Problem 1 - Thomas Calculus

Question

In Exercises \(1-8\) \begin{equation}\begin{array}{l}{\text { a. Identify the equilibrium values. Which are stable and which }} \\ {\text { are unstable? }} \\ {\text { b. Construct a phase line. Identify the signs of } y^{\prime} \text { and } y^{\prime \prime} \text { . }} \\ {\text { c. Sketch several solution curves. }}\end{array}\end{equation} $$\frac{d y}{d x}=(y+2)(y-3)$$

Step-by-Step Solution

Answer

Equilibrium values are \( y = -2 \) (stable) and \( y = 3 \) (unstable).

Step 1: Find Equilibrium Points

Equilibrium points occur where \( \frac{dy}{dx} = 0 \). This happens when either \( y + 2 = 0 \) or \( y - 3 = 0 \). Solving these gives the equilibrium points \( y = -2 \) and \( y = 3 \).

Step 2: Determine Stability of Equilibrium Points

To determine stability, we need to analyze the derivative \( \frac{dy}{dx} = (y+2)(y-3) \) around the equilibrium points. For \( y = -2 \), the sign changes from positive to negative when crossing this point, so it is a stable equilibrium. For \( y = 3 \), the sign changes from negative to positive, making it an unstable equilibrium.

Step 3: Construct the Phase Line

Draw a vertical line representing the \( y \)-axis and mark the equilibrium points \( y = -2 \) and \( y = 3 \). On intervals \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \), check the sign of \( \frac{dy}{dx} \): it is positive on \( (-2, 3) \), and negative on \( (-\infty, -2) \) and \( (3, \infty) \). This shows the flow direction towards \( y = -2 \) and away from \( y = 3 \).

Step 4: Identify Signs of First and Second Derivatives

Since \( \frac{dy}{dx} = (y+2)(y-3) \), for the second derivative, differentiate \( \frac{dy}{dx} \) with respect to \( y \): \( \frac{d^2y}{dx^2} = 2y - 1 \). This helps confirm the behavior around each equilibrium point: \( \frac{d^2y}{dx^2} > 0 \) at \( y = 3 \) (concave up, unstable) and \( \frac{d^2y}{dx^2} < 0 \) at \( y = -2 \) (concave down, stable).

Step 5: Sketch Solution Curves

Sketch curves on a graph to reflect the stability analysis: approaches \( y = -2 \) as \( x \to \infty \) or \( x \to -\infty \), and curves diverging away from \( y = 3 \). Between these lines (from \( y = -2 \) to \( y = 3 \)), curves move upwards, showing growth.

Key Concepts

Equilibrium Points

Equilibrium points in differential equations are values where the derivative is zero, indicating that the system is at rest at these points.
In simpler terms, equilibrium points are where the rate of change, \( \frac{dy}{dx} \), equals zero. This means at those points, the function \( y \) is not changing.
In our exercise, we have the equation \( \frac{dy}{dx} = (y+2)(y-3) \). To find the equilibrium points:

• Set each factor in the equation to zero: \( y + 2 = 0 \) and \( y - 3 = 0 \).
• Solving these, we get \( y = -2 \) and \( y = 3 \).

These are the points where the slope of the function is zero, meaning they are horizontal tangents on the graph. Understanding these points is crucial as they form the foundation for further stability analysis and graphical representations.

Phase Line

A phase line is a simple graphical tool used to understand the behavior of solutions to differential equations around equilibrium points.
It is essentially a one-dimensional diagram that shows these equilibrium points and indicates the direction of movement towards or away from them.
For the given equation \( \frac{dy}{dx} = (y+2)(y-3) \), we create a phase line by:

• Drawing a vertical line, symbolizing the \( y \)-axis.
• Marking the equilibrium points, \( y = -2 \) and \( y = 3 \).
• Determining the sign of \( \frac{dy}{dx} \) in different intervals:
  • For \( y < -2 \) and \( y > 3 \), \( \frac{dy}{dx} < 0 \), indicating the graph moves downwards.
  • Between \( y = -2 \) and \( y = 3 \), \( \frac{dy}{dx} > 0 \), showing upward movement.

The phase line helps visualize the stability: movements towards \( y = -2 \) suggest stability, while movements away from \( y = 3 \) indicate instability.

Stability Analysis

Stability analysis in the context of differential equations examines whether solutions will gravitate towards or diverge from equilibrium points over time.
This is crucial as it determines the long-term behavior of the system's solutions.
To analyze stability, consider the sign changes in \( \frac{dy}{dx} = (y+2)(y-3) \) as you cross each equilibrium point:

• At \( y = -2 \), the derivative changes from positive to negative, indicating that solutions move towards this point, making it a stable equilibrium.
• For \( y = 3 \), the derivative changes from negative to positive, suggesting that solutions move away from this point, marking it as unstable.

Beyond first derivatives, looking at the second derivative \( \frac{d^2y}{dx^2} \) provides more insight:

• If \( \frac{d^2y}{dx^2} < 0 \), the function is concave down, supporting stability, as seen at \( y = -2 \).
• If \( \frac{d^2y}{dx^2} > 0 \), like at \( y = 3 \), the curvature is concave up, reinforcing instability.

Through stability analysis, you understand not just where equilibrium points lie, but how solutions respond around them.