Answer
$$\frac{1}{2}{y^2} = - x\cos x + \sin x + 8$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{x}{y}\sin x,{\text{ }}\left( {0,4} \right) \cr
& {\text{Separate the variables}} \cr
& ydy = x\sin xdx \cr
& {\text{Integrate both sides}} \cr
& \frac{1}{2}{y^2} = \int {x\sin x} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{*Integrate }}\int {x\sin x} dx{\text{ by parts}} \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x \Rightarrow du = dx \cr
& dv = \sin xdx \Rightarrow v = - \cos x \cr
& {\text{Integration by parts formula }} \cr
& \int {x\sin x} dx = - x\cos x + \int {\cos x} dx \cr
& \int {x\sin x} dx = - x\cos x + \sin x + C \cr
& {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr
& \frac{1}{2}{y^2} = - x\cos x + \sin x + C{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Use the intial condition }}\left( {0,4} \right) \cr
& \frac{1}{2}{\left( 4 \right)^2} = - 0\cos \left( 0 \right) + \sin \left( 0 \right) + C \cr
& C = 8 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{2}} \right) \cr
& \frac{1}{2}{y^2} = - x\cos x + \sin x + 8 \cr
& \cr
& {\text{Graph}} \cr} $$
