I need to solve the following ODE:
$$y'+A\sin(2y)+B\sin(y)+C\cos(y)+D=0$$
I want to find y' (explicit/unexplicit) rather than y.
I tryied to use the Weierstrass transform, after substituting the terms of the transformation [$\cos(y)=(1-z^2)/(1+z^2)$ ; $\sin(y)=2z/(1+z^2)$ ; $y'=2z'/(1+z^2)$ ; $\sin(2y)=2\sin(y)\cos(y)$] I got this eqn: $$z'=1/(a*z^4+b*z^3+c*z^2+d*z+e)$$ where a,b,..,e are functions of the scalars A,B,C,D I tried to solve it in a web calculator, and it does give a really long solution and then when substituting back z= tan(y/2) I couldn't extract y and derivate it to find y'.
It's is not mandatory to use the Weierstrass transform, I used it because it worked for me when I try to solve the easier case when $B=C=0$.
Thanks to anyone who devotes his time to this question.
Regards, Daniel.