Solution to the differential equation $\left(x \csc\left(\frac{y}{x}\right)-y\right) dx + xdy$? What is the general solution to the differential equation $\left(x \csc\left(\frac{y}{x}\right)-y\right) dx + xdy$?
My work
If Im going to reaarange the differential equation above, it would look like this:
$$x \csc\left(\frac{y}{x}\right) dx-ydx + xdy$$
$$x \csc\left(\frac{y}{x}\right) dx-(ydx - xdy)$$
I got an idea how to deal with the expression $(ydx - xdy)$, but I don't know how to approach the expression $x \csc\left(\frac{y}{x}\right) dx$ 
to get the general solution of the d.e.
How to get the general solution to the differential equation above?
 A: I guess the differential equation you are trying to solve is :
$$\left(x \csc\left(\frac{y}{x}\right)-y\right) dx + xdy =0$$
Let $y(x) = xv(x)$, which gives : $\frac{dy(x)}{dx} = v(x) + \frac{dv(x)}{dx}$ ::
$$x\csc(v(x)) + x \bigg(x\frac{dv(x)}{dx} + v(x) \bigg) - xv(x) = 0$$
$$\Leftrightarrow$$
$$x\bigg(\csc(v(x)) + x\frac{dv(x)}{dx}\bigg) = 0$$
$$\Leftrightarrow$$
$$\frac{dv(x)}{dx} = - \frac{\csc(v(x))}{x}$$
$$\Leftrightarrow$$
$$\frac{dv(x)}{dx}\sin(v(x)) = -\frac{1}{x}$$
$$\Rightarrow$$
$$\int \frac{dv(x)}{dx}\sin(v(x))dx = \int-\frac{1}{x}dx$$
$$\Leftrightarrow$$
$$\cos(v(x)) = -\ln(x) + c_1$$
Solving for $v(x)$ gives you :
$$v(x) = \begin{cases} -\arccos(\ln(x) - c_1) \\ \arccos(\ln(x) - c_1)\end{cases}$$
Substitute in the initial $y(x) = xv(x)$ that we used to solve the equation and you will get $y(x)$ !
A: Well, we have:
$$\left(x\cdot\csc\left(\frac{\text{y}}{x}\right)-\text{y}\right)\space\text{d}x+x\space\text{d}\text{y}=0\space\Longleftrightarrow\space\text{y}\space'\left(x\right)=\frac{\text{y}\left(x\right)}{x}-\csc\left(\frac{\text{y}}{x}\right)\tag1$$
Let $\text{y}\left(x\right):=x\cdot\text{v}\left(x\right)$:
$$x\cdot\left(\csc\left(\text{v}\left(x\right)\right)+x\cdot\text{v}\space'\left(x\right)\right)=0\space\Longleftrightarrow\space-\frac{\text{v}\space'\left(x\right)}{\csc\left(\text{v}\left(x\right)\right)}=\frac{1}{x}\tag2$$
Let $\text{u}:=\text{v}\left(x\right)$:
$$\int-\frac{\text{v}\space'\left(x\right)}{\csc\left(\text{v}\left(x\right)\right)}\space\text{d}x=\int-\frac{1}{\csc\left(\text{u}\right)}\space\text{d}\text{u}=\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{C}\tag3$$
A: Substitute $y=xu$ like in a homogeneous ODE to find
$$
0=\frac{dx}x+\sin(u)du.
$$
A: We need to reaarange the differential equation so that it can be solved easily...
$$\left( x \csc\left( \frac{y}{x}\right) - y\right) dx + x dy = 0$$
$$x \csc\left( \frac{y}{x}\right)dx - ydx + xdy = 0$$
$$xdy = -x \csc\left( \frac{y}{x}\right)dx + ydx$$
Divide the terms of the differential equation by $x$, getting:
$$dy = - \csc\left( \frac{y}{x}\right)dx + \left( \frac{y}{x}\right)dx$$
Apparently this modified form of differential equation is a homogenous differential equation. So we let $v = \frac{y}{x}$, $y = vx$ and 
$dy = vdx + xdv$. So we get:
$$dy = - \csc\left( \frac{y}{x}\right)dx + \left( \frac{y}{x}\right)dx$$
$$vdx + xdv = - \csc(v)dx + vdx$$
$$ xdv = - \csc(v)dx$$
$$\frac{dv}{-\csc(v)} = \frac{1}{x} dx$$
$$-sin(v)dv = \frac{1}{x} dx$$
The above differential equation is now a variable-separable type...so we get the integral of individual terms.
$$-sin(v)dv = \frac{1}{x} dx$$
$$\int -sin(v)dv = \int \frac{1}{x} dx$$
$$cos(v) + c_1 = ln(x) + c_2$$
$$cos(v)  = ln(x) + c_2 - c_1$$
$$cos(v)  = ln(x) + C$$
Now getting the $v$:
$$cos(v)  = ln(x) + C$$
$$v = cos^{-1}(ln(x) + C)$$
The solution of the given differential equation $\left( x \csc\left( \frac{y}{x}\right) - y\right) dx + x dy = 0$ would be:
$$y = vx$$
$$y = (cos^{-1}(ln(x) + C))x$$
$$y = xcos^{-1}(ln(x) + C)$$
Alternate ways of answering it are encouraged....
