The integrating factor of differential equation \([y(1-x\tan x)+x^2\cos x]\,dx-x\,dy=0\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{x\cos x}\)
Dividing by \(x\,dx\) and rearranging:
\( \displaystyle \frac{dy}{dx} = \frac{y(1-x\tan x)}{x}+x\cos x \)
\( \displaystyle \frac{dy}{dx}+\left(\tan x-\frac{1}{x}\right)y = x\cos x \)
This is a linear differential equation \(\frac{dy}{dx}+Py = Q\) with \(P = \tan x-\frac{1}{x}\).
\( \displaystyle \int P\,dx = \log\sec x-\log x = \log\frac{\sec x}{x} \)
\( \displaystyle \text{I.F.} = e^{\int P\,dx} = \frac{\sec x}{x} = \frac{1}{x\cos x} \)