91Ó°ÊÓ

A useful new product is introduced into an isolated fixed population of \(1,000,000\) people, and 100 of these people adopt this product initially, that is, at time \(t=0\). Suppose the rate at which the product is adopted is proportional to the number of the people who have adopted it already multiplied by the number of them who have not yet done so. If we let \(x\) denote the number of people who have adopted the product at time \(t\), measured in weeks, then we have the initial-value problem $$ \begin{aligned} \frac{d x}{d t} &=k x(1,000,000-x) \\ x(0) &=100 \end{aligned} $$ where \(k\) is the constant of proportionality. (a) Solve this initial-value problem. (b) How many people have adopted the product after two weeks? (c) When will one half of the given population have adopted it?

Step by step solution

01

Identify the type of differential equation

The given differential equation is of the form: \[\frac{dx}{dt} = kx(1{,}000{,}000 - x)\] This is a first-order separable differential equation because we can separate the variables x and t on different sides of the equation.

02

Separate the variables

The goal is to have all x terms on one side of the equation and all t terms on the other side. To do this, we will divide by \(x(1{,}000{,}000 - x)\) and multiply by dt: \[\frac{1}{x(1{,}000{,}000 - x)}dx = k dt\]

03

Integrate both sides of the equation

Now, we need to integrate both sides of the equation. We can use the method of partial fraction decomposition for the integral on the left-hand side, which results in: \[\int{\frac{1}{x} + \frac{1}{1{,}000{,}000-x}}dx = \int k dt\] Now we integrate: \[\ln{x} - \ln{(1{,}000{,}000 - x)} = kt + C\]

04

Simplify and apply initial condition

Recall that some properties of logarithms allow us to rewrite the equation as: \[\ln{\frac{x}{(1{,}000{,}000 - x)}} = kt + C\] Now we need to apply the initial condition \(x(0) = 100\) to find the particular solution: \[\ln{\frac{100}{1{,}000{,}000 - 100}} = k \cdot 0 + C\] \[\ln{\frac{1}{10{,}000}} = C\]

05

Solve for x(t)

Now, we will rewrite the equation in terms of x to solve for x(t): \[\frac{x}{(1{,}000{,}000 - x)} = \frac{1}{10{,}000}e^{kt}\] From here, we can solve for x(t): \(x(t) = \frac{1{,}000{,}000}{1 + 10{,}000e^{-kt}}\)

06

Answer parts (b) and (c)

(b) To find how many people have adopted the product after two weeks, we plug in \(t = 2\) weeks into the particular solution equation: \(x(2) = \frac{1{,}000{,}000}{1 + 10{,}000e^{-2k}}\) (c) To find when half of the population will have adopted the product, we need to solve for t when \(x(t) = 500{,}000\): \(\frac{1{,}000{,}000}{1 + 10{,}000e^{-kt}} = 500{,}000\) Since we don't have an explicit expression for k, we cannot solve for the exact time t algebraically. However, we could use numerical methods (like the bisection method or Newton-Raphson method) to get an approximate value for t.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem

An Initial Value Problem (IVP) involves solving a differential equation with an initial condition. In our exercise, the differential equation models the rate of adoption for a new product in a population. The initial condition specifies that 100 people have already adopted the product at time zero. This initial condition is crucial because it allows us to find the particular solution that fits this specific scenario. Understanding and solving IVPs help us predict future behavior of dynamic systems by using their starting conditions. This allows us to tailor the general solution of the differential equation to meet the conditions provided at the start of the problem.

• Initial condition: Information about the state of a system at the beginning of the problem, e.g., \(x(0) = 100\).
• Particular solution: A solution to a differential equation that satisfies the initial condition.

First-order Differential Equation

A First-order Differential Equation involves derivatives of the first degree. In mathematical terms, it looks like \( \frac{dx}{dt} = f(t, x) \). Our exercise uses a first-order equation to express the rate of change in the number of adopters based on current adopters and non-adopters. The equation is written as \( \frac{dx}{dt} = kx(1,000,000-x) \), which models the growth factored by how many have and have not yet adopted the product.

• First-order: Involves only the first derivative (\( dx/dt \) in this case).
• Real-world application: This type of equation is commonly used for population growth or decay processes.

In practice, such equations help us understand and predict changes in different systems, from economics to ecology.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used in calculus to integrate complex rational expressions. In this problem, the left-hand side of the separated equation \( \frac{1}{x(1{,}000{,}000 - x)} \) was decomposed into simpler fractions. By rewriting it as \( \frac{1}{x} + \frac{1}{1{,}000{,}000 - x} \), we can easily integrate to find a solution. Without this step, integration would be much more complicated.

• Purpose: Simplifies complex fractions to make integration manageable.
• Process: Rewrite complex expressions into sums of simpler fractions.

This is a key step in solving separable differential equations since it allows us to find the integral on one side, making the equation solvable.

Numerical Methods

While analytical solutions to differential equations provide exact solutions, sometimes we rely on Numerical Methods for approximations. These methods help when it's difficult to solve an equation analytically due to complex parameters or missing specific values. In our exercise, without the exact value of the proportionality constant \( k \), we use numerical methods to find when half the population has adopted the product. Methods like bisection or Newton-Raphson can be applied for these approximations.

• Bisection Method: Divides intervals to find roots based on sign changes.
• Newton-Raphson Method: Iteratively finds roots using tangents.

These methods are invaluable, especially in fields like engineering and physics, where estimating outcomes quickly and sufficiently accurately is sometimes more practical than finding an exact solution.