Fundamental matrix solution and commutativity. Please I have a question.
Let $$y'(t) = M(t)y(t)~~~~~~~~~~~(*)$$ where $M(t)$ is a matrix with continuous entries on the interval $(a,b)$.  
Let $Y(t,t_0)$ be its fudamental solution. It is known that if $M(t)$ commutes with $A(t)=\int_{t_0}^t M(s)~ds$, then $$Y(t,t_0) = \exp\left(\int_{t_0}^t M(s)~ds\right)~~t,t_0 \in (a,b)~~~~~~~~~~~~~~~~~~~~~(**)$$
I know that if $M(t)$ and $A(t)$ do not commute, then it is not possible to have $(**)$. The question I want to ask is this:  

Is it possible to for $Y(t,t_0) = \exp(\int M(s))$ to be a fundamental solution of $(*)$ if $M(t)$ do not commute with $A(t)?$

 A: In general, no. The problem is that the chain rule doesn't work the way you would like it to, i.e.
$$
\frac{d}{dt} \exp(A(t)) \neq A'(t) \exp(A(t)) = M(t) \exp(A(t))
$$
in general.
To see why this is, write out the Taylor expansion for $\exp(A(t))$, and when you have to take the derivative of terms of the form
$$
(A(t))^n
$$
you'll see the problem if you proceed carefully. In particular, it is not true that
$$
\frac{d}{dt}(A(t))^n = n (A(t))^{n-1} A'(t); 
$$
really, one must write, e.g.
$$
\frac{d}{dt}(A(t))^3 = \frac{d}{dt}(A(t)A(t)A(t)) = A'(t)A(t)A(t) + A(t)A'(t)A(t) + A(t)A(t)A'(t)\\ = MA^2 + AMA + A^2M.
$$
It would be nice to think of a concrete counterexample; I'll see what I can come up with.
EDIT: (maybe) a counterexample
$$
M(t) = \left(\begin{array}{cc} t & 1 \\ 0 & 1\end{array}\right)
$$
$$
A(t) = \left(\begin{array}{cc} 0.5 t^2 & t \\ 0 & t\end{array}\right)
$$
You can check that $MA \neq AM$.
Wolfram alpha gives me
$$
Y(t) = \exp(A(t)) = \left(\begin{array}{cc} e^{0.5 t^2} & -2\frac{e^t - e^{0.5 t^2}}{t-2}
 \\ 0 & e^t\end{array}\right).
$$
Now we need to see if $Y'(t) \neq M(t)Y(t)$ (hopefully it isn't!). I haven't set down and done the calculation, but a little messing around convinced me this probably works out. In particular, it's the entry in the 1st row, 2nd column that won't be equal.
