What is the solution for y^'sin(y)=1 ?

The solution for y^'sin(y)=1 is y=arccos(-t-c_{1})+2pin,y=-arccos(-t-c_{1})+2pin

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y^'sin(y)=1

y^{^{'}} sin (y) = 1

y = arccos (- t - c_{1}) + 2 πn, y = - arccos (- t - c_{1}) + 2 πn

Solution steps

y^{'} (t) sin (y) = 1

Solve separable ODE:

y = arccos (- t - c_{1}) + 2 πn, y = - arccos (- t - c_{1}) + 2 πn

y = arccos (- t - c_{1}) + 2 πn, y = - arccos (- t - c_{1}) + 2 πn

\frac{d x}{d t} = x^{2} + \frac{1}{16}, x (0) = 6

\frac{d v}{d x} = 9 + 16 v^{2}

y^{^{'}} + \frac{2 x y ^{2} + 2 y}{2 x ^{2} y + 2 x} = 0

\frac{d x}{d y} = \frac{1 + x ^{2}}{1 - y ^{2}}

u^{^{'}} + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

What is the solution for y^'sin(y)=1 ?
The solution for y^'sin(y)=1 is y=arccos(-t-c_{1})+2pin,y=-arccos(-t-c_{1})+2pin