this was a problem from a textbook:
If $x>0$, $y>0$, find the general solution to the differential equation, $$ x \frac{dy}{dx} = y + \frac{x}{\ln y - \ln x }$$ giving your answer in the form $ye^{y/x}=f(x)$
I approached with the substitution $y=vx$ since this is a homogenous equation. Giving,
$$ x\frac{dv}{dx}+v = v+ \frac{1}{\ln v}$$
By separating variables yields,
$$ v \ln v -v = \ln x + C $$
The by taking the power of $e$ $$e^{v \ln v - v } = xA$$
where $e^C= A$. This is equivalent to
$$e^{-v}v^{v} = Ax$$
Now when I substitute the value of $v= \frac{y}{x}$ back, I cannot obtain the equation of the form $ye^{y/x} = f(x)$.
Maybe I have done a mistake in my calculations. May someone explain where? Or how to solve the problem? Thank you so much!!
Here my solution ( maybe someone can find something more simple).
Using the substitution $\log y-\log x=t \iff y=xe^t$ the differential equation become: $$ te^t\,dt=\dfrac{dx}{x} $$ with solution: $$ e^t(t-1)=\log x +c $$ divide by $e$: $$ (t-1)e^{t-1}=\dfrac{\log x+c}{e} $$ and using the Lambert W function we find: $$ t-1=W\left(\dfrac{\log x+c}{e}\right) $$ so we have (see here) $$ e^t=\dfrac{y}{x}=\dfrac{\log x +c}{W\left(\dfrac{\log x+c}{e}\right)}=e\dfrac{\dfrac{\log x +c}{e}}{W\left(\dfrac{\log x+c}{e}\right)}= $$ $$ =ee^{W\left(\dfrac{\log x+c}{e}\right)}=F(x) $$ and $y=xF(x)$ is the solution of the differential equation.
If we want the form requested in OP we can do: $$ ye^{\frac{y}{x}}=xF(x)e^{F(x)} $$