<s>[INST] Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$. Solve the differential equation $\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$ given that $y(1)=0$.
I was able to find the solution to it by the method of solving homogeneous differential equations, which is as follows:
$\frac {x^2}{2y^2} \log \frac{x}{y}-\frac {x^2}{4y^2}+ \log y = C$.
But I am not able to find the value of $C$, I am not getting how to substitute for $y$ because it seems to give not-defined values. Please help.
According to the answer hint, $C=1- \frac{3}{4e^2}$.
Thanks in advance...
EDIT:
Since most of you are asking for the complete steps I followed to get the final solution, I am posting it as follows:
$\{xy\log \frac{x}{y}\}dx+\{y^2-x^2 \log \frac{x}{y}\}dy=0$
Let $x=vy \implies \frac{dx}{dy}=v+y \frac{dv}{dy}$
Therefore, $\frac{dx}{dy}=-(\frac{y^2-x^2 \log \frac{x}{y}}{xy\log \frac{x}{y}})$
$\implies v+y \frac{dv}{dy}= \frac{v^2 \log v-1}{v \log v}$
$\implies y \frac{dv}{dy}= \frac{v^2 \log v-1}{v \log v}-v=\frac{-1}{v \log v}$
$\implies v \log v dv + \frac{1}{y} dy = 0$
Integrating the equation we get,
$\frac{v^2}{2} \log v-\frac{v^2}{4}+\log y=C$
On replacing value of $v=\frac{x}{y}$
$\frac {x^2}{2y^2} \log \frac{x}{y}-\frac {x^2}{4y^2}+ \log y = C$
Now, $y(1)=0$
Therefore, $\frac {1^2}{2\times 0^2} \log \frac{1}{0}-\frac {1^2}{4\times 0^2}+ \log 0 = C$
How do I solve for $C$? Please help...
 [/INST] $$\left(xy\log\frac{x}{y}\right)dx+\left(y^2-x^2\log\frac{x}{y}\right)=0$$
$$\left(xy\log\frac{y}{x}\right)dx=\left(y^2+x^2\log\frac{y}{x}\right)dy$$ $$\left(\frac{y}{x}\log\frac{y}{x}\right)dx=\left(\left(\frac{y}{x}\right)^2+\log\frac{y}{x}\right)dy$$ Let $\frac{y}{x}=u \implies y=ux \implies \frac{dy}{dx}=u+x\frac{du}{dx}$ 
$$\left(u\log u\right)=\left(u^2+\log u\right)\left(u+x\frac{du}{dx}\right)$$ 
$$x\frac{du}{dx}=\frac{u\log u}{u^2+\log u}-u$$ 
$$x\frac{du}{dx}=\frac{u\log u-u^3-u\log u}{u^2+\log u}=\frac{-u^3}{u^2+\log u}$$
$$\frac{(u^2+\log u)du}{u^3}=-\frac{dx}{x}$$  
Can you take it from here?
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