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=-((e+1)(1+cos(t)))/2+1+e

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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 = - \frac{( e + 1 ) ( 1 + cos ( t ) )}{2} + 1 + e

Solution steps

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

Solve separable ODE:

y = - \frac{( e + 1 ) ( 1 + cos ( t ) )}{2} + 1 + e

y = - \frac{( e + 1 ) ( 1 + cos ( t ) )}{2} + 1 + e

(x - 2)^{2} (x - 1)

\frac{d v}{d t} = 0.64 - \frac{v}{29400} \cdot 8, v (0) = 0

cos (2 (t + 1))

\int \frac{x ^{2} - 16 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\frac{\partial}{\partial z} (x y + z^{2})

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=-((e+1)(1+cos(t)))/2+1+e