The integrating factor of the differential equation \((x\log_e x)\displaystyle\frac{dy}{dx}+y=2\log_e x\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log_e x\)
Dividing by \(x\log_e x\) gives the linear form \(\displaystyle\frac{dy}{dx} + Py = Q\):
\( \displaystyle \frac{dy}{dx} + \frac{1}{x\log_e x}\,y = \frac{2}{x} \)
So \(P = \displaystyle\frac{1}{x\log_e x}\). Putting \(t = \log_e x\), \(dt = \displaystyle\frac{dx}{x}\):
\( \displaystyle \int \frac{1}{x\log_e x}\,dx = \int \frac{dt}{t} = \log_e(\log_e x) \)
The integrating factor is:
\( \displaystyle \text{I.F.} = e^{\int P\,dx} = e^{\log_e(\log_e x)} = \log_e x \)