Analytical solution to a nonlinear ODE How might I analytically solve the following differential equation? $$yy'' = y' + y^3$$
I've tried certain substitutions ($y = ux$ etc.) but none of them work. 
 A: $yy''=y'+y^3$
$y\dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}+y^3$
Let $u=\dfrac{dy}{dx}$ ,
Then $\dfrac{d^2y}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{dy}\dfrac{dy}{dx}=u\dfrac{du}{dy}$
$\therefore yu\dfrac{du}{dy}=u+y^3$
$u\dfrac{du}{dy}=\dfrac{u}{y}+y^2$
This belongs to an Abel equation of the second kind.
Let $y=e^t$ ,
Then $\dfrac{du}{dy}=\dfrac{\dfrac{du}{dt}}{\dfrac{dy}{dt}}=\dfrac{\dfrac{du}{dt}}{e^t}=e^{-t}\dfrac{du}{dt}$
$\therefore e^{-t}u\dfrac{du}{dt}=e^{-t}u+e^{2t}$
$u\dfrac{du}{dt}-u=e^{3t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
A: Since the problem has translational symmetry, it should be easier to solve for $x=x(y)$ instead and then invert the relationship. We have $\frac{dx}{dy}=\frac{1}{y'}$, $\frac{d^2x}{dy^2}=\frac{d}{dy}\frac{1}{y'}=\frac{-1}{y'^2}y''\frac{dx}{dy}=\frac{-1}{y'^3}y''=-x'^3y''$, so using the ODE we get $x''=-x'^3(\frac{y'}{y}+y^2)\implies x''=-\frac{x'^2}{y}-x'^3y^2$ and this is first order in $x'$, but I'm not sure it can be reduced, wolframalpha doesn't seem to be able to do it.
