Possible Close-form solution for 2 dimensional $\frac{d\Sigma}{dt} = \mathbf{A}\Sigma + \Sigma \mathbf{A}^T + \mathbf{B}\mathbf{B}^T$ When $\mathbf{A},\mathbf{B}$ are 2x2 matrix and $\Sigma(t)$ are PSD, can we expect a close form solution for the following ODE,
$$
\frac{d\Sigma}{dt} = \mathbf{A}\Sigma + \Sigma \mathbf{A}^T + \mathbf{B}\mathbf{B}^T.
$$
The problem is from solving the covariance of a time-invariant SDE
$$
 x = \mathbf{A}x + \mathbf{B}dw.
$$
When $x$ is in two dimensions, I guess there could be a close-form solution for $\Sigma$. But I am not sure how to solve it.
 A: Notice that \begin{align*}\dfrac{d}{dt}\left[e^{-t\mathbf{A}}\mathbf{\Sigma}e^{-t\mathbf{A}^T}\right] &= -e^{-t\mathbf{A}}\mathbf{A}\mathbf{\Sigma}e^{-t\mathbf{A}^T} + e^{-t\mathbf{A}}\dfrac{d\mathbf{\Sigma}}{dt}e^{-t\mathbf{A}^T} - e^{-t\mathbf{A}}\mathbf{A}^T\mathbf{\Sigma}e^{-t\mathbf{A}^T} 
\\
&= e^{-t\mathbf{A}}\left(-\mathbf{A}\mathbf{\Sigma}+\dfrac{d\mathbf{\Sigma}}{dt} -\mathbf{\Sigma}\mathbf{A}^T \right)e^{-t\mathbf{A}^T}
\\
&= e^{-t\mathbf{A}}\mathbf{B}\mathbf{B}^Te^{-t\mathbf{A}^T}.\end{align*}
Hence, $$e^{-t\mathbf{A}}\mathbf{\Sigma}(t)e^{-t\mathbf{A}^T} - \mathbf{\Sigma}(0) = \int_{0}^{t}e^{-s\mathbf{A}}\mathbf{B}\mathbf{B}^Te^{-s\mathbf{A}^T}\,ds,$$ i.e., $$\mathbf{\Sigma}(t) = e^{t\mathbf{A}}\mathbf{\Sigma}(0)e^{t\mathbf{A}^T} + \int_{0}^{t}e^{(t-s)\mathbf{A}}\mathbf{B}\mathbf{B}^Te^{(t-s)\mathbf{A}^T}\,ds.$$
A: $
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\BR#1{\Big(#1\Big)}
\def\vecc#1{\operatorname{vec}\LR{#1}}
\def\Mat#1{\operatorname{Mat}\LR{#1}}
\def\Reshape#1{\operatorname{Mat}\!\BR{#1}}
\def\diag#1{\operatorname{diag}\LR{#1}}
\def\Diag#1{\operatorname{Diag}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\iff\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
$Vectorize the equation
$$\eqalign{
s &= \vecc{\Sigma} \quad\qiq \Sigma = \Mat{s} \\
b &= \vecc{BB^T} \\
\dot s &= \BR{I\otimes A+A\otimes I}s \;+\; b \\
\dot s &= Cs \;+\; b \\
}$$
Substitute the dependent variable
and solve the resulting ODE by inspection
$$\eqalign{
x &= s + C^{-1}b \\
\dot x &= \dot s \;=\; Cx \qquad\qquad\qquad\qquad\qquad\quad \\
x(t) &= e^{Ct}\,x(0) \\
}$$
Recover the original variable and de-vectorize to matrix form
$$\eqalign{
s(t) &= e^{Ct}s(0) + \LR{e^{Ct}-I}C^{-1}b \qquad\qquad \\
\Sigma(t) &= \Reshape{s(t)} \\
}$$
Update
Since $C$ is the Kronecker sum of $A$ with itself,
its exponential can be written in a factored form
$$C=\LR{A\oplus A} \qiq e^{Ct} = \LR{e^{At}\otimes e^{At}}$$
which allows the solution to be de-vectorized algebraically
$$\eqalign{
V &= \Mat{C^{-1}b} \qiq \vecc{V} = C^{-1}b \\
\Sigma(t)
 &= e^{At}\,\Sigma(0)\,e^{A^Tt}
  + e^{At}V e^{A^Tt} - V \\
}$$
Coincidentally, this provides a solution of the matrix integral in the other answer.
