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Solve the differential equations in Exercises \(1-14\) $$e^{x} \frac{d y}{d x}+2 e^{x} y=1$$

Short Answer

Expert verified
The solution is \(y = \frac{1}{2} e^{-x} + Ce^{-3x}\).

Step by step solution

01

Recognize the Type of Differential Equation

The given differential equation is \(e^x \frac{dy}{dx} + 2e^x y = 1\). Notice that it is a first-order linear differential equation in standard form, \( \frac{dy}{dx} + P(x) y = Q(x) \), where \(P(x) = 2\) and \(Q(x) = e^{-x}\).
02

Find the Integrating Factor

The integrating factor \(\mu(x)\) for a linear differential equation is given by \(\mu(x) = e^{\int P(x) \, dx}\). Here, \(P(x) = 2\), so \(\int P(x) \, dx = 2x\). Thus, \(\mu(x) = e^{2x}\).
03

Multiply the Entire Equation by the Integrating Factor

Multiply through the equation by \(e^{2x}\), yielding:\[ e^{x} e^{2x} \frac{dy}{dx} + 2e^{x} e^{2x} y = e^{2x} \]This simplifies to:\[ e^{3x} \frac{dy}{dx} + 2e^{3x} y = e^{2x} \]
04

Express the Left Side as a Derivative

Notice that the left side can now be written as a derivative of the product of the integrating factor and \(y\), i.e.,\[ \frac{d}{dx} (e^{3x} y) = e^{2x} \]
05

Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(e^{3x} y) \, dx = \int e^{2x} \, dx \]The left side simplifies to \(e^{3x} y\), and the integral of the right side is \(\frac{1}{2}e^{2x}\). Thus:\[ e^{3x} y = \frac{1}{2} e^{2x} + C \]
06

Solve for \(y\)

Divide both sides by \(e^{3x}\) to solve for \(y\):\[ y = \frac{1}{2} e^{-x} + Ce^{-3x} \]

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Key Concepts

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

Integrating Factor
In solving first-order linear differential equations, one essential method is using an integrating factor. An integrating factor is a function that, when multiplied by the original equation, simplifies it into an easily integrable form. For any differential equation of the form
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
the integrating factor \( \mu(x) \) is defined as \( e^{\int P(x) \, dx} \). This clever trick transforms the original equation into one whose left-hand side is the derivative of a product. This can be integrated easily.

In the problem from the exercise, \( P(x) = 2 \), leading to the integrating factor \( e^{2x} \). Once multiplied throughout the equation, it allows the left side to be expressed as a derivative. Integrating factors are powerful because they make the equation's structure simpler and more straightforward to solve.
Standard Form Differential Equation
The standard form of a first-order linear differential equation provides a structured approach to problem-solving. These equations are expressed as:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
This format is not just convenient; it's essential for certain solution techniques, like the integrating factor method. Recognizing this form is the first step in utilizing the method efficiently.

In the given differential equation:
  • \( e^x \frac{dy}{dx} + 2e^x y = 1 \)
we rewrite it into its standard form by dividing through by \( e^x \). This yields \( \frac{dy}{dx} + 2y = e^{-x} \). Identifying and rewriting into this standard form simplifies subsequent steps, paving the way for integration and solution of the equation.
Solving Differential Equations
Solving differential equations, especially the first-order linear type, involves several steps which highlight their underlying structure. Once we've identified the equation's standard form, our primary objective is to transform it into an expression that's straightforward to integrate.

Using the integrating factor, the differential equation is multiplied through to modify it into a form where the left-hand side becomes a derivative of a product in terms of the variable \( y \). For example,
  • after multiplying through by \( e^{2x} \), the left side becomes \( \frac{d}{dx}(e^{3x}y) \).
Once both sides of the equation are integrated with respect to \( x \), it becomes possible to express \( y \) explicitly. This showcases the beauty of integrating factors—they convert a complex, multi-term equation into one that can be easily handled with basic calculus techniques.

Thus, solving these equations is an exercise in understanding the structure, manipulating terms via integrating factors, and applying fundamental integration skills to arrive at a solution expressing \( y \) in terms of \( x \).

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Most popular questions from this chapter

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Solve the differential equations in Exercises \(1-14\) $$x y^{\prime}-y=2 x \ln x$$

Solve the differential equations in Exercises \(1-14\) $$2 y^{\prime}=e^{x / 2}+y$$

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