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2nd order to matrix

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    x'' + 3x' + 2x = 0

    Find fundamental matrix

    2. Relevant equations

    x = x1
    x' = x2 = x1'
    x'' = x3 = x2' = x1''

    3. The attempt at a solution

    Not sure how to convert this to a matrix...

    The eiganvalues should be 1 and 2
     
  2. jcsd
  3. Dec 14, 2013 #2

    vela

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    You want to write the derivatives in terms of non-derivatives. You have, so far,
    \begin{align*}
    x_1' &= x_2 \\
    x_2' &= x''
    \end{align*} Use the differential equation to express x'' in terms of x1 and x2. You can then express this system as a matrix equation
    $$\begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = A \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$ with the appropriate A.
     
  4. Dec 14, 2013 #3
    x1' = 0x2 + 2x1
    x2' = -3x2 -2x1

    [tex]
    \begin{bmatrix}
    0 & 2\\
    -3 & -2
    \end{bmatrix}

    [/tex]

    But the eiganvalues for this do not work.
     
  5. Dec 14, 2013 #4

    Mark44

    Staff: Mentor

    Why did you write them this way?
    I would write the system like this:
    x1' = -2x1 + 0x2
    x2' = -2x1 - 3x2

    That will make a difference in how your matrix appears.
     
  6. Dec 14, 2013 #5

    vela

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    If you multiply the system out with your matrix, you get
    $$\begin{bmatrix}
    0 & 2\\
    -3 & -2
    \end{bmatrix}
    \begin{bmatrix}
    x_1 \\
    x_2
    \end{bmatrix}
    =
    \begin{bmatrix}
    2x_2 \\
    -3x_1-2 x_2
    \end{bmatrix}
    $$ which isn't what you want.
     
  7. Dec 14, 2013 #6
    Okay,

    then

    [tex]

    \begin{bmatrix}
    2 & 0\\
    -3 & -2
    \end{bmatrix}

    [/tex]

    then

    [tex]
    (2-\lambda )(-2-\lambda )

    [/tex]

    I need a lamda = 2 and lamda = 1.

    I will need

    [tex]
    \begin{bmatrix}
    1 & 0\\
    -3 & 2
    \end{bmatrix}
    [/tex]

    but I dont see how i get from the original DE to this matrix
     
  8. Dec 14, 2013 #7

    vela

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    Recheck your equations.
     
  9. Dec 14, 2013 #8
    Is there a proper name for this method that i can lookup a lesson on how to do this? The professor went over this the last few minuets of class and just y = x1, x2 = x1', x3 = x2'.... then just built the matrix mentally but I am not not seeing the process.
     
  10. Dec 14, 2013 #9

    vela

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    No, there's not a name for this because it's trivial to do. You're really overthinking this.

    http://tutorial.math.lamar.edu/Classes/DE/SystemsDE.aspx

    Look at the equation you wrote above and look at the second equation your professor wrote. They're supposed to be the same.
     
  11. Dec 14, 2013 #10
    x'' + 3x' + 2x = 0

    r^2 + 3r + 2
    (r+2)(r+1)

    r= -2 r = -1

    x1' = x2
    x2' = -3x2 -2x1

    [tex]

    \begin{bmatrix}
    0 & 1\\
    -3 & -2
    \end{bmatrix}

    [/tex]

    Then,

    [tex]
    \begin{bmatrix}
    0 &1 \\
    -3&-2
    \end{bmatrix}\begin{bmatrix}
    x_2\\
    x_1
    \end{bmatrix}= \begin{bmatrix}
    x_1 \\
    -3x_2 -2x1
    \end{bmatrix}

    [/tex]

    Okay I see now.

    So if I had:

    x''' + x'' + 3x' + 2x = 0

    x1' = x2
    x2' = x3
    x3' = -x'' - 3x' - 2x
     
  12. Dec 14, 2013 #11
    So to complete the problem

    [tex]

    \begin{bmatrix}
    2 & 1\\
    -3 &0
    \end{bmatrix}\begin{bmatrix}
    a\\
    b
    \end{bmatrix}
    =
    \begin{bmatrix}
    0\\
    0

    \end{bmatrix}
    and
    \begin{bmatrix}
    1 &1 \\
    -3& -1
    \end{bmatrix}\begin{bmatrix}
    a\\
    b
    \end{bmatrix}=\begin{bmatrix}
    0\\
    0
    \end{bmatrix}

    [/tex]


    2a + b = 0
    -3a=0

    and

    a + b = 0
    -3a-b = 0

    Unfortunately, these don't seem correct so I am guessing i still have an error.
     
  13. Dec 14, 2013 #12
    Okay, found the error.

    Should be:

    [tex]

    \begin{bmatrix}
    0 & 1\\
    -2 &-3
    \end{bmatrix}

    [/tex]

    That works out to be

    [tex]

    C_1\begin{bmatrix}
    -1\\
    1
    \end{bmatrix}
    e^{-t}+C_2\begin{bmatrix}
    1\\
    -2
    \end{bmatrix}e^{-2t}


    [/tex]

    Would you agree?
     
  14. Dec 14, 2013 #13

    Dick

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    Now that looks reasonable. Looking back that wasn't so hard, was it?
     
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