Consider the following system of ODEs
$$
Y^{'}(t) = - \left[ A Y(t) + Y(t) A \right] + B(t) 
,
$$
where $Y(t)$,$A$,$B(t)$ are all matrices, with the properties $A=A^T$, $Y=Y^T$. $Y(t)$ is the matrix of functions to be solved. Of course, there's a standard way to solve this system, for example, by rearranging the $n \times n$ matrix $Y(t)$ into a column vector of dimension $n^2$ and finding the eigenvalues of the resulting constant $n^2 \times n^2$ matrix. But my question is, are there any known results concerning this system such that the problem may be reduced from dimension $n^2 \times n^2$ to, say, $n \times n$? Any comment would be appreciated.


Some random observations


1. $A Y(t) + Y(t) A$ looks like an anti-commutator. Is it related to the following fact: "the time-evolution of a quantum variable is governed by its commutator with the Hamiltonian"
2. Is the above system related to the exponential map in Lie Algebras/ Lie Groups?


P.S. Why do I want to consider it as a generic matrix system? It's hard to answer this question. Indeed I just want to express the solution $Y(t)$ in a form that is "natural". For example, in 1-D (physical situation), the matrix $A$ is constant and band-diagonal, and equals $I + c R$, where R is the discretisation of the linear operator $\partial\_x^2$. In 2-D (physical situation), $R$ is the discretisation of $\partial\_x^2 + \partial\_y^2$.



A Non-homogeneous, Linear (Matrix) System of ODEs: What's Known About it?
Well, there won't be solutions unless $B(t)^T = B(t)$ as well. If you have this, then consider $U(t) = e^{At}Y(t)e^{At}$. This will satisfy
$$
U'(t) = e^{At}B(t)e^{At},
$$
so your solution with given initial value $Y(0)$ is going to be 
$$
Y(t) = e^{-At}\left(Y(0) + \int\_0^te^{A\tau}B(\tau)e^{A\tau}\ \mathrm{d}\tau\right)e^{-At}.
$$


By the way, as for 'reducing the dimension of the system' (which one might want to do so as to reduce the complexity of computing the matrix exponential) by 'uncoupling the system', this can be done: Writing $A = RDR^T$ where $R$ is orthogonal and $D$ is diagonal (which can always be done) and setting $Y(t) = RZ(t)R^T$ and $B(t) = RC(t)R^T$, the equation becomes
$$
Z'(t) = -(DZ(t)+Z(t)D) + C(t),
$$
and one recognizes this as $\tfrac12n(n{+}1)$ independent first order ODE
$$
Z\_{ij}'(t) = -(D\_i+D\_j) Z\_{ij}(t) + C\_{ij}(t)\qquad i\le j.
$$


