Differential equation how to continue?

We have the equation $$2x(y') - y=\ln(y') \tag{1}$$ We replace $y'=p$, then
$$2xp-\ln p=y.$$ Taking derivative with respect to x. I have $$p=2p + [(2x)-(1/p)] \frac{dp}{dx}$$

So I have $0=p + (2x) - (1/p)dp/dx$. What to do now?

EDIT : From here, I have to write the equation in such way,that I can replace it in (1) and have a solution?

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You know you are in trouble when none here replies :/ – hgdhg Jun 2 '13 at 19:05
A CAS gives an implicit solution, which I won't post spoilers to here. So there is an algorithmic approach that yields an exact (implicit) solution family. It appears from the solution that the substitution $z=W(-2xe^{-y})$ would help, where $W$ is the Lambert function. – alex.jordan Jun 2 '13 at 20:48

Try differentiating from both sides of the OE with respect to $x$. You'll get: $$2y'+2xy''-y'=\frac{y''}{y'}$$ Now set $y'=p$ and then you have an exact OE. Do the rest by yourself.

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Perfect hint! +1 – Amzoti Jun 2 '13 at 21:24
I agree with Amzoti! +1 – amWhy Jun 3 '13 at 1:36

We have $p = y'$ and

$2xp - \ln{p} = y$
$\implies \frac{d}{dx} (2xp - \ln {p}) = y'$
$\implies 2p - (1/p) y'' = p$
$\implies p = (1/p)y''$
$\implies p^2 = y''$
$\implies p^2 = \frac{d}{dx} y'$
$\implies p^2 dx = dy'$
$\implies (\frac{dy}{dx})^2dx = dy'$
$\implies \frac{dy}{dx} dy = dy'$
$\implies y'dy = dy'$
$\implies \int dy = \int dy' / y'$
$\implies y = \ln{y'}$

I couldn't go further. Hope it helps.

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