The following conjecture is inspired by asymptotic results in generalizations of the secretary problem.
CONJECTURE Consider a sequence of functions {$F_n$} with $F_{n}:[0,n]\cap \mathbb{Z}\rightarrow\mathbb{R}$ defined recursively, for $k\in \{1,\ldots,n\}$, by: $$F_{n}(k)=G_{n}(k)+H_{n}(k)F_{n}(k-1)\text{ and }F_{n}(0)=\mu_n \rightarrow \mu.$$
with $|H_n(k)|< 1$ and $|G_n(k)|<M$ for all $(n,k)$
Let $f_{n}(x):=F_{n}(\lfloor{nx}\rfloor)$, $h_{n}(x):=n(1-H_{n}(\lfloor {nx}\rfloor))$ and $g_{n}(x):=nG_{n}(\lfloor{nx}\rfloor)$. If both $h_{n}(x)$ and $g_{n}(x)$ converge in $(0,1)$ and uniformly in $[\varepsilon ,\varepsilon^{\prime}]$ for all $0<\varepsilon<\varepsilon^{\prime}<1$ to continuous functions in $(0,1)$, $h(x)$ and $g(x)$, respectively, then $f_{n}(x)\rightarrow f(x)$ uniformly in $[\varepsilon ,\varepsilon^{\prime}]$ for all $0<\varepsilon<\varepsilon^{\prime}<1$ with $f\in \mathcal{C}^1(0,1)$ and $f$ satisfies the following differential equation in $(0,1)$ $$f^{\prime}(x)= -f(x)h(x)+g(x).$$
Example :
$$F_{n}(k)=\frac{1}{n}+\frac{n-k}{n-k+1}F_{n}(k-1)\text{ and }F_{n}(0)=0.$$ $$F_{n}(\lfloor{nx}\rfloor) \rightarrow (x-1) \log(1-x) $$ The associated differential equation is: $$1-\frac{f(x)}{1-x}=f'(x)$$
(1) Ribas, J.M.G. A new look at the returning secretary problem. J Comb Optim 37, 1216–1236 (2019). https://doi.org/10.1007/s10878-018-0349-8
(2) Bayón, L., Fortuny Ayuso, P., Grau, J.M. et al. A New Method for Computing Asymptotic Results in Optimal Stopping Problems. Bull. Malays. Math. Sci. Soc. 46, 46 (2023). https://doi.org/10.1007/s40840-022-01436-4
Unfortunately, similarly to the situation with your other question, the uniform convergence in $[\varepsilon,\varepsilon^{\prime}]$ for all $0<\varepsilon<\varepsilon^{\prime}<1$ is a too weak condition for the conclusion. For instance, let $\mu\ne 0$, $M>1$, for each natural $n$ pick $\varepsilon_n>0$ such that $(1-\varepsilon_n)^n\ge 1-\frac 1{n^2}$, and for each natural $k<n$ put $G_n(k)=0$ and $H_n(k)=1-\varepsilon_n$, but the case $H_n(1)=\left(\frac 12+\frac {(-1)^n}4\right) (1-\varepsilon_n)$. Then $F_n(k)=H_n(1)\mu_n(1-\varepsilon_n)^{k-1}$ when $k\ge 2$. Thus both $h_{n}(x)$ and $g_{n}(x)$ converge in $(0,1)$ and uniformly in $[\varepsilon,\varepsilon^{\prime}]$ for all $0<\varepsilon<\varepsilon^{\prime}<1$ to the zero function, but even the sequence $(f_n(1/2))_{n\in\mathbb N}$ diverges.