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Chapter 2: Problem 21

Suppose that the population \(P(t)\) of a country satisfies the differential equation \(d P / d t=k P(200-P)\) with \(k\) constant. Its population in 1940 was 100 million and was then growing at the rate of 1 million per year. Predict this country's population for the year 2000.

Short Answer

Expert verified
The population in the year 2000 is approximately 148 million.

Step by step solution

01

Interpret the Given Information

We are given a differential equation \( \frac{dP}{dt} = kP(200-P) \) describing the population growth, where \( P(t) \) is the population at time \( t \). We know at \( t = 0 \) (year 1940), \( P = 100 \) and \( \frac{dP}{dt} = 1 \) million per year. We need to predict the population at \( t = 60 \) (year 2000).
02

Substitute Initial Conditions to Define Transformation

Substitute \( P = 100 \) and \( \frac{dP}{dt} = 1 \) into the differential equation to find \( k \). Thus, \( 1 = k \times 100 \times (200-100) \). Solving this gives \( k = 0.0001 \).
03

Solve the Differential Equation

We solve the equation \( \frac{dP}{dt} = 0.0001 P (200 - P) \) by separating variables: \( \frac{dP}{P(200-P)} = 0.0001 dt \). Integrate both sides to distinguish \( t \) and \( P \):\[ \int \frac{1}{P(200-P)} \, dP = \int 0.0001 \, dt \].
04

Perform Partial Fraction Decomposition

To solve the integral on the left, express \( \frac{1}{P(200-P)} \) as a sum of partial fractions:\[ \frac{1}{P(200-P)} = \frac{A}{P} + \frac{B}{200-P} \]. Solve for \( A \) and \( B \), giving you \( A = \frac{1}{200} \), \( B = -\frac{1}{200} \). Thus,\[ \int \left( \frac{1}{200P} + \frac{1}{200(200-P)} \right) \, dP = \int 0.0001 \, dt \].
05

Integrate Both Sides

Integrate both sides:\[ \frac{1}{200} \ln |P| - \frac{1}{200} \ln |200-P| = 0.0001 t + C \]. Simplifying, \( \ln \frac{P}{200-P} = 0.02t + C' \).
06

Apply Initial Conditions to Solve for Constant

At \( t = 0 \), \( P = 100 \):\[ \ln \frac{100}{100} = 0.02 \times 0 + C' \Rightarrow C' = 0 \]. So the equation simplifies to: \( \ln \frac{P}{200-P} = 0.02t \).
07

Solve for Population at Year 2000

Substitute \( t = 60 \) into \( \ln \frac{P}{200-P} = 0.02 \times 60 \) resulting in:\[ \ln \frac{P}{200-P} = 1.2 \]. By exponentiating both sides, \( e^{1.2} = \frac{P}{200-P} \). Solve for \( P \) giving:\[ P = \frac{200e^{1.2}}{1 + e^{1.2}} \]. Evaluating this gives \( P \approx 147.5 \).

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Key Concepts

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

Population Growth Modeling
Population growth modeling is a way to understand how a population changes over time. Often, we use mathematical models to represent these changes. A popular model is the logistic growth model which accounts for a carrying capacity of an environment. This capacity represents the maximum population size that the environment can sustain.
In our exercise, the differential equation \( \frac{dP}{dt} = kP(200-P) \) describes how the population \( P(t) \) grows. Here, 200 is the carrying capacity, and \( k \) is a constant that affects the growth rate. As the population approaches this carrying capacity, the growth rate slows down.
To model population growth accurately, it's crucial to include real-world factors like food supply, space, and other resources. This helps us predict future population sizes and understand how populations interact with their environment.
Separable Differential Equations
Separable differential equations are a type of differential equation in which variables can be separated on different sides of the equation. This makes them easier to solve. In our exercise, we have the differential equation \( \frac{dP}{dt} = 0.0001 P (200 - P) \).
To solve it, we rearrange the terms to separate \( P \) and \( t \): \( \frac{dP}{P(200-P)} = 0.0001 dt \). Now, each side of the equation depends only on one variable.
This separation allows us to integrate each side separately. It simplifies finding the function \( P(t) \), which expresses how the population changes over time. Solutions to separable differential equations are common in modeling population growth as they often simplify complex real-world behaviors.
Initial Value Problems
An initial value problem (IVP) involves solving a differential equation with given initial conditions. These conditions specify the value of the function, and sometimes its derivatives, at a particular point.
In our exercise, the initial conditions are \( P(0) = 100 \) and \( \frac{dP}{dt}(0) = 1 \). These specify that the population at the start of our model (year 1940) is 100 million, growing at 1 million per year.
By using these initial values, we determine constants in our general solution. After solving the differential equation, we apply these conditions to find specific solutions.
The entire process ensures that the model closely follows real-world conditions at the start, providing more accurate predictions.

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Most popular questions from this chapter

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. Suppose that the fish population \(P(t)\) in a lake is attacked by a disease at time \(t=0\), with the result that the fish cease to reproduce (so that the birth rate is \(\beta=0\) ) and the death rate \(\delta\) (deaths per week per fish) is thereafter proportional to \(1 / \sqrt{P}\). If there were initially 900 fish in the lake and 441 were left after 6 weeks, how long did it take all the fish in the lake to die?

A programmable calculator or a computer will be useful for Problems 11 through \(16 .\) In each problem find the exact solution of the given initial value problem. Then apply the RungeKutta method twice to approximate (to five decimal places) this solution on the given interval, first with step size \(h=0.2\), then with step size \(h=0.1 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x\). \(y y^{\prime}=2 x^{3}, y(1)=3 ; 1 \leqq x \leqq 2\)

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=2 y, y(0)=\frac{1}{2} ; y(x)=\frac{1}{2} e^{2 x} $$

Find the exact solution of the given initial value problem. Then apply Euler's method twice to approximate (to four decimal places) this solution on the given interval, first with step size \(h=0.01\), then with step size \(h=0.005 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x\). $$ y y^{\prime}=2 x^{3}, y(1)=3 ; 1 \leqq x \leqq 2 $$

A tumor may be regarded as a population of multiplying cells. It is found empirically that the "birth rate" of the cells in a tumor decreases exponentially with time, so that \(\beta(t)=\beta_{0} e^{-\alpha t}\) (where \(\alpha\) and \(\beta_{0}\) are positive constants), and hence $$ \frac{d P}{d t}=\beta_{0} e^{-\alpha t} P, \quad P(0)=P_{0} $$ Solve this initial value problem for $$ P(t)=P_{0} \exp \left(\frac{\beta_{0}}{\alpha}\left(1-e^{-\alpha t}\right)\right) $$ Observe that \(P(t)\) approaches the finite limiting population \(P_{0} \exp \left(\beta_{0} / \alpha\right)\) as \(t \rightarrow+\infty\).
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