Answer
$$\frac{1}{2}{y^2} = - 3\cos x + 5$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{3\sin x}}{y},{\text{ }}\left( {0,2} \right) \cr
& {\text{Separate the variables}} \cr
& ydy = 3\sin xdx \cr
& {\text{Integrate both sides}} \cr
& \int y dy = 3\int {\sin x} dx \cr
& \frac{1}{2}{y^2} = - 3\cos x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0,2} \right) \cr
& \frac{1}{2}{\left( 2 \right)^2} = - 3\cos \left( 0 \right) + C \cr
& 2 = - 3 + C \cr
& C = 5 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{1}{2}{y^2} = - 3\cos x + 5 \cr
& \cr
& {\text{Graph}} \cr} $$
