 Solution

Comparison with the multivariate normal density shows that $D$ is its precision matrix. Since the mean is zero,

Multiplying $a+\frac{1}{2}bD=0$ on the right by $C=D^{-1}$ gives $aC=-b/2$. Taking the transpose and using the symmetry of $b$ and $C$ gives $C{a}^T=-b/2$. Hence the stationary covariance obeys the Continuous Lyapunov equationThe same equation follows directly by setting the time derivative of the second-moment equation from part (c) to zero.