Question:

The integrating factor of the differential equation \( \frac{dy}{dx} + \frac{2}{x} y = 0 \), \( (x \neq 0) \) is:

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For linear first-order differential equations, the integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} \). This factor helps to simplify the equation for easier integration.
  • \( \frac{2}{x} \)
  • \( x^2 \)
  • \( e^{\frac{2}{x}} \)
  • \( e^{\log(2x)} \)
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The Correct Option is B

Solution and Explanation

The given differential equation is: \[ \frac{dy}{dx} + \frac{2}{x} y = 0 \] This is a linear first-order differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] where \( P(x) = \frac{2}{x} \) and \( Q(x) = 0 \). 
Step 1: Find the integrating factor
The integrating factor \( \mu(x) \) is given by the formula: \[ \mu(x) = e^{\int P(x) \, dx} \] Substitute \( P(x) = \frac{2}{x} \) into the equation: \[ \mu(x) = e^{\int \frac{2}{x} \, dx} \] The integral of \( \frac{2}{x} \) is \( 2 \ln |x| \), so: \[ \mu(x) = e^{2 \ln |x|} = |x|^2 \] Since \( x \neq 0 \), we can write: \[ \mu(x) = x^2 \] 
Step 2: Conclusion
Thus, the integrating factor is \( x^2 \), which corresponds to option (B). 
 

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