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Euler's method for coupled ODE's

  1. Aug 14, 2014 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    Consider the following pair of coupled first order ODEs

    [itex]\dot{y_{1}} = y_{2}[/itex] with ##y_{1}(0) = 1##
    [itex]\dot{y_{2}} = -y_{1}[/itex] with ##y_{2}(0) = 1##

    Use the Euler integration method with a step-size ##h = 1## and fill out the entries in the table below

    [itex]\begin{bmatrix}
    t_{k}&y_{1}(t_{k})&y_{2}(t_{k})\\
    0 & &\\
    1 & &\\
    2 & &\\
    3 & & -4\\
    \end{bmatrix}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    Normally I understand how to do Euler's method, but of course now it's a coupled ODE so I am very confused how to do it. I know in general you do

    ##y_{k+1} = y_{k} + f(t_{k},y_{k})h##

    But with the coupled ODE I am lost
     
    Last edited: Aug 14, 2014
  2. jcsd
  3. Aug 14, 2014 #2

    pasmith

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    Homework Helper

    In this equation:
    [itex]y[/itex] can be (and normally is) a vector, and [itex]f[/itex] can be (and normally is) a vector-valued function.
     
  4. Aug 14, 2014 #3

    Maylis

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    Gold Member

    How does it look?

    [itex]\begin{bmatrix}
    t_{k}&y_{1}(t_{k})&y_{2}(t_{k})\\
    0 & 1 &1\\
    1 & 2 &0\\
    2 & 2 & -2\\
    3 & 0 & -4\\
    \end{bmatrix}[/itex]

    [itex] \begin{bmatrix} y_{{1},t_{k}=1} \\ y_{{2},t_{k}=1}\\ \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} + \begin{bmatrix}1\\-1\\ \end{bmatrix}*1 = \begin{bmatrix} 2 \\ 0 \\ \end{bmatrix} [/itex]

    [itex] \begin{bmatrix} y_{{1},t_{k}=2} \\ y_{{2},t_{k}=2}\\ \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ \end{bmatrix} + \begin{bmatrix}0\\-2\\ \end{bmatrix}*1 = \begin{bmatrix} 2\\ -2 \\ \end{bmatrix}[/itex]

    [itex] \begin{bmatrix} y_{{1},t_{k}=3} \\ y_{{2},t_{k}=3}\\ \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \\ \end{bmatrix} + \begin{bmatrix}-2\\-2\\ \end{bmatrix}*1 = \begin{bmatrix} 0\\ -4 \\ \end{bmatrix}[/itex]
     
    Last edited: Aug 14, 2014
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