Question

Consider the differential equation given below.
\[ \frac{dy}{dx} + \frac{x}{1 - x^2} y = x \sqrt{y}. \] The integrating factor of the differential equation is

• A. \( (1 - x^2)^{-\frac{3}{4}} \)
• B. \( (1 - x^2)^{-\frac{1}{4}} \)
• C. \( (1 - x^2)^{-\frac{3}{2}} \)
• D. \( (1 - x^2)^{-\frac{1}{2}} \)

Hint:

The integrating factor for a linear first-order differential equation is found by taking the exponential of the integral of the coefficient \( P(x) \).

Answer 1

The Correct Option is B

The given equation is: \[ \frac{dy}{dx} + \frac{x}{1 - x^2} y = x \sqrt{y}. \] This is a linear first-order differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) y^n, \] where \( P(x) = \frac{x}{1 - x^2} \) and \( Q(x) = x \) with \( n = \frac{1}{2} \). The integrating factor \( I(x) \) for a linear differential equation is given by: \[ I(x) = e^{\int P(x) \, dx}. \] To find the integrating factor, we compute the integral of \( P(x) \): \[ \int \frac{x}{1 - x^2} dx. \] The integral of \( \frac{x}{1 - x^2} \) is \( -\frac{1}{2} \ln(1 - x^2) \). Therefore, the integrating factor is: \[ I(x) = e^{-\frac{1}{2} \ln(1 - x^2)} = (1 - x^2)^{-\frac{1}{4}}. \] Final Answer: \( (1 - x^2)^{-\frac{1}{4}} \)