Ordinary Differential Equation System with Variable Coefficients

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Homework Statement


For t\in\mathbb{R}, let:

A=\left[\begin{array}{ccc}<br /> -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 2 &amp; 2t\\<br /> 0 &amp; 0 &amp; 2\end{array}\right]

Get the solution for the general equation: X&#039;=A(t)X

Homework Equations



The Attempt at a Solution


I done many of these problems, all with constant coefficients, but I don't know how to do in this case.
 
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Did you know that the solution to
x&#039;(t) = f(t) x(t)
is
x(t) = x_0 \exp\left( \int_{t_0}^t f(\xi) \, d\xi \right)
?
 
Yes, I know. That's the formula I use after getting the exponential matrix, by "diagonalizing" the matrix A. My problem is that I'm not sure if I can do it as I do with constant coefficients, because supposedly it should be different.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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