Is there any known closed-form solutions to $y''yx+y'y-(y')^2x=x^3$? The title says it all... 
The equation $y''yx+y'y+(y')^2x=x^3$ belongs to a known DE class (listed on EqWorld).
The equation $y''yx+y'y-(y')^2x=x^3$ appears to be way more difficult to solve...
Is the analytic solution to this equation known at all?
 A: I've solved this. There indeed is a closed-form solution that is obtained as follows:
The equation is a generalized homogeneous equation, so we start by substituting 
$$x \rightarrow kx, \qquad  y \rightarrow k^my, \qquad y' \rightarrow k^{m-1}y', \qquad y'' \rightarrow k^{m-2}y''$$ and obtain $m=2$ because $x^3$ becomes $k^3x^3$ on the right side, and all summands on the left side have the common factor of $k^{2m-1}$.
Next, we substitute $x$ for $e^t$, and $y$ for $ze^{mt}$, i.e. $y=ze^{2t}$
Keeping in mind that $t_x'=\frac{1}{x}=e^{-t}$ and $(e^t)_x'=1$, we write
$$ 
\begin{equation}
y_x'=z_t'e^{2t}t_x'+2ze^{2t}t_x'=e^{2t}(z_t'+2z)e^{-t}=e^t(z_t'+2z)\\
y_{xx}''=(z_t'+2z)+e^t(z_{tt}''t_x'+2z_t't_x')=z_t'+2z+z_{tt}''+2z_t'=z_{tt}''+3z_t'+2z\\
ze^{3t}(z'+2z)+ze^{3t}(z''+3z'+2z)-e^{3t}(z'^2+4zz'+4z^2)=e^3t\\
z(z'+2z)+z(z''+3z'+2z)-(z'^2+4zz'+4z^2)=1\\
zz''-z'^2=1
\end{equation}
$$
The equation $zz''-z'^2=1$ allows for order reduction by subbing $z'=p(z)$. 
I'll skip the routine, and just borrow the general solution from wolfram:
$$
\begin{equation}
z_1(t) = \frac{1}{2} e^{-e^{c_1}t-2c_1-e^{c_1}c_2}(e^{2e^{c_1}(c_2 + t)}+e^{2c_1}) \\
z_2(t) = \frac{1}{2} (e^{-e^{c_1}t-2c_1-e^{c_1}c_2}+e^{e^{c_1}t+e^{c_1}c_2})
\end{equation}
$$
All that is left is to perform the backward substitutions, $z\rightarrow y e^{-2t}$ and $e^t \rightarrow x$, to get our closed forms:
$$
\begin{equation}
y(x) = \frac{1}{2}( x^{2-e^{c_1}}e^{-2c_1-c_2e^{c_1}}+ x^{2+e^{c_1}}e^{c_2e^{c_1}} )\\
y(x) = \frac{1}{2}( x^{2+e^{c_1}}e^{-2c_1+c_2e^{c_1}}+ x^{2-e^{c_1}}e^{-c_2e^{c_1}} )
\end{equation}
$$
We can rewrite it further if we want:
$$
y(x) = \frac{1}{2}( x^{2\pm e^{c_1}}e^{-2c_1\pm c_2e^{c_1}}+ x^{2\mp e^{c_1}}e^{\mp c_2e^{c_1}} )
$$
There may be typos in the solution, but I've verified the obtained closed forms in Mathcad, and they indeed satisfy the initial equation.
A: There is no closed-form solution to the differential equation mentioned and generally not even a solution in terms of standard functions.
Sampling some initial values for $y$ and $y'$, without loss of generality, one can receive a general idea about the function $y(x)$, where it seems to exist an asymptotic "duality" between them and thus the inability of describing the solution in closed-form terms is (in an essence) derived :
$ \qquad \qquad \qquad \qquad $
The existence of a solution though, it may be simply derived by standard theorems or just in a functional analysis approach.
