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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. $$ \begin{array}{l} y^{\prime}=80-2 y \\ y(0)=0 \end{array} $$

Short Answer

Expert verified
The solution is \( y(t) = 40(1 - e^{-2t}) \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = 80 - 2y \). This is a first-order linear ordinary differential equation and can be restructured as \( y' + 2y = 80 \). It resembles a limited growth model, where growth depends on the current state, inclined to stabilize at a constant value in the long term.
02

Solve the Differential Equation

To solve \( y' + 2y = 80 \), calculate the integrating factor \( \mu(t) = e^{\int 2 \, dt} = e^{2t} \). Multiply throughout by the integrating factor: \( e^{2t} y' + 2e^{2t} y = 80e^{2t} \). The left side is the derivative of \( e^{2t} y \), so: \( \frac{d}{dt}(e^{2t} y) = 80e^{2t} \). Integrate both sides: \[ e^{2t} y = 40e^{2t} + C \], where \( C \) is a constant of integration.
03

Find the General Solution

Solve for \( y \) by dividing through by \( e^{2t} \): \( y = 40 + Ce^{-2t} \). This is the general solution of the differential equation.
04

Apply Initial Conditions

Use the initial condition \( y(0) = 0 \) to find \( C \). Substitute into the general solution: \( 0 = 40 + Ce^{0} \). Hence, \( 0 = 40 + C \). Solving for \( C \) gives \( C = -40 \).
05

Write the Particular Solution

Substitute \( C = -40 \) back into the general solution: \( y = 40 - 40e^{-2t} \). This is the particular solution satisfying the initial condition.

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Key Concepts

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

Linear Differential Equations
A linear differential equation is a type of differential equation where the unknown function and its derivatives appear linearly. This means that the function or its derivatives are not raised to any power other than one, and they are not multiplied by each other. These equations can often be written in the form \( y' + p(t) y = g(t) \), where \( y' \) is the first derivative of \( y \), and \( p(t) \) and \( g(t) \) are functions of \( t \).

This simplicity allows for a methodical approach in solving them, using a tool called the integrating factor.
To solve such an equation, we assume an integrating factor \( \mu(t) = e^{\int p(t) \, dt} \). This is used to multiply through the equation to make the left-hand side a complete derivative.
  • The solution process hinges on the ability to identify this structure clearly and use the integrating factor effectively.
  • Understanding this solves the differential equation step by step, leading us to the general and particular solutions.
This systematic approach aids in predicting the behavior of dynamic systems governed by such equations.
Limited Growth Model
The limited growth model is a concept where growth depends proportionately on the difference between the current state and a maximum limit.
It is often described by a linear differential equation like \( y' = r(M - y) \), where \( y' \) is the rate of change, \( r \) is the growth rate, and \( M \) is the limiting value.

For the equation \( y' = 80 - 2y \), you can see that \( 80 \) is akin to the maximum carrying capacity or limit towards which the solution tends.
  • As time progresses, the term \( -2y \) causes the rate of increase to slow down, making \( y \) stabilize at \( y = 40 \) in the absence of other forces.
  • This behavior mirrors numerous real-world processes where growth isn't unbounded and naturally reaches a plateau.
The limited growth model's power is its applicability to biological, economic, and social systems, making it a vital concept in differential equations.
Initial Conditions
In solving differential equations, initial conditions are vital in pinning down particular solutions from general solutions.
Differential equations can yield an infinite set of functions as their solutions. Initial conditions provide specific values that the function must satisfy at certain points.

For example, if we have a solution like \( y = 40 + Ce^{-2t} \), where \( C \) is an arbitrary constant, the initial condition \( y(0) = 0 \) is crucial.
  • Substituting \( t = 0 \) into the solution and using the initial condition, we identify the constant \( C \).
  • In this case, it led to finding \( C = -40 \), thus giving the unique particular solution \( y = 40 - 40e^{-2t} \).
Initial conditions convert general solutions into specific equations that accurately describe a system's unique behavior under given circumstances.

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Most popular questions from this chapter

The air in a 100,000 -cubic-foot room contains \(483 \mathrm{pCi}\) of radon per cubic foot. A ventilation system is turned on that each hour will bring in 7250 cubic feet of fresh air containing \(8 \mathrm{pCi}\) of radon per cubic foot, while an equal amount of air leaves the room. Assuming that the air mixes thoroughly, in the long run, what will be the total amount of radon in the room?

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.

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}+4 y=10 x \\ y(1)=0 \end{array} $$

Is a differential equation of the form \(u(x) y^{\prime}+v(x) y=w(x)\) first- order linear? If so, what is its standard form?

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(y^{\prime}=x e^{-y}\) \(y(1)=0.5\) Approximate the solution at \(x=3\)

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