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Differential equation/rational function

  1. Jan 31, 2012 #1

    nos

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    Hi all,

    Three forces are acting on an object.
    Gravity, mg
    Lorentz force, av(a is a constant)
    Air resistance, kv^2( k is a constant)
    v is the velocity of the object

    I figured that this differential equation must be the right one:

    M(total)[itex]\frac{dv}{dt}[/itex]= mg-av-kv^2

    Where m and M(total) are different masses

    [itex]\frac{dv}{kv^2+av-mg}[/itex] = -[itex]\frac{dt}{M(total)}[/itex]

    Integrate both sides:

    -[itex]\frac{2}{\sqrt{a^2+4kmg}}[/itex] artanh([itex]\frac{2kv+a}{\sqrt{a^2+4kmg}}[/itex]) = -[itex]\frac{T}{M(total)}[/itex] + C

    So I want to find the expression for velocity v.
    How do I begin? And should the integration constant even be there? When I put limits on the integral(v goes from 0 to v and t goes from 0 to t) then there is no constant of integration.
    Thanks so much.
     
  2. jcsd
  3. Jan 31, 2012 #2

    nos

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    I'm actually starting to think that maybe the whole anti-derivative is wrong.
     
  4. Jan 31, 2012 #3

    nos

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    So I figured using the limits on the integral gives me:

    arctanh([itex]\frac{2kv+a}{\sqrt{a^2+4kmg}}[/itex])-arctanh([itex]\frac{0+a}{\sqrt{a^2+4kmg}}[/itex])=[itex]\frac{\sqrt{a^2+4kmg}}{2M(total)}T[/itex]

    Taking the second term arctanh tot the other side and taking the tanh of both sides:

    2kv+a = [itex]\sqrt{a^ 2+4kmg}[/itex] tanh([itex]\frac{\sqrt{a^2+4kmg}}{2M(total)}T[/itex]
    +arctanh([itex]\frac{a}{\sqrt{a^2+4kmg}}[/itex]))

    which will finally give v in terms of t:

    v(t) = [itex]\frac{-a}{2k}[/itex]+[itex]\sqrt{a^ 2+4kmg}[/itex] tanh([itex]\frac{\sqrt{a^2+4kmg}}{2M(total)}T[/itex]
    +arctanh([itex]\frac{a}{\sqrt{a^2+4kmg}}[/itex]))/2k
     
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