Problem 15 - Calculus for Biology and Medicine

Question

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d y}{d t}=4-y^{2} $$

Step-by-Step Solution

Answer

Equilibria are \( y = 2 \) (stable) and \( y = -2 \) (unstable).

Step 1: Identify Equilibria

An equilibrium occurs when the rate of change is zero. Set \( \frac{d y}{d t} = 0 \), leading to the equation \( 4 - y^2 = 0 \). Solving for \( y \), we find the equilibria at \( y = 2 \) and \( y = -2 \).

Step 2: Analyze Stability Using Derivatives

To determine the stability of each equilibrium, evaluate the derivative of the function \( f(y) = 4 - y^2 \). Calculate \( \frac{df}{dy} = -2y \). At \( y = 2 \), \( \frac{df}{dy} = -4 < 0 \), indicating the equilibrium is stable. At \( y = -2 \), \( \frac{df}{dy} = 4 > 0 \), indicating the equilibrium is unstable.

Step 3: Plot Vector Field

Create a vector field plot by evaluating the differential equation at various values of \( y \). For example, when \( y < -2 \), \( \frac{dy}{dt} > 0 \), indicating upward arrows. When \( -2 < y < 2 \), \( \frac{dy}{dt} > 0 \), indicating upward arrows toward \( y = 2 \). Beyond \( y > 2 \), \( \frac{dy}{dt} < 0 \), indicating downward arrows back toward \( y = 2 \). This visually confirms stability at \( y = 2 \) and instability at \( y = -2 \).

Step 4: Classify Equilibria

Using the vector field plot and derivative analysis, classify the equilibria. \( y = 2 \) is a stable equilibrium, as nearby solutions converge towards it. Conversely, \( y = -2 \) is an unstable equilibrium, as nearby solutions diverge away.

Key Concepts

Equilibrium Analysis

Equilibrium points in a differential equation are crucial as they provide insight into the system's behavior over time. To find these points, set the rate of change to zero, essentially solving for when the system is in a state of rest. In the given equation \( \frac{dy}{dt} = 4 - y^2 \), equating to zero leads to \( 4 - y^2 = 0 \). Solving this, you find the roots \( y = 2 \) and \( y = -2 \), which are the equilibrium points. Equilibria tell us where the system might settle. These points can be stable, drawing nearby solutions in, or unstable, repelling them. Recognizing this distinction helps predict long-term behavior.

Vector Field Plots

A vector field plot is a valuable visual tool in understanding differential equations. It shows the direction of change for the system at various points. Construct this by evaluating the differential equation at a spectrum of \( y \) values. For example, in \( \frac{dy}{dt} = 4 - y^2 \), check how \( dy/dt \) behaves over different intervals:

• For \( y < -2 \), \( \frac{dy}{dt} > 0 \), indicating movement upwards.
• For \( -2 < y < 2 \), \( \frac{dy}{dt} > 0 \), arrows point toward \( y = 2 \).
• For \( y > 2 \), \( \frac{dy}{dt} < 0 \), showing downward movement.

This plotting effectively lets one see where each equilibrium lies and how the system behaves near these points. It offers a straightforward way to confirm equilibrium stability and system behavior.

Stability of Equilibria

Stability analysis involves determining whether solutions are attracted to or repelled from equilibrium points as time progresses. The derivative concept comes in handy here. By taking the derivative of the governing function, \( f(y) = 4 - y^2 \), you get \( \frac{df}{dy} = -2y \). Plugging equilibria into this derivative:

• At \( y = 2 \), \( \frac{df}{dy} = -4 \), which is negative, implying a stable equilibrium as nearby solutions are pulled towards it.
• At \( y = -2 \), \( \frac{df}{dy} = 4 \), indicating positivity, and an unstable equilibrium since nearby solutions are driven away from it.

Understanding stability allows predictions of system behavior over time, which can be vital in various scientific and engineering applications.

Phase Line Analysis

Phase line analysis is another powerful tool in studying differential equations. This approach effectively distills a system's behavior into a simple linear diagram, highlighting equilibrium points and general solution trends. By marking \( y = -2 \) and \( y = 2 \) on a line and adding arrows based on \( \frac{dy}{dt} = 4 - y^2 \), you visually display system dynamics:

• Before \( y = -2 \), upward arrows show positive \( dy/dt \).
• Between \( y = -2 \) and \( y = 2 \), the positive \( dy/dt \) continues, with arrows towards \( y = 2 \).
• Beyond \( y = 2 \), downward arrows signal negative \( dy/dt \), suggesting movement back to \( y = 2 \).

The phase line provides an intuitive overview of how solutions evolve, complementing vector field plots and solidifying understanding of equilibrium stability.