What is the solution for (1+cos(x))y^'=(1+e^{-y})sin(x),y(0)=0 ?

The solution for (1+cos(x))y^'=(1+e^{-y})sin(x),y(0)=0 is y=-ln((-cos(x)-1)/(cos(x)-3))

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(1+cos(x))y^'=(1+e^{-y})sin(x),y(0)=0

(1 + cos (x)) y^{^{'}} = (1 + e^{- y}) sin (x), y (0) = 0

y = - ln (\frac{- cos ( x ) - 1}{cos ( x ) - 3})

Solution steps

(1 + cos (x)) y^{'} (x) = (1 + e^{- y}) sin (x)

Solve separable ODE:

y = - ln (\frac{- cos ( x ) - 1}{cos ( x ) - 3})

y = - ln (\frac{- cos ( x ) - 1}{cos ( x ) - 3})

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}

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

e^{y} \frac{d y}{d x} - x^{2} = 0

\frac{d x}{d y} = e^{x + y}

\frac{d y}{d x} = (\frac{5 y}{x})

What is the solution for (1+cos(x))y^'=(1+e^{-y})sin(x),y(0)=0 ?
The solution for (1+cos(x))y^'=(1+e^{-y})sin(x),y(0)=0 is y=-ln((-cos(x)-1)/(cos(x)-3))