The integrating factor of the differential equation $(x\log_ex)\frac{dy}{dx}+y=2\log_ex$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\log_ex$
$ (x\log_{e}x)\frac{dy}{dx}+y=2\log_{e}x $
$\text{Rewrite in standard form: }$
$ \frac{dy}{dx}+\frac{1}{x\log_{e}x}\,y=\frac{2}{x} $
$\text{Here } P(x)=\frac{1}{x\log_{e}x}. $
$\text{Integrating factor } IF=e^{\int P(x)\,dx} =e^{\int \frac{1}{x\log_{e}x}\,dx } $
$ \int \frac{1}{x\log_{e}x}\,dx =\log(\log_{e}x) $
therefore $ IF=e^{\log(\log_{e}x)}=\log_{e}x $
The integrating factor is $\log_{e}x$.