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Magazine Chapter 4: Problem 230
Solve the following differential equations by using integrating factors. $$ y^{\prime}=y+e^{x} $$
Short Answer
Expert verified
The solution is \( y = e^x(x + C) \).
Step by step solution
01
Identify the Standard Form
First, identify if the given differential equation can be expressed in the standard linear form of a first-order differential equation: \[ y' + P(x)y = Q(x) \] In this case, rewrite the equation \[ y' - y = e^x \] where, \[ P(x) = -1 \] and \[ Q(x) = e^x \] This is the standard form of a linear first-order differential equation.
02
Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by the formula: \[ \mu(x) = e^{\int P(x) \, dx} \]Compute this integrating factor for our equation:\[ \mu(x) = e^{-\int 1 \, dx} = e^{-x} \]
03
Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( e^{-x} \):\[ e^{-x}(y' - y) = e^{-x} \, e^x \]Simplifying the right side gives:\[ e^{-x}y' - e^{-x}y = 1 \]
04
Recognize the Left Side as a Product Derivative
The left side of the equation can now be recognized as the derivative of a product:\[ \frac{d}{dx}(e^{-x}y) = 1 \]
05
Integrate Both Sides
Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{-x}y) \, dx = \int 1 \, dx \]This gives:\[ e^{-x}y = x + C \]where \( C \) is the constant of integration.
06
Solve for \( y \)
Finally, solve for \( y \) by multiplying both sides by \( e^x \) to isolate \( y \):\[ y = e^x(x + C) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear First-Order Differential Equations
Understanding linear first-order differential equations is key to solving many real-world problems. These equations have the general form: \[ y' + P(x)y = Q(x) \]Where:
- \( y' \) is the derivative of \( y \) with respect to \( x \).
- \( P(x) \) and \( Q(x) \) are functions of \( x \), which can often be constants.
Derivative of a Product
Recognizing a derivative of a product can simplify complex expressions. When working with differential equations, this concept often arises after applying an integrating factor. Recall that the derivative of a product can be expressed as:\[ \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) \]For example, in the equation:\[ e^{-x}y' - e^{-x}y = 1 \]The left side can be condensed into a product derivative expression:\[ \frac{d}{dx}(e^{-x}y) = 1 \]This realization allows us to integrate simplistically, easing the solution path.
Constant of Integration
The constant of integration, represented as \( C \), emerges when integrating indefinite integrals. In the context of differential equations, it accounts for the family of solutions that differ by a constant shift. Upon integrating an expression like:\[ e^{-x}y = x + C \]The constant \( C \) is necessary to capture solutions that meet variable initial conditions, reflecting different specific scenarios.
Thus, if provided with initial conditions, you substitute them here to find the exact value of \( C \), tailoring the generic solution to meet particular requirements.
Thus, if provided with initial conditions, you substitute them here to find the exact value of \( C \), tailoring the generic solution to meet particular requirements.
Integrating Factor Formula
The integrating factor formula is a method to solve linear first-order differential equations. It involves multiplying through by a strategically chosen function. The integrating factor \( \mu(x) \) is defined as:\[ \mu(x) = e^{\int P(x) \, dx} \]where \( P(x) \) is part of the standard form.For the equation\[ y' - y = e^x \]we calculated\[ \mu(x) = e^{\int -1 \, dx} = e^{-x} \]This technique transforms the equation into one that can be written as the derivative of a product, allowing us to integrate both sides easily. Applying the integrating factor method is key for efficiently resolving first-order linear differential equations.
