TITLE: Solve $ f'(x)f(-x)=f(x)$
QUESTION [7 upvotes]: I want to find all differentiable functions $ f$ from $ \mathbb{R}$ to $ \mathbb{R}$ such that
$$ 
 f'(x)f(-x)=f(x)
$$
for all $ x \in \mathbb{R}$.
It's easy to check that this equation admits $f(x)=0$ and $ f(x)=1+e^x$ as solutions. The main difficulty comes from the appearance of $f(-x) $. For the case in which $f(x) \neq 0 $ for all $ x$, I've tried differentiating both sides of the equation and using the relation $ f'(-x) f'(x) = 1 $ to get rid of the term $ f'(-x) $ and obtained
$$
f'' f = f'^2 + f'
$$
But this equation seems to be more complicated.

REPLY [2 votes]: $$f'' f = f'^2 + f'$$
Simply rewrite the ODE as:
$$\dfrac {f'' }{f'+1} = \dfrac {f'}{f}$$
$$\dfrac {(f'+1)' }{f'+1} = \dfrac {f'}{f}$$
Note that $\dfrac {f'}{f}=(\ln f)'$
$$(\ln (f'+1))'=(\ln f)'$$
Integrate both sides.