Answer
$${y^2} = \sin {x^2} + 4$$
Work Step by Step
$$\eqalign{
& yy' - x\cos {x^2} = 0,{\text{ }}y\left( 0 \right) = - 2 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& y\frac{{dy}}{{dx}} - x\cos {x^2} = 0 \cr
& {\text{Add }}x\cos x{\text{ to both sides }} \cr
& y\frac{{dy}}{{dx}} = x\cos {x^2} \cr
& {\text{Separate the variables}} \cr
& ydy = x\cos {x^2}dx \cr
& {\text{Integrate both sides}} \cr
& \int y dy = \int {x\cos {x^2}} dx \cr
& \frac{1}{2}{y^2} = \frac{1}{2}\sin {x^2} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 0 \right) = - 2 \cr
& \frac{1}{2}{\left( { - 2} \right)^2} = \frac{1}{2}\sin {\left( 0 \right)^2} + C \cr
& 2 = C \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{1}{2}{y^2} = \frac{1}{2}\sin {x^2} + 2 \cr
& {y^2} = \sin {x^2} + 4 \cr} $$
