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

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.

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 $$f'-Cf=-1\implies f(x)=De^{Cx}+\frac1C.$$
• It should be noted that $f$ is indeed twice differentiable here as $f(x)=f'(x)f(-x)$ and $f(x)$ and $f(-x)$ are differentiable. In general, it is possible that the product of two functions is differentiable but that one of the functions itself is not differentiable (take for example $g(x)=0$ and $h(x)=|x|$). Apr 28, 2021 at 6:33
• @Mathematician42 (and for others interested) To expand on their attempt, differentiating the original equation gives $f''(x)f(-x)-f'(x)f'(-x)=f'(x)$. The OP made the acute observation that $f'(x)f'(-x)=1$ and using $f(-x)=f(x)/f'(x)$ we obtain $f''(x)f(x)/f(x)=1+f'(x)$. Apr 28, 2021 at 6:41
$$f'' f = f'^2 + f'$$ 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.