Problem 17 - Advanced Engineering Mathematics

Question

In Problems 1-22, solve the given differential equation by separation of variables. $$ \frac{d P}{d t}=P-P^{2} $$

Step-by-Step Solution

Answer

The solution is \( P(t) = \frac{Ae^t}{1+Ae^t} \), where \( A \) is a constant.

Step 1: Identify the Differential Equation Type

The given differential equation is \( \frac{dP}{dt} = P - P^2 \). It is a first-order separable ordinary differential equation.

Step 2: Rewrite the Equation in Separable Form

To separate variables, rewrite the equation as \( \frac{dP}{dt} = P(1 - P) \). This allows us to express it as \( \frac{dP}{P(1 - P)} = dt \).

Step 3: Integrate Both Sides

Integrate both sides: \( \int \frac{1}{P(1 - P)} \, dP = \int dt \). The left-hand side requires partial fraction decomposition.

Step 4: Partial Fraction Decomposition

Express \( \frac{1}{P(1-P)} \) as \( \frac{1}{P} + \frac{1}{1-P} \). Thus, \( \int \left( \frac{1}{P} + \frac{1}{1-P} \right) \, dP = \int dt \).

Step 5: Solve the Integrals

Integrate both sides: \( \ln |P| - \ln |1-P| = t + C \), where \( C \) is the constant of integration.

Step 6: Simplify the Solution

Combine the logarithms on the left: \( \ln \left( \frac{P}{1-P} \right) = t + C \). Exponentiate both sides to solve for \( P \): \( \frac{P}{1-P} = e^{t+C} = ke^t \), where \( k = e^C \).

Step 7: Solve for P

Rearrange \( \frac{P}{1-P} = ke^t \) to solve for \( P \): \( P = \frac{ke^t}{1+ke^t} \).

Step 8: Express General Solution

The general solution to the differential equation is \( P(t) = \frac{Ae^t}{1+Ae^t} \), where \( A \) is a constant determined by initial conditions.

Key Concepts

Separation of Variables

In solving ordinary differential equations (ODEs), separation of variables is a powerful technique that helps us handle equations where terms can be separated based on the variables involved. This method is particularly useful for first-order separable equations.

• To apply it, rewrite the differential equation such that all terms involving one variable are on one side and terms involving another variable are on the opposite side.
• An example is the equation \( \frac{dP}{dt} = P - P^2 \), which can be rewritten as \( \frac{dP}{P(1 - P)} = dt \).

Once the variables are separated, you can integrate both sides to find the solution. This approach makes it easier to handle complexities by breaking them down into simpler parts.

Ordinary Differential Equations

Ordinary differential equations, or ODEs, are equations involving functions of one independent variable and their derivatives. Solving them often involves finding an unknown function based on its relationship to its derivative.

• ODEs are classified based on their order, which is determined by the highest derivative present in the equation. For example, an equation involving \( \frac{dP}{dt} \) is a first-order equation.
• These equations can describe a wide range of phenomena, from population growth to mechanical vibrations.

The goal in solving an ODE is to find a function that satisfies the equation, often requiring initial conditions to find a specific solution. Understanding the types and methods of solving ODEs, such as separable ODEs, is crucial in calculus and applied mathematics.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in algebra to simplify the integration of rational expressions. It breaks down a complex fraction into simpler fractions that are easier to integrate.

• When faced with an equation like \( \frac{1}{P(1-P)} \), it can be expressed as \( \frac{1}{P} + \frac{1}{1-P} \), making it easier to integrate each component separately.
• This method is particularly useful when dealing with the integration of expressions that appear in solving differential equations.

By using partial fraction decomposition, integrating complex rational expressions becomes more manageable, leading to a clearer path to the solution of the differential equation.