Question
Consider a job shop that consists of three identical machines and two technicians. Suppose that the amount of time to breaking down for each machine is exponentially distributed with parameter $0.1$ and the amount of time taken by a technician to fix a broken machine is exponentially distributed with parameter $0.4$. Suppose also that all these times to breakdown and repair are independent random variables and let $X(t)$ be the number of machines operating at time $t$.
Write the Kolmogorov forward equations involving the $P'_{0j}(t)\ \forall\ j \in \{0, 1, 2, 3\}$ in terms of $p_j(t)$, where $$\begin{aligned} p_j(t) & = P_{0j}(t)\\[1 mm] & = \mathbb{P}[X(t) = j \mid X(0) = 0] \end{aligned}$$ and obtain the equilibrium probabilities $$p_j = \lim\limits_{t \to \infty} p_j(t).$$
My working
First, I have determined the generator matrix to be $$Q = \begin{pmatrix} -0.8 & 0.8 & 0 & 0\\ 0.1 & -0.9 & 0.8 & 0\\ 0 & 0.2 & -0.6 & 0.4\\ 0 & 0 & 0.3 & -0.3 \end{pmatrix}.$$
Next, I have obtained the forward equations to be $$\begin{aligned} p'_0(t) & = -0.8p_0(t) + 0.1p_1(t),\\[1 mm] p'_1(t) & = 0.8p_0(t) - 0.9p_1(t) + 0.2p_2(t),\\[1 mm] p'_2(t) & = 0.8p_1(t) - 0.6p_2(t) + 0.3p_3(t),\\[1 mm] p'_3(t) & = 0.4p_2(t) - 0.3p_3(t). \end{aligned}$$
However, this is where I am stuck. In order to find the $p_j$'s, I thought I had to solve the equations above, but that seems very complicated for me. Instead, I am thinking that the $p_j$'s would also be the components of the unique stationary distribution $\pi$ that solves $\pi Q = 0$. Would my line of reasoning be correct?
Moreover, if my generator matrix or forward equations are wrong, please also point out my mistakes. As this is my first time trying to solve the forward equations, any help will be greatly appreciated :)