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Elementary Differential equations : seperable

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Tan∅ dr + 2r d∅=0

    2. Relevant equations



    3. The attempt at a solution

    ∫1/2r dr +∫1/tan∅ d∅=0

    1/2ln(2r)+ln(tan∅)=c
    ln[r(tan∅)]=ln(c)

    r(tan∅)=c

    the solution in the book says the answer is rsin^2∅=c

    Where did I go wrong?

    Thank you
     
  2. jcsd
  3. Sep 15, 2012 #2
    Here's your mistake:
    ∫1/tan∅ d∅≠ln(tan∅)
     
  4. Sep 15, 2012 #3
    Wow I see it would be ∫cos∅/sin∅ d∅ and with some u substitution it would give us ln(sin∅) correct?
     
  5. Sep 15, 2012 #4
    Yep, that's right but you have done one more mistake.
    ∫1/2r dr≠(1/2)ln(2r). It is equal to (1/2)ln(r).
     
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