What is the solution for y^'=(e^{-x})/(sin(y)) ?

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

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y^'=(e^{-x})/(sin(y))

y^{^{'}} = \frac{e ^{- x}}{sin ( y )}

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

Solution steps

y^{'} (x) = \frac{e ^{- x}}{sin ( y )}

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{x ^{2}}{1 - 2 y}

\frac{d y}{d x} = \frac{3 ( 1 + x )}{x y}

x y^{^{'}} + 4 y = 0

\frac{d y}{d x} = \frac{1 + y}{2 + x}

\frac{d y}{x} + \frac{d x}{y} = 0

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