What is the solution for y^'=e^{-y}cos(t),y(pi/6)=1 ?

The solution for y^'=e^{-y}cos(t),y(pi/6)=1 is y=ln(sin(t)+e-1/2)

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y^'=e^{-y}cos(t),y(pi/6)=1

y^{^{'}} = e^{- y} cos (t), y (\frac{π}{6}) = 1

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} + \frac{3}{x} y = \frac{3}{x}

(6 \frac{d y}{d x}) - 36 y^{2} = 36

y^{^{'}} = 4 y^{- 2} (x^{6} + x^{4} + x^{2})

y^{^{'}} = (8 x - 5) (y + 2), y (0) = 7

cos (y) y^{^{'}} = sin^{2} (t) cos (t), y (0) = \frac{π}{6}

What is the solution for y^'=e^{-y}cos(t),y(pi/6)=1 ?
The solution for y^'=e^{-y}cos(t),y(pi/6)=1 is y=ln(sin(t)+e-1/2)