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A seperation of variables intergration

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dV}{dt} = 20 - kV[/tex]

    By solving this show that

    [tex]V = A + Be^{-kt}[/tex]

    2. Relevant equations

    Well im guessing there is a ln coming into play somewhere during the intergration
    if the diff the bottom = the top then you get a ln(bottom)

    3. The attempt at a solution

    seperate the variables

    [tex]\int{\frac{1}{20-kV}dv} = \int{1dt}[/tex]

    i dont know how to form the ln part out of the LHS, cus you cant just take a constant out here?

    Thanks :)
     
  2. jcsd
  3. Nov 24, 2008 #2

    gabbagabbahey

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    Hint: what is [tex]\frac{d}{dx} \ln(20-kV)[/tex]?
     
  4. Nov 24, 2008 #3

    Mark44

    Staff: Mentor

    Let w = 20 - kV. Then dw = -k dV
     
  5. Nov 24, 2008 #4
    ahh take a constant of -1/k out

    so -1/k ln (20-kV) = t
    multiply through by -k to give

    ln(20-kV) = -tk
    then e it man
    gives you
    20-kV = e^(-tk)
    V = 20/k -e^(-tk)/k
    but the stupid mark scheme says
    V = 20/k -20e^(-tk)/k

    who is right?

    Thanks :)
     
  6. Nov 24, 2008 #5

    gabbagabbahey

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    Remember, your integrals are indefinite integrals; so you need to include a constant of integration:

    [tex]\Rightarrow \frac{-1}{k} \ln(20-kV) = t +C[/tex]
     
  7. Nov 24, 2008 #6
    Ooops missed that.
    however how does that effect the co-efficient of e being 20?

    Thanks :)
     
  8. Nov 24, 2008 #7

    gabbagabbahey

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    Well, what do you get when you solve it with the constant C?
     
  9. Nov 24, 2008 #8
    using v=0 and t=0
    (as im told the container is empty)
    so that means c = -1/k ln 20

    which means i can combine the ln s!!!

    cheers :)
     
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