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

Question

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

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

Answer 1

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.

After doing this, you might well start plugging in series solutions and devising creative coefficient recursions...