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More linear eqn trouble

  1. Jun 20, 2004 #1

    Math Is Hard

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    My teacher gave us this problem in class and then when she went to show us how to solve, she got stuck too!

    [tex] 1+xy = xy' [/tex]

    after some rearranging and dividing we have

    [tex] y' - y = 1/x [/tex]

    we used an integrating factor of [tex] e^ {-x} [/tex]

    and got the result of [tex] (e^ {-x}y)' = e^ {-x}/x [/tex]

    integrating gives [tex] (e^ {-x}y) = {\int e^ {-x}/x \dx}[/tex]

    and that's about where we got stuck. Integration by parts just took us in a loop.
    Any help is appreciated. Thanks!
     
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  3. Jun 20, 2004 #2

    Gokul43201

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    I think you/she got stuck there because the integral of [tex]{e^ {-x}/x \dx}[/tex] has no analytic form - it diverges at x=0
     
    Last edited: Jun 20, 2004
  4. Jun 20, 2004 #3

    Math Is Hard

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    Thanks, Gokul.

    so as far as the solution for y , do you just throw up your hands and say "OK, divergent, can't be solved since the integral won't converge" or can you express y as something like this and call that the solution?

    [tex] y = {e^ {x}\int e^ {-x}/x \dx}[/tex]
     
  5. Jun 20, 2004 #4

    Tom Mattson

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    You are correct about both of those statements, but I think I should clarify by saying that one has nothing to do with the other. The integral of exp(-x2) has no analytic form, and it doesn't diverge anywhere. Conversely, the integral of tan(x) does have an analytic form, and it diverges at (2n+1)π/2, for any integer n.

    Math Is Hard,

    Yes, you can leave an integral as part of a solution.
     
  6. Jun 20, 2004 #5

    Math Is Hard

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    Thanks, Tom.
     
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