How to solve the differential equation $xx'' = (x')^2$? There is an differential given:
$$xx'' = (x')^2,$$
where $x' \neq 0.$
I tried to solve this problem but I cannot see the proper substitution.
 A: The comment seen here uses a "trick". Here is a more standard solution.
The independent variable is not rendered explicitly so you have an autonomous equation.  Given this, put in $u=x'$.  Then 
$$x''=\frac{d}{dt}u(x(t))=\frac{du}{dx}\, x'=u\,\frac{du}{dx}$$
using the Chain Rule.  So
$$x\cdot u\cdot \frac{du}{dx}=u^2$$
Divide by $u$ and solve the separable equation that results, then substitute $x'$ for $u$ into that first integration to get another first order equation that is easily solved.
A: With the substitution $$v(x)=\frac{dx}{dt},$$ we get
$$x\frac{dv(x)}{dx}v(x)=(v(x))^2.$$
This can be written as
$$-v(x)\left(-x\frac{dv(x)}{dx}+v(x)\right)=0,$$
so either $v(x)=0$ or
$$\frac{dv(x)}{dx}=\frac{v(x)}{x}$$ and this is
$$\int\frac{\frac{dv(x)}{dx}}{v(x)}dx=\int\frac{1}{x}dx.$$
Can you finish?
A: $$xx'' = (x')^2,$$
Divide by $x^2 \,\, x\neq 0)$
$$\frac {x''}x = \frac {(x')^2}{x^2}$$
Note that 
$$(\frac {x'}x)'= \frac {x''}x-\frac {x'^2}{x^2}$$
Therefore
$$\frac {x''}x = \frac {(x')^2}{x^2} \implies (\frac {x'}x)' =0$$
$$\frac {x'}x =K_1 \implies \ln|x|=K_1t+K_2 \implies x=K_2e^{K_1t} $$
A: Think that
$$
\frac{\ddot x}{\dot x} = \frac{\dot x}{x}\Rightarrow \frac{d}{dt}\ln(\dot x) = \frac{d}{dt}\ln (x)
$$
so
$$
\ln(\dot x) = \ln(x) + C \Rightarrow \dot x = C_1 x\Rightarrow x = C_2e^{C_1 t}
$$
