Integrating factor of $(x\, log_ex)\frac{dy}{dx}+y = 2log_e x $ is :
Answer & explanation
Correct answer: option 3
The correct answer is option (3) → $\log_ex$
dividing eq. by $x\log x$
so $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x}$
$I.F.=e^{\int\frac{1}{x\log x}}dx=e^{\log\log x}=\log x$