The integrating factor of the differential equation (y log_ey)dy/dx+x…

The correct answer is Option (3) → $\log_ey$

$(y \log_ey)\frac{dy}{dx}+x=2\log_ey$

$⇒\frac{dy}{dx}+\frac{1}{y\log_ey}x=\frac{2}{y}$

$⇒\frac{dy}{dx}=\frac{2\log_ey-x}{y\log_ey}$

$⇒\frac{dx}{dy}=\frac{y\log_ey}{2\log_ey-x}$

$⇒\frac{dx}{dy}+\frac{1}{y\log_ey}x=\frac{2}{y}$

$e^{\int\frac{1}{y\log_ey}dy}=e^{\log_e(\log_ey)}+C$

$I.F.=e^{\log_e(\log_ey)}$

$=\log_ey$