What is the solution for 2xdx+sin(y)dy=0 ?

The solution for 2xdx+sin(y)dy=0 is y=arccos(x^2-c_{1})+2pin,y=-arccos(x^2-c_{1})+2pin

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2xdx+sin(y)dy=0

2 x d x + sin (y) d y = 0

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

Solution steps

2 x d x + sin (y) d y = 0

Solve separable ODE:

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

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

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

y^{^{'}} = 2 y + 1 cos (x)

2 x d x - y d y = 0

\frac{d y}{d x} = \frac{20 x ^{4}}{y}, y > 0, x > 0

\frac{d y}{d x} = \frac{x ^{2} e ^{x^{3} - y^{2}}}{y}

What is the solution for 2xdx+sin(y)dy=0 ?
The solution for 2xdx+sin(y)dy=0 is y=arccos(x^2-c_{1})+2pin,y=-arccos(x^2-c_{1})+2pin