The integrating factor of the differential equation, $x^2\frac{dy}{dx}+xy = \log_ex$ is equal to

$x$

The correct answer is Option (3) → $x$

The given equation is

$x^{2}\frac{dy}{dx}+xy=\log x$

Divide by $x^{2}$:

$\frac{dy}{dx}+\frac{1}{x}y=\frac{\log x}{x^{2}}$

This is linear of the form

$\frac{dy}{dx}+P(x)y=Q(x)$ where $P(x)=\frac{1}{x}$

Integrating factor is

$IF=e^{\int \frac{1}{x}dx}$

$=e^{\log x}$

$=x$

Integrating factor is $x$.