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Integral question

  1. Nov 9, 2004 #1
    say a ball falling through shampoo had a drag force of F=6.5v

    the differential equation for the objects motion is:

    mdv/dt=6.5v

    to solve it, I get all of the v value on one side and integrate them:

    dv/v=(6.5/m)(dt)
    integral[dv/v]=integral[(6.5/m)(dt)]
    I know that integral[(6.5/m)(dt)] becomes 6.5t/m+c, but what will happen to the dv/v?
     
  2. jcsd
  3. Nov 9, 2004 #2
    How about mg ?
     
  4. Nov 9, 2004 #3
    ignoring the effects of gravity, sry
     
  5. Nov 9, 2004 #4
    the indefinite integral of dx/x is ln|x|+C

    i guess if you are looking at this as a seperable differential equation you'd have

    ln|v|=(6t/m)+C
    |v|=e^((6t/m)+C)
    |v|=e^(6t/m) * e^C
    if K = +,- e^C
    v=Ke^(6t/m)

    i don't know, is that what you were looking for?
     
    Last edited: Nov 9, 2004
  6. Nov 9, 2004 #5
    [tex]\int \frac{1}{v}dv=ln |v|[/tex]
     
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