Consider the linear matrix differential equation

$\def\diag{\mathrm{diag}}$ \begin{align} U(0) &= I\\ \frac{\mathrm{d}U}{\mathrm{d}t}(t) &= U(t) \phantom{.} Q(t) & & \quad(1) \end{align}

where $Q(t),U(t)$ are $n\times n$ real valued matrices and $Q(t)$ is a transition rate matrix, which means that the off diagonal entries are nonnegative and each row sums to zero. Unfortunately, in general $Q(t_1)Q(t_2)\neq Q(t_2)Q(t_1)$ so the tempting equality $U(t)=\exp\left(\int_0^t Q(s)\,\mathrm{d}s\right)$ is false in general.

For some $\delta>0$, consider a "magnified" process $V^\delta$

\begin{align} V^\delta(0) &= I\\ \frac{\partial V^\delta}{\partial t}(t) &= V^\delta(t) \phantom{.} (1+\delta)Q(t) & & \quad(2) \end{align}

Suppose one can compute the solution of (1) explicitly. Is there a simple expression for the solution of (2) in terms of the solution of (1)?

Actually I am only interested in calculating $\left. \frac{\partial V^\delta (t)}{\partial \delta}\right|_{\delta=0}$, which may be easier.

UPDATE: Let me clarify the time dependence of $Q$. Let be $S$ be a symmetric transition rate matrix with unit speed (i.e., all diagonal entries are -1). Consider the process \begin{align*} H(0)&=I\\ \frac{\mathrm{d}H}{\mathrm{d}t}(t) &= H(t)\phantom{.} S. \end{align*} In this case, we have $H(t) = \exp(tS)$. Let $f\in[n]$ be a state and $T>0$ be a real. Let $U(t)$, $t\in[0,T]$ be the distribution function for of the process obtained by conditioning that the $H$ process ends up at state $f$ at time $T$.

By Bayes' rule we must have \begin{align*} U(t) &= \diag^{-1}(H_T 1_f) H_t\diag(H_{T-t}1_f). &(3) \end{align*} By differentiation with respect to $t$, we see that $U$ satisfies the differential equation (1) with $Q(t)$ being as follows \begin{align*} Q(t) = \diag^{-1}(H_{T-t}1_f)\phantom{.}S\phantom{.}\diag(H_{T-t}1_f) - \diag^{-1}(H_{T-t}1_f)\diag(S H_{T-t}1_f). \end{align*}

Note that (3) gives an explicit solution for (1). I am wondering if $(2)$ has a simple solution in terms of (3).

Thanks