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Thread: Particular integral, Matrix method

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    142

    Particular integral, Matrix method

    Hi

    I have found the complementary function but I am not sure how to work out the particluar integral. Here are the equations

    $\displaystyle
    \dot{x}=3x+8y-2t
    $
    $\displaystyle
    \dot{y}=-5x+-10y+1
    $

    Which is

    $\displaystyle
    \left[\begin{array}{c} \dot{x}\\
    \dot{y}\end{array}\right]=\left[\begin{array}{cc}
    3 & 8\\
    -5 & -10\end{array}\right]\cdot\left[\begin{array}{c}
    x\\
    y\end{array}\right]+\left[\begin{array}{c}
    -2t\\
    1\end{array}\right]
    $

    I have solved the complementary function

    $\displaystyle
    \left[\begin{array}{c}
    x\\
    y\end{array}\right]=\alpha\left[\begin{array}{c}
    1\\
    -5/8\end{array}\right]e^{-2t}+\beta\left[\begin{array}{c}
    1\\
    -1\end{array}\right]e^{-5t}

    $

    But I am not sure how to solve for the particular integral, any ideas

    Thanks
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Use variation of parameter by substituting:

    $\displaystyle \left(\begin{array}{c} x \\ y\end{array}\right)_p=\alpha(t)\left(\begin{array} {c} 1 \\-5/8\end{array}\right)e^{-2t}+\beta(t)\left(\begin{array}{c} 1 \\ -1\end{array}\right)e^{-5t} $

    into:

    $\displaystyle \left(\begin{array}{c} x \\ y\end{array}\right)^{'}=\left(\begin{array}{cc} 3 & 8 \\ -5 & -10\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)+\left(\begin{array}{c} -2t \\ 1\end{array}\right)$

    obtaining a system of two equations in the unknowns $\displaystyle \alpha'(t)$ and $\displaystyle \beta'(t)$ which you can solve for each by elementary means.
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  3. #3
    Member
    Joined
    Jan 2009
    Posts
    142
    Please excuse me for being dumb, but could you elaborate.

    James
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  4. #4
    Member
    Joined
    Jan 2009
    Posts
    142
    HI

    Sorted it out, I was been dim. Thanks
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