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H value in runge kutta method 2nd order

  1. Oct 27, 2005 #1
    hi can anyone explain to me how to get the H value for runge kutta second method? ive searched everywhere online but i just dont understand it.

    if found h = tn - to/n??

    i know what value of "to" is but no clue what values to put in for n and tn?

    thanks
     
  2. jcsd
  3. Oct 27, 2005 #2

    Integral

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    If I am interpreting what you are saying correctly, h is your time step, the errors of a 2nd order RK method are on the order of h5. It looks to me like your t0 is the starting time, therefore tn is the time end point, n is the number of time steps you are taking so your h is the time interval divided by the number of time steps.

    for example let:
    t0=0
    tn=100
    n=1000

    [tex] h = \frac {t_n - t_0} n = \frac {100 - 0} {1000} = .1 [/tex]
     
    Last edited: Oct 27, 2005
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