Problem 17 - Advanced Engineering Mathematics

Question

From \(\frac{1}{P-P^{2}} d P=\left(\frac{1}{P}+\frac{1}{1-P}\right) d P=d t\) we obtain \(\ln |P|-\ln |1-P|=t+c\) so that \(\ln \left|\frac{P}{1-P}\right|=t+c\) \(\frac{P}{1-P}=c_{1} e^{t} .\) Solving for \(P\) we have \(P=\frac{c_{1} e^{t}}{1+c_{1} e^{t}}\).

Step-by-Step Solution

Answer

\( P = \frac{c_1 e^t}{1+c_1 e^t} \)

Step 1: Understand the Given Equation

We start with the equation \( \frac{1}{P-P^2} dP = \left(\frac{1}{P} + \frac{1}{1-P}\right) dP = dt \). This equation suggests we have a differential form that needs to be simplified or integrated.

Step 2: Simplify the Expression

The left-hand side \( \frac{1}{P-P^2} \) can be factored as \( \frac{1}{P(1-P)} \). So, the equation becomes \( \left( \frac{1}{P} + \frac{1}{1-P} \right) dP = dt \). It shows the integrable form has already been simplified.

Step 3: Integrate Both Sides

Integrating both sides gives us \( \int \left( \frac{1}{P} + \frac{1}{1-P} \right) dP = \int dt \). This leads to \( \ln |P| - \ln |1-P| = t + c \) where \( c \) is the constant of integration.

Step 4: Combine Logarithmic Terms

The expression \( \ln |P| - \ln |1-P| \) can be combined using properties of logarithms: \( \ln \left| \frac{P}{1-P} \right| = t + c \). This step condenses the expression to a single logarithmic form.

Step 5: Exponentiate to Solve for \( \frac{P}{1-P} \)

To remove the logarithm, exponentiate both sides: \( \left| \frac{P}{1-P} \right| = e^{t+c} = c_1 e^t \) where \( c_1 = e^c \) is another constant.

Step 6: Solve for \( P \)

We express \( \frac{P}{1-P} = c_1 e^t \). Solving for \( P \), multiply both sides by \( 1-P \): \( P = c_1 e^t (1-P) \).

Step 7: Express \( P \) Explicitly

Rearranging gives \( P = c_1 e^t - c_1 e^t P \) or \( P + c_1 e^t P = c_1 e^t \). Factor \( P \) out: \( P(1+c_1 e^t) = c_1 e^t \). Divide both sides by \( 1+c_1 e^t \) to solve for \( P \): \( P = \frac{c_1 e^t}{1+c_1 e^t} \).

Key Concepts

Integration

Integration is a fundamental concept in calculus that helps in finding the antiderivative or the area under a curve. In the context of differential equations, integration is crucial for solving equations where derivatives are involved. In this exercise, the equation given is \[\left(\frac{1}{P} + \frac{1}{1-P}\right) dP = dt\] Here, each side of the equation needs to be integrated separately to progress towards a solution.
- On the left side, we integrate terms involving the variable \(P\). - On the right side, we integrate the term involving \(t\).
After integration, we combine these results to find an expression that relates \(P\) and \(t\). At this step, it is also crucial to add the constant \(c\) to account for the indefinite nature of integration, which represents the family of solutions.

Exponential Functions

Exponential functions appear often in solutions to differential equations, especially when dealing with growth and decay problems. Here, the relationship \[\ln \left|\frac{P}{1-P}\right| = t + c\]can be converted to an exponential form by exponentiating both sides.
- Exponentiating helps remove the logarithm, transforming the equation to a clearer exponential expression.
The exponential form derived is\[\left| \frac{P}{1-P} \right| = e^{t+c} = c_1 e^t\]where \(c_1 = e^c \). This simplification using the exponential function assists in isolating and eventually solving for the variable \(P\), making the function exponentially dependent on \(t\). Exponential functions are characterised by their constant rate of growth, which is reflected in how \(P\) changes with respect to \(t\).

Logarithmic Equations

Logarithmic equations are vital when manipulating expressions involving exponential terms or solving for a variable trapped inside a logarithm. They simplify multiplication into addition or division into subtraction, a useful property in the context of this problem. In this step-by-step solution, a crucial point was the consolidation of the logarithmic terms:\[\ln |P| - \ln |1-P| = t + c\]Applying logarithmic rules, specifically the quotient rule, these can be combined into a single term:\[\ln \left| \frac{P}{1-P} \right|\]Logarithms help in transitioning between multiplicative and additive forms, simplifying the manipulation of complex equations. This is why understanding logarithmic properties is essential for solving differential equations that involve exponential functions.

Separation of Variables

The separation of variables is a technique used to solve ordinary differential equations (ODEs), where you literally separate the variables into different sides of the equation. In this exercise, the separation process is subtly already in place by presenting a differential form \[\frac{1}{P-P^2} dP = dt\]We separate the variables by ensuring that all terms containing \(P\) are on one side, while dividing the differential elements accordingly, and all terms containing \(t\) are on the other.
- Once separated, each side is independently integrated. This method simplifies the process of solving ODEs by transforming them into manageable algebraic forms.
It also allows the integration process to be applied effectively, leading towards a solution where the variable \(P\), the population in this context, can be expressed explicitly in terms of \(t\), time. This separation makes it easier to see how \(P\) evolves over time based on the given differential equation.