The Question:
Consider the following boundary-value problem for the function $f: \mathbb R^2 \rightarrow \mathbb R$: $$\frac{\partial}{\partial x}f(x,A)=A \ g(x) f(x,A)$$ $$f(0,A)=1$$ It's easy to show that the solution is: $$f(x,A) = e^{A\int_0^x dt \ g(t)}$$ which can be rewritten as: $$f(x,A)=(e^{\int_0^x dt \ g(t)})^A$$ In other words: $$f(x,A)=(f(x,1))^A \qquad (*)$$ This means that if I had the solution to the differential equation for some value of $A$, I would be able to find the solution for any other $A$ from it. Now let's generalize this problem to a matrix differential equation: $$\frac{\partial}{\partial x}F(x,A)=A \ G(x) F(x,A)$$ with $F$ and $G$ being two $n\times n$ square-matrix-valued functions; and the boundary condition being $F(0,A)\equiv \mathbb I_{n\times n}$.
My question is: is there any straightforward generalization of the relation ($*$) for a general matrix-valued function $G$? In other words, given the solution of such a DE for some specific $A$, is there a relation one can use to calculate its solution for any other $A$?
Some of my own thoughts:
1- For the special case of a matrix $G(x)$ which commutes with itself at any two $x$ values, our $(*)$ relation also trivially works for the matrix DE. But this of course breaks down for the general case.
2- Another attempt of mine was to first use the change of variables $t\equiv x A$, which by defining $\mathcal F(t,A) \equiv F(t/A,A)=F(x,A)$ gives:
$$\frac{\partial}{\partial t} \mathcal F(t,A) = G(\frac{t}{A}) \mathcal F(t,A)$$ with the boundry condition $\mathcal F(0,A)= \mathbb I_{n\times n}$. But even though we absorbed the $A$, this isn't exactly what I want because the function $G(t)$ has been changed to $\tilde{G}(t) \equiv G(t/A)$.