Problem 59 - 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). \(\frac{d y}{d t}=y(1-y)\) (a) \(y(0)=0\), (b) \(y(0)=1 / 2\), (c) \(y(0)=1 / 4\), (d) \(y(0)=2\).

Step-by-Step Solution

Answer

Sketch solution curves: (a) stays at 0, (b) and (c) decrease to 0, (d) decreases to 1.

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{d y}{d t} = y(1-y) \). This equation represents a first-order autonomous differential equation.

Step 2: Analyze Critical Points

Set the right-hand side of the equation to zero to find critical points: \[ y(1-y) = 0 \]Solving this, we get critical points at \( y = 0 \) and \( y = 1 \). These are equilibrium solutions where the rate of change is zero.

Step 3: Determine Stability of Critical Points

Check the stability of these critical points by considering the sign of \( \frac{d y}{d t} \). - For \( y < 0 \) or \( y > 1 \), \( \frac{d y}{d t} > 0 \), so solutions increase.- For \( 0 < y < 1 \), \( \frac{d y}{d t} < 0 \), so solutions decrease.This tells us \( y = 0 \) is unstable and \( y = 1 \) is stable.

Step 4: Draw the Vector Field

On a \( (t, y) \) graph, draw arrows indicating the direction of change.- Arrows will point upwards (positive slope) when \( y < 0 \) or \( y > 1 \).- Arrows will point downwards (negative slope) when \( 0 < y < 1 \).

Step 5: Sketch Solution Curves with Initial Conditions

Using the initial conditions, sketch the trajectory for each case:- (a) \( y(0) = 0 \): Stays at zero since \( y = 0 \) is an equilibrium.- (b) \( y(0) = \frac{1}{2} \): Moves downwards towards \( y = 0 \).- (c) \( y(0) = \frac{1}{4} \): Also decreases towards \( y = 0 \).- (d) \( y(0) = 2 \): Decreases towards \( y = 1 \), as \( y = 1 \) is stable.

Key Concepts

Vector Field Plots

When you graph a differential equation like \( \frac{d y}{d t} = y(1-y) \), drawing a vector field plot gives you a visual representation of how solution curves behave over the plane. In our specific case, the plane we refer to is defined by the coordinates \((t, y)\). Here's how you can create this plot:

• On the plot, the horizontal axis represents the independent variable, \(t\), and the vertical axis represents the dependent variable, \(y\).
• A vector field consists of arrows indicating the slope of the solution curve at various points. These arrows give a glimpse into the direction and speed at which \(y\) changes over time.
• For this equation, when \(y < 0\) or \(y > 1\), the arrows point upwards indicating a positive slope. Likewise, when \(0 < y < 1\), the arrows point downwards to show a negative slope.

This vector field helps us visualize how solutions converge or diverge towards equilibrium points. By sketching this field, we can easily see the behavior of the system for different initial conditions.

Equilibrium Solutions

Equilibrium solutions are key points where a differential equation has no change, meaning where \( \frac{d y}{d t} = 0 \). These are also called critical points. They are vital because they help us understand where a system stabilizes or remains constant.
How do we find them? For our differential equation \( \frac{d y}{d t} = y(1-y) \), you set the expression \(y(1-y)\) to zero. Solving gives \( y = 0 \) and \( y = 1 \). At these points, the rate of change is zero, so the system stays put.

• \(y = 0\) is an equilibrium solution. If a solution starts here, it'll remain.
• \(y = 1\) is another equilibrium, but we'll explore its nature in the stability section.

Equilibrium solutions are the 'resting points' in the system. Whether they are stable or unstable is determined through further analysis, known as stability analysis.

Stability Analysis

Stability analysis tells us whether a system returns to equilibrium after a disturbance. For our differential equation, it helps determine if an equilibrium solution like \(y = 0\) or \(y = 1\) is stable or unstable.
To perform this analysis, you observe the sign of \( \frac{d y}{d t} = y(1-y) \):

• For \( y < 0 \) or \( y > 1 \), \( \frac{d y}{d t} > 0 \). This means \(y\) is increasing, suggesting that \(y = 0\) is not a point to which solutions return.
• For \( 0 < y < 1 \), \( \frac{d y}{d t} < 0 \). Thus, \(y\) decreases, indicating solutions might move towards \(y = 1\).
• This makes \(y = 0\) an unstable equilibrium, as solutions move away. On the other hand, \(y = 1\) acts as a stable one because solutions nearby tend to move towards it.

By understanding stability, we learn whether solutions to our differential equation will settle at an equilibrium or continue to change. This knowledge helps in predicting the long-term behavior of dynamical systems modeled by differential equations.