Problem 6 - Advanced Engineering Mathematics

Question

Solving \(P(5-P)-\frac{25}{4}=0\) for \(P\) we obtain the equilibrium solution \(P=\frac{5}{2} .\) For \(P \neq \frac{5}{2}, d P / d t<0 .\) Thus, if \(P_{0}<\frac{5}{2},\) the population becomes extinct (otherwise there would be another equilibrium solution.) Using separation of variables to solve the initial-value problem, we get \\[ P(t)=\left[4 P_{0}+\left(10 P_{0}-25\right) t\right] /\left[4+\left(4 P_{0}-10\right) t\right] \\] To find when the population becomes extinct for \(P_{0}<\frac{5}{2}\) we solve \(P(t)=0\) for \(t .\) We see that the time of extinction is \(t=4 P_{0} / 5\left(5-2 P_{0}\right).\)

Step-by-Step Solution

Answer

The population becomes extinct at \( t = \frac{4P_0}{5(5 - 2P_0)} \) for \( P_0 < \frac{5}{2} \).

Step 1: Understanding the Given Expression

The exercise provides the function \( P(t) = \frac{4P_0 + (10P_0 - 25)t}{4 + (4P_0 - 10)t} \) and states we need to find when \( P(t) = 0 \). This will determine when the population becomes extinct.

Step 2: Setting the Function to Zero

To find when the population becomes extinct, we must set the given expression for population \( P(t) \) to zero and solve for \( t \): \[ \frac{4P_0 + (10P_0 - 25)t}{4 + (4P_0 - 10)t} = 0 \]

Step 3: Solving the Numerator

A fraction is zero when its numerator is zero (while the denominator is non-zero). This means we need to solve: \( 4P_0 + (10P_0 - 25)t = 0 \).

Step 4: Isolating \( t \)

Rearrange the equation to solve for \( t \):\[ 4P_0 = 25t - 10P_0t \] leads to \[ 4P_0 = t(25 - 10P_0) \].

Step 5: Expressing \( t \) in Terms of \( P_0 \)

Solving for \( t \), we get: \[ t = \frac{4P_0}{25 - 10P_0} = \frac{4P_0}{5(5 - 2P_0)} \].

Step 6: Conclusion on Extinction Time

Thus, the population becomes extinct at \( t = \frac{4P_0}{5(5 - 2P_0)} \) when \( P_0 < \frac{5}{2} \).

Key Concepts

Equilibrium Solutions

In the realm of differential equations, equilibrium solutions play a fundamental role. An equilibrium solution refers to a steady state solution where the rate of change

• typically, represented by derivatives,
• is zero over time.

In simpler terms, it's a stable condition where the system doesn't change.

When we solve a differential equation, the equilibrium solution gives us insights on the behavior of the system. For instance, in the given problem, the equilibrium solution is calculated by setting the derivative to zero. Solving for the variable, in this case, gives us the value of \( P = \frac{5}{2} \). This indicates a state where the population remains constant.

It's also important to note that if the current state is not at equilibrium, indicated by \( P eq \frac{5}{2} \), the system (population) will naturally seek to return to its equilibrium state. Here, for values less than the equilibrium, the derivative is negative leading the population towards extinction. If \( P_0 < \frac{5}{2} \), the population decreases until it becomes extinct as there aren't enough individuals to sustain growth.

Separation of Variables

Separation of variables is a robust mathematical technique used to solve ordinary differential equations. It's especially handy when dealing with equations where variables can be separated and individually addressed.

• This method essentially involves rearranging the equation so that one variable and its derivative is on one side, while the remaining terms are on the other.
• This simplifies the integration of the equation.

In our exercise, applying separation of variables helps break down a complex problem into a more manageable form.

Initially, the differential equations express relationships involving variables interlinked with their derivatives. By using this technique, one can disentangle these relations and solve the integral by considering each variable independently. For example, in the population problem, solving \( P(t) = \left[4 P_{0} +\left(10 P_{0} - 25\right) t\right] /\left[4+\left(4 P_{0} - 10\right) t\right] \) demonstrates how the population formula changes over time by isolating time and population terms.

Initial Value Problems

Initial value problems (IVPs) are a common way to study differential equations by providing initial conditions.

• This narrows down solutions by setting the starting point of the equation at \( t = 0 \).
• The initial condition is often denoted \( P_0 \), representing the starting value of the function.

For the problem provided, recognizing \( P_0 \) as the initial population allows us to find how the population evolves over time.

The goal is to determine the function \( P(t) \) that fulfills the equation and meets the initial criterion. With \( P_0 \) set, solving the differential equation reveals how the system progresses from this initial state. Importantly, understanding IVPs allows us to predict future behavior which is essential in applications like population studies, as it dictates when this population will reach zero or become extinct. Recognizing changes starting from a specific point ensures precise modeling of dynamic systems.