Differential equation with homoegeneous coefficient, solution other than in book I have a  differential equation:
$$ x \frac{dy}{dx} - y - x\sin\left(\frac{y}{x}\right) = 0. $$
I'm multiplying both sides by $dx$ and I'm obtaining:
$$ x\,dy - y\,dx - x \sin\left(\frac{y}{x}\right)\, dx = 0. $$
Next, after simplification I have:
$$ x\,dy - \left(y+\sin\left(\frac{y}{x}\right)\right)\,dx = 0.$$
This is a homogeneous differential equation with homogeneous functions of order $1$ right?
So I use substitution:
$$ y = ux, dy = u\,dx + x\,du $$
and I'm obtaining the equation:
$$ x(u\,dx + x\,du ) -\left(ux + \sin\left(\frac{y}{x}\right)\right)\,dx = 0.$$
After simplification I'm obtaining:
$$ x^{2}\, du - \sin(u)\, dx = 0.$$  
So, next I'm dividing equation boths sides by: $\sin(u)x^{2}$:
$$ \frac{du}{\sin(u)} - \frac{dx}{x^{2}}  = 0.$$
Because :
$$ \int \frac{dx}{x^{2}} = \frac{-1}{x} + C $$ 
and
$$ \int\frac{du} {\sin(u)} = \ln \left| \tan\left(\frac{u}{2}\right)\right| + C. $$ 
So:
$$ \ln \left|\tan\left(\frac{u}{2}\right)\right| +  \frac{1}{x}  = C.$$
Next:
$$ \ln \left| \tan\left(\frac{y}{2x}\right)\right| = C - \frac{1}{x}$$
$$ e^{C-\frac{1}{x}} = \left|\tan\left(\frac{y}{2x}\right)\right|  $$
$$ \pm e^{c} e^{\frac{-1}{x}} = \tan\left(\frac{y}{2x}\right). $$   Now I'm substituting $d = \pm e^{e^{c}} $
and in consequence I have:
$$ de^{\frac{-1}{x}} = \tan\left(\frac{y}{2x}\right) $$
$$ \arctan\left(d e^{\frac{-1}{x}} \right) = \frac{y}{2x} $$ 
$$ y = 2x \cdot \arctan\left(de^{\frac{-1}{x}}\right).$$
When I look on the answer from the book there is:
$$ y = 2x \cdot \arctan(cx).$$
Why here is $ x $ instead $e^{\frac{-1}{x}} $ ? I don't know. Is my answer wrong? I will be greatfull for help.
Best regards.

The book's solution is correct, as you can easily check by substituting it in to the differential equation: note that
$$ \eqalign{\sin(2 \arctan(cx)) &= 2 \sin(\arctan(cx)) \cos(\arctan(cx))\cr &= 2 \tan(\arctan(cx)) \cos^2(\arctan(cx))\cr & = \frac{2 \tan(\arctan(cx)}{1+\tan^2(\arctan(cx))}\cr &= \frac{2 c x}{1 + c^2 x^2}}$$
