y^'=((2sin(x))/(sin(y))),y(0)= pi/2

解答

y^{^{'}} = (\frac{2 sin ( x )}{sin ( y )}), y (0) = \frac{π}{2}

y = arccos (2 cos (x) - 2) + 2 πn, y = - arccos (2 cos (x) - 2) + 2 πn

求解步骤

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

求解可分离方程:

y = arccos (2 cos (x) - 2) + 2 πn, y = - arccos (2 cos (x) - 2) + 2 πn

y = arccos (2 cos (x) - 2) + 2 πn, y = - arccos (2 cos (x) - 2) + 2 πn