/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find the solution \(y(t)\) by re... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 29

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. $$ \begin{array}{l} y^{\prime}=0.25 y(0.5-y) \\ y(0)=0.1 \end{array} $$

Short Answer

Expert verified
The solution is \(y(t) = \frac{0.5}{1 + 4e^{-0.25t}}\).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(y' = 0.25y(0.5-y)\). By recognizing the terms, you can see this is a logistic growth model, which is of the form \(y' = ry(1-\frac{y}{K})\) where \(r\) is the growth rate and \(K\) is the carrying capacity.
02

Determine the Constants

For the logistic equation \(y' = ry(1-\frac{y}{K})\), comparing with \(y' = 0.25y(0.5-y)\), we identify that \(r = 0.25\) and \(K = 0.5\). These represent the growth rate and carrying capacity, respectively.
03

Write the Logistic Solution Formula

The general solution for a logistic growth equation is \(y(t) = \frac{K}{1 + Ae^{-rt}}\), where \(A\) is a constant determined by initial conditions.
04

Substitute Known Values and Initial Condition

Substitute \(K = 0.5\) and \(r = 0.25\) into the logistic formula: \(y(t) = \frac{0.5}{1 + Ae^{-0.25t}}\). Use the initial condition \(y(0) = 0.1\) to find \(A\).
05

Solve for the Constant A

At \(t = 0\), \(y(0) = 0.1\): \[ 0.1 = \frac{0.5}{1 + A} \]Solving for \(A\), we get:\[ 1 + A = \frac{0.5}{0.1} \rightarrow 1 + A = 5 \rightarrow A = 4 \]
06

Write the Final Solution

Substituting the value of \(A\) back into the logistic formula gives:\[ y(t) = \frac{0.5}{1 + 4e^{-0.25t}} \] This is the complete solution for \(y(t)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Logistic Growth Model
The logistic growth model is an essential concept used to describe how quantities grow in a restricted or limited environment. Unlike unlimited exponential growth, which assumes infinite resources and environment, logistic growth recognizes practical limits. This model is frequently used in biology but also appears in other fields addressing population growth, resource consumption, and even spread phenomena like viruses.
To represent logistic growth mathematically, we use the differential equation:
  • \(y' = ry(1-\frac{y}{K})\)
Here, \(y\) is the observed variable typically representing population, \(r\) is the intrinsic growth rate, and \(K\) represents the carrying capacity or maximum sustainable limit.
The solution form for such an equation is structured to balance growth and limitation effects. By resolving this equation, we can derive a solution that adjusts to the initial conditions and incorporates the maximum sustainable level variable, known as the carrying capacity.
Carrying Capacity
Carrying capacity, usually symbolized by \(K\), is a fundamental aspect of the logistic growth model. It represents the maximum limit of the environment to sustain a certain population or quantity over time.
  • Think of it like the maximum number of individuals that can survive indefinitely in a given environment, considering available resources, habitat space, and other ecological constraints.
For the equation \(y' = 0.25y(0.5-y)\), the carrying capacity \(K\) is identified as \(0.5\). This means the model predicts the population will asymptotically approach 0.5, stabilizing at this level barring any changes in environmental constraints. Carrying capacity ensures the logistic model mirrors real-world situations where populations level off as they exhaust resources.
Understanding \(K\) helps in long-term planning and sustainability assessments since it defines the deceleration points beyond which growth balances out with resource distribution.
Growth Rate
Growth rate, often denoted by \(r\), is another critical factor in understanding logistic growth. It indicates how quickly the population or quantity can grow per unit time, considering each additional individual can impact future growth.
In our context, with \(y' = 0.25y(0.5-y)\), we have a growth rate \(r\) of \(0.25\). This value suggests a moderate pace for population increase when resources and space are initially abundant. However, as the population approaches the carrying capacity, the effective growth slows down due to decreasing available resources.
  • The logistic model incorporates \(r\) to ensure the growth is initially steep when far from carrying capacity, becoming progressively less steep as limits are reached.
Understanding the growth rate is crucial for projections and interventions, ensuring processes stay within sustainable parameters while assessing various scenarios and strategic planning.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 500-gallon tank is filled with water containing 0.2 ounce of impurities per gallon. Each hour, 200 gallons of water (containing 0.01 ounce of impurities per gallon) is added and mixed into the tank, while an equal volume of water is removed. a. Write a differential equation and initial condition that describe the total amount \(y(t)\) of impurities in the tank after \(t\) hours. b. Solve this differential equation and initial condition. c. Use your solution to find when the impurities will reach 0.05 ounce per gallon, at which time the water may be used for drinking. d. Use your solution to find the "long-run" amount of impurities in the tank.

The following problems extend and augment the material presented in the text. BIOMEDICAL: Fick's Law 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 \(C_{0}\).) b. Find the particular solution that satisfies the initial condition \(y(0)=y_{0},\) where \(y_{0}\) is the initial concentration inside the cell.

Solve each differential equation with the given initial condition. $$ \begin{array}{l} y^{\prime}+2 x y=4 x \\ y(0)=0 \end{array} $$

BIOMEDICAL: Heart Function In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T\), \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. (K and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself. \()\) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$ \frac{d p}{d t}=K I_{0}-\frac{K}{R} p $$ Find the general solution of this differential equation. (I \(_{0}\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example \(7 .\) d. Find the particular solution that satisfies the condi- $$ \text { tion } p\left(t_{0}\right)=p_{0} $$ e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right) .\) Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0) .\) Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$.

A 12,000 -cubic-foot room has 500 smoke particles per cubic foot. A ventilation system is turned on that each minute brings in 600 cubic feet of smoke-free air, while an equal volume of air leaves the room. Also, during each minute, smokers in the room add a total of 10,000 particles of smoke to the room. Assume that the air in the room mixes thoroughly. a. Find a differential equation and initial condition that govern the total number \(y(t)\) of smoke particles in the room after \(t\) minutes. b. Solve this differential equation and initial condition. c. Find how soon the smoke level will fall to 100 smoke particles per cubic foot.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.