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How to solve $\dot{X} = MX+XN$?

  1. $M,N,X\in \mathbb{R}^{n\times n}$

I know in the scalar case:

$$y' + p(x)y=q(x)$$

the solution is $$\mu = e^{\int p(x)dx}$$ so $$y=\frac{\int \mu(x)q(x)dx + c}{\mu(x)}$$

The mapping is : $y\rightarrow X$, $x \rightarrow t$, $p(x) = -(MX+XN)$, $q(x) = 0$

However, I have no idea how to deal with that in the matrix case.

  • 1
    One way would be to write $X$ as an element of $\mathbb{R}^{n^2}$ and solve the resulting $n^2$ system. – 2017-02-25

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