# 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!