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?

$$yy''+ (y')^2-(y´)^3\ln (y)=0$$
Consider $x'$
And note that
$$y''=\frac 1{x'}\frac {d}{dy} \frac 1 {x'}=-\frac {x''}{x'^3}$$
$$yy''+ (y')^2-(y´)^3\ln (y)=0$$
The differential equation becomes:
$$-x''y+ x'=\ln |y|$$
Substitute  $z=x'$ it's a first order linear equation:
$$z'y-z=-\ln |y|$$
Or multiply by $y$
$$y^2x''- yx'=-y\ln |y|$$
Then it's Cauchy-Euler's equation.
