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Chapter 9: Problem 4

Solve and check each first-order linear differential equation. $$ y^{\prime}+3 y=e^{3 x} $$

Short Answer

Expert verified
The solution to the differential equation is \( y = \frac{1}{6} e^{3x} + Ce^{-3x} \).

Step by step solution

01

Recognize the Standard Form

The given differential equation is \( y^{\prime} + 3y = e^{3x} \). This is already in the standard form of a first-order linear differential equation: \( y^{\prime} + P(x) y = Q(x) \) where \( P(x) = 3 \) and \( Q(x) = e^{3x} \).
02

Determine the Integrating Factor

For first-order linear differential equations, we use the integrating factor method. The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \). Here, \( P(x) = 3 \), so \( \mu(x) = e^{\int 3 \, dx} = e^{3x} \).
03

Multiply the Entire Equation by the Integrating Factor

Multiply each term in the equation by the integrating factor \( e^{3x} \): \[ e^{3x} y^{\prime} + 3 e^{3x} y = e^{3x} e^{3x} \]This simplifies to: \[ e^{3x} y^{\prime} + 3 e^{3x} y = e^{6x} \].
04

Recognize the Left-hand Side as a Product Derivative

The left-hand side of the equation \( e^{3x} y^{\prime} + 3 e^{3x} y \) can be written as the derivative of a product: \[ \frac{d}{dx}(e^{3x} y) = e^{6x} \].
05

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{3x} y) \, dx = \int e^{6x} \, dx \].This yields:\[ e^{3x} y = \frac{1}{6} e^{6x} + C \], where \( C \) is the constant of integration.
06

Solve for \( y \)

Solve for \( y \) by dividing each side by \( e^{3x} \):\[ y = \frac{1}{6} e^{3x} + C e^{-3x} \].
07

Check the Solution

Plug \( y = \frac{1}{6} e^{3x} + C e^{-3x} \) into the original equation to confirm it satisfies it:Calculate \( y' = \frac{1}{2} e^{3x} - 3C e^{-3x} \). Substitute \( y \) and \( y' \) back into the original equation:\[ (\frac{1}{2} e^{3x} - 3C e^{-3x}) + 3(\frac{1}{6} e^{3x} + C e^{-3x}) = e^{3x} \]Simplifying verifies equality, confirming the solution is correct.

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Key Concepts

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

Integrating Factor
When faced with a first-order linear differential equation, one powerful technique to solve it is the integrating factor method. This method involves a particular function that transforms the equation into a more easily solvable form. For our equation, identified as \( y^{\prime} + 3y = e^{3x} \), the integrating factor \( \mu(x) \) is given by the formula \( e^{\int P(x) \, dx} \). Here, \( P(x) \) is the coefficient of \( y \), which is 3.
So, \( \mu(x) = e^{\int 3 \, dx} = e^{3x} \).
  • The purpose of an integrating factor is to rewrite the differential equation so that its left-hand side becomes the derivative of a product. This simplifies the process of solving the equation.
  • It essentially harmonizes the equation, making it much easier to integrate.
In our example, multiplying the entire equation \( y^{\prime} + 3y = e^{3x} \) by the integrating factor \( e^{3x} \) helps convert the left side into an exact derivative.
Product Derivative
The beauty of the integrating factor is that it often turns the left-hand side of the differential equation into a product derivative. This is key to solving first-order linear differential equations.
In our equation, after multiplying by the integrating factor \( e^{3x} \), we obtained:\[ e^{3x} y^{\prime} + 3e^{3x} y = e^{6x} \] This can be restated as the derivative of a product, specifically:\[ \frac{d}{dx}(e^{3x} y) = e^{6x} \]
  • A product derivative combines two individual functions into a single differential expression.
  • Recognizing this form is crucial as it indicates that the left-hand side has been successfully prepared for integration.
Applying the product rule in reverse allows us to simplify the equation, significantly easing the task of integration on both sides.
Constant of Integration
When we integrate a differential equation, a constant of integration, denoted as \( C \), often appears. This constant represents a family of solutions for the differential equation, reflecting the infinite possibilities of initial conditions that might specify a unique solution.
In our integrated result:\[ e^{3x} y = \frac{1}{6} e^{6x} + C \]
The \( C \) emerges from the integration process. This term is critical because it accounts for all possible solutions of the differential equation based on different initial values.
  • Without the constant of integration, we would have a specific, rather than general, solution.
  • The constant allows us to encapsulate the general case of all possible solutions.
  • When initial or boundary conditions are provided, the exact value of \( C \) can be determined, reducing the general solution to a particular one.
This is why the constant of integration is essential in expressing the complete solution to differential equations.

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Most popular questions from this chapter

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 four times the one-half power of the sales. b. Write an initial condition that says that at time \(t=0\) sales were 10,000 c. Solve this differential equation and initial value. d. Use your solution to predict the sales at time \(t=12\) months.

A medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body \(t\) hours after the murder, and if the room temperature is \(70^{\circ},\) then \(y\) satisfies $$ \begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &=98.6 \text { (body temperature initially } \left.98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the \text { body when it was discovered was } 80^{\circ} .

A 10,000 -cubic-foot-room has an initial radon level of \(800 \mathrm{pCi}\) (picocuries) per cubic foot. A ventilation system is installed that each hour brings in 500 cubic feet of outside air (containing \(5 \mathrm{pCi}\) per cubic foot), while an equal volume of air leaves the room. Assume that the air in the room mixes thoroughly. a. Find a differential equation and initial condition that govern the total amount \(y(t)\) of radon in the room after \(t\) hours. b. Solve this differential equation and initial condition. c. Graph the solution on a graphing calculator and find how soon the radon level will fall to the EPA safety level of \(112 \mathrm{pCi}\) per cubic foot.

Solve each differential equation with the given initial condition. $$ \begin{array}{l} x y^{\prime}=2 y+x^{2} \\ y(1)=3 \end{array} $$

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=\sqrt{y} e^{x}-\sqrt{y} \\ y(0)=1 \end{array}\right.$$
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