Problem 67 - Calculus and its applications

Question

Exponential growth. a) Use separation of variables to solve the differentialequation model of uninhibited growth, \(\frac{d P}{d t}=k P\) b) Rewrite the solution of part (a) in terms of the condition \(P_{0}=P(0)\)

Step-by-Step Solution

Answer

The solution is \( P(t) = P_0 e^{kt} \).

Step 1: Separate Variables

The given differential equation is \( \frac{dP}{dt} = kP \). To apply the separation of variables technique, we rearrange this equation to have all \( P \) terms on one side and \( t \) on the other side: \( \frac{1}{P} dP = k \, dt \).

Step 2: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{P} dP = \int k \, dt \). The left side integrates to \( \ln |P| \) and the right side integrates to \( kt + C \) where \( C \) is the constant of integration. So we have: \( \ln |P| = kt + C \).

Step 3: Solve for P

To solve for \( P \), exponentiate both sides to remove the logarithm: \( e^{\ln |P|} = e^{kt + C} \). This simplifies to: \( |P| = e^{kt} \cdot e^{C} \). Let \( e^{C} = C' \) where \( C' \) is another constant, so \( P = C' e^{kt} \).

Step 4: Apply Initial Condition

Replace \( C' \) by finding it in terms of the initial condition \( P(0) = P_0 \). Substitute this into the equation: \( P_0 = C' \cdot e^{k \cdot 0} \). Therefore, \( C' = P_0 \).

Step 5: Write Final Solution

Substitute \( C' = P_0 \) into the equation from Step 3: \( P = P_0 e^{kt} \). This is the solution to the differential equation representing uninhibited growth with the initial condition \( P(0) = P_0 \).

Key Concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations. It's particularly effective when you can rewrite an equation so all terms involving one variable are on one side, and all terms involving the other variable are on the other side.

In the context of the exponential growth problem, we start with the differential equation \( \frac{dP}{dt} = kP \). Our goal is to separate the variables, placing all \( P \) terms on one side and \( t \) terms on the other. After rearranging, we get \( \frac{1}{P} dP = k \, dt \).

By doing this, each side of the equation only contains one type of variable. This separation makes it possible to integrate both sides easily, helping us to find the general solution. Once you've separated the variables, the next logical step is to perform integration on both sides. This is where the magic happens, allowing us to solve for the desired function.

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. It describes the relationship between a function and its rates of change.

In the exponential growth model, the differential equation \( \frac{dP}{dt} = kP \) describes how a quantity \( P \) changes over time. Here, \( k \) represents a constant growth rate. The form of this equation is simple yet powerful, representing many real-world growth scenarios like populations, radioactive decay, and even simple interest growth.

Solving differential equations often requires finding a function \( P(t) \) that satisfies the original equation. After separating variables in our problem, we integrated both sides with respect to their independent variables, yielding the solution \( P = Ce^{kt} \). This represents the general solution, where \( C \) is determined by initial conditions.

Initial Condition

An initial condition is a specific value that gives a starting point for solving differential equations. It allows us to find particular solutions from general solutions by specifying conditions that the solution must satisfy at a certain point.

In our problem, the initial condition is \( P(0) = P_0 \). This means that when time \( t = 0 \), the population or amount \( P \) is \( P_0 \).

Applying the initial condition to our solution \( P = Ce^{kt} \), we substitute \( t = 0 \) and \( P = P_0 \), resulting in \( P_0 = Ce^{k\cdot0} = C \). This lets us determine the constant \( C \) to be \( P_0 \). Thus, the specific solution satisfying the initial condition becomes \( P = P_0 e^{kt} \), perfectly modeling uninhibited exponential growth starting from the initial state \( P_0 \).

The use of initial conditions is crucial, as it adapts the general nature of differential equation solutions to fit particular, real-world scenarios.