Answer
$y(x)=e^{-1 / x}(x^2+C)$
Work Step by Step
$\begin{aligned} & y^{\prime}-\frac{1}{x^2} y=\frac{2 x^3 e^{-1 / x}}{x^2} \\ & \mu(x)=e^{\int \frac{-1}{x^2} d x} \\ & \mu(x)=e^{1 / x} \\ & \therefore y(x)=\frac{\int 2 e^{1 / x} * e^{-1 / x} * x}{e^{1 / x}}dx \\ & y(x)=e^{-1 / x}(x^2+C)\end{aligned}$
