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Given a matrix $X(t)=e^{tA}$, we know that $X(t)$ is the solution of the following matrix differential equation: $$ \frac{dX(t)}{dt} =X(t) \cdot A .$$

Now could anyone help to construct a matrix differential equation in terms of $Y(t)$, such that $Y(t)=e^{tA} \cdot e^{tB}$ is its solution?

(NOTE: the matrices $A$ and $B$ do not commute, meaning that $e^{A+B} \neq e^A \cdot e^B.$ )

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What did you try? Did you compute $Y'(t)$? –  martini May 10 '12 at 6:20
    
I want to construct a matrix differential equation such that its solution is $Y(t) = e^{tA} \cdot e^{tB}$. –  John Smith May 10 '12 at 6:23
    
The IMO easiest way to do so is computing $Y'(t)$ (as you KNOW the solution). –  martini May 10 '12 at 6:24
    
@JohnSmith: martini is right. Take the derivative and you're done. –  user26872 May 10 '12 at 6:25
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Notice that $$\begin{eqnarray*} \frac{d}{dt} Y &=& \left(\frac{d}{dt} e^{t A}\right) e^{t B} + e^{t A} \left(\frac{d}{dt}e^{t B}\right) \\ &=& A e^{t A} e^{t B} + e^{t A}e^{t B} B \\ &=& A Y + Y B. \end{eqnarray*}$$ Since $A$ commutes with itself, it also commutes with $e^{t A}$, so $\frac{d}{dt} e^{t A} = A e^{t A} = e^{t A}A$. If you are unfamiliar with matrix calculus, here is not a bad place to start.

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Is $\frac{d}{dt} e^{tA} = A e^{tA}$ or $\frac{d}{dt} e^{tA} = e^{tA} A $ ? –  John Smith May 10 '12 at 6:41
    
@JohnSmith: Notice that $A$ commutes with itself, so they are the same thing. –  user26872 May 10 '12 at 6:42
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@JohnSmith oenamen's argument become obvious, when you write down the Taylor series for $e^{tA}$. –  draks ... May 10 '12 at 6:52
    
Yeah, I see. Thanks! The key point is that $e^{A}$ and $A$ are commutative. –  John Smith May 10 '12 at 6:55
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