I have a functional differential equation $$\Big(\epsilon x \frac{d}{dx} + a x^2 - b x + 1\Big) f(x) = f(q x).$$ Notice that this function involves both $f(x)$ and $f(q x)$, where $q$ is some real number with $0 \leq q < 1$. I want to solve this equation on the condition that $f(0)=1$.
In the limit $\epsilon=0$:
I have solved similar equations that lack a derivative term ($\epsilon = 0$) by noting that the equation constitutes a recursion relation: $$ f(x) = \frac{f(q^N x)}{(a x^2 - b x + 1)^N}.$$ As $N \rightarrow \infty$, we have $q^N \rightarrow 0$, since $0\leq q < 1$, providing $$ f(x) = \lim_{N\rightarrow\infty} \frac{1}{(a x^2 - b x + 1)^N}$$ if we note the condition $f(0)=1$. This result is actually enough to solve the problem with $\epsilon = 0$, since expansion in partial fractions gives the result as an infinite series.
With $\epsilon\neq 0$:
In this case a similar recursion approach can be made, but it provides an infinite order ODE: $$ \lim_{N\rightarrow \infty} \Big(\epsilon x\frac{d}{dx}+ a x^2 - b x + 1\Big)^N f(x) = 1.$$ This is now very difficult. Does anyone have a recommendation to make progress on the problem with $\epsilon \neq 0 $ ? One solution is to solve the problem $L^N f(x) = 1$ for arbitrary $N$, where $L$ has the obvious meaning. Limiting $N\rightarrow \infty$ would then provide the desired solution of the functional ODE.