A general solution for $\frac{f'\left(x\right)}{f'\left(1-x\right)}=-\frac{x}{1-x}$

Asked Nov 01 '21 at 19:08

Active Nov 02 '21 at 16:34

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I am trying to understand the equation (for $x\neq \frac{1}{2}$): $$\frac{f'\left(x\right)}{f'\left(1-x\right)}=\frac{x}{1-x}$$

I can see that, $f(x)=ax^2 + c$ solves this, but I want to know if this is all that solves it.

Is there a name for functions of this type? It is not quite a delay differential equation. Where can I look for guidance on solving equations like this?

edited Nov 02 '21 at 16:34

asked Nov 01 '21 at 19:08

CommonerG

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2

Hint: simplify to $g(x)=-g(1-x)$ with $g:=f^\prime/x$, then define $h(x):=g(1/2+x)$. – J.G. Nov 01 '21 at 19:52
Is the message here that the derivative of of f at x is proportional to x’s (signed) distance from 1/2? – CommonerG Nov 01 '21 at 20:51
1

It's that $h$ has to be of a certain parity. – J.G. Nov 01 '21 at 20:53
Yes, I see. h is odd – CommonerG Nov 01 '21 at 20:57
There's a sign issue in your proposed solution that makes me think by $f^\prime(1-x)$ you might have meant $df(1-x)/dx=-f^\prime(1-x)$. If that's the intended problem, $h$ actually ends up being even. – J.G. Nov 01 '21 at 21:03
You’re absolutely right. – CommonerG Nov 01 '21 at 21:52

A general solution for $\frac{f'\left(x\right)}{f'\left(1-x\right)}=-\frac{x}{1-x}$

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