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I came across $f'''(x)=f(x)f'(x)f''(x)$ but I don't know how to solve it.

I tried

$\frac{f'''(x)}{f''(x)}=f(x)f'(x)$

$\ln|f''(x)|=\frac{1}{2}f(x)^2+c_{1}$

But from there I have no idea how to proceed.

Please help me solve this if possible.

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After

$$ f'' = C_0e^{\frac 12 f^2} $$

we have

$$ f'' f' = C_0e^{\frac 12 f^2}f'\Rightarrow \frac 12(f')^2=C_0\sqrt{\frac{\pi}{2}}\phi\left(\frac{f}{\sqrt 2}\right)+C_1 $$

with

$$ \phi\left(x\right)=\int_0^x e^{\zeta^2}d\zeta $$

and finally we arrive to the solution after integrating

$$ \frac{df}{\sqrt{2\left(C_0\sqrt{\frac{\pi}{2}}\phi\left(\frac{f}{\sqrt 2}\right)+C_1\right)}} = dx $$

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  • $\begingroup$ Wow. That's pretty elegant. Thank you. $\endgroup$ – clathratus Sep 29 '18 at 22:21
  • $\begingroup$ Just a clarity question: with respect to what variable does one integrate the last expression? Edit: It's x, right? $\endgroup$ – clathratus Sep 29 '18 at 22:28
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    $\begingroup$ @clathratus The last expression was obtained due to the variable separation property. So the result should be $F(f) = x + C_2$ which gives an implicit solution for $f(x)$ $\endgroup$ – Cesareo Sep 29 '18 at 22:35
  • $\begingroup$ Okay thank you. $\endgroup$ – clathratus Sep 30 '18 at 1:19

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