# Differential Equation with inverse function $\frac{1-f^{-1}\left(\frac{f(x)}{x}\right)}{1-x} = 1- \frac{f(x)}{xf'(x)}$

$$\frac{1-f^{-1}\left(\frac{f(x)}{x}\right)}{1-x} = 1- \frac{f(x)}{xf'(x)}$$

I know $$f(x) = ax+b$$ is a solution. How can I find other solutions?

• Have you checked if $f(x)=ax+b \ \ \forall \ \ a,b\neq 0$ is a solution? This is the first what I would do if I´m facing this problem. – callculus May 21 at 15:21
• Yes. That's a solution. I want to find other solutions. Or prove that is the only solution. – ftor May 21 at 15:28
• I have to admit that the solution doesn´t work for me. Maybe I´m wrong, but maybe I´m right. A calculation that shows that $f(x)=ax+b$ is a solution would indicate that you are really interested in that exercise. – callculus May 21 at 15:30
• I checked again. That is a solution. – ftor May 21 at 15:36
• Yes. That is a critical equation in my research project. – ftor May 21 at 15:38

This is an extended comment, not an answer, but I'd need the space... Inverse functions are stressful, so you might eliminate yours by $$f^{-1}\left(\frac{f(x)}{x}\right) =1 -\left (1- \frac{f(x)}{xf'(x)}\right ) (1-x),$$ so that $$\frac{f(x)}{x} =f\left (1 -\left (1- \frac{f(x)}{xf'(x)}\right ) (1-x)\right ),$$ so that $$f(x)=xf\left( \frac{f(x)/f’(x)}{ x}+\frac{xf'(x) -f(x)}{f'(x)} \right),$$ manifestly possessing your leading binomial solution. Near the origin, only the leading term in the r.h.side argument is singular.