Answer
$$y = 2x - \frac{1}{3}\cos 3x - \frac{{2\pi + 1}}{3}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 2 + \sin 3x \cr
& {\text{distributing the differentials}} \cr
& dy = \left( {2 + \sin 3x} \right)dx \cr
& {\text{integrating both sides of the equation}} \cr
& \int {dy} = \int {\left( {2 + \sin 3x} \right)} dx \cr
& y = 2x - \frac{1}{3}\cos 3x + C \cr
& {\text{using the initial value problem }}y\left( {\pi /3} \right) = 0 \cr
& 0 = 2\left( {\frac{\pi }{3}} \right) - \frac{1}{3}\cos 3\left( {\frac{\pi }{3}} \right) + C \cr
& 0 = \frac{{2\pi }}{3} - \frac{1}{3}\cos \pi + C \cr
& - \frac{{2\pi }}{3} = - \frac{1}{3}\left( { - 1} \right) + C \cr
& - \frac{{2\pi + 1}}{3} = C \cr
& C = - \frac{{2\pi + 1}}{3} \cr
& {\text{substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = 2x - \frac{1}{3}\cos 3x - \frac{{2\pi + 1}}{3} \cr} $$
