Consider the differential equation:
$$xf'(x) = f(x-1)$$
for $f : \mathbb{R} \to \mathbb{R}$. This has a solution given by $(1+x)$. However, this solution does not work for me, as I require a (non-zero) solution which satisfies $\lim_{x \to \infty} f(x) = 0$. This is due to the fact that $f$ here represents the derivative of an ansatz for a probability density function.
Which are the other solutions to this differential equation, and/or how can I solve to find them?
Edit: I now remembered the existence of the Fourier transform. If we apply this to both sides, we get:
$$i \frac{d}{dk} \hat{f'}(k) = e^{-ik} \hat{f}(k)$$
$$- \frac{d}{dk} \bigg( k \hat{f}(k) \bigg)= e^{-ik} \hat{f}(k)$$
$$\implies \hat{f} + k \frac{d \hat{f}}{dk} = -e^{-ik} \hat{f}$$
$$\implies \frac{d \hat{f}}{dk} = \frac{-(1 + e^{-ik})}{k} \hat{f}$$
I recall that the Fourier transform is a bijection if $f, \hat{f}$ are integrable. Of course, $1+x$ is not integrable.