Question:

The integrating factor of the differential equation \[ x \frac{dy}{dx} + 2y = x e^x \] is:

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For first-order linear differential equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is given by \( \mu(x) = e^{\int P(x) \, dx} \), which can be used to solve the equation.
Updated On: Mar 12, 2025
  • \( \log_e x \)
  • \( \log_e 2x \)
  • \( x \)
  • \( x^2 \)
  • \( 2x \)
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The Correct Option is D

Solution and Explanation

We are given the first-order linear differential equation: \[ x \frac{dy}{dx} + 2y = x e^x \] Step 1: Rewrite in standard form
First, we rewrite the equation in standard linear form: \[ \frac{dy}{dx} + \frac{2}{x} y = e^x \] Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} \] where \( P(x) = \frac{2}{x} \). Thus, the integrating factor is: \[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \log x} = x^2 \] Step 3: Conclusion
Thus, the integrating factor is \( x^2 \).
Thus, the correct answer is option (D), \( x^2 \).
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