Solve $x \ddot {x} + {\dot {x}}^2=0$. I solved a physics problem and I got this equation, but I don't know how to proceed. Could you solve for $x (t)$ this equation:
$x \ddot {x} + {\dot {x}}^2=0$

$(xx')'=xx''+x'^2=0$ and $2xx'=2C$ then $(x^2)'=2C$ so $x^2=2Ct+D$.
