Problem 27 - Calculus: Early Transcendental Functions

Question

Find the equilibrium solutions and determine which are stable and which are unstable. $$y^{\prime}=y^{2}-y^{4}$$

Step-by-Step Solution

Answer

The equilibrium solutions are \( y = 0, +1, -1 \). From the stability analysis, \( y = -1, 1 \) are stable equilibrium points and \( y = 0 \) is an unstable equilibrium point.

Step 1: Finding the Equilibrium Solutions

In order to find the equilibrium solutions, set the derivative equal to zero and solve for y. So, \( y^{2}-y^{4} = 0 \). This can be simplified by factoring out y², we get \( y^2 (1 - y^2) = 0 \). Hence the equilibrium solutions are \( y = 0, +1, -1 \).

Step 2: Determining Stability

To determine stability, consider the sign of the derivative on either side of these equilibrium points. The derivative, \( y^{2}-y^{4}\), can be simplified as \( y^2 (1-y^2) \), so consider the intervals \( y < -1, -1 < y < 0, 0 < y < 1 \), and \( y > 1 \). It can be observed that \( y^{2}-y^{4} > 0 \) when either \( -1 < y < 0 \) or \( 0 < y < 1 \), and \( y^{2}-y^{4} < 0 \) when either \( y < -1 \) or \( y > 1 \). Therefore, \( y = -1, 1 \) are stable equilibrium points and \( y = 0 \) is an unstable equilibrium point.

Key Concepts

Stability Analysis in Differential Equations

In the context of differential equations, stability analysis helps us determine if solutions will remain close to an equilibrium point when slightly disturbed. Equilibrium solutions are points where the derivative of the function equals zero, meaning there is no change at these points. For the equation \( y' = y^2 - y^4 \), we found the equilibrium solutions to be \( y = 0, +1, -1 \).

Understanding the stability of these solutions involves looking at how \( y' \) behaves just before and after these points:

• If \( y' > 0 \), the function is increasing.
• If \( y' < 0 \), the function is decreasing.

For \( y = 0 \), the expression becomes negative for \( y < 0 \) and positive for \( y > 0 \), indicating it is unstable. Slight changes will cause the system to move away from this equilibrium.

For \( y = 1 \) and \( y = -1 \), the derivative moves toward these points in adjacent intervals. Thus, they are stable since they attract nearby solutions.

Basics of Differential Equations

Differential equations involve equations that hold a relationship between a function and its derivatives. They are essential for modeling various physical systems where the change of a quantity is related to the amount of the quantity itself.

The given differential equation \( y' = y^2 - y^4 \) is a first-order nonlinear differential equation. The order of a differential equation refers to the highest derivative present, and the term 'nonlinear' states that the relationship is not a straight line. Such equations often require specific techniques like factoring or substituting to solve.

Understanding these tools not only helps in solving the equations, but it also provides insights into the behavior of dynamic systems governed by these equations.

Factoring Polynomials for Solving Equations

Factoring polynomials is a critical skill in algebra that helps with solving equations, including differential equations. The process involves breaking down a polynomial into simpler 'factor' components that are easier to solve.

In the equation \( y' = y^2 - y^4 \), we factored it as \( y^2 (1 - y^2) \) to find equilibrium points. Recognizing patterns, such as the difference of squares formula, can simplify complex expressions and aid in finding solutions effectively.

• Differentiation is setting the derivative equal to zero to find critical points, which implies considering each factor separately equal to zero.
• This helps us quickly identify solutions like \( y = 0, +1, -1 \) without complicated calculations.

By mastering factoring techniques, you gain a powerful tool for analyzing polynomial expressions and finding key points in equations.