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

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

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

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

y = ln (\frac{c _{1}}{1 + cos ( x )} - 1)

Solution steps

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

Solve separable ODE:

y = ln (\frac{c _{1}}{1 + cos ( x )} - 1)

y = ln (\frac{c _{1}}{1 + cos ( x )} - 1)

\int \frac{1}{( 2 x - 1 ) ^{2}} d x

\frac{\partial}{\partial x} (arctan (x^{2} y))

n \to \infty lim (\frac{10}{n})

\frac{d}{d t} (\frac{x}{t})

\int_{- \infty}^{- 1} 0 d x

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