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Math Help - Urgent Again

  1. #1
    Junior Member
    Joined
    Sep 2008
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    47

    Unhappy 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
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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Quote Originally Posted by Jimmy_W View Post
    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: <br />
\left( {\begin{array}{*{20}c}<br />
   {4 \cdot t^{ - 1}  - 1} & {2 - 4 \cdot t^{ - 1} }  \\<br />
   {2 \cdot t^{ - 1}  - 1} & {2 - 2 \cdot t^{ - 1} }  \\<br /> <br />
 \end{array} } \right) \cdot X\left( t \right) = X'\left( t \right)<br />
where <br />
X\left( t \right) = \left( {\begin{array}{*{20}c}<br />
   {x_1 \left( t \right)}  \\<br />
   {x_2 \left( t \right)}  \\<br /> <br />
 \end{array} } \right)<br />
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