Answer
$$y = - \frac{{18}}{{37}}{e^{ - x/3}}\cos 2x - \frac{3}{{35}}{e^{ - x/3}}\sin 2x$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {e^{ - x/3}}\sin 2x,{\text{ }}\left( {0, - \frac{{18}}{{37}}} \right) \cr
& {\text{Separate the variables}} \cr
& dy = {e^{ - x/3}}\sin 2xdx \cr
& {\text{Integrate both sides}} \cr
& y = \int {{e^{ - x/3}}\sin 2x} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate by parts}} \cr
& {\text{ }}u = {e^{ - x/3}},{\text{ }}du = - \frac{1}{3}{e^{ - x/3}}dx; \cr
& dv = \sin 2xdx \Rightarrow v = - \frac{{\cos 2x}}{2} \cr
& {\text{Using the Integration - by - Parts Formula gives}} \cr
& \int {{e^{ - x/3}}\sin 2x} dx = \left( {{e^{ - x/3}}} \right)\left( { - \frac{{\cos 2x}}{2}} \right) - \int {\left( { - \frac{{\cos 2x}}{2}} \right)\left( { - \frac{1}{3}{e^{ - x/3}}} \right)} dx \cr
& \int {{e^{ - x/3}}\sin 2x} dx = - {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{6}\int {{e^{ - x/3}}} \cos 2xdx \cr
& {\text{Use integration by parts again for }}\int {{e^{ - x/3}}} \cos 2xdx \cr
& {\text{ }}u = {e^{ - x/3}},{\text{ }}du = - \frac{1}{3}{e^{ - x/3}}dx; \cr
& dv = \cos 2xdx \Rightarrow v = \frac{{\sin 2x}}{2} \cr
& {\text{Therefore,}} \cr
& \int {{e^{ - x/3}}\sin 2x} dx = \cr
& = - {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{6}\left( {{e^{ - x/3}}\frac{{\sin 2x}}{2} - \int {\left( {\frac{{\sin 2x}}{2}} \right)\left( { - \frac{1}{3}{e^{ - x/3}}} \right)dx} } \right) \cr
& = - {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{6}\left( {\frac{1}{2}{e^{ - x/3}}\sin 2x + \frac{1}{6}\int {{e^{ - x/3}}\sin 2xdx} } \right) \cr
& = - {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{{12}}{e^{ - x/3}}\sin 2x - \frac{1}{{36}}\int {{e^{ - x/3}}\sin 2xdx} \cr
& \int {{e^{ - x/3}}\sin 2x} dx + \frac{1}{{36}}\int {{e^{ - x/3}}\sin 2xdx} = {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{{12}}{e^{ - x/3}}\sin 2x \cr
& \frac{{37}}{{36}}\int {{e^{ - x/3}}\sin 2xdx} = - {e^{ - x/3}}\frac{{\cos 2x}}{2} - \frac{1}{{12}}{e^{ - x/3}}\sin 2x \cr
& {\text{Solve for }}\int {{e^{ - x/3}}\sin 2xdx} \cr
& \int {{e^{ - x/3}}\sin 2xdx} = \frac{{36}}{{37}}\left( { - {e^{ - x/3}}\frac{{\cos 2x}}{2}} \right) - \frac{{36}}{{37}}\left( {\frac{1}{{12}}{e^{ - x/3}}\sin 2x} \right) + C \cr
& \int {{e^{ - x/3}}\sin 2xdx} = - \frac{{18}}{{37}}{e^{ - x/3}}\cos 2x - \frac{3}{{35}}{e^{ - x/3}}\sin 2x + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& y = - \frac{{18}}{{37}}{e^{ - x/3}}\cos 2x - \frac{3}{{35}}{e^{ - x/3}}\sin 2x + C \cr
& {\text{Use the intial condition }}\left( {0, - \frac{{18}}{{37}}} \right) \cr
& - \frac{{18}}{{37}} = - \frac{{18}}{{37}}{e^0}\cos 0 - \frac{3}{{35}}{e^{ - x0/3}}\sin 0 + C \cr
& C = 0,{\text{ then}} \cr
& y = - \frac{{18}}{{37}}{e^{ - x/3}}\cos 2x - \frac{3}{{35}}{e^{ - x/3}}\sin 2x \cr
& \cr
& {\text{Graph}} \cr} $$
