The integrating factor of the differential equation \((y\log_e y)\displaystyle\frac{dx}{dy} + x = 2\log_e y\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\log_e y\)
The equation is linear in x (x as a function of y). Dividing by \(y\log_e y\):
\( \displaystyle \frac{dx}{dy} + \frac{1}{y\log_e y}\,x = \frac{2}{y} \)
This is of the form \(\frac{dx}{dy} + P\,x = Q\) with
\( \displaystyle P = \frac{1}{y\log_e y} \)
The integrating factor is \(e^{\int P\,dy}\). Putting \(t = \log_e y\), \(dt = \frac{dy}{y}\):
\( \displaystyle \int\frac{dy}{y\log_e y} = \int\frac{dt}{t} = \log_e(\log_e y) \)
\( \displaystyle \text{I.F.} = e^{\log_e(\log_e y)} = \log_e y \)