Find $\displaystyle x_1 (t) $ and $\displaystyle x_2 (t) $ by writing a system of differential eqns as a matrix eqn and diagonalizing the matrix.

The system of eqns is

$\displaystyle \frac{dx_1}{dt} = (4t^{-1} - 1)x_1 + (2 - 4t^{-1})x_2 $

$\displaystyle \frac{dx_2}{dt} = (2t^{-1} - 1)x_1 + (2 - 2t^{-1})x_2 $

with IC's $\displaystyle x_1(1) = 2 $ and $\displaystyle x_2(1) = 1$

Im normally fine with these, and its porbably something really obvious thats making me stuck.

Anyway,

I need to put it in the form Ax = b.

Something like $\displaystyle A = \begin{bmatrix} (4t^{-1} - 1) & (2 - 4t^{-1}) \\ (2t^{-1} - 1) & (2 - 2t^{-1}) \end{bmatrix} $ and $\displaystyle b = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $

anyway, i know A isnt correct, im just really having trouble concentrating. Need

....even had to check this post about 10 times to make sure it was right