Identify equilibrium solutions

Solutions that stay put

The field shows \(\dfrac{dy}{dx} = (y - 1)(y + 3)\). The segments on two rows are flat, and the dashed lines mark these rows, \(y = 1\) and \(y = -3\).

Press "Start at (0, 1)". The path follows the dashed line, because every segment on that line is flat, so the solution is the horizontal line \(y = 1\) itself.

Press "Start at (0, 0)". The slope at \((0, 0)\) is \((0 - 1)(0 + 3) = -3\), so this path falls to the right and is not a horizontal line.

A line that follows its segments

Move the orange line up and down. The right side of the equation contains only \(y\), and \(y\) is the same along the line, so the orange segments all have one slope.

If that slope is 0, a path on the line follows its flat segments and stays on it, so the line is a solution.

At any other height the segments tilt, so a path that starts on the line leaves it, and the line is not a solution. Press "Move to a flat row" to jump to the nearest such height.

Notation

1. A solution with the same \(y\)-value for every \(x\) is a constant solution, also called an equilibrium solution. It has the form \(y = c\), where \(c\) is a number, so its graph is a horizontal line.

   \[\cy{y = c}\]

2. The derivative of a constant is 0, so the right side of the equation, \(f(x, y)\), must be 0 at every point of the line. The orange tangent slides along the line and reads slope 0 everywhere.

   \[\begin{aligned} &\frac{dy}{dx} = 0 \\ &\Rightarrow\; \cy{f(x, c) = 0} \text{ for every } x \end{aligned}\]

3. To find the numbers \(c\), set the right side equal to 0 and solve for \(y\).

   \[\begin{aligned} &(y - 1)(y + 3) = 0 \\ &\Rightarrow\; y = 1 \text{ or } y = -3 \end{aligned}\]

   Test an answer. If you put \(y = 1\) into the right side, the result is 0 for every \(x\), so \(y = 1\) is a solution.

When it fails

Press the button. The equation changes to \(\dfrac{dy}{dx} = x - 3\), and the start moves to \((3, 1)\). Every flat segment lies on the dashed orange line \(x = 3\), so the path through \((3, 1)\) is flat at that one point only. The slope \(x - 3\) is negative to the left of that line and positive to the right, so the path falls and then rises.

No value of \(y\) makes \(x - 3\) equal 0 for every \(x\), so this equation has no equilibrium solution.

Example 1: Two equilibria

Find the equilibrium solutions of \(\dfrac{dy}{dx} = (y + 2)(y - 5)\).

1. The orange row of segments moves through the window and is flat at only two heights. To find them, set the right side equal to 0.

   \[(y + 2)(y - 5) = 0\]

2. A product is 0 when a factor is 0, so \(y = -2\) or \(y = 5\). The dashed orange lines mark these two rows, where every segment is flat.

3. Test \(y = 5\). The derivative of the constant 5 is 0, and the right side is 0 on the line, so \(y = 5\) is a solution. The same holds for \(y = -2\).

   \[\begin{aligned} &\frac{d}{dx}(5) = 0 \\ &(5 + 2)(5 - 5) = 0 \end{aligned}\]

Example 2: A flat point, not a flat line

The equation is \(\dfrac{dy}{dx} = (x - 2)(y + 1)\). The segment at \((2, 3)\) is flat. Find the equilibrium solutions.

1. At \((2, 3)\) the slope is 0, so the solution through this point is flat there.

   \[(2 - 2)(3 + 1) = 0\]

2. A tempting wrong move: The segment at \((2, 3)\) is flat, so \(y = 3\) is an equilibrium solution.

   \[y = 3\]

   The red line \(y = 3\) is not a solution, because a constant solution needs flat segments along its whole length. At \((4, 3)\) the slope is \((4 - 2)(3 + 1) = 8\), so the orange segment there rises steeply, and a path along the line would leave it.

3. A number \(c\) gives an equilibrium only if the right side is 0 for every \(x\). The factor \(x - 2\) is 0 only when \(x = 2\), so the factor \(c + 1\) must be 0, which gives \(c = -1\).

   \[\begin{aligned} &(x - 2)(c + 1) = 0 \text{ for every } x \\ &\Rightarrow\; c = -1 \end{aligned}\]

4. The dashed orange lines hold every flat segment. The line \(y = -1\) is the only equilibrium solution, and on the vertical line \(x = 2\) each solution is flat at one point only.

Practice

Find the equilibrium solutions of \(\dfrac{dy}{dx} = y^3 - 9y\).

Move the orange line up and down. At which heights are all the segments on the line flat?

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = y(y - 3)\).

1. Which horizontal lines are equilibrium solutions?

   • \(y = 0\) and \(y = 3\)
   • \(y = 1\) and \(y = 2\)
   • \(y = 0\) only
   Answer

   \(y = 0\) and \(y = 3\)

2. What is the solution through the orange point \((0, 3)\)?

   • The horizontal line \(y = 3\)
   • A curve that is flat at \((0, 3)\) and then falls
   • A curve that rises from \((0, 3)\)
   Answer

   The horizontal line \(y = 3\)

3. What are all the equilibrium solutions of \(\dfrac{dy}{dx} = y^2(y - 1)\)?

   • \(y = 0\) and \(y = 1\)
   • \(y = 1\) only
   • \(y = 0\) only
   • \(y = -1\), \(y = 0\) and \(y = 1\)
   Answer

   \(y = 0\) and \(y = 1\)

Summary

You set the right side of the equation equal to 0 and solve for \(y\), and each value \(c\) gives the constant solution \(y = c\).

This works because a constant has derivative 0, so a horizontal line is a solution when the right side is 0 along all of it.

You use it when a problem asks for the equilibrium solutions, and a curve that is flat at one point only is not one.

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