how to solve $(y')^4 x -2 y (y')^3 + 12 x^3 = 0$ Is there a smart way to solve this first order ode
$$
    (y')^4 x -2 y (y')^3 + 12 x^3 = 0
$$
This is problem from Ordinary differential equations and their solutions. By George Moseley Murphy. 1960.
I solved it myself, but my method is a brute force. First solved for $y'$, which gave 4 solutions. This generated 4 ode's to solve.
Each one of these ode's turned out to be isobaric. After applying the isobaric transformation, the ode becomes separable. However, the integrals are so complicated and could not solve them even on the computer.  So the solutions are left with unevaluated integrals. But they were verified correct by Maple. But will not post them here, as the integrals are too large.
Maple solves this and gives these simple 5(!) solutions


The Maple trace says it used

trying 1st order ODE linearizable_by_differentiation

Which I do not know how. Here is the full trace
ode:=x*diff(y(x),x)^4-2*y(x)*diff(y(x),x)^3+12*x^3 = 0; infolevel[dsolve]:=5; 
dsolve(ode)

 trying 1st order ODE linearizable_by_differentiation
 -> Solving 1st order ODE of high degree, Lie methods, 1st trial
 -> Computing symmetries using: way = 2
 -> Solving 1st order ODE of high degree, 2nd attempt. Trying parametric methods
  *** Sublevel 3 ***
  Methods for first order ODEs:
  --- Trying classification methods ---
  trying homogeneous types:
  trying exact
  Looking for potential symmetries
  trying an equivalence to an Abel ODE
  trying 1st order ODE linearizable_by_differentiation
  -> Calling odsolve with the ODE diff(y(x) x) = (3*(x^4+12*y(x)^2)*y(x)/x- 4*y(x)*x^3)/(-x^4+36*y(x)^2) y(x)
  *** Sublevel 3 ***
  Methods for first order ODEs:
  --- Trying classification methods ---
  trying a quadrature
  trying 1st order linear
  <- 1st order linear successful
  <- 1st order, parametric methods successful

My question is, how did Maple obtain these simple solutions? From trace it says it used Lie symmetries which I am still learning.
Does anyone see a "simple" method to solve this ode using some smart transformation and be able to obtain the solutions found by Maple?
 A: Set $y(x)=\frac12u(x^2)$, then $y'(x)=xu'(x^2)$ and
$$
x^5[u'(x^2)]^4-x^3u(x^2)[u'(x^2)]^3+12x^3=0.
$$
Cancel $x^3$, divide by $u'(x^2)^3$, substitute $x^2=s$ and reorder to give the transformed equation
$$
u(s)=su'(s)+12u'(s)^{-3}.
$$
This now is a Clairaut DE with linear solution family
$$
u(s)=Cs+12C^{-3},~~y(x)=\frac{C}2x^2 +6C^{-3}
$$
You get one of the Maple solutions with $C=\frac1{C_1}$.
The singular solution or envelope of $u(s)=su'(s)+f(u'(s))$ is obtained from $0=s+f'(u'(s))$, here
$$
0=s-36[u'(s)]^{-4}
$$
The real solutions are
$$
u'(s)=\pm\left(\frac{36}{s}\right)^{1/4}
$$
so that
$$
u(s)=u'(s)\left(s+12[u'(s)]^{-4}\right)=\pm\frac43s\left(\frac{36}{s}\right)^{1/4}
\\~\\
y(x)=\pm\frac23x\sqrt{6|x|}
$$
This should work for  $x>0$ as well as for $x<0$.
As usual, one can switch branches where solutions from the regular family touch the singular solution.
A: Here's a continuation of what you were doing. The ODE is generalized homogeneous (which you've called isobaric, same thing), so under the transformation $y=x^{3/2}u$, $\ln|x|=\xi$ the ODE becomes
\begin{align}
\left(u'+\frac{3u}{2}\right)^4-2u\left(u'+\frac{3u}{2}\right)^3+12=0.
\end{align}
Solving this for $u'$ doesn't work out well, but one can use the method of 'integration by differentiation' outlined in Handbook of exact solutions for ordinary differential equations / Andrei D. Polyanin Valentin F. Zaitsev 2ed. Taking $u'=t$ the equation can be written as
\begin{align}\tag{*}\label{1}
F(\xi,u,t)=\left(t+\frac{3u}{2}\right)^4-2u\left(t+\frac{3u}{2}\right)^3+12=0.
\end{align}
Taking the differential of \ref{1} and using the relationship $\mathrm du/\mathrm d\xi=t$ one arrives at the two equations
\begin{align}
(F_{\xi}+tF_u)\xi'_t+F_t=0,\quad\quad (F_{\xi}+tF_u)u'_t+tF_t=0.
\end{align}
In our case this yields that
\begin{align}
\frac{t}{4}(2t+3u)^2(4t-3u)\xi'_t+t(2t+3u)^2&=0 \quad\text{and}\\
\frac{t}{4}(2t+3u)^2(4t-3u)u'_t+t^2(2t+3u)^2&=0.
\end{align}
(The two equations are identical, via the chain rule we see that $\xi'_t=u'_t/t$). For the second we get that
\begin{align}
t(2t+3u)^2\big[(4t-3u)u'_t+4t\big]=0.
\end{align}
The first solution, $t=0$, yields that $y=cx^{3/2}$. Substituting this into the original equation we find that this solves the equation for $c^2=\pm 8/3$. The four different $c$'s correspond to the four particular solutions Maple has given you.
For $2t+3u=0$ we find that $y=c$, which does not solve the equation, I'm not sure why this occurs. For
\begin{align}
(4t-3u)u'_t+4t=0
\end{align}
we can undo some of the substitutions (using $u'_t=u'_\xi/u''_{\xi\xi})$ to get a constant coefficient linear equation
\begin{align}
u''_{\xi\xi}+u'_{\xi}-\frac{3}{4}u=0.
\end{align}
This has the solution
\begin{align}
u(\xi)=c_1e^{\xi/2}+c_2e^{-3\xi/2}\quad\rightarrow\quad
y(x)=c_1x^2+c_2.
\end{align}
Substitution into the original equation yields our general solution
\begin{align}
y(x)=c_1x^2+\frac{3}{4c_1^3}.
\end{align}
Plot the general solution along with the two real particular solutions and you'll see that by changing $c_1$ the general solution arcs along the particular one! Very nice.
My feeling is that this could be extended into the complex domain, but my skills stop at solving ODE's. Perhaps a skilled individual could do so.
A: Let's look for polynomial solutions. If $y = c x^d + \ldots$, then $(y')^4 x - 2 y (y')^3 = c^4 (d^4-2 d^3) x^{4d-3} + \ldots$.  This $x^{4d-3}$ term can only cancel if $d=2$.  So now try $y = c_2 x^2 + c_1 x + c_0$ and it's not hard to see we must have
$c_1 = 0$ and $c_0 c_2^3 = 3/4$.
