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Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. $$ \begin{array}{l} y^{\prime}=2(100-y) \\ y(0)=0 \end{array} $$

Short Answer

Expert verified
The solution is \( y(t) = 100(1-e^{-2t}) \).

Step by step solution

01

Identify the Type of Growth

The differential equation given is \( y' = 2(100-y) \). This equation resembles the form \( P' = r(K-P) \) which is indicative of an unlimited growth scenario. In this form, the solution follows an exponential or logistic-like pattern where growth slows as the population approaches a carrying capacity of \( K \).
02

Identify Constants

From the equation \( y' = 2(100-y) \), we can extract the parameters: the growth rate \( r = 2 \) and the carrying capacity \( K = 100 \).
03

Solve the Differential Equation

The equation is separable, so we rearrange and integrate: \[ \int \frac{dy}{100-y} = \int 2dt \]. The left side integrates to \(-\ln |100-y|\) and the right side integrates to \(2t + C\).
04

Find the Particular Solution

Solving \(-\ln |100-y| = 2t + C \) gives \(|100-y| = e^{-2t-C} = Ae^{-2t}\), where \(A = e^{-C}\). From the initial condition \( y(0) = 0 \), we have \(|100-0| = A \Rightarrow A = 100\). So, \(100-y = 100e^{-2t}\).
05

Express \( y(t) \)

Solving for \( y(t) \), we obtain \( y = 100 - 100e^{-2t} \). Therefore, the solution is \( y(t) = 100(1-e^{-2t}) \).

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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 concept in populations where resources limit repeated exponential growth. This model is particularly significant in ecologies when a species initially grows quickly, but as resources like food and space become constrained, the growth rate slows down. The logistic growth equation is generally expressed as \( P' = r(K-P) \), where:
  • \( P \) stands for the population size at any given time.
  • \( r \) is the intrinsic growth rate.
  • \( K \) is the carrying capacity of the environment.
Initially, when the population size is small \( P \ll K \), the term \( K-P \) remains large, allowing for a quick increase in population size, similar to exponential growth.
However, as \( P \) nears \( K \), the factor \( K-P \) reduces, slowing growth and eventually stabilizing the population at the carrying capacity.
Logistic growth is realistic in nature as it considers the limitation and competition caused by finite resources.
Carrying Capacity
Carrying capacity is a critical concept within the idea of logistic growth. It refers to the maximum population size that an environment can support indefinitely without significant negative impacts.Think of the carrying capacity, denoted as \( K \), as the equilibrium point for a population given its resources.
In the equation \( y' = 2(100-y) \) from the exercise, the carrying capacity is 100. This means, in practical terms, that the population \( y(t) \) will grow until it stabilizes at 100 as resources become the limiting factor for further growth.
Carrying capacity helps in understanding environmental limits and resource consumption, highlighting the balance required for sustainable population levels.
Separable Differential Equation
Separable differential equations are a class of differential equations that can be solved by separating the variables. This means you can rearrange the equation such that all terms involving the dependent variable \( y \) are on one side of the equation and all terms involving the independent variable \( t \) are on the other.For example, in the equation given: \( y' = 2(100-y) \), we rearranged to get \( \int \frac{dy}{100-y} = \int 2dt \).
This setup allows us to integrate each side independently, where:
  • The left side, \( \int \frac{dy}{100-y} \), simplifies to \(-\ln |100-y| \).
  • The right side, \( \int 2dt \), becomes \( 2t + C \) after integrating.
This process leads to a general solution, which can then be refined using boundary conditions or initial values, such as \( y(0) = 0 \), to find the particular solution.
Separable differential equations are user-friendly because they allow for straightforward integration, simplifying the solution process.

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Most popular questions from this chapter

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}=3 y+x^{4} \\ y(1)=7 \end{array} $$

When you swallow a pill, the medication passes through your stomach lining into your bloodstream, where some is absorbed by the cells of your body and the rest continues to circulate for future absorption. The amount \(y(t)\) of medication remaining in the bloodstream after \(t\) hours can be modeled by the differential equation $$ \frac{d y}{d t}=a b e^{-b t}-c y $$ for constants \(a, b,\) and \(c\) (respectively the dosage of the pill, the dissolution constant of the pill, and the absorption constant of the medication). For the given values of the constants: a. Substitute the constants into the stated differential equation. b. Solve the differential equation (with the initial condition of having no medicine in the bloodstream at time \(t=0)\) to find a formula for the amount of medicine in the bloodstream at any time \(t\) (hours). c. Use your solution to find the amount of medicine in the bloodstream at time \(t=2\) hours. d. Graph your solution on a graphing calculator and find when the amount of medication in the bloodstream is maximized. \(a=10 \mathrm{mg}, \quad b=3, \quad c=0.2\)

Suppose that you meet 30 new people each year, but each year you forget \(20 \%\) of all of the people that you know. If \(y(t)\) is the total number of people who you remember after \(t\) years, then \(y\) satisfies the differential equation \(y^{\prime}=30-0.2 y .\) (Do you see why?) Solve this differential equation subject to the condition \(y(0)=0\) (you knew no one at birth).

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For the initial value problem \(\left\\{\begin{array}{l}y^{\prime}=4 x y \\\ y(1)=3\end{array}\right.\) state the initial point \(\left(x_{0}, y_{0}\right)\) and calculate the slope of the solution at this point.

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