differential equation help please?

Question

find the general solution :

dy/dt = (2^y)*(sin t)^2

Answer 1

Separating variables

2^(-y)dy = (sin t)^2dt Remembering that (sin t)^2=1/2-(cos 2t)/2
and integrating both sides

-2^(-y)/ln 2 = 1/2 *t - (sin 2t) / 4 +C