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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.

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    $\begingroup$ $(x/x')'=((x')^2-xx'')/(x')^2$ may help. $\endgroup$ – Gerry Myerson May 11 '18 at 9:43
  • $\begingroup$ Worth noting the next time you see a problem like this is that the closed-form solutions to second-degree differential equations of one variable often involve $c_1 \sin k_1 x + c_2 \cos k_2 x$, as the first and second derivatives have the same form. The $\sinh$ and $\cosh$ functions also often appear, for similar reasons. $\endgroup$ – Davislor May 16 '18 at 1:41
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Think that

$$ \frac{\ddot x}{\dot x} = \frac{\dot x}{x}\Rightarrow \frac{d}{dt}\ln(\dot x) = \frac{d}{dt}\ln (x) $$

so

$$ \ln(\dot x) = \ln(x) + C \Rightarrow \dot x = C_1 x\Rightarrow x = C_2e^{C_1 t} $$

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  • $\begingroup$ Thanks LutzL for the hint. $\endgroup$ – Cesareo May 11 '18 at 10:59
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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.

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  • $\begingroup$ Which works in this case because $u = x' \neq 0$. $\endgroup$ – Davislor May 16 '18 at 1:49
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    $\begingroup$ Agreed. Ironically, the solution gotten by assuming $u\ne 0$ also covers $u=0$ given the proper value of one integration constant. Dumb luck strikes again! $\endgroup$ – Oscar Lanzi May 16 '18 at 1:55
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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?

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$$xx'' = (x')^2,$$ 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 $$\frac {x''}x = \frac {(x')^2}{x^2} \implies (\frac {x'}x)' =0$$ $$\frac {x'}x =K_1 \implies \ln|x|=K_1t+K_2 \implies x=K_2e^{K_1t} $$

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