Finding exact solutions to a nonlinear ODE I've come across the differential equation
$$ 2 x^2 f(x)^2 f^{(3)}(x)+12 x f(x)^2 f''(x)+3 x^2 f'(x)^3-12 x f(x) f'(x)^2+12 f(x)^2 f'(x)-6 x^2 f(x) f'(x) f''(x)=0 $$
and I would like to find the most general exact solution possible.
Plugging in $f(x)=cx^\alpha$, I see that the powers $c,c x^2,cx^{-2}$ are solutions. I couldn't find anything more general than these. Can more general solutions be found for this DE?
Thanks
 A: Maple code 
 $$infolevel[dsolve] := 5:  dsolve(2*x^2*f(x)^2*(diff(f(x), x, x, x))+12*x*f(x)^2*(diff(f(x), x, x))+$$ $$3*x^2*(diff(f(x), x))^3-12*x*f(x)*(diff(f(x), x))^2+
  12*f(x)^2*(diff(f(x), x))-$$ $$6*x^2*f(x)*(diff(f(x), x))*(diff(f(x), x, x)) = 0)$$
  produces
  $$f \left( x \right) =2\,{\it WeierstrassP} \left( -1/2\,{\frac {\sqrt [
3]{-2\,{\it \_C2}}}{x}}+{\it \_C3},-{\frac {{\it \_C1}\, \left( -2\,{
\it \_C2} \right) ^{2/3}}{{\it \_C2}}},0 \right) {\frac {1}{\sqrt [3]{
-2\,{\it \_C2}}}}, $$ indicating the use of computation of integrating factors. See the pdf file.
A: Just an accompanying comment:
Note that the ODE is such that, whenever $f$ is a solution, then $cf$ will be
a solution as well, for any constant factor $c$.
That $c$ is one parameter of the general solution, and by eliminating it,
the order of the ODE should be reduced by one.
In order to do this, set
$$v(x) = (\ln f)'(x) = \frac{f'(x)}{f(x)}$$
Plugging in, I get
$$2x^2 v''(x) + 12x v'(x) + 12 v(x) - x^2 v(x)^3 = 0$$
which should then lead to the Weierstrass $\wp$ solution given by user64494.
Note: The original ODE thus can be written as
$$\frac{\mathrm{d}^3}{\mathrm{d}x^3}\left(2x^2\ln f(x)\right) =
x^2 \left((\ln f)'(x)\right)^3$$
which comes close to the hint user1337 has given.
