Why solving trigonometry equation for $xy' = y\cos\ln \frac{y}{x}$? I have the equation which is homogeneous, but I am concerned of missing one solution, what I have and what I have tried so far:
$$xy' = y\cos\ln \frac{y}{x}$$
$$y = tx \Rightarrow y' = t'x+t$$
$$t'x = t\cos \ln t - t$$
after some answers exploration I figured out that author also solves
$$\cos\ln t = 1$$
Which with the substitution can be represented as $\cos u = 1$ and $u = 2\pi k$ and therefore $\ln \frac{y}{x} = 2\pi k$. Only then the author goes back to the original equation and solves it as an equation with separable variables.
But my question is: do we really have to do this,  and why? It looks like the equation in the equation...

solution in picture, disregard comments in russian:
1.) shows trigonometry equation
2.) the whole equation
http://i.imgur.com/LAt7KHf.png
 A: With $u=\ln y-\ln x$, 
$$
u'=\frac{y'}y-\frac1x=\frac1x(\cos u-1)
$$
This has stationary solutions $u=c=const$, $y=e^cx$ for the roots of $\cos u-1=0$ which are at $c=2k\pi$. For all other points you can separate the equation to get
$$
\cot(\frac u2)=\ln x+c.
$$ 
A: Well, we have
$$x\cdot\text{y}\space'\left(x\right)=\text{y}\left(x\right)\cdot\cos\left(\ln\left(\frac{\text{y}\left(x\right)}{x}\right)\right)\tag1$$
Substitute $\text{y}\left(x\right)=x\cdot\mathscr{P}\left(x\right)$:
$$x\cdot\left(x\cdot\mathscr{P}\space'\left(x\right)+\mathscr{P}\left(x\right)\right)=x\cdot\mathscr{P}\left(x\right)\cdot\cos\left(\ln\left(\mathscr{P}\left(x\right)\right)\right)\tag2$$
So, we also get:
$$\frac{\mathscr{P}\space'\left(x\right)}{\mathscr{P}\left(x\right)\cdot\cos\left(\ln\left(\mathscr{P}\left(x\right)\right)\right)-\mathscr{P}\left(x\right)}=\frac{1}{x}\tag3$$
Integrate both sides with respect to $x$:
$$\int\frac{\mathscr{P}\space'\left(x\right)}{\mathscr{P}\left(x\right)\cdot\cos\left(\ln\left(\mathscr{P}\left(x\right)\right)\right)-\mathscr{P}\left(x\right)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\tag4$$
So:
$$\cot\left(\frac{\ln\left(\mathscr{P}\left(x\right)\right)}{2}\right)=\ln\left|x\right|+\text{C}\tag5$$
We also now know:
$$\cot\left(\frac{\ln\left(\frac{\text{y}\left(x\right)}{x}\right)}{2}\right)=\ln\left|x\right|+\text{C}\tag6$$
