solving a certain homogeneous ODE I want to solve the homogeneous first-order ODE
$$y'=\sin(y/x)+(y/x)$$
Using substitution $v=y/x$ we separate variables to get
$$\csc(v)\;dv=1/x\;dx$$
and hence
$$|\csc(v)+\cot(v)|^{-1}=A|x|,\;\;\;A>0.$$
But this is nasty and I don't see how to give a closed-form general solution.  Maybe I made a mistake in my work above.
Ideas?
Thanks!
 A: You can rewrite
$$\csc v + \cot v = \cot\left(\frac{v}{2}\right)$$
so after separation of variables

$$\csc(v)\;dv=1/x\;dx$$

You get (rewriting the constant):
$$-\log\left(\cot\left(\frac{v}{2}\right)\right)=\log x + c \iff \log\left(\cot\left(\frac{v}{2}\right)\right)=\log x^{-1} - c\iff \cot\left(\frac{v}{2}\right)= \frac{a}{x}$$
This allows to explicitly solve for $v$ and via $v=y/x$ for y as well:
$$\frac{v}{2}=\cot^{-1}\left(\frac{a}{x}\right)\iff
y=2x\cot^{-1}\left(\frac{a}{x}\right)$$
A: The trick is to use some kind of trigonometric identity, for instance
$$
\frac{dx}{x}=\frac{dv}{\sin v}=\frac{\sin(v)\,dv}{1-\cos^2v}.
$$
Now perform partial fraction decomposition and integrate
$$
2\frac{dx}{x}=\frac{\sin v\,dv}{1-\cos v}+\frac{\sin v\,dv}{1+\cos v},
\\
c+2\ln |x|=\ln|1-\cos v|-\ln|1+\cos v|,
$$
to get an implicit form of the solution. There are now multiple ways to solve, one could extract $\cos v$ and then apply the inverse cosine,
$$
Cx^2=\frac{1-\cos v}{1+\cos v},~~C\ge0,
\\
 \cos v=\frac{1-Cx^2}{1+Cx^2},
\\
v=\cos^{-1}\frac{1-Cx^2}{1+Cx^2}.
$$
Or use the double angle identities to get
$$
Cx^2=\tan^2\frac v2,
\\
v=2\tan^{-1}(C_1x).
$$
