Question

Find the integrating factor of the differential equation: \[ (3 \sin x \cos x) \, dy = (1 + 3y \sin^2 x) \, dx, \quad {where} \quad 0<x<\frac{\pi}{2} \] is

• A. \( \sec x \)
• B. \( \sin x \)
• C. \( \tan x \)
• D. \( \cos x \)
• E. \( \cot x \)

Hint:

The integrating factor for a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x) y = Q(x) \) is given by \( I(x) = e^{\int P(x) dx} \).

Answer 1

The Correct Option is D

Rewriting the given differential equation: \[ \frac{dy}{dx} - \frac{1 + 3y \sin^2 x}{3 \sin x \cos x} = 0 \] Rearranging: \[ \frac{dy}{dx} - \frac{3y \sin^2 x}{3 \sin x \cos x} = -\frac{1}{3 \sin x \cos x} \] Simplifying: \[ \frac{dy}{dx} - \frac{y \sin x}{\cos x} = -\frac{1}{3 \sin x \cos x} \] Recognizing the integrating factor \( I(x) \): \[ I(x) = e^{\int -\frac{\sin x}{\cos x} dx} = e^{-\ln |\cos x|} = \cos x \] Thus, the integrating factor is \( \cos x \).