How to solve $ xyy'' − 2x(y')^2 + yy' = 0$ How can I solve the given IVP 
$$ xyy'' − 2x(y')^2 + yy' = 0 $$
$$y(1) = 1, \\
y''(1) = 2.$$
Using substitution $u = \ln{x}$ is given as hint.
Yes this is a homework problem but I have no clue to solve it. I wish had some and tell you about it. I appreciate any help, even a little bit.

The "how on Earth did you spot that?!" solution:
Consider the derivative of $xy^n y'$. Why? Because I thought the original equation looked a bit like a product rule, what with the $yy''$ and the $y'^2$, which look like $(yy')'$. This doesn't quite work by itself, but
$$ (xy^n y')' = y^n y' + nxy^{n-1}y'^2 + xy^n y'', $$
and if we set this equal to zero and cancel off a $y^{n-1}$, we find
$$ yy'+nxy'^2+xyy''=0, $$
and taking $n=-2$ gives us the original equation. Hence,
$$ \left( \frac{xy'}{y^2} \right)' = 0 \\
 \frac{xy'}{y^2} = A \\
\frac{y'}{y^2} = \frac{A}{x}, $$
after one integration. Integrating again,
$$ -\frac{1}{y} = A\log{x}+B, $$
and you can go on from here to find the values of $A$ and $B$ (note that if $x=1$, $-1/1 = B$, so $B=-1$, for example).
