Let $T >0$ and let $x(t)$ with $t \in [0,T]$ be the solution of a differential equation of the form
$$\dot{x}(t) + v(t) x(t) = 1, \quad x(0) = A \in \mathbb{R}$$
Here $v(\cdot)$ is a control input. Do we know if we can find $v(\cdot)$ such that $x(T) = B$ for some fixed $T$? Thanks.
Without any additional requirements there can be infinitely many solutions. But there can only be solutions if $A$ and $B$ have the same sign or when $A$ negative and $B$ positive. This is because when $x(t)=0$, then $\dot{x}(t)=1$ independently of what you choose for $v(t)$. So $x(t)$ can't become negative, because that would require a negative time derivative at $x(t)=0$.
Assuming that $x(t)$ does not have to cross zero then an solution would be to choose an input which would result in a linear path from $A$ to $B$,
$$ \dot{x}(t) = \frac{x(T) - x(0)}{T} = \frac{B - A}{T}. $$
Or if you want to have some disturbance rejection/feedback,
$$ \dot{x}(t) = \frac{x(T) - x(t)}{T - t} = \frac{B - x(t)}{T - t}. $$
Solving this last equation for $v(t)$ yields,
$$ v(t) = \frac{x(t) - B + T - t}{(T - t)\, x(t)}. $$
If $A$ negative and $B$ positive then you would have to adjust the path such that $\dot{x}(t)=1$ near $x(t)=0$.