Problem 61 - Calculus for Biology and Medicine

Question

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d t}=(y+3)(1-y)\) (a) \(y(0)=-1\), (b) \(y(0)=-1 / 2\), (c) \(y(0)=-2\), (d) \(y(0)=2\).

Step-by-Step Solution

Answer

The solutions approach the stable equilibrium at \( y = -3 \) for \( y(0) = -1, -1/2 \), and \( y(0) = -2 \); it diverges upwards for \( y(0) = 2 \).

Step 1: Understanding the Differential Equation

The differential equation provided is \( \frac{dy}{dt} = (y + 3)(1 - y) \). This is a first-order ordinary differential equation (ODE) which represents the rate of change of \( y \) with respect to \( t \). The expression \((y + 3)(1 - y)\) describes how this rate of change varies with \( y \).

Step 2: Identify Equilibrium Points

To find equilibrium points, set \( \frac{dy}{dt} = 0 \), giving us \((y + 3)(1 - y) = 0\). Solving this gives the equilibrium points \( y = -3 \) and \( y = 1 \). These are the values of \( y \) where the system does not change.

Step 3: Analyzing Stability of Equilibrium Points

Evaluate the stability by considering the sign of \( \frac{dy}{dt} \) around the equilibrium points. If \( y < -3 \), then \( (y+3) \) is negative and \( (1-y) \) is positive, making \( \frac{dy}{dt} \) negative, and \( y \) decreases, showing \( y = -3 \) is a stable equilibrium. If \( y > 1 \), both terms are negative, making the product positive, so \( y \) increases, showing \( y = 1 \) is unstable.

Step 4: Plotting the Vector Field

Create a vector field by plotting \( \frac{dy}{dt} \) as arrows on the graph. Each arrow points in the direction of \( \frac{dy}{dt} \) and is scaled to represent the magnitude of \( \frac{dy}{dt} \). Use software or graphing tools to visualize this.

Step 5: Sketch Solution Curves Using Initial Conditions

Use the initial conditions: (a) \(y(0)=-1\), (b) \(y(0)=-1/2\), (c) \(y(0)=-2\), (d) \(y(0)=2\). Starting from each initial \( y \) value, trace out the solution on the vector field plot to identify how \( y \) changes over time. Each solution curve will show how \( y \) evolves for the given initial value.

Step 6: Drawing the Graphs of the Solutions

On the graph of \( y \) against \( t \), draw solution curves starting from each initial condition. For \( (a) \) and \( (b) \), \( y \) will approach the stable equilibrium at \( y = -3 \). For \( (c) \), it will diverge towards \( y = -3 \). For \( (d) \), \( y \) will continue increasing indefinitely.

Key Concepts

Differential Equations

Differential equations serve as a powerful tool for expressing the relationship between a function and its derivatives. In simple terms, a differential equation connects how a quantity changes with respect to another. In our exercise, the equation \( \frac{dy}{dt} = (y + 3)(1 - y) \) is a first-order ordinary differential equation (ODE).

• This particular equation captures how the variable \( y \) changes over time \( t \).
• The right-hand side of the equation, \((y + 3)(1 - y)\), determines the rate at which \( y \) changes.

Each component of the expression affects this rate differently. Comprehending these expressions within differential equations aids us in predicting and describing various phenomena, from physics to finance. Thus, understanding these equations can unravel the nature of dynamic systems.

Equilibrium Points

Equilibrium points in differential equations represent conditions where the system remains unchanged over time. These points occur when the derivative or rate of change, \( \frac{dy}{dt} \), is equal to zero. For the given equation, equilibrium points are found by solving \((y + 3)(1 - y) = 0\).

• This yields the equilibrium points \( y = -3 \) and \( y = 1 \).
• At these values, the system is in a steady state, meaning there's no net change in \( y \).

Equilibrium points are essential for understanding the long-term behavior of dynamical systems. They tell us about the stability and potential states where systems tend to settle. Recognizing these points is fundamental when analyzing any system described by differential equations.

Stability Analysis

Analyzing the stability of equilibrium points is crucial for predicting how systems behave near these points. Once we know the equilibrium points, like \( y = -3 \) and \( y = 1 \), we determine their stability by examining the sign of \( \frac{dy}{dt} \) around them.

• If the rate of change is positive just before the equilibrium, the system tends to move away, indicating instability.
• If it's negative just before the point, the system moves towards the equilibrium, suggesting stability.

In our scenario:

• \( y = -3 \) is a stable equilibrium, as nearby values tend to return to \( -3 \).
• \( y = 1 \) is unstable, and nearby values tend to move away from \( 1 \).

Understanding stability helps in predicting the likelihood of a system to return to equilibrium after a disturbance, a critical insight for real-world applications.

Solution Curves

Solution curves are an insightful way to visualize how the dependent variable \( y \) evolves over time \( t \) given different initial conditions. For our equation, solution curves are sketched starting from initial conditions like \( y(0) = -1 \), \( y(0) = -1/2 \), \( y(0) = -2 \), and \( y(0) = 2 \).

• These curves on a graph illustrate the path each solution takes in the \( y \)-\( t \) plane.
• For example, starting conditions \( y(0) = -1 \) and \( y(0) = -1/2 \) show a trajectory moving towards the stable point \( y = -3 \).
• The initial condition \( y(0) = -2 \) diverges towards \( y = -3 \), showcasing the stability of this equilibrium.
• Meanwhile, \( y(0) = 2 \) results in \( y \) moving away indefinitely, affirming the instability at \( y = 1 \).

Sketching these solution curves not only aids in understanding the qualitative behavior of solutions but also in anticipating the system's response to various conditions. This process is essential in systems models, providing a clear picture of possible future states.