall above the curve, the broken line starts catching up, and it can end on either side.

❌ Too strong. Beyond

x \approx 2.05

this solution is concave, and there the struts run above it instead.

❌ A sign at a point is not a sign everywhere.

y''(0) > 0

fixes the bend at the start; here

y''

turns negative near

x \approx 2.05

❌ Too pessimistic.

y''

is continuous along the solution, so its sign at the start persists on a neighbourhood — one that comfortably contains

[0,\,0.2]

❌ Not quite. Ask what a single evaluation of

y''

can certify, and over what stretch of

x

Show solution

The computation produced one number: $y_1''(0) = 2$ . Because $y'' = 2x - 2y\,y'$ is continuous along the solution, a positive value at $x = 0$ stays positive on an interval around $0$ — so the solution is convex there and the broken line runs below it. That interval is exactly what the argument certifies.

Why the caveat matters. If the solution changes from convex to concave, the tangents past the inflection run above the curve, the broken line begins closing the gap, and it can cross to the other side. Nothing in the starting-point calculation announces where that happens; here a numerical solution locates the inflection near $x \approx 2.05$ — ten times beyond the interval $[0,\,0.2]$ that was computed.

The recipe, stated generally. For any $y' = f(x,y)$ , differentiating the equation along the solution gives $y''$ in terms of $x$ , $y$ and $y'$ ; its sign at the starting point tells you which side Euler lands on nearby — without ever calculating the solution.

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