1. ## Urgent Again

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

The system of eqns is

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

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

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

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

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

2. Originally Posted by Jimmy_W
Find $x_1 (t)$ and $x_2 (t)$ by writing a system of differential eqns as a matrix eqn and diagonalizing the matrix.

The system of eqns is

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

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

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

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

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
Well, $A$ is okay, however the others are not.

You should get: $
\left( {\begin{array}{*{20}c}
{4 \cdot t^{ - 1} - 1} & {2 - 4 \cdot t^{ - 1} } \\
{2 \cdot t^{ - 1} - 1} & {2 - 2 \cdot t^{ - 1} } \\

\end{array} } \right) \cdot X\left( t \right) = X'\left( t \right)
$
where $
X\left( t \right) = \left( {\begin{array}{*{20}c}
{x_1 \left( t \right)} \\
{x_2 \left( t \right)} \\

\end{array} } \right)
$