What is the solution for y^'=cos^2(t)cos^2(2y) ?

The solution for y^'=cos^2(t)cos^2(2y) is y=(arctan(t+1/2 sin(2t)+c_{1}))/2+(pin)/2

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y^'=cos^2(t)cos^2(2y)

y^{^{'}} = cos^{2} (t) cos^{2} (2 y)

y = \frac{arctan ( t + \frac{1}{2} sin ( 2 t ) + c _{1} )}{2} + \frac{πn}{2}

Solution steps

y^{'} (t) = cos^{2} (t) cos^{2} (2 y)

Solve separable ODE:

y = \frac{arctan ( t + \frac{1}{2} sin ( 2 t ) + c _{1} )}{2} + \frac{πn}{2}

y = \frac{arctan ( t + \frac{1}{2} sin ( 2 t ) + c _{1} )}{2} + \frac{πn}{2}

x \to 0 lim (1 - cot (x))

\frac{t}{( t - 7 ) ^{2}}

\frac{d y}{d x} = sin (x + y)

f (x) = (4 x - 1)^{- 2}

e^{x} \cdot (\frac{1}{1 - x})

What is the solution for y^'=cos^2(t)cos^2(2y) ?
The solution for y^'=cos^2(t)cos^2(2y) is y=(arctan(t+1/2 sin(2t)+c_{1}))/2+(pin)/2