Question

Create a direction field for the differential equation \(y^{\prime}=(y-3)^{2}\left(y^{2}+y-2\right)\) and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

Short answer

Equilibrium: \(y = 3\) (semi-stable), \(y = 1\) (unstable), \(y = -2\) (unstable).

Step-by-step solution

Identify Equilibrium Solutions

Equilibrium solutions occur where the right side of the differential equation equals zero. We set \((y-3)^{2}(y^{2}+y-2) = 0\) and solve for \(y\). The equation is zero if either factor is zero. So, first solve \((y-3)^{2} = 0\), giving \(y = 3\). Second, solve \(y^{2}+y-2 = 0\) using the quadratic formula: \(y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}\), resulting in \(y = 1\) and \(y = -2\). Thus, the equilibrium solutions are \(y = 3\), \(y = 1\), and \(y = -2\).

Classify Stability of Equilibrium Solutions

To classify stability, analyze the sign of the derivative around each equilibrium point:- For \(y = 3\): Slight deviations result in \(y' = (y-3)^{2}(y^{2}+y-2)\). Since \((y-3)^{2} > 0\) for all \(y eq 3\), the behavior depends on \(y^{2} + y - 2\). At \(y = 3\), \(y' = 0\). For \(y > 3\), \(y^{2} + y - 2 > 0\), and for \(y < 3\), it is also positive. This points to \(y = 3\) being semi-stable.- For \(y = 1\): Near \(y = 1\), signs of \(y^{2}+y-2 = 0\) change — negative to positive as you move away from 1. Thus, \(y = 1\) is unstable.- For \(y = -2\): \(y^{2}+y-2 = 0\) also changes from negative to positive; hence, \(y = -2\) is unstable.

Key concepts

Equilibrium Solutions

In the context of differential equations, equilibrium solutions are the values of the dependent variable, commonly denoted as \( y \), for which the derivative \( y' \) equals zero. This means that at these values, the system is in a state of balance or rest. To find the equilibrium solutions for any given differential equation, you need to set the equation equal to zero and solve for \( y \).

Let's take the example equation from the original problem: \( y^{\prime} = (y-3)^{2}(y^{2}+y-2) \). By setting \((y-3)^{2}(y^{2}+y-2) = 0\), we find the values of \( y \) that make this expression zero.

• Firstly, \((y-3)^{2} = 0\) provides the solution \( y = 3 \) because squaring any non-zero number results in a positive value.
• Secondly, solving \( y^{2}+y-2 = 0 \) is done using the quadratic formula \( y = \frac{-1 \pm \sqrt{1 + 8}}{2} \), resulting in \( y = 1 \) and \( y = -2 \).

So, the equilibrium solutions for the given differential equation are \( y = 3 \), \( y = 1 \), and \( y = -2 \). These points mark where the system does not experience change in \( y \) over time.

Direction Field

A direction field is a graphical representation which helps us to visualize solutions of a differential equation without actually solving it. It consists of small line segments or arrows drawn at a grid of points. Each segment's direction and steepness represents the slope given by the differential equation at that point.

For a differential equation like \( y^{\prime} = (y-3)^{2}(y^{2}+y-2) \), the direction field will illustrate how solutions may behave as you vary \( y \) and the independent variable, typically \( x \) or time \( t \). By examining the direction field, you can quickly identify equilibrium solutions as they will appear where the arrows have zero slope.

• A direction field can reveal whether solutions tend to move towards or away from equilibrium points.
• This undertstanding can guide further analysis in studying the stability of equilibrium solutions.

Hence, using a direction field is crucial for getting an intuitive grasp of the dynamics modeled by a differential equation.

Stability Analysis

Stability analysis involves examining the behavior of solutions near the equilibrium points to determine whether these points are stable or unstable.

When assessing stability for the differential equation given, you check the sign of the derivative (\( y' \)) around each equilibrium point:

• For an equilibrium solution to be **stable**, any small deviation should result in the system returning to the equilibrium point.
• For **unstable** solutions, small deviations lead to the system moving away from the equilibrium.
• A solution is **semi-stable** if it exhibits stability from one side and instability from the other side.

In our original example, at \( y = 3 \), \( y' \) evaluates to zero at the point itself, and the behavior is the same on either side, indicating semi-stability. At \( y = 1 \) and \( y = -2 \), the sign of \( y' \) changes as \( y \) passes through the point, making them unstable equilibria.

Quadratic Equation

The quadratic equation is a fundamental tool used to find the roots of certain second-degree polynomials, which in turn helps in determining the equilibrium solutions for some differential equations. A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \).

The roots can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula is derived from completing the square and is fundamental in algebra and calculus.

• For the term \( y^{2}+y-2 = 0 \) in our differential equation, identify \( a = 1 \), \( b = 1 \), and \( c = -2 \).
• Plugging these values into the quadratic formula gives the roots as \( y = 1 \) and \( y = -2 \).

Understanding the application of the quadratic formula is essential for solving a wide range of mathematical problems and is particularly useful in differential equations for finding equilibrium solutions.