A integrating factor of the differential equation $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x^2};(x > 0)$ is equal to

$x$

The correct answer is Option (1) → $x$

$\text{Differential equation: }\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^2},\; x>0$

$\text{Integrating factor } \mu(x)=\exp\!\left(\int\frac{1}{x}\,dx\right)=\exp(\ln x)=x$

The integrating factor is $x$.