I have a system
$$X_1'(\tau)+p (e^{X_2(\tau)-X_1(\tau)}-1)=v_1$$
$$X_2'(\tau)+q (e^{X_1(\tau)-X_2(\tau)}-1)=v_2$$
First, how can I solve this system and obtain $X_i(\tau)$?
Second, consider a two state continuous time Markov chain with transition probabilities $p$ and $q$. In the first state it takes a value $v_1$ and in the second the value $v_2$. Can we relate the solution to the properties of the Markov chain?
My intuition tells that $X_i(\tau)$ is related to $\mathbb{E}_{t,i}\int_{t}^{t+\tau}v(s)ds$, the expectation over the integral of values of the Markov-chain conditional on currently being in state $i$. One can see that this is true in the special case that the transition probabilities are zero $p=q=0$. But is this generally correct?