What is the solution for (1+cos(t))y^'=(1+e^{-y})sin(t),y(0)=0 ?

The solution for (1+cos(t))y^'=(1+e^{-y})sin(t),y(0)=0 is y=-ln((-cos(t)-1)/(cos(t)-3))

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(1+cos(t))y^'=(1+e^{-y})sin(t),y(0)=0

(1 + cos (t)) y^{^{'}} = (1 + e^{- y}) sin (t), y (0) = 0

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

Solution steps

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

Solve separable ODE:

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

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

(x^{2} + 5) (8 - x)

\int \frac{2 x}{1 + 4 x ^{2}} d x

\frac{6}{x ^{3}}

\int \frac{3 x ^{2}}{4 5 - x ^{3}} d x

f (x) = 3 x^{2} - 3 x, at x = 3

What is the solution for (1+cos(t))y^'=(1+e^{-y})sin(t),y(0)=0 ?
The solution for (1+cos(t))y^'=(1+e^{-y})sin(t),y(0)=0 is y=-ln((-cos(t)-1)/(cos(t)-3))