Particular Solution of $xy' = y(lnx - lny)$ I was solving for the differential equation $xy' = y(lnx - lny)$, $y(1) = 4$ and $x>0$
My work 
The equation, I think, can't be categorized into variable-seprable, exact, homogenous, or others, unless modified. 
Modifying the differential equation above: 
$$xy' = y(lnx - lny)$$
$$xy' = y(ln\left(\frac{x}{y}\right))$$
$$y' = \frac{y}{x}ln\left(\frac{x}{y}\right)$$
Noticing that the modified differential equation $y' = \frac{y}{x}ln\left(\frac{x}{y}\right)$ is a homogenous one. To get the solution of a 
homogenous differential equation, make the substitution $v = \frac{y}{x}$, then $y = vx$ and $y' = v + xv'$. Substituting these appropriate substitution 
to the modified differential equation, we get:
$$y' = \frac{y}{x}ln\left(\frac{x}{y}\right)$$
$$v + xv' = vln\left(\frac{1}{v}\right)$$
$$xv' = vln\left(\frac{1}{v}\right) - v$$
$$xv' = v\left(ln\left(\frac{1}{v}\right) - 1\right)$$
$$\frac{v'}{v\left(ln\left(\frac{1}{v}\right) - 1\right)} = \frac{1}{x} dx$$
$$\frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)} = \frac{dx}{x}$$
The heavily-modified differential equation becomes a variable-separable one.
Now integrating the left and the right-side of the variable-separable differential equation:
$$\frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)} = \frac{dx}{x}$$
$$\int \frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)} =\int \frac{dx}{x}$$
To evaluate $\int \frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)}$, we let $u = ln\left(\frac{1}{v}\right) - 1$ and $du = v dv$
With that in mind....oh wait...I'm stuck. Why? If I do it.....this will happen:
$$\int \frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)} = \int \frac{du}{vu}$$
Evaluating $\int \frac{du}{vu}$ isn't allowed. 
Now I'm stuck. How to evaluate $\int \frac{dv}{v\left(ln\left(\frac{1}{v}\right) - 1\right)}$ properly, and then, ultimately, the 
particular solution of the given differential equation posed on the problem.
 A: Almost as you wrote$$y' = -\frac{y}{x}\,\log\left(\frac{y}{x}\right)$$ To get rid of the logarithms, let $$y=x e^v \implies y'=e^{v} \left(x v'+1\right)$$ which makes $$e^{v} \left(x v'+1\right)=-e^v v$$ that is to say $$x v'+v+1=0$$ which is separable and easy to integrate.
A: Alternatively you’re very close. Remember $\ln\left(\frac{1}{v}\right)=-\ln(v)$ so the integral in question becomes $\int\frac{1}{v(-\ln(v)-1)}\,\mathrm{d}v$ then making the substitution $u=\ln(v)\implies\mathrm{d}u=\frac{1}{v}\mathrm{d}v$ changes our integral into $-\int\frac{1}{u+1}\,\mathrm{d}u=-\ln|u+1|=-\ln|\ln(v)+1|$. Pairing this with your $\int\frac{\mathrm{d}x}{x}=\ln|x|+C$ we get
\begin{align*}
&\implies-\ln|\ln(v)+1|=\ln|x|+C\\
&\implies \ln(v)=\frac{C}{x}-1\\
&\implies v=e^{\frac{C}{x}-1}\\
&\implies\boxed{y=xe^{\frac{C}{x}-1}}
\end{align*}
A: With writing
$$y'=\dfrac{y}{x}\ln\dfrac{x}{y}$$
you see that the function $f(x,y)=\dfrac{y}{x}\ln\dfrac{x}{y}$ is homogeneous of degree $1$. with $\dfrac{x}{y}=u$ then this equation simplifies to $u-u'x=u\ln u$ and then
$$\dfrac{du}{u(1-\ln u)}=\dfrac{dx}{x}$$
integration gives $1-\ln u=\dfrac{1}{Cx}$ and the answer will simplify to $\color{blue}{y=xe^{\frac{1}{Cx}-1}}$.
A: An alternative way : 
Change of variables $\begin{cases}
x=e^X\\
y=e^Y
\end{cases}\quad \begin{cases}
dx=e^XdX\\
dy=e^YdY
\end{cases}$
$$xy'=y\left(\ln(x)-\ln(y)\right)=e^X\left(\frac{e^Y}{e^X}\frac{dY}{dX}\right)=e^Y(X-Y)$$
$$\boxed{Y'=X-Y}$$
$Y'+Y=X\quad\implies\quad Y=c\:e^{-X}+X-1\quad\implies\quad y=x\exp(\frac{c}{x}-1)$
