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

[Wildcat04]

Homework Statement

Find v_z(r)

Boundary Conditions:

1. v_z(R_o) = 0
2. v_z(R_i) = W

Homework Equations

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

Let v = dv_z / dr

d/dr [r v] = 0

r v = c₁

v = c₁ / r

v_z = c₁ ln r + c₂

BC 1:

0 = c₁ ln R_o + c₂

c₂ = - c₁ ln R_o

BC 2:

W = c₁ ln R_i - c₁ ln R_o

W = c₁ (ln R_i - ln R_o)

W = c₁ ln(R_i/R_o)

c₁ = W / ln(R_i/R_o)

My Soultion:

v_z = (W ln r) / (ln(R_i/R_o) - (W ln R_o)/ln(R_i/R_o)

v_z = W[ln (r / R_o) / ln(R_i/R_o)]


Unfortunately, the answer key says it should be:

v_z = W [ln (R_o / r) / ln(R_o/R_i)]

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?

 

[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}}​

 

[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!