<s>[INST] Solve the differential equations: $$y' = {y \over {\sin x}}+\tan{x \over 2}$$
I was trying to do this by substitution $u=y/x $ and it did not work and also with
$$y' - {y \over {\sin x}} = 0$$
$${dy \over dx} = {y \over {\sin x}} $$
$${\ln(y) = \ln\left|\tan\frac{x}{2}\right|+c } $$
$${y = c\cdot\tan\frac{x}{2} } $$
but then when im trying to calculate $y'$ I have a problem and I have too many equations. Is there some easier way or am I making some mistakes.
 [/INST] I don't think there "too many equations" 
$${ y=C\left( x \right) \tan  \frac { x }{ 2 }  }\\ y'=C'\left( x \right) \tan { \frac { x }{ 2 }  } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 }  }  }{ 2 } \\ C'\left( x \right) \tan { \frac { x }{ 2 }  } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 }  }  }{ 2 } =\frac { C\left( x \right) \tan  \frac { x }{ 2 }  }{ \sin { x }  } +\tan { \frac { x }{ 2 }  } \\ \\ C'\left( x \right) \tan { \frac { x }{ 2 }  } =\tan { \frac { x }{ 2 }  } \\ C\left( x \right) =x+{ C }_{ 1 }\\ { y=C\left( x \right) \tan  \frac { x }{ 2 }  }={ \tan  \frac { x }{ 2 }  }\left( x+{ C }_{ 1 } \right) \\ \\ $$
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