Better method for solving ODE $yy''+ (y')^2-(y')^3\ln (y)=0$ What's a better way to solve this equation?
$$yy''+ (y')^2-(y')^3\ln (y)=0 $$
so far, I've tried:


*

*reducing the order of the equation ($p=y'$):  $yp'+ p^2-p^3ln(y)=0 $

*dividing everything by $p^3$: $y\frac{p'}{p^3} + \frac{1}{p}=ln(y)$

*solving the homogeneous equation: $y\frac{p'}{p^3} + \frac{1}{p}=0 \implies p=(\frac{1}{3ln(cy)})^\frac{1}{3}$
but using constant variation to solve this from this point on seems unnecessarily complicated. Is there a better method I'm not seeing?

As I stated in the comments, use the fact that
$$y y'' + y'^2 = \frac{d}{dx} (y y')$$
Then
$$\frac{d}{dx} (y y') = y'^3 \log{y} $$
This may be rewritten as
$$\frac{d(y y')}{(y y')^2} = \frac{\log{y}}{y^2} y' dx = \frac{\log{y}}{y^2} dy$$
This equation may be integrated to produce
$$-\frac1{y y'} + C_1 = -\frac{1+\log{y}}{y} $$
Rearrange to get
$$y' = \frac1{1+\log{y}-C_1 y}$$
which may be expressed as
$$(1+\log{y}-C_1 y) dy = dx$$
which may be integrated again to produce

$$x = y \log{y} - \frac12 C_1 y^2 + C_2$$

which is about as far as I can take this.
