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Diff Eq - Should be easy, but my answer is upside down!

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Find vz(r)

    Boundary Conditions:

    1. vz(Ro) = 0
    2. vz(Ri) = W


    2. Relevant equations

    (1/r) d/dr [r dvz/dr] = 0 (from previous problem)

    Let v = dvz / dr

    d/dr [r v] = 0

    r v = c1

    v = c1 / r

    vz = c1 ln r + c2

    BC 1:

    0 = c1 ln Ro + c2

    c2 = - c1 ln Ro

    BC 2:

    W = c1 ln Ri - c1 ln Ro

    W = c1 (ln Ri - ln Ro)

    W = c1 ln(Ri/Ro)

    c1 = W / ln(Ri/Ro)

    My Soultion:

    vz = (W ln r) / (ln(Ri/Ro) - (W ln Ro)/ln(Ri/Ro)

    vz = W[ln (r / Ro) / ln(Ri/Ro)]


    Unfortunately, the answer key says it should be:

    vz = W [ln (Ro / r) / ln(Ro/Ri)]

    So I am close but no cigar. I have recompleted this problem several times and keep arriving at the same solution. Can anyone point out my mistake?
     
  2. jcsd
  3. Oct 27, 2009 #2
    Actually, the two answers are equivalent! Recall that
    [tex]\frac{\ln\left(\frac{1}{a}\right)}{\ln\left(\frac{1}{b}\right)} = \frac{\ln\left(a^{-1}\right)}{\ln\left(b^{-1}\right)} = \frac{-\ln{a}}{-\ln{b}} = \frac{\ln{a}}{\ln{b}}[/tex]​
     
  4. Oct 27, 2009 #3
    Hrmmm...I always forget about identities, weither it be sin / cos or in this case natural logs.

    Thank you foxjwill, I thought this was an easy one but I couldn't figure out how to get the correct answer, when, all along, I had it!
     
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