Homogeneous D.E show that $ x\frac{dy}{dx} = y + \frac{x}{lny-lnx} $ is a homogeneous D.E.
Find the general solution in the form $ye^{y/x} = f(x) $
The answer given is $ye^{y/x} = Ax^2 $ but I am getting $ (\frac{y}{x})^{y/x} = Axe^{y/x} $
Is the given answer correct?

Consider the given answer $\quad ye^{y/x} = Ax^2 \quad\to\quad \ln(y)+\frac{y}{x}=\ln(A)+2\ln(x) \quad$ and differentiate it :
$$\frac{dy}{y}+\frac{dy}{x}-\frac{ydx}{x^2}=2\frac{dx}{x}$$
$$\left( \frac{1}{y}+\frac{1}{x}\right)dy=\left(\frac{2}{x}+\frac{y}{x^2} \right)dx$$
$$\frac{dy}{dx}=\frac{\frac{2}{x}+\frac{y}{x^2}}{\frac{1}{y}+\frac{1}{x}} = \frac{y}{x}\left(\frac{2x+y}{x+y}\right)$$
$$x\frac{dy}{dx}= y\left(\frac{2x+y}{x+y}\right)=y\left(1+\frac{x}{x+y}\right)$$
$$x\frac{dy}{dx}= y+\frac{x}{\frac{x}{y}+1}\tag 2$$
Compared to the given ODE 
$$x\frac{dy}{dx}= y+\frac{x}{\ln(y)-\ln(x)}\tag 1$$
we see that $\quad ye^{y/x} = Ax^2\quad$ should be the solution of $(1)$ if $\quad \frac{x}{y}+1=\ln(y)-\ln(x)\quad$ 
Except for particular values of $x$ and $y$, this isn't the case in general.
So, in general $\quad ye^{y/x} = Ax^2\quad$ isn't solution of $(1)$.
The general solution of $(1)$ is $ (\frac{y}{x})^{y/x} = Axe^{y/x} $
Probably there is a typo or a mistake in the wording of the problem. May be the question isn't to find the general solution of $(1)$, but only to find a particular solution on a specified form.
