Answer
$${\text{false}}$$
Work Step by Step
$$\eqalign{
& x\frac{{dy}}{{dx}} + 5y = {e^{2x}} \cr
& {\text{divide both sides by }}x \cr
& \frac{{dy}}{{dx}} + \left( {\frac{5}{x}} \right)y = \frac{{{e^{2x}}}}{x} \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + \left( {\frac{5}{x}} \right)y = \frac{1}{{{x^2}}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }}\frac{5}{x}{\text{ }} \cr
& {\text{the integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {\frac{5}{x}dx} }} \cr
& I\left( x \right) = {e^{5\ln x}} = {e^{\ln {x^5}}} \cr
& I\left( x \right) = {x^5} \cr
& {\text{then, the statement is false}} \cr} $$
