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: $x \ddot {x} + {\dot {x}}^2=0$

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    $\begingroup$ $$x(t)=c_2 \sqrt{2 t-c_1}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 13 '17 at 10:23
  • $\begingroup$ Was that a guess or is there an analytical way to do this. $\endgroup$ – Piyush Divyanakar Oct 13 '17 at 10:24
  • $\begingroup$ Wolfram alpha and the textbook solution give $x(t)= c_2 \sqrt {c_1 + 2t}$ $\endgroup$ – Flaffo Oct 13 '17 at 10:27
  • $\begingroup$ Divide by $x$ and also by $x'$. Then it is easy to integrate. $\endgroup$ – user121049 Oct 13 '17 at 10:32

$$x\ddot{x} + \dot{x}^2=\frac{d}{dt}(x\dot{x})=0\\ x\frac{dx}{dt}=c_1\\ \int xdx=\int c_1 dt\\ \frac{x^2}{2}=c_1t+c_2\\ x=\sqrt{2c_1t+c_2}$$ which may be rewritten as $$x=c_1\sqrt{2t+c_2}$$


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


Note that $x'' = \dfrac{\mathrm dx'}{\mathrm dt} = \dfrac{\mathrm dx'}{\mathrm dx} \dfrac{\mathrm dx}{\mathrm dt} = \dfrac{\mathrm dx'}{\mathrm dx} x'$.

Therefore: $$x \dfrac{\mathrm dx'}{\mathrm dx} x' + (x')^2 = 0$$

Then: $$x \dfrac{\mathrm d(x')^2}{\mathrm dx} + 2(x')^2 = 0$$

Let $v=(x')^2$: $$x \dfrac{\mathrm dv}{\mathrm dx} + 2v = 0$$

Separating variables: $$\dfrac1{-2v} \dfrac{\mathrm dv}{\mathrm dx} = \dfrac1x$$

Integrating both sides: $$-\dfrac12\ln v = \ln x + C$$

Therefore: $$v = k_0 x^{-2}$$

Substituting the definition of $v$: $$\dfrac{\mathrm dx}{\mathrm dt} = k_1 x^{-1}$$

Taking reciprocal of both sides: $$\dfrac{\mathrm dt}{\mathrm dx} = k_2 x$$

Integrating both sides: $$t=Ax^2+B$$

  • $\begingroup$ @samjoe solved. $\endgroup$ – Kenny Lau Oct 13 '17 at 10:53

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