Suppose $X_t$ satisfies the following time inhomogenous SDE: $$dX_t=f(t,X_t)dt+g(t,X_t)dW_t,$$ and define $u(s,x;t_1)=E(v(X_{t_1})\mid X_s=x)$ for $s\in [t_0,t_1]$ for a good enough fixed function $v$, then Kolmogorov backward equation shows $$ u_t(s,x)+Lu(s,x)=0,\quad u(t_1,x)=v(x),$$ for $(s,x)\in [t_0,t_1]\times \mathbb{R}$, where $L$ denotes the generator of $X_t$.
My questions is in the definition of $u(s,x)$, do we assume we know the information before $s$? The answer is yes, if $X_t$ is time homogeneous based on the Markov property. But I am not sure about the time inhomogenous case.
For example, if $dX_s=f(X_{t_0})ds+g(X_{t_0})dW_s, s\in [t_0,t_1]$, we know $$X_{t_1}=X_s+f(X_{t_0})(t_{t_1}-s)+g(X_{t_0})(W_{t_1}-W_s).$$
Then if we want to calculate $u(s,x;t_1)$, do we need to consider the randomness of $X_{t_0}$?