This is the three-state continuous-time Markov chain in which the transition rates are given by:
$$Q = \left[ \begin{matrix} 0 & 2\lambda & 0 \\ \lambda & 0 & \lambda \\ 0 & 2\lambda & 0 \\ \end{matrix} \right] $$
I need to find the probability that of the state starting $i$, will be in $j$ state at time $t$?I am suppose to use differential equations but I am not sure how to. The hint given is:
Hint: Whenever confronted with an ordinary differential equation of the form x′(t) = ax(t)+b(t), it might be beneficial to consider the function y(t) = $e^{−at}x(t)$.
The generator matrix is $$ G = \left[ \begin{matrix} -2\lambda & 2\lambda & 0 \\ \lambda & -2\lambda & \lambda \\ 0 & 2\lambda & -2\lambda \\ \end{matrix} \right]. $$ Now, if $P(t) = \mathbb P(X(t)=j\mid X(0)=i)$, then $P$ satisfies the backward Kolmogorov equation $$ P'(t) = GP(t), $$ which has unique solution $e^{tG}$. Now, $G$ has eigenvalues $-4\lambda, -2\lambda, 0$ with corresponding eigenvectors $(1, -1, 1), (-1, 0, 1), (1, 1, 1)$. It follows that $$ G = ADA^{-1} $$ where $$ A = \left( \begin{array}{ccc} 1 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} \right),\quad D =\left( \begin{array}{ccc} -4 \lambda & 0 & 0 \\ 0 & -2 \lambda & 0 \\ 0 & 0 & 0 \\ \end{array} \right). $$ For each nonnegative integer $n$ we have $G^n = AD^nA^{-1}$, and hence $$ e^{tG} = \sum_{n=0}^\infty \frac{(tG)^n}{n!} = \left( \begin{array}{ccc} \frac{1}{4} e^{-4 \lambda t} \left(e^{2 \lambda t}+1\right)^2 & \frac{1}{2}-\frac{1}{2} e^{-4 \lambda t} & e^{-2 \lambda t} \sinh ^2(\lambda t) \\ \frac{1}{4}-\frac{1}{4} e^{-4 \lambda t} & \frac{1}{2} \left(e^{-4 \lambda t}+1\right) & \frac{1}{4}-\frac{1}{4} e^{-4 \lambda t} \\ e^{-2 \lambda t} \sinh ^2(\lambda t) & \frac{1}{2}-\frac{1}{2} e^{-4 \lambda t} & \frac{1}{4} e^{-4 \lambda t} \left(e^{2 \lambda t}+1\right)^2 \\ \end{array} \right). $$