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Euler forward equation

  1. May 15, 2009 #1
    Hi all, I'm having trouble understanding a basic concept introduced in one of my lectures. It says that:

    To solve the DE
    [tex]y(t) + \frac{dy(t)}{dt} = 1[/tex] where [tex]y(t) = 0[/tex],

    using the Euler (forward) method, we can approximate to:

    [tex]y[n+1] = T + (1-T)y[n] [/tex] where [tex]T[/tex] is step size and [tex]y[0] = 0[/tex].

    I have no idea how this result is obtained, the only thing they say is that in general for

    [tex]\frac{dx_1}{dt} = \frac{x_1[n+1] - x_1[n]}{T}[/tex] for [tex]t = nT[/tex].

    Can anyone please help me understand how they arrived at the solution for [tex]y[n+1][/tex]? Thanks!
     
  2. jcsd
  3. May 15, 2009 #2
    Bah, it is simple plug-and-chug. Should have known! Thanks!
     
  4. May 15, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Four minutes! You didn't even give us a chance to explain!
     
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