Problem 5 - Advanced Engineering Mathematics

Question

(a) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1

Step-by-Step Solution

Answer

The equilibrium solutions are \( P = 1 \) and \( P = 4 \). Population becomes extinct at \( t = -\frac{1}{3} \ln \frac{4(P_0-1)}{P_0-4} \).

Step 1: Identifying Equilibrium Solutions

First, solve the equation \( P(5-P) - 4 = 0 \) to find the equilibrium solutions. Factoring yields \( (P-1)(P-4) = 0 \), so the solutions are \( P = 1 \) and \( P = 4 \). These are the equilibrium solutions where the population remains constant.

Step 2: Deriving the Differential Equation

The given differential equation is \( \frac{dP}{dt} = P(5-P) - 4 = -(P-4)(P-1) \). This shows how the rate of change of \( P \) is related to \( P \) itself and its distance from the equilibrium points.

Step 3: Separation of Variables

To solve the differential equation, separate variables: \( \frac{dP}{(P-4)(P-1)} = -dt \). This allows us to integrate each side independently.

Step 4: Integrating Both Sides

Integrate the left side using partial fractions \( \frac{1/3}{P-4} - \frac{1/3}{P-1} \) and the right side with respect to \( t \). This results in \( \frac{1}{3} \ln \left| \frac{P-4}{P-1} \right| = -t + c \).

Step 5: Solving for the Constant and Function of \( P \)

Set \( t = 0 \) and \( P = P_0 \) to find that the constant \( c_1 = \frac{P_0 - 4}{P_0 - 1} \). Solve for \( P(t) \) to obtain the solution: \( P(t) = \frac{4(P_0 - 1) - (P_0 - 4) e^{-3t}}{(P_0 - 1) - (P_0 - 4) e^{-3t}} \).

Step 6: Finding Time of Extinction

To determine when the population becomes extinct (\( P = 0 \)), set \( P = 0 \) in the expression \( \frac{P-4}{P-1} = \frac{P_0-4}{P_0-1} e^{-3t} \). Solving for \( t \) gives \( t = -\frac{1}{3} \ln \frac{4(P_0-1)}{P_0-4} \). This gives the time at which extinction occurs, provided \( P_0 < 1 \).

Key Concepts

Equilibrium Solutions

Equilibrium solutions are points where the rate of change in a system, such as a differential equation, becomes zero. In simple terms, they represent the values at which the system remains constant over time.
In our problem, the differential equation is given as \( \frac{dP}{dt} = P(5-P) - 4 \). We solve the equation \( P(5-P) - 4 = 0 \) to find the equilibrium solutions. Factoring the expression gives \((P-1)(P-4) = 0\), leading to the solutions \( P = 1 \) and \( P = 4 \).
These points are called equilibrium points because when \( P \) is equal to 1 or 4, the population \( P \) neither increases nor decreases; it stays the same over time. Hence:

• For \( P = 1 \): This represents a stable equilibrium, as any deviation from this point results in a system that returns back to equilibrium.
• For \( P = 4 \): This is another equilibrium solution where the system neither grows nor diminishes.

Separation of Variables

Separation of Variables is a powerful method used to solve differential equations. It involves rearranging an equation so that each variable appears on a different side of the equation, allowing for independent integration.
In the differential equation \( \frac{dP}{dt} = -(P-4)(P-1) \), our goal is to isolate \( P \) on one side and \( t \) on the other. We rearrange it to \( \frac{dP}{(P-4)(P-1)} = -dt \).
Now, both sides can be integrated:

• The left side involves partial fraction decomposition: \( \frac{1/3}{P-4} - \frac{1/3}{P-1} \)
• The right side is simpler, as it integrates to \(-t + c\)

By integrating both sides, we find the relationship between \( P \) and \( t \), which is essential to understand how \( P \) changes over time given an initial condition.

Phase Portrait

A phase portrait is a visual representation that helps us understand the behavior of differential equations. It consists of arrows that indicate the direction of change in a given system.
For our differential equation \( \frac{dP}{dt} = -(P-4)(P-1) \), the phase portrait visually depicts how the population \( P \) will change over time for different initial conditions. By analyzing the arrows' directions:

• For \( P_0 > 4 \), the population decreases towards 4, illustrating stability.
• For \( 1 < P_0 < 4 \), the population increases towards 4, also illustrating stability.
• For \( 0 < P_0 < 1 \), the population decreases to zero, showing extinction.

The phase portrait provides an intuitive understanding of possible future states of the system based on current conditions, helping predict long-term behavior and stability points.

Integration of Differential Equations

Integration of differential equations is a method used to find a solution to equations that involve derivatives. In our context, it helps to determine the relationship between the variables \( P \) and \( t \) over time.
After separating variables in the equation \( \frac{dP}{(P-4)(P-1)} = -dt \), we integrate both sides:

• The left-hand side, after decomposition, becomes \( \int \left(\frac{1/3}{P-4} - \frac{1/3}{P-1}\right) dP \).
• The right-hand side simplifies to integration over time \( - \int dt \), resulting in \(-t + c\).

After integration, solving for \( P \) yields a function that helps express \( P \) in terms of \( t \).
This solution function, \( P(t) = \frac{4(P_0-1) - (P_0-4)e^{-3t}}{(P_0-1) - (P_0-4)e^{-3t}} \), describes the population behavior over time, revealing how the initial population \( P_0 \) and time \( t \) influence \( P \). This mathematical integration enables predicting how populations evolve based on initial starting points.