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Math Help - show vector satisfies given DE

  1. #1
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    show vector satisfies given DE

    Hi,

    I am very rusty with my math, and need help understanding how to approach this type of problem.

    (problem)
    verify that the given vector satisfies the given differential equation:

    x'=Ax
    where A is a 3x3 matrix

    A = {1,1,1;2,1,-1;0,-1,1}

    and x = Be^(-t) + 2Ce^(2t)
    where B and C are 3x1 matrices

    B = {6;-8;4}
    C = {0;1;-1}
    (\problem)

    I differentiated the x with respect to t, but I dont know how to show that the given vector x' fits the DE.

    Thanks,
    OkashiiKen
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  2. #2
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    nm, solved the problem

    feeling rusty.....

    x' = Ax, so A^(-1)x' = x

    from there I can show the given vector satisfies the DE provided.
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  3. #3
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    Quote Originally Posted by OkashiiKen View Post
    Hi,

    I am very rusty with my math, and need help understanding how to approach this type of problem.

    (problem)
    verify that the given vector satisfies the given differential equation:

    x'=Ax
    where A is a 3x3 matrix

    A = {1,1,1;2,1,-1;0,-1,1}

    and x = Be^(-t) + 2Ce^(2t)
    where B and C are 3x1 matrices

    B = {6;-8;4}
    C = {0;1;-1}
    (\problem)

    I differentiated the x with respect to t, but I dont know how to show that the given vector x' fits the DE.

    Thanks,
    OkashiiKen
    We have a proposed solution x:
    x = \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    So
    x' = - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    Thus we need to verify the equation:
    x' = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) x

    - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

    The RHS becomes:
    \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} \right )  + \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

     = \left ( \begin{array}{c} 2 \\ 0 \\ 12 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 2 \\ -2 \end{array} \right ) e^{2t}

    We can factor a 2 from each of the 3 x 1 matrices, giving:
     = 2 \left ( \begin{array}{c} 1 \\ 0 \\ 6 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    which is not x'. This shows that the given x is NOT a solution to the differential equation.

    -Dan
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  4. #4
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    We have a proposed solution x:
    x = \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    So
    x' = - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    Thus we need to verify the equation:
    x' = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) x

    - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} =\left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

    The RHS becomes:
    \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} \right )+ \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

    = \left ( \begin{array}{c} 2 \\ 0 \\ 12 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 2 \\ -2 \end{array} \right ) e^{2t}

    We can factor a 2 from each of the 3 x 1 matrices, giving:
    = 2 \left ( \begin{array}{c} 1 \\ 0 \\ 6 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

    which is not x'. This shows that the given x is NOT a solution to the differential equation.

    -Dan
    Thanks for your solution, and you are correct. I mistyped the problem 3x1 matrix so that the given vector does not satisfy the give DE. Vector B should be {6;-8;-4} and NOT {6;-8;4}. Using the correct vector, and following your methodology:

    A = {1,1,1;2,1,-1;0,-1,1}

    x' = Ax
    = A{6;-8;-4}e^(-t) + 2{0;1;-1}e^(2t)
    = {-6;8;4}e^(-t) + 2{0;2;-2}e^(2t)
    = {-6;8;4}e^(-t) + {0;4;-4}e^(2t)

    furthermore, by deriving x to get x' we see that
    x' = -{6;-8;-4}e^(-t) + 2*2{0;1;-1}e^(2t)
    = {-6;8;4}e^(-t) + {0;4;-4}e^(2t)

    I initially did this using A^-1, which was not needed, but still worked. Wasnt really sure what I was doing. Thanks for your help!
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