What is the solution for 2xdx+sin(y)dy=0,y(1)=0 ?

The solution for 2xdx+sin(y)dy=0,y(1)=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,y(1)=0

2 x d x + sin (y) d y = 0, y (1) = 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

q^{^{'}} (p) = \frac{5400 p ^{2} q ( p )}{( p ^{3} + 2 ) ^{2}}

\frac{d y}{d x} = 4 x^{3} + x^{3} y^{2}

\frac{d x}{d t} = - x^{2}

\frac{d y}{d t} = - y - y^{2}

x^{2} y^{^{'}} = 1 + y, y (1) = 3

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