Consider the following DE:
$$x' = f(x) + g(x)z'$$ with initial condition: $$x(0) = x_0$$
where $f,g$ are infinitely differentiable functions of $x$ and $z'$ is the derivative (in the distributional sense) of some given locally integrable function z(t).
By distribution, I mean the definition given here on wikipedia.
I would like to know if there exists a unique solution $x$ (a function) such that $x$ is a solution for all t in $[-d,d]$ for some constant $d>0$.
Details
By the expression $x'$, I denote the distributional derivative of $x$ when thought of as a distribution, so $x'$ is a distribution rather than a function. Because the product of a function and a distribution is also a distribution, $g(x)z'$ is a distribution. $f$ can be interpreted as a distribution via the usual inclusion map from functions to distributions, thus we can consider the sum $f(x)+g(x)z'$ and therefore also consider the equation of distributions $x'=f(x)+g(x)z'$.
No; without further assumptions there can be many solutions even in senses much stronger than distributional. The key idea is that the equation is singular around any common roots of $f$ and $g$ and the forcing from $z'$ can push onto (or off) those singularities.
Let, for all $x$, $f(x)=0$ and $g(x)=x$; for all $t$, $z(t)=-\frac{2}{\sqrt{t}}$; and $x_0=0$. Then we have in distribution $$x'=-\frac{x}{t^{3/2}}\tag{1}$$ For any $C$, the function $$x(t)=Ce^{-\frac{2}{\sqrt{t}}}$$ solves (1) on all of $\mathbb{R}$.
(Inspired by the solution to this question.)