# Can $\boldsymbol{u}' - \boldsymbol{A}(t) \boldsymbol{u} = \boldsymbol{f}(t)$ be solved using an matrix exponential integrating factor?

Can

$$\boldsymbol{u}' - \boldsymbol{A}(t) \boldsymbol{u} = \boldsymbol{f}(t)$$

be solved using an integration factor method analogous to that used in the non-system case? (i.e. $u'-a(t)u=f(t)$). Giving the following solution:

$$\boldsymbol{u} = e^{-\int^t A(k)dk}\left (\int^t e^{\int^s A(k)dk}\boldsymbol{f}(s) ds + \boldsymbol{c}\right )$$

where the matrix exponential defined in the usual way. If you rearrange the solution and differentiate you find $$\frac {\text{d}e^{-\int^t A(k)dk}} {\text{d}t} \boldsymbol{u} + e^{-\int^t A(k)dk} \boldsymbol{u}' = e^{-\int^t A(k)dk} \boldsymbol{f}$$

If this is indeed the solution, it would then imply that $$\frac {\text{d}e^{-\int^t A(k)dk}} {\text{d}t} = - e^{-\int^t A(k)dk} \boldsymbol{A}$$

However this doesn't seem to be the case from Eq. 2.1 of http://aip.scitation.org/doi/pdf/10.1063/1.1705306 referenced on the Wiki page for matrix exponentials.

EDIT: Disproved on page 6 here.

• If your $A$ is time-dependent, then there is no simple closed formula. If $A$ is constant instead then there is a closed form solution. – MrYouMath Feb 19 '18 at 9:34
• @MrYouMath: Yes, I believe you are correct. – Freeman Feb 19 '18 at 9:35
• – Lutz Lehmann Feb 19 '18 at 10:34