In order to find a prior probability distribution I have to solve the following functional equation:
$$af\left(\frac{a\theta}{1-\theta-a\theta}\right)=(1-\theta+a\theta)^2f(\theta)$$
the solution of this equation must be a function $f(\theta)$.
I am new to functional equations, and my question is: could someone demonstrate me that $$f(\theta)=A \theta^{a-1}(1-\theta)^{a-1}$$ is a solution of this equation?
EDIT: I asked to my university teacher and he said that the problem was badly posed. We have that for $a=0$ the function $f(\theta)=\frac{A}{ \theta(1-\theta)}$ is a solution of the previous functional eq. (the demonstration is easy).
Then $f(\theta)=A \theta^{a-1}(1-\theta)^{a-1}$ is a "generalization of the solution".
So a better question is, given that $f(\theta)=A \theta^{a-1}(1-\theta)^{a-1}$ is a solution of the previous functional eq. for $a=0$ , are there values of $a\neq0$ for which this function is still a solution?