What is the solution for y^'=(sin(2x))/(cos(3y)),y(0)=1 ?

The solution for y^'=(sin(2x))/(cos(3y)),y(0)=1 is

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y^'=(sin(2x))/(cos(3y)),y(0)=1

y^{^{'}} = \frac{sin ( 2 x )}{cos ( 3 y )}, y (0) = 1

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

Solution steps

y^{'} (x) = \frac{sin ( 2 x )}{cos ( 3 y )}

Solve separable ODE:

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

2 y t (d y) + y^{2} (d t) = 0

y^{^{'}} + y = y (t e^{t^{2}} + 1), y (0) = 1

y^{^{'}} = \frac{y ^{2}}{x ^{3}}, y (2) = 4

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

What is the solution for y^'=(sin(2x))/(cos(3y)),y(0)=1 ?
The solution for y^'=(sin(2x))/(cos(3y)),y(0)=1 is