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

The solution for y^'sin(y)=cos(x),y(0)=0 is y=arccos(-sin(x)+1)+2pin,y=-arccos(-sin(x)+1)+2pin

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

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

y = arccos (- sin (x) + 1) + 2 πn, y = - arccos (- sin (x) + 1) + 2 πn

Solution steps

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

Solve separable ODE:

y = arccos (- sin (x) + 1) + 2 πn, y = - arccos (- sin (x) + 1) + 2 πn

y = arccos (- sin (x) + 1) + 2 πn, y = - arccos (- sin (x) + 1) + 2 πn

\frac{d x}{d t} + 1 = x

e^{y} (1 + x^{2}) \frac{d y}{d x} - 2 (1 + e^{y}) = 0

\frac{d y}{d x} = e^{x} y^{3}

y^{^{'}} = (5 x + 1) (y + 11), y (0) = 0

x^{2} sin (x d x) + x y d y = 0

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