What is the solution for y^'=(-0.8)/(cos(y)),y(0)= pi/2 ?

The solution for y^'=(-0.8)/(cos(y)),y(0)= pi/2 is y=arcsin(-0.8t+1)+2pin,y=pi+arcsin(0.8t-1)+2pin

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y^'=(-0.8)/(cos(y)),y(0)= pi/2

y^{^{'}} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

Solution steps

y^{'} (t) = \frac{- 0.8}{cos ( y )}

Solve separable ODE:

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

f^{^{'}} (x) = f (x) (x + 1)

y^{^{'}} = 0.2877 y, y (0) = 150

y^{^{'}} = 5 x y^{2}

y^{^{'}} = x^{2} y

y^{^{'}} - \frac{x}{5 y ^{4}} = 0, y (0) = 2

What is the solution for y^'=(-0.8)/(cos(y)),y(0)= pi/2 ?
The solution for y^'=(-0.8)/(cos(y)),y(0)= pi/2 is y=arcsin(-0.8t+1)+2pin,y=pi+arcsin(0.8t-1)+2pin