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.


Thanks



Solving a matrix ODE
Let's suppose all the $V^\delta(t)$ have a common initial condition
$V^\delta(0) = U\_0$. If $W^\delta(t) = \dfrac{\partial}{\partial \delta} V^\delta(t)$, then we have 
$$ \dfrac{dW^\delta}{dt} = \dfrac{\partial}{\partial \delta} \dfrac{dV^\delta}{dt} = \dfrac{\partial}{\partial \delta} (1+\delta) V^\delta(t) Q(t)
= V^\delta(t) Q(t) + (1+\delta) W^\delta(t) Q(t)$$
with $W^\delta(0) = 0$. In particular, for $\delta = 0$ we have
$$ \dfrac{dW^0}{dt} = (U(t) + W^0(t)) Q(t)$$
The solution to this is $$W^0(t) = Z(t) U(t)$$
where 
$$ Z(t) = \int\_0^t U(s) Q(s) U(s)^{-1}\; ds $$


