Differential equation problem, how to solve such types? $$y'=\frac {y(x-y\log(y))}{x(x\log(x)-y)}$$
Its a high school mathematics question.
I tried solving by many methods but failed. It is not a homogeneous equation, so not in parametric form. I think it can be solved by inspection.....but when i tried I was stuck in a loop and couldn't progress forward in the question. How to do ?
Options(4):
$(log(x)÷x) + or -(log(y)÷y)=c$
$(x.logx +or - ylogy)÷xy=c$
 A: $$y(x-y\log(y))dx+x(y-x\log(x))dy=0$$
$$p(x,y)dx+q(x,y)dy=0\quad\begin{cases}
p=y(x-y\log(y))\\
q=x(y-x\log(x))
\end{cases}$$
We look for an integrating factor $\mu(x,y)$ to make it an exact differential of a function $F(x,y)$.
$$\begin{cases}
\frac{\partial F}{\partial x}=\mu(x,y)y(x-y\log(y))\\
\frac{\partial F}{\partial y}=\mu(x,y)x(y-x\log(x))
\end{cases}$$
See http://mathworld.wolfram.com/OrdinaryDifferentialEquation.html , Eq.$(12)$.
$$\frac{\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}}{xp-yq}=
\frac{(y-2x\log(x)-x)-(x-2y\ln(y)-y)}{xy(x-y\log(y))-yx(y-x\log(x))}=\frac{2}{xy}$$
Thus $\mu$ is a function of $xy$. 
$$\begin{cases}
\frac{\partial F}{\partial x}=\mu(xy) y(x-y\log(y))\\
\frac{\partial F}{\partial y}=\mu(xy) x(y-x\log(x))
\end{cases}$$
$\frac{\partial \mu}{\partial x}=y\mu'(xy)$ and $\frac{\partial \mu}{\partial y}=x\mu'(xy)$
$$\frac{\partial^2 F}{\partial x \partial y}=\\=x\mu'y(x-y\log(y))+\mu(x-2y\ln(y)-y)=y\mu'x(y-x\log(x))+\mu(y-2x\ln(x)-x)$$
After simplification : $\quad\mu'yx +2\mu=0$
Not forgetting that $\mu$ is function of $xy$, let $X=xy$.
$$\frac{d\mu}{\mu}=-2\frac{dX}{X}\quad\to\quad \mu(X)=\frac{c}{X^2}\quad\to\quad \mu(x,y)=\frac{c}{(xy)^2}$$
Doesn't matter the value of $c$. Any one is sufficient as integrating factor.
$$\begin{cases}
\frac{\partial F}{\partial x}=\frac{1}{(xy)^2} y(x-y\log(y))\\
\frac{\partial F}{\partial y}=\frac{1}{(xy)^2} x(y-x\log(x))
\end{cases}$$
$$F(x,y)=\int\frac{1}{x^2y}(x-y\log(y))dx = \int \frac{1}{y^2x} (y-x\log(x))dy$$
$y$ is constant parameter in the first integral and $x$ is constant parameter in the second integral.
$$F(x,y)=\frac{\ln(x)}{y}+\frac{\ln(y)}{x}$$
Coming back to the initial ODE with integrating factor :
$$\mu y(x-y\log(y))dx+\mu x(y-x\log(x))dy=dF=0$$
Thus $F=$constant. The solution of the ODE expressed on implicit form is :
$$\frac{\ln(x)}{y}+\frac{\ln(y)}{x}=c$$
Search for explicit form of solution :
$$\frac{x\ln(x)}{y}+\ln(y)=cx\quad\to\quad y\exp\left(\frac{x\ln(x)}{y}\right)=e^{cx}$$
$$\frac{1}{y}\exp\left(-\frac{x\ln(x)}{y}\right)=e^{-cx} $$
$$\left(-\frac{x\ln(x)}{y}\right)\exp\left(-\frac{x\ln(x)}{y}\right)=-e^{-cx}x\ln(x)$$
From the definition of the Lambert's W function : $\quad W(\xi)e^{W(\xi)}=\xi$
and with $\xi=-e^{-cx}x\ln(x)$
$$-\frac{x\ln(x)}{y}=W\left(-e^{-cx}x\ln(x) \right)$$
$$y(x)=-\frac{x\ln(x)}{W\left(-e^{-cx}x\ln(x) \right)}$$
