Say we consider a process $X_t$ with dynamics $$dX_t=\alpha X_tdt+\sigma dW_t$$ and $X_0=x_0$ where $W_t$ is a standard brownian motion. It has the solution $$X_t=e^{\alpha t}x_0+\sigma \int_0^te^{\alpha (t-s)}dW_s$$
But what if we for $T>t$ wish to know the distribution of $X_T$ conditional on $X_t=x_t$. My textbook says (as I understand it) without going into depth that $$X_T=e^{\alpha (T-t)}x_t+\sigma \int_t^Te^{\alpha (T-s)}dW_s$$ but how do we get here? Thanks!
So if you understand how going from the initial condition $X_0 = x_0$ to the solution for $X_t$ at time $t$, then notice that a SDE in general is of the form \begin{align*} dX_t = f(t, X_t )dt + g(t, X_t) dW_t, \end{align*} now the linear SDE you are interested in is the special case where the functions drift function $f$ and diffusion function $g$ do not depend on $t$, therefore considering the solution at time $t$ starting from $0$ is the same as considering the solution at time $T - t$ starting from $t$, in both cases it is the length of the time interval you solve over that matters, and so the solution is the same, the second solution is just a relabelling of the first with initial time $0 \mapsto t$ and terminal time $t \mapsto T - t$
This would not be the case if your drift and diffusion functions had been time dependent.
Just integrate your process from $t$ to $T$ instead of $0$ to $t$.