#### Answer

$${\text{linear and separable}}$$

#### Work Step by Step

$$\eqalign{
& \frac{{dy}}{{dx}} + x = xy \cr
& {\text{subtract }}xy + x{\text{ from each side}} \cr
& \frac{{dy}}{{dx}} - xy = - x \cr
& {\text{The equation can be written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\,\,\,\left( {{\text{linear form}}} \right), \cr
& {\text{the given equation is linear}} \cr
& \cr
& and \cr
& \frac{{dy}}{{dx}} = xy - x \cr
& \frac{{dy}}{{dx}} = x\left( {y - 1} \right) \cr
& \frac{{dy}}{{dx}} = \frac{x}{{{{\left( {y - 1} \right)}^{ - 1}}}} \cr
& {\text{The equation can be written in the form }}\frac{{dy}}{{dx}} = \frac{{p\left( x \right)}}{{q\left( y \right)}} \cr
& {\text{ then}}{\text{, the given equation separable}} \cr} $$