= Solution
At the stationary mean,
$$
a=\begin{pmatrix}-\gamma&0\\\gamma&-\beta\end{pmatrix},\qquad
b=\lambda\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.
$$
The $(1,1)$ entry of the <Continuous Lyapunov equation> gives $-2\gamma C_{11}+2\lambda=0$, so $C_{11}=\lambda/\gamma$. Its $(1,2)$ entry gives $\gamma C_{11}-(\gamma+\beta)C_{12}-\lambda=0$, hence $C_{12}=0$. Finally $(2,2)$ gives $2\gamma C_{12}-2\beta C_{22}+2\lambda=0$. Thus
$$
\boxed{\operatorname{var}(x_1)=\lambda/\gamma,\qquad\operatorname{var}(x_2)=\lambda/\beta,\qquad\operatorname{cov}(x_1,x_2)=0.}
$$
Both variances equal their means; the absence of stationary covariance does not mean that the maturation jumps are independent events in the two populations.
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