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.

(I think this is probably different enough, at least at the beginning, to warrant a separate answer.)
We're going to look for an integrating factor Euler-style: that is, if we write the equation as
$$ 0=xyd(y')+yy' dx-2xy'dy, $$
where we define the operator $d$ to be linear and satisfy $d(uv)=v \, du+ u \, dv$ (don't do this in public, but it looks like "multiplying the equation by $dx$" from one of the $dy/dx$s: $dy = y' \, dx$).
Then the first two terms are $y \, d(xy')$, so we seek and integrating factor $z$ so that
$$ 0= d(zxy') = z \, d(xy') + z' xy' \, dx = z\left(d(xy') + \frac{z'}{z} xy' \, dx \right). $$
In other words, we have at present
$$ 0 = y \left( d(xy') + \frac{-2y'}{y} xy' \, dx \right). $$
Dividing the above equations by $z$ and $y$ and equating coefficients, we conclude that such a $z$ has to satisfy
$$ \frac{z'}{z} = -2\frac{y'}{y}. $$
Integrating,
$$ \log{(z/A)} = -2\log{y} = \log{y^{-2}}, $$
so $z=Ay^{-2}$. The constant is obviously irrelevant, and we end up with
$$ 0 = d\left(\frac{xy'}{y^2}\right) = \left(\frac{xy'}{y^2}\right)' \, dx, $$
and then proceed as in my other answer.
