What is the solution for sin(y)y^'=2xcos(y) ?

The solution for sin(y)y^'=2xcos(y) is y=arccos(e^{-x^2-c_{1}})+2pin,y=-arccos(e^{-x^2-c_{1}})+2pin

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sin(y)y^'=2xcos(y)

sin (y) y^{^{'}} = 2 x cos (y)

y = arccos (e^{- x^{2} - c_{1}}) + 2 πn, y = - arccos (e^{- x^{2} - c_{1}}) + 2 πn

Solution steps

sin (y) y^{'} (x) = 2 x cos (y)

Solve separable ODE:

y = arccos (e^{- x^{2} - c_{1}}) + 2 πn, y = - arccos (e^{- x^{2} - c_{1}}) + 2 πn

y = arccos (e^{- x^{2} - c_{1}}) + 2 πn, y = - arccos (e^{- x^{2} - c_{1}}) + 2 πn

\frac{1}{2} \frac{d y}{d x} = y + 1 cos (x), y (π) = 0

\frac{d y}{d x} = 6 x sec (y)

y^{^{'}} = - 6 x y, y (0) = 7

\frac{d y}{d x} = \frac{x ^{2}}{y ^{3}}

y^{^{'}} + x^{3} y = 0

What is the solution for sin(y)y^'=2xcos(y) ?
The solution for sin(y)y^'=2xcos(y) is y=arccos(e^{-x^2-c_{1}})+2pin,y=-arccos(e^{-x^2-c_{1}})+2pin