# Solving $f'(x)=f(x)f(-x)$

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.

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.
• 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. Commented Nov 9, 2021 at 6:05
• Couldn't $f$ be an odd function plus a constant? Commented Nov 9, 2021 at 6:05
• @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. Commented Nov 9, 2021 at 6:08