What is the solution for y^'=(2sin(2x)cos(2x))y^2,y(0)=2 ?

The solution for y^'=(2sin(2x)cos(2x))y^2,y(0)=2 is y= 4/(cos(4x)+1)

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y^'=(2sin(2x)cos(2x))y^2,y(0)=2

y^{^{'}} = (2 sin (2 x) cos (2 x)) y^{2}, y (0) = 2

y = \frac{4}{cos ( 4 x ) + 1}

Solution steps

y^{'} (x) = (2 sin (2 x) cos (2 x)) y^{2}

Solve separable ODE:

y = \frac{4}{cos ( 4 x ) + 1}

y = \frac{4}{cos ( 4 x ) + 1}

What is the solution for y^'=(2sin(2x)cos(2x))y^2,y(0)=2 ?
The solution for y^'=(2sin(2x)cos(2x))y^2,y(0)=2 is y= 4/(cos(4x)+1)