Define a matrix differential equation $\dot{X}=A(t)X(t)$ with initial condition $X(0)$, where $X=[x_1,x_2,\ldots]^T$ is a 1D vector and $A(t)$ is a complex-valued time-dependent matrix.
This system can be solved by $X(t)= \mathcal{T}{\rm exp}(\int_0^t A(t')dt') X(0)$, where we introduce a time-ordered exponential. In general, this is an arduous task. For my specific case, I know how to solve a similar, but simpler limit.
Assume $A(t)=A_s+A_I(t)$, then $\dot{X}=(A_s+A_I(t))X(t)$.
For the time-independent part, I know the answer $X_{s}(t)=e^{A_0 t}X(0)$.
Given these conditions, and knowing that the effect $A_I(t)$ has on the system is small, i.e. the new solution shouldn't be too much different from $X_{s}(t)$, is there anything I can do? Any approximation?
I am expecting I could do some transformation of the sort $X(t)=U(t)X_s(t)$, with $U(t)=U(t,A_I)$. Given the time-ordered exponential, is there any perturbation that would give me this transformation?
The sort of transformation you are looking for is given by $$ U(t):=e^{A_st}U_I(t)e^{-A_st}, \tag{1} $$ where $U_I(t)$ is the solution to the IVP \begin{align} \dot{U_I}(t)&=e^{-A_st}A_I(t)e^{A_st}U_I(t), \tag{2} \\ U_I(0)&=\mathbb{1}. \tag{3} \end{align} To prove this, we have to show that $$ X(t)=U(t)X_s(t)=U(t)e^{A_st}X_0 \tag{4} $$ satisfies the differential equation $$ \dot{X}(t)=(A_s+A_I(t))X(t) \tag{5} $$ with initial condition $X(0)=X_0$. Let's begin with the latter: $$ X(t)=U(t)X_s(t)\overset{(1,4)}{=}e^{A_st}U_I(t)e^{-A_st}e^{A_st}X_0=e^{A_st}U_I(t)X_0 \tag{6} $$ $$ \implies X(0)=U_I(0)X_0\overset{(3)}=X_0. \tag{7} $$ Now let's show that (4) satisfies $(5)$: \begin{align} \dot{X}(t)&\overset{(6)}{=}A_se^{A_st}U_I(t)X_0+e^{A_st}\dot{U_I}(t)X_0 \\ &\overset{(2)}{=}A_sX(t)+e^{A_st}e^{-A_st}A_I(t)e^{A_st}U_I(t)X_0 \\ &\overset{(6)}{=}A_sX(t)+A_I(t)X(t). \quad{\square} \tag{8} \end{align}