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.
 A: Providing you're well versed with the concept of solving via an integrating factor.

Rewrite in a linear form
\begin{equation}
    \frac{dy}{dx}-y\left(\frac{1}{sin(x)}\right)=tan(\frac{x}{2}).
\end{equation}
We identify the integrating factor $\mu (x)$ as (I'll leave it as an exercise for you to verify this):
\begin{equation}
\mu (x)= exp\left(\int(\frac{-1}{sin(x)})\right) \Rightarrow \mu (x) = exp\left(-\int(cosec(x)dx\right) \Rightarrow \mu(x)= cot\left(\frac{x}{2}\right).
\end{equation}
Then, if we multiply our differential equation through by our integrating factor we get
$$cot(\frac{x}{2})\frac{dy}{dx}-ycot\left(\frac{x}{2}\right)\frac{1}{sin(x)}=tan\left(\frac{x}{2}\right)cot\left(\frac{x}{2}\right).$$
Now by virtue of the product rule (and I suggest you check this out for yourself to verify), we have that
$$\left(cot\left(\frac{x}{2}\right)y\right)'=cot\left(\frac{x}{2}\right)\frac{dy}{dx}-ycot\left(\frac{x}{2}\right)\frac{1}{sin(x)},$$
which is equivalent to the left hand side of our equation. Then making the appropriate replacement and observing the right hand side cancels out to $1$ we have
$$\left(cot\left(\frac{x}{2}\right)y\right)'=1,$$
and it should be trivial from here
$$cot\left(\frac{x}{2}\right)y=x+c_1,$$
that is
$$y=\frac{x+c_1}{cot\left(\frac{x}{2}\right)}=tan\left(\frac{x}{2}\right)(x+c_1),$$
where $c_1 \in \mathbb{R}.$
A: I don't think there "too many equations" 
$${ y=C\left( x \right) \tan  \frac { x }{ 2 }  }\\ y'=C'\left( x \right) \tan { \frac { x }{ 2 }  } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 }  }  }{ 2 } \\ C'\left( x \right) \tan { \frac { x }{ 2 }  } +C\left( x \right) \frac { \sec ^{ 2 }{ \frac { x }{ 2 }  }  }{ 2 } =\frac { C\left( x \right) \tan  \frac { x }{ 2 }  }{ \sin { x }  } +\tan { \frac { x }{ 2 }  } \\ \\ C'\left( x \right) \tan { \frac { x }{ 2 }  } =\tan { \frac { x }{ 2 }  } \\ C\left( x \right) =x+{ C }_{ 1 }\\ { y=C\left( x \right) \tan  \frac { x }{ 2 }  }={ \tan  \frac { x }{ 2 }  }\left( x+{ C }_{ 1 } \right) \\ \\ $$
