Answer
$${\text{true}}$$
Work Step by Step
$$\eqalign{
& x\frac{{dy}}{{dx}} = \left( {x + 1} \right)\left( {y + 1} \right) \cr
& {\text{divide both sides of the equation by }}x\left( {y + 1} \right) \cr
& \frac{x}{{x\left( {y + 1} \right)}}\frac{{dy}}{{dx}} = \frac{{\left( {x + 1} \right)\left( {y + 1} \right)}}{{x\left( {y + 1} \right)}} \cr
& {\text{simplifying}} \cr
& \frac{1}{{y + 1}}\frac{{dy}}{{dx}} = \frac{{x + 1}}{x} \cr
& or \cr
& \frac{1}{{y + 1}}dy = \frac{{x + 1}}{x}dx \cr
& {\text{the differential equation can be written in the form}} \cr
& q\left( y \right)dy = p\left( x \right)dx \cr
& {\text{then, the statement is true}}{\text{. It is possible solve it using the method of}} \cr
& {\text{separation of variables}}{\text{.}} \cr} $$
