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

Reading Stability Off the Arrows: Why a Tangency Is Half Stable, and What a Constant Harvest Adds

IKey Formulas

Formula	Name	What it tells you
$\dfrac{dy}{dt} = f(y), \quad f(y_0) = 0$	Critical point $y_0$	A single level on the $y$ -axis, carrying the constant solution $y \equiv y_0$
signs of $f$ just below and just above $y_0$ : $(+,\,-)$ , $(-,\,+)$ , or the same sign twice	Stable, unstable, semi-stable	Arrows point in from both sides; out on both sides; the same way on both sides
$ay - by^2 = y\,(a - by)$	Logistic right-hand side	$y = 0$ is unstable, $y = a/b$ is stable
$\dfrac{dy}{dt} = ay - by^2 - h$	Logistic model with constant harvesting	$h$ is a fixed weight per day, in lb/day; a term $hy$ would only lower the growth rate from $a$ to $a - h$

Key Insight: Classifying a critical point never needs a formula for the solution, only the sign of $f$ on each side of it. A graph that crosses the axis gives a stable or an unstable point; a graph that touches the axis without crossing keeps one sign on both sides, and that gives a semi-stable point: stable from one side, unstable from the other.

IIThe Arrows Predict the Curves

The sign of $f$ at the starting level is the direction the solution moves, so the arrows predict every curve.

Right-hand side Start y(0) 1

the graph of the right-hand side

the solution curves

For a smooth $f$ two different solution curves never meet, so no solution can cross a constant one: a curve heading for a critical level can only flatten against it.

IIIThree Kinds of Critical Point

Stable, unstable or semi-stable: the signs of $f$ just left and just right of each root decide.

Graph

IVWhere Growth Meets the Harvest

Critical points sit where growth $ay - by^2$ meets the harvest: a flat line for $h$ , a sloped line for $hy$ .

Harvest Stock y 1500 lb

Raising $h$ lifts the flat line and slides its two crossings toward each other, until the line just touches the growth curve: a tangency, and so a semi-stable point. How large $h$ can be before that happens is the next video's question.

VQuiz Questions

Problem 1 · Read the Signs

Given: $\dfrac{dy}{dt} = y^2 - 3y$ — classify its two critical points.

✅ Correct!

f(y) = y\,(y - 3)

is positive for

y < 0

, negative between the roots and positive for

y > 3

, so the arrows run toward

0

from both sides and away from

3

on both sides.

❌ That is the logistic pattern, but this parabola opens upward. Between the roots

f(1) = -2 < 0

, so solutions there decrease: toward

0

and away from

3

❌ An upward parabola does not push everything away. Just below

0

f(-1) = 4 > 0

, so solutions there rise toward

0

; the arrows face

0

from both sides.

❌ Not quite. Find the sign of

f(y) = y\,(y - 3)

in each of the three regions, then draw the arrows.

Show solution

Factor the right-hand side: $f(y) = y^2 - 3y = y\,(y - 3)$ , so the critical points are $y = 0$ and $y = 3$ . Test one point in each region:

f(-1) = 4 > 0, \qquad f(1) = -2 < 0, \qquad f(4) = 4 > 0

At $y = 0$ : $f > 0$ below and $f < 0$ above, so the arrows point in from both sides, and $y = 0$ is stable.
At $y = 3$ : $f < 0$ below and $f > 0$ above, so the arrows point out on both sides, and $y = 3$ is unstable.

This is the logistic picture turned upside down: the graph is an upward parabola, so every arrow reverses and the two roots swap roles.

Problem 2 · The Unstable Side of a Tangency

Given: $\dfrac{dy}{dt} = -(y - 2)^2$ with $y(0) = 1.9$ — what does $y$ do as $t$ increases?

✅ Correct!

f(1.9) = -(1.9 - 2)^2 = -0.01 < 0

, and

f < 0

for every

y < 2

, so the solution keeps falling. The level

y = 2

is semi-stable, stable from above and unstable from below, and

1.9

sits on the unstable side.

❌ Only a start above $2$ , on the stable side, approaches $2$ , and it gets there by falling. The graph touches the axis from below, so

f < 0

on both sides of

2

: below

2

the arrow points away from it, toward decreasing

y

❌ Two things go wrong.

f(1.9) < 0

, so

y

starts by decreasing; and no solution can cross the constant solution

y \equiv 2

❌ $f(1.9) = -0.01$ is small, but it is not zero. Only the critical level

y = 2

carries a constant solution.

❌ Not quite. Find the sign of

f

on each side of

y = 2

Show solution

The right-hand side $f(y) = -(y - 2)^2$ is zero only at $y = 2$ and negative everywhere else: its graph touches the axis from below. So $y = 2$ is semi-stable, stable from above and unstable from below.

The start $y(0) = 1.9$ is below $2$ , where $f(1.9) = -(1.9 - 2)^2 = -0.01 < 0$ . The solution decreases, and since $f < 0$ for every $y < 2$ it keeps decreasing, moving away from $2$ .

Explicitly, $u = y - 2$ satisfies $u' = -u^2$ with $u(0) = -0.1$ , so

u(t) = \frac{u(0)}{1 + u(0)\,t} = \frac{-0.1}{1 - 0.1t},

which runs off to $-\infty$ as $t \to 10$ . A start at $2.1$ would instead give $u(t) = \dfrac{0.1}{1 + 0.1t} \to 0$ : the same level attracts from above.

Problem 3 · Setting Up the Harvest

Given: salmon growing logistically with $a = 0.3$ per day and $a/b = 12{,}000$ lb. Harvest (i): a cannery takes $400$ lb every day, whatever the stock. Harvest (ii): a net takes 5% of the stock every day.

