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Homework Help: Quick DE fix

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm looking to solve a DE for a problem I have but I haven't done so in a while and must be forgetting something. I've checked the answer in Mathematica and what I'm getting is wrong.

    u''(r)+1/r u'=G/m

    3. The attempt at a solution

    homogeneous solution:

    a^2+1/r a = 0 -> a = 0, -1/r

    u1 = Ae^(-1)+B

    try u2 = c r (as a constant B already exists as part of the solution, multiply by r)

    u' = c u'' = 0

    c/r = G/m
    c = Gr/m

    u2 = Gr^2/m

    u = Ae^(-1)+B+Gr^2/m

    Whereas the actual solution is

    u = A Ln(r)+B+Gr^2/4m

    Thanks for any help
     
  2. jcsd
  3. Apr 22, 2010 #2

    Mark44

    Staff: Mentor

    You are tacitly assuming that r is a constant, but it's not. It's the independent variable.

    Your DE is u''(r) + (1/r)u'(r) = G/m
    Since there's no u(r) term, I would make this substitution.
    Let w(r) = u'(r) ==> w'(r) = u''(r)
    With this substitution you can rewrite the DE as
    w'(r) + (1/r)w(r) = G/m
    Now you can use the standard technique of finding an integrating factor, v(r) = e^{integral(1/r).
    That's what I would do.
     
  4. Apr 22, 2010 #3
    Thanks, that worked out good.

    Could you point out in my initial work where I implied r was constant? because I didn't mean to.
     
  5. Apr 22, 2010 #4

    Mark44

    Staff: Mentor

    You treated the equation u'' + (1/r)u' = G/m as if it were a constant coefficient differential equation.

    From the homogeneous equation, u'' + (1/r)u' = 0, you got a characteristic equation of a^2 + (1/r)a = 0, and then factored the left side. This would have worked if r were a constant, but r is the independent variable, or so I gather by your use of u''(r).

    It's the difference between, say y' + 2y = 0 (with general solution y = Ae-2x) and y' + xy = 0 (with general solution y = Ce-(1/2)x2).
     
  6. Apr 22, 2010 #5
    Cheers, I understand now
     
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