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=-e^{-c_{1}}-e^{-c_{1}}cos(t)+1+e

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

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

y = - e^{- c_{1}} - e^{- c_{1}} cos (t) + 1 + e

Solution steps

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

Solve separable ODE:

y = - e^{- c_{1}} - e^{- c_{1}} cos (t) + 1 + e

y = - e^{- c_{1}} - e^{- c_{1}} cos (t) + 1 + e

y^{^{''}} + 4 y = 8 tan (2 x)

y^{^{''}} + 2 y = - 18 x^{2} e^{2 x}

\int_{1}^{0} - 9 x d x

y^{^{''}} - 4 y^{^{'}} - 12 y = t e^{4 t}

\int_{- 2}^{2} \frac{1}{2} (4^{2} - x^{4}) d x

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=-e^{-c_{1}}-e^{-c_{1}}cos(t)+1+e