Question

The integrating factor of the differential equation \[ (1 - x^2) \frac{dy}{dx} + xy = ax, \quad -1 < x < 1, \] is:

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

Hint:

To find the integrating factor, use \( e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \) in a linear differential equation.

Answer 1

The Correct Option is D

Step 1: Rewrite the given differential equation. \[ (1 - x^2) \frac{dy}{dx} + xy = ax \] Dividing throughout by \( (1 - x^2) \) gives: \[ \frac{dy}{dx} + \frac{xy}{1 - x^2} = \frac{ax}{1 - x^2} \] Step 2: Identify the integrating factor (IF). The standard form of a linear differential equation is: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \( P(x) = \frac{x}{1 - x^2} \). 
Step 3: Compute the integrating factor. \[ \text{IF} = e^{\int P(x) \, dx} \] Substituting \( P(x) \): \[ \int \frac{x}{1 - x^2} \, dx \] Letting \( u = 1 - x^2 \), \( du = -2x \, dx \): \[ = -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln |1 - x^2| \] Step 4: Simplify the integrating factor. \[ \text{IF} = e^{-\frac{1}{2} \ln |1 - x^2|} = (1 - x^2)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 - x^2}} \] Final Answer: \[ \boxed{\frac{1}{\sqrt{1 - x^2}}} \]