Answer
$${y^2} + \sin y = {x^3} + {\pi ^2}$$
Work Step by Step
$$\eqalign{
& y' = \frac{{3{x^2}}}{{2y + \cos y}},\,\,\,\,\,\,\,\,{\text{initial condition }}y\left( 0 \right) = \pi \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{2y + \cos y}} \cr
& \cr
& {\text{separating the variables}} \cr
& \left( {2y + \cos y} \right) = 3{x^2}dx \cr
& {\text{integrate both sides of the equation}} \cr
& \int {\left( {2y + \cos y} \right)dy} = \int {3{x^2}} dx \cr
& {y^2} + \sin y = {x^3} + C \cr
& \cr
& {\text{use the initial condition }}y\left( 0 \right) = \pi \cr
& {\left( \pi \right)^2} + \sin \pi = {\left( 0 \right)^3} + C \cr
& {\pi ^2} + 0 = C \cr
& C = {\pi ^2} \cr
& \cr
& Then,{\text{ the particular solution of the differential equaion is}} \cr
& {y^2} + \sin y = {x^3} + {\pi ^2} \cr} $$
