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I think it is connected to the logistic differential equation,

$$f(x)=f(x)(1-f(x))$$

Because the logistic function has the property that $f(-x)=1-f(x)$.

Perhaps these differential equations are equivalent? I'm not really sure. T

This is not for school, I'm just curious.

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Well, it must be that $f'(-x) = f(-x) f(x) = f'(x)$ also, which means that $f(x)$ is an odd function plus a constant $f_0$. That is, $f(-x) = f_0 - f(x)$ and so we have the equivalent differential equation $f'(x) = f(x) (f_0 - f(x))$, i.e. recover the Logistic equation.

Edit: Fixed basic error (eek), thanks Ninad.

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    $\begingroup$ Your final $f$ contradicts your first assumption that $f$ must be odd. The logical error is assuming that the antiderivative of an even function must be an odd one, which is not correct. $\endgroup$ Commented Nov 9, 2021 at 6:05
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    $\begingroup$ Couldn't $f$ be an odd function plus a constant? $\endgroup$ Commented Nov 9, 2021 at 6:05
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    $\begingroup$ @Hans that would be the best way to fix this, then the real differential equation simplifies to $$f'=f(f_0-f)$$ so OP was on the money about the relationship to the logistical differential equation. $\endgroup$ Commented Nov 9, 2021 at 6:08
  • $\begingroup$ Thanks everyone! I'll accept an answer that reflects the conclusion of these comments. $\endgroup$ Commented Nov 9, 2021 at 9:08

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