What is the solution for (1+cos(t))y^'=(1+e^{-y})sin(t) ?

The solution for (1+cos(t))y^'=(1+e^{-y})sin(t) is y=ln((c_{1})/(1+cos(t))-1)

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(1+cos(t))y^'=(1+e^{-y})sin(t)

(1 + cos (t)) y^{^{'}} = (1 + e^{- y}) sin (t)

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

Solution steps

(1 + cos (t)) y^{'} (t) = (1 + e^{- y}) sin (t)

Solve separable ODE:

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

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

\frac{d}{d x} (lo g_{4} (2 x + 3))

f (x) = (5 t^{2} - 2 t + 1)^{\frac{3}{2}}

\int_{1}^{2} \frac{1}{( x + 1 ) ^{2}} d x

f (x) = ∣ x - 1 ∣

What is the solution for (1+cos(t))y^'=(1+e^{-y})sin(t) ?
The solution for (1+cos(t))y^'=(1+e^{-y})sin(t) is y=ln((c_{1})/(1+cos(t))-1)