What is the solution for y^'= t/(sin(y)) ?

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

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y^'= t/(sin(y))

y^{^{'}} = \frac{t}{sin ( y )}

y = arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn, y = - arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn

Solution steps

y^{'} (t) = \frac{t}{sin ( y )}

Solve separable ODE:

y = arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn, y = - arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn

y = arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn, y = - arccos (- \frac{t ^{2} + c _{1}}{2}) + 2 πn

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