#### Answer

$${\text{true}}$$

#### Work Step by Step

$$\eqalign{
& \frac{{dy}}{{dx}} + 3\frac{y}{x} = \frac{1}{{{x^2}}} \cr
& or \cr
& \frac{{dy}}{{dx}} + \left( {\frac{3}{x}} \right)y = \frac{1}{{{x^2}}} \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + \left( {\frac{3}{x}} \right)y = \frac{1}{{{x^2}}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }}\frac{3}{x}{\text{ }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {\frac{3}{x}dx} }} \cr
& I\left( x \right) = {e^{3\ln x}} = {e^{\ln {x^3}}} \cr
& I\left( x \right) = {x^3} \cr
& {\text{then, the statement is true}} \cr} $$