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Chapter 6: Problem 31

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. \(y^{\prime}=3 y(10-y)\) \(v(0)=20\)

Short Answer

Expert verified
The solution is \( y(t) = \frac{10}{1 - 0.5e^{-3t}} \).

Step by step solution

01

Identify the Type of Growth

The given differential equation is \[ y^{\prime} = 3y(10-y) \].This equation is in the form of a logistic growth model, as it resembles the general structure \[ y^{\prime} = ky(M-y) \],where \( M \) is the carrying capacity and \( k \) is the growth rate. Here, \( k = 3 \) and \( M = 10 \).
02

Set Up the Logistic Equation

The logistic growth equation can be defined as\[ y(t) = \frac{M}{1 + \left( \frac{M - y_0}{y_0} \right)e^{-kt}} \],where \( y_0 = v(0) \) is the initial value. According to the problem, \( y_0 = 20 \), and the carrying capacity \( M = 10 \).
03

Substitute Given Values

Substitute \( M = 10 \), \( y_0 = 20 \), and \( k = 3 \) into the logistic growth formula:\[ y(t) = \frac{10}{1 + \left( \frac{10 - 20}{20} \right)e^{-3t}} \].
04

Simplify the Equation

Calculate the fraction inside the exponential term:\( \frac{10 - 20}{20} = -0.5 \). Substitute back to the equation:\[ y(t) = \frac{10}{1 - 0.5e^{-3t}} \].
05

Write the Solution

The final solution for the differential equation is\[ y(t) = \frac{10}{1 - 0.5e^{-3t}} \].

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Key Concepts

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

Logistic Growth
Logistic growth is a common model in biology that describes how populations grow in an environment with limited resources. Unlike exponential growth, which assumes unlimited resources, logistic growth includes the impact of a carrying capacity, which is the maximum population size that the environment can sustain. This model is represented by a specific type of differential equation.
  • The general form of a logistic growth differential equation is: \( y' = ky(M-y) \)
  • In this equation, \( M \) represents the carrying capacity, \( k \) is the growth rate, and \( y \) is the population size.
  • The solution to the logistic growth equation is a function that shows how the population size changes over time.
When you solve a logistic growth equation, as seen in the original exercise, you determine the constants and use them to express a population's growth functionally over time. This provides a realistic view of population dynamics, including the slowing growth as it nears the carrying capacity.
Carrying Capacity
Carrying capacity is a key component in understanding logistic growth. It is crucial to comprehending the limits a population faces due to environmental constraints like food, space, and other resources.
  • Carrying capacity, often denoted as \( M \), is the maximum number of individuals that an environment can support without degradation.
  • In logistic growth equations, \( M \) affects how quickly the population approaches stability.
  • As the population size, \( y \), gets closer to \( M \), the growth rate decreases, indicating resource limitations.
This concept helps explain why populations do not grow indefinitely. In our example exercise, the carrying capacity \( M \) is 10, meaning, in theory, the population of interest will not surpass this number as it reaches equilibrium.
Initial Value Problem
An initial value problem in the context of differential equations involves finding a specific solution that passes through a given initial condition. This requires both an equation and an initial value from which the solution will begin.
  • An initial value problem specifies the value of the solution at a starting point, typically expressed as \( y(0) = y_0 \).
  • In logistic growth, the initial value \( y_0 \) is essential for determining the particular solution that describes the population's growth from a known starting size.
  • The solution not only solves the differential equation but also satisfies this initial condition.
For the original exercise, the initial value given is \( v(0) = 20 \). This initial condition provides the starting point for the population size, ensuring that the particular solution reflects the real-world scenario modeled by the differential equation.

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Most popular questions from this chapter

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. Contamination is leaking from an underground waste-disposal tank at the rate of \(4 \ln t\) thousand gallons per month, where \(t\) is the number of months since the leak began. Find the total leakage from the end of month 1 to the end of month \(4 .\)

Fick's Law governs the diffusion of a solute across a cell membrane. According to Fick's Law, the concentration \(y(t)\) of the solute inside the cell at time \(t\) satisfies \(\frac{d y}{d t}=\frac{k A}{V}\left(C_{0}-y\right)\), where \(k\) is the diffusion constant, \(A\) is the area of the cell membrane, \(V\) is the volume of the cell, and \(C_{0}\) is the concentration outside the cell. a. Find the general solution of this differential equation. (Your solution will involve the constants \(k, A, V\), and \(\left.C_{0} .\right)\) b. Find the particular solution that satisfies the initial condition \(y(0)=y_{0}\), where \(y_{0}\) is the initial concentration inside the cell.

Integration by parts often involves finding integrals like the following when integrating \(d v\) to find \(v\). Find the following integrals without using integration by parts (using formulas 1 through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure. $$ \int \sqrt{x} d x $$

Integration by parts often involves finding integrals like the following when integrating \(d v\) to find \(v\). Find the following integrals without using integration by parts (using formulas 1 through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure. $$ \int x^{5} d x $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int x \sqrt{x+1} d x $$
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