Predict long-term behavior

Where paths end up

The field shows \(\dfrac{dy}{dx} = 4 - y^2\). The rows \(y = 2\) and \(y = -2\) are flat, and the dashed lines mark them.

Press "Start at (0, 3)". Above \(y = 2\), the slope \(4 - y^2\) is negative, so the path falls, and the segments flatten as the path nears the row \(y = 2\).

Press "Start at (0, 0)". Between the two rows, the slope is positive, so the path rises toward the row \(y = 2\).

Both paths come closer to \(y = 2\) as \(x\) grows, so the limit of \(y\) as \(x\) grows without bound is 2.

Toward or away

A flat row is an equilibrium, which is a constant solution.

Press "Toward the row". The segments above and below the row tilt toward it, so the faint paths approach the row and the limit of \(y\) is its height.

Press "Away from the row". The segments tilt away from the row, so the paths leave it. The orange label names each case: an equilibrium that paths approach is stable, and one that paths leave is unstable.

The slopes next to the row decide the long-term behavior.

Notation

1. Find the equilibria. Set the right side equal to 0 and solve for \(y\).

   \[4 - y^2 = 0 \;\Rightarrow\; y = 2 \text{ or } y = -2\]

2. The equilibria divide the \(y\)-values into three intervals. For \(y > 2\) and for \(y < -2\), \(y^2 > 4\), so the slope \(4 - y^2\) is negative. For \(-2 < y < 2\), \(y^2 < 4\), so the slope is positive. Red marks a negative slope and green a positive one.

3. A path cannot cross an equilibrium, because two solutions never cross. Between two equilibria the slope keeps one sign, so a path that starts there moves in one direction and approaches one of them.

4. The limit of \(y\) is the equilibrium that the slopes point the path toward. From \((0, 0)\) the slope is positive, so the path rises to \(y = 2\).

   \[\lim_{x \to \infty} y = \cy{2}\]

When it fails

Press the button to start at \((0, -3)\). The start is below \(y = -2\), where the slope is negative, so the path falls away from the row \(y = -2\). No equilibrium lies below it to stop the fall, so the path leaves the window and \(y\) does not approach any number.

The nearest equilibrium is not the answer in every case. The sign of the slope on the side where the path starts decides it.

Example 1: One equilibrium

A solution of \(\dfrac{dy}{dx} = 6 - 2y\) has \(y(0) = 8\). Find \(\lim_{x \to \infty} y\).

1. Find the equilibrium. Set the right side equal to 0.

   \[6 - 2y = 0 \;\Rightarrow\; y = 3\]

2. At the start \(y = 8\), so the slope is \(-10\) and the orange tangent falls steeply. For every \(y > 3\), \(2y > 6\), so \(6 - 2y\) is negative and the path continues to fall.

   \[6 - 2(8) = -10\]

3. The path falls and cannot cross \(y = 3\), so it approaches 3.

   \[\lim_{x \to \infty} y = \cy{3}\]

Example 2: Two equilibria

A solution of \(\dfrac{dy}{dx} = y(y - 5)\) has \(y(0) = 3\). Find \(\lim_{x \to \infty} y\).

1. Find the equilibria.

   \[y(y - 5) = 0 \;\Rightarrow\; y = 0 \text{ or } y = 5\]

2. A tempting wrong move: The start \(y = 3\) is closest to 5, so the limit is 5.

   \[\lim_{x \to \infty} y = 5\]

   The orange segment at \((0, 3)\) has slope \(3(3 - 5) = -6\), so the path falls there, away from 5. The red path rises, but every segment between the rows \(y = 0\) and \(y = 5\) falls.

3. Find the sign of the slope between the equilibria. For \(0 < y < 5\), \(y\) is positive and \(y - 5\) is negative, so the slope is negative on the whole interval. The path falls and cannot cross \(y = 0\), so it approaches 0.

   \[\lim_{x \to \infty} y = \cy{0}\]

Practice

A solution of \(\dfrac{dy}{dx} = (y + 1)(3 - y)\) has \(y(0) = 0\). Find \(\lim_{x \to \infty} y\).

Find the two flat rows. Is the start between them? Which way do the segments tilt there?

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = (y - 1)(4 - y)\) and its two flat rows.

1. A solution starts at the orange point \((0, 2)\). What is \(\lim_{x \to \infty} y\)?

   • 4
   • 1
   • 2
   Answer

   4

2. A solution starts at \((0, 0.5)\). What does it do as \(x\) grows?

   • It falls and does not approach any number.
   • It approaches 1.
   • It approaches 4.
   Answer

   It falls and does not approach any number.

3. A solution of \(\dfrac{dy}{dx} = 12 - 3y\) has \(y(0) = 10\). What is \(\lim_{x \to \infty} y\)?

   • 4
   • 12
   • 0
   • 10
   Answer

   4

Summary

You find the equilibria and the sign of the slope on each side, and that sign tells you which equilibrium the path moves toward.

This works because a path cannot cross an equilibrium, so it moves in one direction toward the next equilibrium, if there is one.

You use it when a problem asks for the limit of a solution, or for its behavior over a long time.

This is one of five skills in topic 7.4 of Which & Why: AP Calculus.