How to reduce the differential equation to first order? How to reduce the following equation to first order differetial equation?
$$y^2 (y'y''' - 2{y''}^2) = y{y'}^2y'' + 2{y'}^4$$
I tried to change variable $y' = p(y)$, then $y'' = p'p$, $y''' = p({p'}^2 + pp'')$, but resulting equation $$y^2 (p^3p'' - p^2{p'}^2) = yp'p^3 + 2p^4$$ isn't homogeneous, so I don't know what should I do now.
 A: One observe that the change of $y$ to $-y$ doesn't change the ODE. This draw us to a change of function : $Y(x)=\left(y(x)\right)^2$.
This change simplifies a lot the equation which reduces to $Y'''Y'-2Y''^2=0$.
Then it is easy to solve it for $Y(x)$ and finally :
$$y(x)=\pm\sqrt{c_3\ln|c_1x+c_2|}$$

A: I managed to get it down to order 2, maybe you can use some of the ideas.
I noticed that after dividing by $y^2\dot{y}^2$ one has
$$
\frac{\dot{y}\dddot{y} - 2\ddot{y}^2}{\dot{y}^2} = \frac{y\ddot{y} + 2\dot{y}^2}{y^2}
$$
Where the left side is the same as the right side with 1 dot more over each y and one sign flipped. Here we can use
$$
\frac{d^2}{dt^2}\log(y) = \frac{d}{dt}\frac{\dot{y}}{y} = \frac{y\ddot{y} - \dot{y}^2}{y^2} 
\qquad\text{and}\qquad
\frac{d^2}{dt^2}\log(\dot{y}) = \frac{d}{dt}\frac{\ddot{y}}{\dot{y}} = \frac{\dot{y}\dddot{y} - \ddot{y}^2}{\dot{y}^2}
$$
To rewrite the equation as
$$
\frac{d}{dt}\frac{\ddot{y}}{\dot{y}} - \left(\frac{\ddot{y}}{\dot{y}}\right)^2 = \frac{d}{dt}\frac{\dot{y}}{y} + 3  \left(\frac{\dot{y}}{y}\right)^2
$$
Now use the substitutes $z = \frac{\dot{y}}{y}$ and $w = \frac{\ddot{y}}{\dot{y}}$ to rewrite as
$$
\frac{d}{dt}(w-z) = w^2 + 3z^3 = (w-z)(w+z) + 4z^2
$$
And now note that $w-z = \frac{d}{dt}(\log(\dot{y}) - \log(y)) = \frac{d}{dt}\log(z) = \frac{\dot{z}}{z}$ and similarly $ w+z = \frac{\dot{z}}{z} + 2z $
Which gives us 
$$
\frac{d}{dt}\left(\frac{\dot{z}}{z}\right) = \frac{\dot{z}}{z}\bigg(\frac{\dot{z}}{z}+2z\bigg) + 4z^2
$$
Or equivalently 
$$
 z \ddot{z} - 2\dot{z}^2 - 2\dot{z}z^2 - 4z^4 = 0
$$
