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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 9, Problem 9.PE.10

In Exercises 1–22, solve the differential equation.
y'/2 + y = e⁻ˣ sin x

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1
Rewrite the given differential equation to isolate \( y' \). Starting with \( \frac{y'}{2} + y = e^{-x} \sin x \), multiply both sides by 2 to get \( y' + 2y = 2 e^{-x} \sin x \).
Recognize that this is a first-order linear differential equation of the form \( y' + P(x) y = Q(x) \), where \( P(x) = 2 \) and \( Q(x) = 2 e^{-x} \sin x \).
Find the integrating factor \( \mu(x) \) using the formula \( \mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x} \).
Multiply the entire differential equation by the integrating factor \( e^{2x} \) to obtain \( e^{2x} y' + 2 e^{2x} y = 2 e^{2x} e^{-x} \sin x \), which simplifies to \( \frac{d}{dx} \left( e^{2x} y \right) = 2 e^{x} \sin x \).
Integrate both sides with respect to \( x \): \( \int \frac{d}{dx} \left( e^{2x} y \right) dx = \int 2 e^{x} \sin x \, dx \). This gives \( e^{2x} y = \int 2 e^{x} \sin x \, dx + C \), where \( C \) is the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Linear Differential Equations

A first-order linear differential equation has the form y' + P(x)y = Q(x). Solving it involves finding an integrating factor to simplify the equation into an exact derivative, allowing integration to find the general solution.
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