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What is the integrating factor of the equation \( \frac{d y}{d x}-\frac{y}{x}=1 \) ? A \( e^{\int-x d x} \) B \( e^{\int-\frac{1}{x} d x} \) C \( e^{\int \frac{1}{x} d x} \) D \( e^{\int x d x} \)

Ask by Newton Marsh. in Malaysia
Mar 08,2025

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Answer

The integrating factor is \( e^{\int -\frac{1}{x} \, dx} \), which is option B.

Solution

The differential equation is \[ \frac{dy}{dx} - \frac{y}{x} = 1. \] This is a linear first-order differential equation in the form \[ \frac{dy}{dx} + P(x)y = Q(x), \] where in our case \[ P(x) = -\frac{1}{x} \quad \text{and} \quad Q(x) = 1. \] The integrating factor is given by \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx}. \] So, the integrating factor is \[ e^{\int -\frac{1}{x} \, dx}, \] which corresponds to option B.

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Beyond the Answer

To find the integrating factor for the differential equation \( \frac{dy}{dx} - \frac{y}{x} = 1 \), we use the standard form of a linear first-order differential equation, which can be written as \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \( P(x) = -\frac{1}{x} \). The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} \). So, the correct answer is: **B \( e^{\int -\frac{1}{x} \, dx} \)**. This integrating factor allows us to solve the differential equation more easily, transforming it into an exact equation where we can gather \( y \) and \( x \) terms together. Once the equation is multiplied by this integrating factor, you can integrate both sides to get a particular solution involving \( y \) and \( x \).

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