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

1. Oct 27, 2009

Wildcat04

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. Oct 27, 2009

foxjwill

Actually, the two answers are equivalent! Recall that
$$\frac{\ln\left(\frac{1}{a}\right)}{\ln\left(\frac{1}{b}\right)} = \frac{\ln\left(a^{-1}\right)}{\ln\left(b^{-1}\right)} = \frac{-\ln{a}}{-\ln{b}} = \frac{\ln{a}}{\ln{b}}$$​

3. Oct 27, 2009

Wildcat04

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!