Problem 29 - Calculus: Early Transcendental Functions

Question

Find the equilibrium solutions and determine which are stable and which are unstable. $$y^{\prime}=(1-y) \sqrt{1+y^{2}}$$

Step-by-Step Solution

Answer

The equilibrium solutions are \(y = 1\) (stable) and \(y = -1\) (unstable).

Step 1: Find the equilibrium solutions

The equilibrium solutions are the solutions for which the derivative of the function is zero. For the given function, this is when \(y'=0\), or \((1-y) \sqrt{1+y^{2}} = 0\). Solving this equation, we find that \(y=1\) and \(y=-1\) meet the criteria and could be equilibrium solutions.

Step 2: Determine their stability

A solution is stable if it returns to the equilibrium position after being disturbed slightly; unstable if it continues to move away from the equilibrium after a slight disturbance. To determine stability, we need to find the sign of the derivative in the vicinity of these equilibrium solutions; this will show whether the solutions are attracting (negative derivative) or repelling (positive derivative). In this case, we see that for \(y < -1\) or \(1 < y\), \(y'>0\), and for \(-1 < y < 1\), \(y'<0\). Therefore, \(y = 1\) is a stable equilibrium and \(y = -1\) is an unstable equilibrium.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In many scientific fields, these equations are crucial as they model the behavior of dynamic systems.
The differential equation presented in this exercise, \(y^{\text{'}}=(1-y) \sqrt{1+y^{2}}\), involves a first-order derivative \(y^{\text{'}}\), indicating it is a first-order differential equation. In the context of our equation, the term \(1-y\) controls the growth rate, while \(\sqrt{1+y^{2}}\) adds complexity, representing a non-linear behavior.
Understanding how to solve such equations is essential for predicting system behavior over time, which can represent a variety of real-world phenomena, like population growth, heat conduction, or even the trajectory of a moving object under various forces.

Stability Analysis

Stability analysis in the realm of differential equations is a method used to determine whether, over time, a system will return to an equilibrium state after a slight perturbation.
An equilibrium solution is where the system can rest indefinitely unless acted upon by an external force; identifying these points is the first step in stability analysis. This exercise requires finding where the derivative \(y^{\text{'}}\) is equal to zero, as these points represent potential equilibria.
Once the equilibrium solutions are identified, like \(y=1\) and \(y=-1\), the next step is to discern their stability. We look at the sign of the derivative near these points. A negative derivative near an equilibrium implies stability since it indicates that the system returns to equilibrium when disturbed. On the contrary, a positive derivative suggests instability, as the system moves away from equilibrium under perturbation. In our example, \(y=1\) is stable, while \(y=-1\) is unstable.

Calculus

Calculus is the branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It provides tools for analyzing functions and enables us to solve problems involving motion, area, volume, and growth among many others.
The exercise in question utilizes differential calculus to explore the behavior of solutions to a differential equation at different values of \(y\). To solve the problem, calculus concepts such as derivatives, critical points, and intervals of increase and decrease are applied. These concepts are not only foundational in higher mathematics but also have practical applications in fields such as physics, engineering, economics, and biology. Equilibrium solutions are identified by setting derivatives equal to zero, which corresponds to finding the critical points of a function within calculus.