Which equation models harvest (i)?

Under harvest (ii), the nonzero critical point is

Under harvest (ii), $y = 0$ is

✅ Correct! A cannery of fixed capacity subtracts a constant,

h = 400

lb/day. A net that takes a fraction subtracts

hy

with

h = 0.05

per day, and

ay - by^2 - hy = y\,(a - h - by)

is the logistic right-hand side again with growth rate

a - h = 0.25

y = 0

stays an unstable critical point and the stock settles at

(a - h)/b = 10{,}000

lb.

❌ $400t$ is the total canned by day $t$ . The equation needs the rate of removal, and that is the same

400

lb every day: a constant.

❌ $400y$ would take $400$ times the whole stock every day. A term

hy

takes a fraction of the stock per day, which is harvest (ii), not a fixed weight.

❌ Harvesting removes fish, so its term is subtracted from the growth.

❌ Check the units.

dy/dt

is in lb per day, and a cannery of fixed capacity removes a fixed number of lb per day.

❌ $12{,}000$ lb is the carrying capacity with no harvest. Taking 5% a day lowers the growth rate from

a

a - h

, and the level moves with it.

❌ The 5% comes off the growth rate, not off the carrying capacity. The level is

(a - h)/b = (a/b)(1 - h/a)

, and

h/a = 0.05/0.3 = 1/6

, so it drops by a sixth, not by 5%.

❌ Sign slip. The net removes fish, so the growth rate drops to

a - h = 0.25

; it does not rise to

a + h

❌ Factor out $y$ .

ay - by^2 - hy = y\,(a - h - by)

; set the bracket to zero.

❌ That is harvest (i), where $f(0) = -400 \ne 0$ . Here an empty pen loses

h \cdot 0 = 0

, so

f(0) = 0

still.

❌ Just above $0$ , $f(y) = y\,(0.25 - by) > 0$ , so a small stock grows away from

0

: a 5% net is too gentle to hold it down.

❌ Not quite. Check the sign of

y\,(0.25 - by)

just below and just above

y = 0

Show solution

Harvest (i). The cannery removes the same weight every day, so it subtracts a constant with the units of $dy/dt$ :

\frac{dy}{dt} = ay - by^2 - h, \qquad h = 400 \text{ lb/day}.

Harvest (ii). Taking a fraction of the stock subtracts $hy$ with $h = 0.05$ per day, and $y$ factors out:

ay - by^2 - hy = y\,(a - h - by).

This is the logistic right-hand side with growth rate $a - h = 0.3 - 0.05 = 0.25$ per day. Since $a/b = 12{,}000$ lb, $b = 0.3/12{,}000 = 0.000025$ per lb per day, and the nonzero critical point is

y = \frac{a - h}{b} = \frac{0.25}{0.000025} = 10{,}000 \text{ lb}.

The factor $y$ keeps $y = 0$ a critical point, and just above it $y\,(0.25 - by) > 0$ , so $y = 0$ is unstable, exactly as in the unharvested logistic model. Under harvest (i), by contrast, $f(0) = -400 \ne 0$ : a fixed weight is charged even to an empty pen, so $0$ is no longer a critical point at all.

Problem 4 · Tangent but Crossing

Given: $\dfrac{dy}{dt} = (y - 1)^3\,(y - 4)^2$ . At both critical points the graph of the right-hand side is tangent to the $y$ -axis. Classify each.

$y = 1$ is

$y = 4$ is

✅ Correct! Tangency alone decides nothing; the signs do. The odd power

(y - 1)^3

changes sign at

1

, so the graph crosses there, flat, from

-

+

: unstable. The even power

(y - 4)^2

does not change sign, so

f > 0

on both sides of

4

: semi-stable, stable from below and unstable from above.

❌ Check the signs.

f(0) = (-1)^3(-4)^2 = -16 < 0

and

f(2) = 1^3 \cdot (-2)^2 = 4 > 0

, so the arrows point away from

1

on both sides.

❌ The graph is tangent to the axis at $1$ , but it also crosses it.

(y - 1)^3

is an odd power, so

f

changes sign there. Touching without crossing is what makes a point semi-stable; a flat crossing is still a crossing.

❌ The arrows need only the sign of $f$ on each side, and that is known everywhere.

f'(1) = 0

just says the graph is flat as it crosses.

❌ Not quite. Test one point on each side of

1

f(0)

and

f(2)

❌ That is the tangent-from-below case. Here

f(3) = 2^3 \cdot (-1)^2 = 8 > 0

and

f(5) = 4^3 \cdot 1^2 = 64 > 0

: both arrows point toward increasing

y

, toward

4

from below and away from it above.

❌ Below $4$ the arrow does point toward it, but above

4

we have

f(5) = 64 > 0

, so solutions there move away.

❌ Not quite. Compare the sign of

f

y = 3

and at

y = 5

Show solution

The critical points are the roots, $y = 1$ and $y = 4$ . At both, $f'(y_0) = 0$ : the graph is tangent to the axis, so tangency cannot be what decides. Test the sign in each region:

f(0) = (-1)^3(-4)^2 = -16, \qquad f(2) = 1^3 \cdot (-2)^2 = 4, \qquad f(5) = 4^3 \cdot 1^2 = 64

At $y = 1$ the sign goes from $-$ to $+$ . The odd power $(y - 1)^3$ changes sign, so the graph crosses the axis, flat, and the arrows point away on both sides: unstable.
At $y = 4$ the sign is $+$ on both sides. The even power $(y - 4)^2$ never changes sign, so the graph touches without crossing: semi-stable. Below $4$ the arrow points toward increasing $y$ , toward $4$ (stable from below); above $4$ it points toward increasing $y$ again, away from $4$ (unstable from above).

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