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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.

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You forgot $x$ in the third step it should be $$x\,dy - \left(y+\ x\sin\left(\frac{y}{x}\right)\right)\,dx = 0.$$

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The book's solution is correct, as you can easily check by substituting it in to the differential equation: note that

$$ \eqalign{\sin(2 \arctan(cx)) &= 2 \sin(\arctan(cx)) \cos(\arctan(cx))\cr &= 2 \tan(\arctan(cx)) \cos^2(\arctan(cx))\cr & = \frac{2 \tan(\arctan(cx)}{1+\tan^2(\arctan(cx))}\cr &= \frac{2 c x}{1 + c^2 x^2}}$$

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