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Chapter 9: Problem 1

Solve and check each first-order linear differential equation. $$ y^{\prime}+2 y=8 $$

Short Answer

Expert verified
The solution to the differential equation is \( y = 4 + Ce^{-2x} \).

Step by step solution

01

Recognize the Equation Form

The given differential equation is a first-order linear differential equation, which has the standard form \( y' + p(x)y = q(x) \). In this case, \( p(x) = 2 \) and \( q(x) = 8 \).
02

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int p(x) \, dx} \). Since \( p(x) = 2 \), we have:\[ \mu(x) = e^{\int 2 \, dx} = e^{2x}. \]
03

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( e^{2x} \):\[ e^{2x} y' + 2e^{2x} y = 8e^{2x}. \]
04

Write the Left Side as a Derivative

Notice that the left side of the equation is the derivative of \( e^{2x}y \):\[ \frac{d}{dx}(e^{2x} y) = 8e^{2x}. \]
05

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{2x} y) \, dx = \int 8e^{2x} \, dx. \]On the left side, integration gives \( e^{2x} y \), and on the right side, we find:\[ e^{2x} y = 4e^{2x} + C, \] where \( C \) is the constant of integration.
06

Solve for \( y \)

Divide both sides by \( e^{2x} \) to solve for \( y \):\[ y = 4 + Ce^{-2x}. \]
07

Verify the Solution

Differentiate \( y = 4 + Ce^{-2x} \) to find \( y' \):\[ y' = -2Ce^{-2x}. \]Substitute \( y = 4 + Ce^{-2x} \) and \( y' = -2Ce^{-2x} \) into the original equation:\[ -2Ce^{-2x} + 2(4 + Ce^{-2x}) = 8. \]Simplifying:\[ -2Ce^{-2x} + 8 + 2Ce^{-2x} = 8. \]The terms with \( Ce^{-2x} \) cancel, and we are left with \( 8 = 8 \), which is true. Therefore, the solution satisfies the original differential equation.

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Key Concepts

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

Integrating Factor
An integrating factor is a tool used to solve first-order linear differential equations. It makes these equations easier to solve by transforming them into a form that can be easily integrated. For any equation of the form \( y' + p(x)y = q(x) \), the integrating factor \( \mu(x) \) is found using the formula: \[ \mu(x) = e^{\int p(x) \, dx} \]In our exercise, we have \( p(x) = 2 \). Thus, the integrating factor becomes:
  • \( \mu(x) = e^{\int 2 \, dx} = e^{2x} \)
The purpose of this integrating factor is to multiply it through the entire differential equation, aiding in simplifying it for further calculations. The resulting equation will allow us to write the left side as a derivative, preparing it for integration.
Standard Form of Linear Differential Equations
First-order linear differential equations hold an important place in differential equations. Their standard form is given by \( y' + p(x)y = q(x) \). This form is crucial because it sets up the equation for subsequent steps like finding the integrating factor.In our example, the differential equation \( y' + 2y = 8 \) fits perfectly into this standard form:
  • \( p(x) = 2 \)
  • \( q(x) = 8 \)
Recognizing the standard form helps in systematically approaching the solution by identifying \( p(x) \) and \( q(x) \), which are necessary for computing the integrating factor and integrating the transformed equation. The structure of the standard form creates a straightforward path towards solving the equation.
Solution Verification
Verifying a solution ensures that it satisfies the original differential equation. After obtaining the potential solution \( y = 4 + Ce^{-2x} \), one must substitute this back into the original differential equation to check its validity.We do this by first differentiating the solution:
  • \( y' = -2Ce^{-2x} \)
Substituting \( y = 4 + Ce^{-2x} \) and \( y' = -2Ce^{-2x} \) into the original equation \( y' + 2y = 8 \) yields:
  • \( -2Ce^{-2x} + 2(4 + Ce^{-2x}) = 8 \)
This simplifies to \( 8 = 8 \), confirming that the terms involving \( Ce^{-2x} \) cancel out, thereby validating the solution. Successful verification assures the correctness and consistency of the solution method used.

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Most popular questions from this chapter

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)? $$ \left\\{\begin{array}{l} \frac{d y}{d x}=\frac{6 x^{2}}{y^{4}} \\ y(0)=2 \end{array}\right. $$

For each initial value problem, calculate the Euler approximation for the solution on the interval [0,1] using \(n=4\) segments. Draw the graph of your approximation. (Carry out the calculations "by hand" with the aid of a calculator, rounding to two decimal places. Answers may differ slightly, depending on when you do the rounding.) $$ \begin{array}{l} y^{\prime}+2 y=e^{4 x} \\ y(0)=2 \end{array} $$

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=x y-5 x \\ y(0)=4 \end{array}\right.$$

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?

The water in a 100,000 -gallon reservoir contains 0.1 gram of pesticide per gallon. Each hour, 2000 gallons of water (containing 0.01 gram of pesticide per gallon) is added and mixed into the reservoir, and an equal volume of water is drained off. a. Write a differential equation and initial condition that describe the amount \(y(t)\) of pesticide in the reservoir after \(t\) hours. b. Solve this differential equation and initial condition. c. Graph your solution on a graphing calculator and find when the amount of pesticide will reach 0.02 gram per gallon, at which time the water is safe to drink. d. Use your solution to find the "long-run" amount of pesticide in the reservoir.
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