/*! 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 58 For more than 75 years the Flexf... [FREE SOLUTION] | 91Ó°ÊÓ

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For more than 75 years the Flexfast Rubber Company in Massachusetts discharged toxic toluene solvents into the ground at a rate of 5 tons per year. Each year approximately \(10 \%\) of the accumulated pollutants evaporated into the air. If \(y(t)\) is the total accumulation of pollution in the ground after \(t\) years, then \(y\) satisfies \(y^{\prime}=5-0.1 y \quad\) (Do you see why?) \(y(0)=0 \quad\) (initial accumulation zero) Solve this differential equation and initial condition to find a formula for the accumulated pollutant after \(t\) years.

Short Answer

Expert verified
The accumulated pollution after \( t \) years is given by \( y(t) = 50 - 50e^{-0.1t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = 5 - 0.1y \), which is a first-order linear differential equation. This is classified as a linear differential equation because it can be expressed in the standard form \( y' + P(x)y = Q(x) \). In our equation, \( P(t) = 0.1 \) and \( Q(t) = 5 \).
02

Calculate the Integrating Factor

For a first-order linear differential equation, the integrating factor \( \mu(t) \) is given by \( \mu(t) = e^{\int P(t) \, dt} \). Here, \( P(t) = 0.1 \), so \( \mu(t) = e^{0.1t} \).
03

Transform the Differential Equation

Multiply the entire differential equation \( y' = 5 - 0.1y \) by the integrating factor \( e^{0.1t} \). This yields:\[e^{0.1t}y' + 0.1e^{0.1t}y = 5e^{0.1t}\] Notice that the left-hand side is the derivative of \( e^{0.1t}y \).
04

Integrate Both Sides

Integrate both sides with respect to \( t \):\[\int e^{0.1t}y' \, dt = \int 5e^{0.1t} \, dt\]The left side simplifies to \( e^{0.1t}y \), and the right side integrates to \( 50e^{0.1t}+C \). Thus, the equation is: \( e^{0.1t}y = 50e^{0.1t} + C \).
05

Solve for y(t)

Divide through by \( e^{0.1t} \) to solve for \( y(t) \): \[y(t) = 50 + Ce^{-0.1t}\].
06

Apply the Initial Condition

Use the initial condition \( y(0) = 0 \) to determine the constant \( C \). Substitute \( t = 0 \) into the equation: \[0 = 50 + Ce^{-0.1 \times 0}\]This simplifies to \( 0 = 50 + C \), so \( C = -50 \).
07

Write the Final Solution

Substitute \( C = -50 \) back into the expression for \( y(t) \): \( y(t) = 50 - 50e^{-0.1t} \). This is the formula for the accumulated pollutant after \( t \) years.

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Key Concepts

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

First-Order Linear Differential Equations
A first-order linear differential equation is a type of differential equation that involves first derivatives of a function. Such equations can be expressed in the general form \( y' + P(t)y = Q(t) \). Here, \( y' \) signifies the derivative of \( y \) with respect to \( t \), and both \( P(t) \) and \( Q(t) \) are functions of \( t \). These equations are called "linear" because the function \( y \) and its derivative appear to the first power only, without being multiplied together or raised to any other power.

To solve a first-order linear differential equation, we seek a function \( y(t) \) that satisfies this linear relationship between its derivative \( y' \) and itself. The solution must satisfy any given initial conditions, like \( y(0) = 0 \) in this exercise.
  • This particular equation involves constants instead of functions for \( P(t) \) and \( Q(t) \): \( P(t) = 0.1 \) and \( Q(t) = 5 \).
  • The linearity and simplicity of these equations make them quite common in fields such as engineering and the physical sciences.
Integrating Factor Method
The integrating factor method is a powerful technique for solving first-order linear differential equations. It cleverly simplifies the equation into an easily integrable form. The integrating factor, commonly denoted as \( \mu(t) \), is crucial to this process.

To calculate the integrating factor, we use the formula \( \mu(t) = e^{\int P(t) \, dt} \). For the given equation, since \( P(t) = 0.1 \), the integrating factor becomes \( e^{0.1t} \).
  • Once the integrating factor is calculated, it is used to multiply through the entire differential equation, transforming it into a form where the left side is the derivative of a product \( \mu(t) y \).
  • This transformation simplifies the equation, allowing us to integrate both sides with respect to \( t \).
  • In this context, transforming the differential equation with an integrating factor sets the stage for resolving it by standard integration, which then is followed by applying given initial conditions.

This method is not only systematic but also highlights the elegance in solving linear differential equations, giving insight into how changes in systems accumulate over time.
Initial Value Problem
An initial value problem is a specific type of differential equation problem that, in addition to the equation itself, includes a condition specifying the value of the unknown function at a given point, usually \( t = 0 \). This initial condition is vital as it helps us find a precise solution rather than a general family of solutions.

In this exercise, the initial condition is \( y(0) = 0 \), meaning the pollutant level starts at zero. This initial caveat helps us determine the constant of integration that naturally arises when we solve the differential equation through integration.
  • Solving begins by first finding a general solution from the differential equation, which typically includes an arbitrary constant.
  • Applying the initial condition allows us to solve for this constant, thereby giving the specific solution to the problem at hand.

Initial value problems are prevalent across various scientific contexts, especially in modeling scenarios where an initial state of a system is known, and future states need predicting.

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Most popular questions from this chapter

Solve each first-order linear differential equation. $$ y^{\prime}-2 x y=0 $$

Your company has developed a new product, and your marketing department has predicted how it will sell. Let \(y(t)\) be the (monthly) sales of the product after \(t\) months. a. Write a differential equation that says that the rate of growth of the sales will be six times the twothirds power of the sales. b. Write an initial condition that says that at time \(t=0\) sales were 1000 c. Solve this differential equation and initial value. d. Use your solution to predict the sales at time \(t=12\) months.

A patient's ability to absorb a drug sometimes changes with time, and the dosage must therefore be adjusted. Suppose that the number of milligrams \(y(t)\) of a drug remaining in the patient's bloodstream after \(t\) hours satisfies $$ y^{\prime}=-\frac{1}{t} y+t \quad \text { (for } \left.t \geq 1\right) $$ $$ y(3)=5 $$ Solve this differential equation and initial condition to find the amount remaining in the bloodstream after \(t\) hours.

A company grows in value by \(10 \%\) each year, and also gains \(20 \%\) of a growing market estimated at \(100 e^{0.1 t}\) million dollars, where \(t\) is the number of years that the company has been in business. Therefore, the value \(y(t)\) of the company (in millions of dollars) satisfies $$ \begin{aligned} y^{\prime} &=0.1 y+20 e^{0.1 t} \\ y(0) &=5 \end{aligned} $$ a. Solve this differential equation and initial condition to find a formula for the value of the company after \(t\) years. b. Use your solution to find the value of the company after 2 years.

The following exercises require the use of a slope field program. For each differential equation and initial condition: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5] b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the point (0,2) c. Solve the differential equation and initial condition. d. Use your slope field program to graph the slope field and the solution that you found in part (c). How good was the sketch that you made in part (b) compared with the solution graphed in part (d)? $$ \left\\{\begin{array}{l} \frac{d y}{d x}=\frac{x^{2}}{y^{2}} \\ y(0)=2 \end{array}\right. $$

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