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Difficult for me

  1. May 15, 2005 #1
    please try to solve: dv/dx={kx/v}[e^{-x/vRC}]
     
  2. jcsd
  3. May 15, 2005 #2

    dextercioby

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    It doesn't look good.I'll let you know why

    [tex]-\frac {1}{2xk}v^2 e^{\frac {x}{vRC}}-\frac {1}{2RCk}v e^{\frac {x}{vRC}}-\allowbreak \frac {1}{2}\frac {x}{R^2C^2k}\mbox{Ei}\left( 1,-\frac{x}{vRC}\right) +v=C [/tex]

    Daniel.
     
  4. May 15, 2005 #3
    sorry but what is Ei?
     
  5. May 15, 2005 #4

    dextercioby

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  6. May 15, 2005 #5

    saltydog

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    Nishant, would you kindly supply some reasonable values for k, R, C, and a reasonable initial condition so I could plot it to see what it looks like.

    Thanks,
    Salty

    Edit: I mean numerically (via Runge-Kutta). I dont' think I could plot it using the implicit solution.
     
    Last edited: May 15, 2005
  7. May 15, 2005 #6
    K,R,C are constants
     
  8. May 15, 2005 #7

    dextercioby

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    He knew that,he was asking for numbers.

    Daniel.
     
  9. May 15, 2005 #8

    saltydog

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    You know Nishant, I shouldn't give you the impression that I "need" reasonable values to plot it. Really, I can just pull them right out of thin air to get a plot: Wait . . . .1,1,1, and another one. See, got a plot. Really though, might be interesting to study how the solution varies as the constants change unless you have a particular set up in mind.

    Edit: Here it is, see, 1, 1, 1 and . . . well, you know.
     

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    Last edited: May 15, 2005
  10. May 15, 2005 #9
    I am not able to understand Ei,how do u write it in mathematical form?
     
  11. May 15, 2005 #10

    dextercioby

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    Well,how did u run into that equation in the first place...?

    Daniel.
     
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