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

Every Zero of f Is a Barrier: Critical Points and the Two-Step Method

IKey Formulas

Formula	Name	What it gives
$\dfrac{dy}{dt} = f(y)$	Autonomous equation	The slope depends on $y$ alone, never on $t$
$f(y_0) = 0 \iff y \equiv y_0 \text{ is a solution}$	Critical point (Step 1)	Every horizontal solution, found by algebra with no calculus
$\dfrac{d(y_0)}{dt} = f(y_0) \iff 0 = f(y_0)$	The one-line proof	On the left $y_0$ is a constant function; on the right it is a number
$f(y) > 0 \Rightarrow y \text{ increasing}, \quad f(y) < 0 \Rightarrow y \text{ decreasing}$	The sign of $f$ (Step 2)	The direction of every solution between, above and below the barriers

Key Insight: The roots of $f$ cut the $(t, y)$ plane into bands. Each root is a constant solution, and no integral curve may cross or even touch another, so a solution stays in the band it starts in, and the sign of $f$ on that band says which way it moves.

💡 The fine print: the barrier is the uniqueness theorem at work, so it needs $f'(y)$ continuous, as it is for every polynomial on this page. Without that a barrier can leak: for $dy/dt = 3y^{2/3}$ from Unit 1, both $y \equiv 0$ and $y = t^3$ pass through $(0, 0)$ , and $y = t^3$ runs straight through the constant solution.

IIWhich Horizontal Lines Are Solutions?

A horizontal line $y \equiv c$ solves $dy/dt = f(y)$ exactly when $f(c) = 0$ , because its own slope is always zero.

Height c 2

💡 Challenge: find every height at which the field lies flat along the line, then confirm the list by solving $(c - 1)(3 - c) = 0$ .

IIITwo Steps Fix Every Curve

No solving needed: the roots of $f$ place the barriers, and the sign of $f$ gives every curve its direction.

dy/dt = y(0) 1.6

IVRead the Arrows Off the Graph

From the graph of $f$ alone, which way does $y$ move on each interval between the critical points?

Graph

VQuiz Questions

Problem 1 · A Constant Solution on Demand

Given: $\dfrac{dy}{dt} = y^2 - ay + 2$ , with $a$ a constant — find the value of $a$ for which the constant function $y \equiv 2$ is a solution.

✅ Correct! Plugging in

y \equiv 2

makes the left side

0

and the right side

f(2) = 6 - 2a

, so the equation holds exactly when

a = 3

. Then

f(y) = (y - 1)(y - 2)

, and

y \equiv 1

is a second constant solution.

❌ The constant term was dropped.

f(2) = 2^2 - 2a + 2 = 6 - 2a

, not

4 - 2a

. With

a = 2

f(2) = 2 \neq 0

❌ Check the sign of the middle term.

-a \cdot 2 = -2a

, so

f(2) = 6 - 2a

, which vanishes at

a = +3

. With

a = -3

f(2) = 4 + 6 + 2 = 12

❌ That checks only the left side.

\dfrac{d(2)}{dt} = 0

whatever

a

is, but the equation needs the right side to match:

f(2) = 6 - 2a

must be

0

as well.

❌ Not quite. A constant

y_0

is a solution exactly when

f(y_0) = 0

. Evaluate

f(2)

and set it to zero.

Show solution

To say $y \equiv 2$ is a solution is to say that plugging it in works:

\frac{d(2)}{dt} = 2^2 - a(2) + 2

The left side is $0$ , because the derivative of a constant is zero. The right side is the number $f(2) = 6 - 2a$ . So the equation holds exactly when

6 - 2a = 0 \quad\Longrightarrow\quad a = 3

Check: with $a = 3$ , $f(y) = y^2 - 3y + 2 = (y - 1)(y - 2)$ , so $f(2) = 0$ ✅. The equation then has two critical points, $1$ and $2$ , and two horizontal solutions, $y \equiv 1$ and $y \equiv 2$ .

Problem 2 · Following One Solution

Given: $\dfrac{dy}{dt} = 9 - y^2$ with $y(0) = 1$ — which statement describes $y(t)$ for $t \ge 0$ ?

✅ Correct!

f(1) = 9 - 1 = 8 > 0

, so

y

increases, and the constant solution

y \equiv 3

is a barrier just above it: the curve climbs toward

3

and never gets there.

❌ That reads the slope of the graph of $f$ , not its sign.

f

is indeed decreasing at

y = 1

, but

dy/dt

is the value

f(1) = 8 > 0

, so

y

increases.

❌ $y = 3$ is a critical point, since

9 - 3^2 = 0

. So

y \equiv 3

is itself an integral curve, and no other integral curve can cross it.

❌ Reaching $3$ would mean touching the integral curve $y \equiv 3$ , putting two integral curves through one point. Integral curves cannot even touch, so the curve only approaches

3

t \to \infty

❌ Not quite. Find the critical points, then read the sign of

f

at the starting value.

Show solution

Step 1 — critical points. $9 - y^2 = (3 - y)(3 + y) = 0$ gives $y = -3$ and $y = 3$ : two horizontal solutions, two barriers.

Step 2 — sign of $f$ . $(3 - y)(3 + y) > 0$ for $-3 < y < 3$ , and $< 0$ outside. The start $y(0) = 1$ lies between the barriers, where $f > 0$ , so $y$ increases, and since it cannot cross or touch $y \equiv 3$ it flattens out toward $3$ .

