# y'=6x^2/(2y+cosy))

${y}^{\prime }=6\frac{{x}^{2}}{2y+\mathrm{cos}y}$

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Nathalie Redfern
This is a separable equation, which can be written as $\left(2y+\mathrm{cos}y\right)dy=6{x}^{2}dx$
Integrate: $\int \left(2y+\mathrm{cos}y\right)dy=\int 6{x}^{2}dx$
LHS: $\int \left(2y+\mathrm{cos}y\right)dy={y}^{2}+\mathrm{sin}y+C1,$ where C1 is some constant.
RHS: $\int 6{x}^{2}dx=2{x}^{3}+C2$ where C2 is some constant.
So, from (1) it follows that ${y}^{2}+\mathrm{sin}y=2{x}^{3}+C$ where we defined C as C2-C1