Is this a valid Substitution Shortcut for differential equation? Given $\frac{y}{x}\cos\frac{y}{x}dx-(\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x})dy = 0$
This is a homogeneous differential equation. 
Is it valid to substitute $u=\frac{y}{x}$ and $dy=udx+xdu$ into the equation?
If so, after substitution, simplification, and separation of variables I got the following equation $-\log|x|=\log|u|+\int\frac{1}{u}\cot{u}du$ 
Singling out $\int\frac{1}{u}\cot{u}du$ to be integrated through integration by parts technique
I got $\int\frac{1}{u}\cot{u}du = \cot{u}\log|u|+\int \log|u|\csc^2{u}du$
So, is my initial substitution valid? If so, I'm not sure what substitutions to use next in integrating $\int \log|u|\csc^2{u}du$..
 A: Yes, it is valid but the expression can be simplified in an easier way.
Let $u=\frac yx$. Then, $\frac{dy}{dx}=u+x\frac{du}{dx}$
$$\frac{y}{x}\cos\frac{y}{x}dx-\left(\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}\right)dy = 0$$
$$\frac{dy}{dx}=\frac{\frac{y}{x}\cos\frac{y}{x}}{\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}}$$
$$u+x\frac{du}{dx}=\frac{u\cos u}{\frac{1}{u}\sin u+\cos u}$$
$$u+x\frac{du}{dx}=\frac{u^2\cos u}{\sin u+u\cos u}$$
$$x\frac{du}{dx}=\frac{u^2\cos u-u\sin u-u^2\cos u}{\sin u+u\cos u}$$
$$x\frac{du}{dx}=\frac{-u\sin u}{\sin u+u\cos u}$$
$$\int\frac{(\sin u+u\cos u)du}{u\sin u}=\int\frac{-dx}{x}$$
$$\int\left(\frac1u+\cot u\right)du=\int\frac{-dx}{x}$$
$$\ln|u|+\ln|\sin u|=-\ln|x|+c$$
A: First write $$\frac{dy}{dx}=\frac{\frac{y}{x}\cos {\frac{y}{x}}}{\frac{x}{y}\sin\frac{y}{x}+\cos\frac{y}{x}}$$
Then put $y=vx$ to get, $\frac {dy}{dx}= v+ x\frac{dv}{dx}$
$$\implies x\frac{dv}{dx}=\frac{v^2 \cos v}{\sin v+v\cos v}-v$$
$$\implies \int \frac{(\sin v+v \cos v) dv}{v\sin v}= -\int \frac{dx}{x}$$
$$\implies \ln |v| +\ln |\sin v|= -\ln x+C$$
Now simplify and substitute $v=\frac{y}{x}$, to get the answer.
