Second order homogenous non-linear DE: $3(x')^2 - 2x''x=0$ How do I solve this for $x$?
$$3\dot{x}^2-2\ddot{x}x=0$$
$$\Leftrightarrow$$
$$3(x')^2 - 2x''x=0 $$
Note: This comes from my working here(on stack exchange meta sandbox[newest activity])
List of methods is acceptable as an answer. I can do the research with helpful direction.
There is likely an approach to this that I don't know, otherwise I may have improperly obtained this DE. I will overlook my working again soon, and see if this possibility is (non) negligible.
 A: Let $\dfrac{dx}{dt} = u$. But, $x^{\prime \prime}(t) = \dfrac{du}{dt} = \dfrac{du}{dx}\dfrac{dx}{dt}=u\dfrac{du}{dx}$. Hence,
$$
3(x^{\prime})^2 - 2x^{\prime \prime}x = 0 \quad \Rightarrow \quad 3u^2 - 2xu\dfrac{du}{dx} =0
$$
If $u = 0$, then $x(t) = k$ is solution. If $u \neq 0$, then,
$$
2x\dfrac{du}{dx} = 3u \quad \Rightarrow \quad \dfrac{du}{u} = \dfrac{3dx}{2x} \quad \Rightarrow \quad \ln u = \dfrac{3}{2}\ln x + \ln C_1 \quad \Rightarrow
$$
$$
u = C_1x^{3/2} \quad \Rightarrow \quad \dfrac{dx}{dt} = C_1x^{3/2} \quad \Rightarrow \quad \int \dfrac{dx}{x^{3/2}} = \int C_1 dt \quad \Rightarrow 
$$
$$
x(t) = \dfrac{4}{(C_1t + C_2)^2}
$$
A: Let $y = x^m$. Then, $\dot{y} = \dfrac{d}{dt}[x^m] = mx^{m-1}\dot{x}$ and $\ddot{y} = \dfrac{d}{dt}[\dot{y}] = \dfrac{d}{dt}[mx^{m-1}\dot{x}] = m(m-1)x^{m-2}\dot{x}^2+mx^{m-1}\ddot{x} = mx^{m-2}\left[(m-1)\dot{x}^2+x\ddot{x}\right]$. 
If we set $m = -\dfrac{1}{2}$, then $\ddot{y} = -\dfrac{1}{2}x^{-5/2}\left[-\dfrac{3}{2}\dot{x}^2+x\ddot{x}\right] = \dfrac{x^{-5/2}}{4}\left[3\dot{x}^2-2x\ddot{x}\right] = 0$. 
Since $\ddot{y} = 0$, we have $\dot{y} = D$, and thus, $x^{-1/2} = y = Ct+D$ for some constants $C,D$. 
Therefore, $x(t) = \dfrac{1}{(Ct+D)^2}$ for some constants $C,D$. 
EDIT: The assumption that we can let $y = x^{-1/2} = \dfrac{1}{\sqrt{x}}$ implicitly assumes that $x > 0$. If $x < 0$, then we should set $y = \dfrac{1}{\sqrt{-x}}$. Also, $x \equiv 0$ is a trivial solution. Thus, the complete solution set is $x = \dfrac{A}{(t+B)^2}$ for any real constants $A,B$. 
A: Suppose that we define $$x(t)=\frac{1}{y^2(t)}$$ $$x'(t)=-\frac{2 y'(t)}{y(t)^3}$$ $$x''(t)=\frac{6 y'(t)^2-2 y(t) y''(t)}{y(t)^4}$$ and then the differential equation reduces to $$\frac{4 y''(t)}{y(t)^5}=0$$ which is quite simple $$y''(t)=0$$ $$y'(t)=c_1$$ $$y(t)=c_1t+c_2$$ and finally $$x(t)=\frac{1}{(c_1t+c_2)^2}$$
A: HINT
I would say 
$\displaystyle 3\dot{x}^2-2\ddot{x}x=0\Rightarrow 2(\dot{x}^2-\ddot{x}x)+\dot{x}^2=0\Rightarrow 2\frac{\dot{x}^2-\ddot{x}x}{\dot{x}^2}+1=0\Rightarrow $
$2\dot{\left(\frac{x}{\dot{x}}\right)}+1=0\Rightarrow 2\frac{x}{\dot{x}}+t=C\Rightarrow \cdots \Rightarrow x=\frac{C_1}{(C-t)^2}$
