Answer
$$2\sqrt y = x\sin x + \cos x + 3$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = x\sqrt y \cos x,{\text{ }}\left( {0,4} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{1}{{\sqrt y }}dy = x\cos xdx \cr
& {y^{ - 1/2}}dy = x\cos xdx \cr
& {\text{Integrate both sides}} \cr
& \int {{y^{ - 1/2}}} dy = \int {x\cos x} dx \cr
& \frac{{{y^{1/2}}}}{{1/2}} = \int {x\cos x} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{*Integrate }}\int {x\cos x} dx{\text{ by parts}} \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x \Rightarrow du = dx \cr
& dv = \cos xdx \Rightarrow v = \sin x \cr
& {\text{Integration by parts formula }} \cr
& \int {x\cos x} dx = x\sin x - \int {\sin x} dx \cr
& \int {x\cos x} dx = x\sin x + \cos x + C \cr
& {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr
& \frac{{{y^{1/2}}}}{{1/2}} = x\sin x + \cos x + C{\text{ }} \cr
& 2\sqrt y = x\sin x + \cos x + C,{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Use the initial condition }}\left( {0,4} \right) \cr
& 2\sqrt 4 = \left( 0 \right)\sin \left( 0 \right) + \cos \left( 0 \right) + C \cr
& 4 = 1 + C \cr
& C = 3 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{2}} \right) \cr
& 2\sqrt y = x\sin x + \cos x + 3 \cr
& \cr
& {\text{Graph}} \cr} $$
