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Linear initial-value question

  1. Sep 8, 2009 #1
    Im doing questions out of a book, and im stuck on one question, and it dosnt give any answers

    dy/dx*(x) = 2y/x + X^2.......when y(1) = 3 ... = ..... dy/dx - 2y/x^2 = x
    i have trouble intergrating...

    I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

    e^(2/x)*y = (intergral of) xe^(2/x)

    im just having trouble with this ^^

    can ya help?
    thanks
     
  2. jcsd
  3. Sep 9, 2009 #2

    LCKurtz

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    Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere.
     
  4. Sep 9, 2009 #3

    HallsofIvy

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    On the contrary, that has a simple polynomial solution.

    Your difficulty is in your very first step: your equation is dy/dx= 2y/x+ x2 and then you rewrote it as dy/dx- 2y/x2= x!

    It should be dy/dx- 2y/x= x2. Now the integrating factor is
    [tex]e^{\int \frac{-2}{x}dx}= x^{-2}[/tex].
     
  5. Sep 9, 2009 #4

    LCKurtz

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    But aren't you missing the x in the first term of the original DE: dy/dx*(x)?
    I would have written it as x*dy/dx but anyway you must divide through by that before computing the integrating factor.
     
  6. Sep 9, 2009 #5
    im only having trouble intergrating xe^(2/x).. if i can do that i can solve the problem..
     
  7. Sep 10, 2009 #6

    LCKurtz

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    In my previous reply I said:

    "Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere."

    There *is* no elementary anti-derivative for that function. None of your usual techniques can ever work.
     
  8. Sep 10, 2009 #7
    oh ok.. thanks. but how could i do it... is there a way of doing it?
     
  9. Sep 11, 2009 #8

    LCKurtz

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    There is no elementary way. If you are familiar with the exponential integral:

    http://en.wikipedia.org/wiki/Exponential_integral

    there is a way to express the answer in terms of that. Like I said before, if this is a typical homework problem I suspect there is a misprint. As another poster has pointed out, if the x factor in the leading term was missing, it would be straightforward. I can give you the expression Maple gives for an answer if you want it.
     
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