The slope of the graph of $f$ plays no role: $f'(1) = -2 < 0$ only says the speed $dy/dt$ shrinks as $y$ grows, which is why the curve flattens.

Check by solving (not needed): $y = 3\tanh(3t + C)$ with $\tanh C = \tfrac{1}{3}$ satisfies $y' = 9\,\mathrm{sech}^2(3t + C) = 9 - y^2$ , and it rises toward $3$ without ever reaching it.

Problem 3 · Trapped in a Band

Given: $\dfrac{dy}{dt} = y^3 - 4y$ with $y(0) = 1$ — find the band that traps the solution and what it does in that band.

Between which constant solutions is $y(t)$ trapped?

What does $y(t)$ do as $t$ increases?

✅ Correct! The critical points are

-2

0

and

2

, so

y(0) = 1

sits in the band

0 < y < 2

. There

f(1) = -3 < 0

, so the curve falls toward the barrier

y \equiv 0

and never reaches it.

❌ Dividing $y^3 = 4y$ by $y$ throws away the root $y = 0$ . Factor instead:

y^3 - 4y = y(y - 2)(y + 2)

, so

y \equiv 0

is a barrier too, and it lies between

-2

and the start

1

❌ $1$ is where the solution starts, not a critical point:

f(1) = 1 - 4 = -3 \neq 0

. Barriers are roots of

f

, never initial values.

❌ $y = 2$ is a root as well:

2^3 - 4(2) = 0

. So

y \equiv 2

fences the curve in from above too.

❌ Check the sign.

f(1) = 1^3 - 4(1) = -3 < 0

, so

dy/dt < 0

and

y

decreases.

❌ $y \equiv 0$ is a constant solution, since

0^3 - 4 \cdot 0 = 0

, so it is a barrier the curve cannot cross. Falling on toward

-2

is what you get after dividing by

y

and losing that root.

❌ Reaching $0$ at a finite time would put two integral curves, the solution and

y \equiv 0

, through one point. The curve gets arbitrarily close to

0

but never arrives.

❌ Not that band. The barriers are the roots of

y^3 - 4y

; factor it completely.

❌ Not quite. Read the sign of

f

at the starting value, and remember that barriers cannot be crossed or touched.

Show solution

Step 1 — critical points. Factor completely, never divide by $y$ :

y^3 - 4y = y(y^2 - 4) = y(y - 2)(y + 2) = 0 \quad\Longrightarrow\quad y = -2,\ 0,\ 2

Step 2 — sign of $f$ , one factor at a time:

Interval	Signs of $y,\ y - 2,\ y + 2$	$f(y)$	$y(t)$
$y < -2$	$-,\ -,\ -$	$< 0$	decreasing
$-2 < y < 0$	$-,\ -,\ +$	$> 0$	increasing
$0 < y < 2$	$+,\ -,\ +$	$< 0$	decreasing
$y > 2$	$+,\ +,\ +$	$> 0$	increasing

The start $y(0) = 1$ lies in the band $0 < y < 2$ , trapped between the barriers $y \equiv 0$ and $y \equiv 2$ . There $f < 0$ (indeed $f(1) = -3$ ), so $y$ decreases, and since it cannot cross or touch $y \equiv 0$ it flattens out toward $0$ without reaching it.

Problem 4 · Reading Backwards From a Solution

Given: a solution of $\dfrac{dy}{dt} = f(y)$ , with $f$ a polynomial, has $y(0) = 1$ and $y(5) = 4$ — what must be true of $f$ on $[1, 4]$ ?

✅ Correct! The curve passes through every height from

1

4

, so none of those heights can be a critical point: that barrier would be crossed or touched. So

f

never vanishes on

[1, 4]

, keeps one sign there, and since

y

went up, that sign is

+

❌ The endpoints cannot be critical points either. If

f(1) = 0

, then

y \equiv 1

is a solution through the point

(0, 1)

, and so is this curve: two integral curves through one point. The curve would have to be

y \equiv 1

and could never reach

4

❌ A zero $c$ in $(1, 4)$ makes $y \equiv c$ a barrier, and a curve going from

1

4

must cross height

c

. Solutions of an autonomous equation never turn around: a turning point would need

dy/dt = f(y) = 0

there, and that height is a constant solution.

❌ To turn negative after being positive at $1$ , the polynomial

f

must pass through

0

in between, and that zero would be a barrier lying across the curve's path.

❌ Not quite. Ask which heights in

[1, 4]

could possibly be critical points, given that the curve passes through all of them.

Show solution

No critical point in $[1, 4]$ . Suppose $f(c) = 0$ for some $c$ with $1 \le c \le 4$ . The solution is continuous and runs from $y(0) = 1$ to $y(5) = 4$ , so it takes the value $c$ at some time $t^*$ in $[0, 5]$ . Then the solution and the constant solution $y \equiv c$ both pass through the point $(t^*, c)$ . Two integral curves cannot touch, so the solution is $y \equiv c$ , which contradicts $y(0) = 1 \neq 4 = y(5)$ .

One sign on $[1, 4]$ . A continuous $f$ with no zero on $[1, 4]$ cannot change sign there.

That sign is $+$ . $y$ rose from $1$ to $4$ , so at some moment it was climbing at a height inside $[1, 4]$ , and there $dy/dt = f(y) > 0$ . Hence

f(y) > 0 \quad \text{for every } y \in [1, 4].

This is the barrier argument run backwards: a solution that climbs from $1$ to $4$ proves there are no barriers in between and that the arrows there all point toward increasing $y$ .

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