Answer
$${\text{false}}$$
Work Step by Step
$$\eqalign{
& y\frac{{dy}}{{dx}} + xy = 2500{e^y} \cr
& {\text{divide both sides of the equation by }}y \cr
& \frac{y}{y}\frac{{dy}}{{dx}} + \frac{{xy}}{y} = \frac{{2500{e^y}}}{y} \cr
& \frac{{dy}}{{dx}} + x = \frac{{2500{e^y}}}{y} \cr
& {\text{subtract }}x + \frac{{2500{e^y}}}{y}{\text{ from each side}} \cr
& \frac{{dy}}{{dx}} - \frac{{2500{e^y}}}{y} = - x \cr
& {\text{this equation cannit be written in the form of a linear differential equation}} \cr
& \frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{the statement is false}}{\text{.}} \cr} $$
