solve the diff.equation dy/dx -(1/x)y =2x^2​

Dear student, 

This is a first order linear differential equation. 

Let us compare this equation with the standard first order linear differential equation which is:

$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y} = \mathrm{Q}\left(\mathrm{x}\right)$
From this we can say, 

$\mathrm{P}\left(\mathrm{x}\right) = -\frac{1}{\mathrm{x}} \mathrm{and} \mathrm{Q}\left(\mathrm{x}\right) = 2{\mathrm{x}}^2$
Now, 
$$\begin{gathered} \mathrm{Integrating} \mathrm{Factor}(\mathrm{I}.\mathrm{F}.) = {\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right) \mathrm{dx}} = {\mathrm{e}}^{\int -\frac{1}{\mathrm{x}}\mathrm{dx}}= {\mathrm{e}}^{-\ln \left(\mathrm{x}\right)} = {\mathrm{e}}^{\ln (\mathrm{x}{)}^{-1}} = {\mathrm{x}}^{-1} = \frac{1}{\mathrm{x}} \\ \mathrm{The} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation} \mathrm{is} \\ \mathrm{y} \times \mathrm{IF} = \int \left[\mathrm{Q}\times \mathrm{IF}\right] \mathrm{dx} + \mathrm{C} \\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}} = \int 2{\mathrm{x}}^2 \times \frac{1}{\mathrm{x}} \mathrm{dx} + \mathrm{C} \\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}} = {\mathrm{x}}^2 + \mathrm{C} \\ \Rightarrow \mathrm{y} = {\mathrm{x}}^3 + \mathrm{Cx} \end{gathered}$$

Regards