Solve $ f'(x)f(-x)=f(x)$

Question

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.

Answer 1

You have made a very good attempt, as the ODE actually has a simple solution.

Notice that taking $g=f'$ yields $f''=g'=gg^*$ where $g^*=dg/df$ so $$ff''=f'(1+f')\implies fgg^*=g(1+g)\implies fg^*=1+g.$$ This is a separable equation with solution $$\int\frac{dg}{1+g}=\int\frac{df}f\implies \log(1+g)=\log f+c\implies g=Cf-1$$ for some constant $C$. Since $g=f'$ we have $$f'-Cf=-1\implies f(x)=De^{Cx}+\frac1C.$$

Answer 2

$$f'' f = f'^2 + f'$$ 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.