Does there exist $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$?

Solution 1:

I cannot find a general solution for this. But I may find one: The given DE eqn. is equivalent to $$ \frac{dg^{-1}}{dx} = g$$ setting $g = f^{-1}$. And this is in turn equivalent to $$ g^{-1}(x) = \int g(x)dx$$ or $$x = g(\int g(x)dx). $$ Try $g = ax^n$. Then you will find $$ x = a(\frac{a}{n+1}x^{n+1})^{n} = \frac{a^{n+1}}{(n+1)^{n}}x^{n(n+1)}.$$ Then by the root formula you will find two $n$ from the equation $n(n+1)=1$, and for these $n$ you also find suitable $a$.

And try if a linear combination of this solution works, i.e., if the solution space is linear.

If the above works, probably you can extend this idea even further: you man consider any function that is expressible as an infinite sum, i.e., the polynomials of $\infty$ power.

The above two processes, though they may indeed work, will not be that easy I guess.