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ODE question - Integrating factor

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of the equation:

    y' = (2y)/(t.logt) = 1/t, t > 0


    2. Relevant equations
    Integrating factor I = exp([tex]\int[/tex]p(x)dx)

    where y' + p(x)y = q(x)


    3. The attempt at a solution
    Hi everyone, here's what I've done so far:

    Let p(t) = -2/(t.logt)

    I = exp([tex]\int[/tex]((-2)/(t.logt))dt)

    Factoring out the -2, consider [tex]\int[/tex]1/(t.logt)dt

    Use integration by parts:
    u = 1/logt
    du = -(logt)^-2.(1/t).dt

    dv = (1/t)dt
    v = logt

    [tex]\int[/tex]u.dv = uv - [tex]\int[/tex]v.du

    I end up with:

    [tex]\int[/tex]1/(t.logt)dt = 1 + [tex]\int[/tex]1/(t.logt)dt

    which gives me 0 = 1, which is clearly wrong.

    But I can't see where I'm going wrong! I've done it five times now and I keep getting the same answer. Can anyone see where I'm going wrong or suggest another way of solving the problem?

    Thanks for any help
     
  2. jcsd
  3. Feb 3, 2010 #2

    vela

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    Try the substitution u=log t.
     
  4. Feb 3, 2010 #3

    vela

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    You didn't do anything wrong when you integrated by parts. If you were evaluating definite integrals, the first term would drop out since you'd get 1-1=0. With the indefinite integrals, the paradoxical result arises because you've neglected the constant of integration.
     
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