This question comes from the book Continuous Time Markov Processes: An Introduction by Thomas Milton Liggett. It is exercise 2.8 and goes as follows:
Consider a process $X(t)$ with state space {0,1} that evolves as follows. If it starts at $0$, then the chain stays at $0$ for an exponential time with parameter $\beta$ and then moves to $1$. It then stays at $1$ for an exponential time with parameter $\delta$, and then moves back to $0$, and repeats, with analogous rules if it starts at $1$. Let $p_t(i,j)$ be the probability that $X(t)=j$ if $X(0)=i$.
a) Show that $$ p_t(0,1)=\int_0^t\beta e^{-\beta s}p_{t-s}(1,1)ds.$$
b) Use Laplace transforms to check that $$p_t(0,1)=\frac \beta {\beta+\delta}\left( 1-e^{-t(\beta+\delta)} \right).$$
I'm stuck on problem a). I tried to describe $p_t(0,1)=P(X(t)=1 \mid X(0)=0) $ as the probability of an odd number of "hits" (for a lack of a better word) of our (independent) exponential variables on $[0,t]$. But I don't know how to move forward with this probability. Because of our two (possibly different) parameters $\beta$ and $\delta$ I don't know how to compute this probability.
In addition, I'm not entirely sure this sort of "direct" computing is even necessary or if an identity of a Markov chain's transition function gives us a quicker solution.
Any help would be greatly appreciated. Thanks!