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Integral question

  1. Nov 12, 2006 #1
    Just wondering if you can help me out, I have the following integral that I go to in some differential equation I came up with to figure something out here:

    Integrate: dy/(4-y^0.5)

    The root y screws me up.
     
  2. jcsd
  3. Nov 12, 2006 #2

    Hurkyl

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    Then get rid of it. (With some sort of substitution)
     
  4. Nov 12, 2006 #3
    Yah, how?
    That's my question.
     
  5. Nov 12, 2006 #4

    Pyrrhus

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    Try

    [tex] z^{2} = y^{0.5} [/tex]

    [tex] dy = 4z^3 dz [/tex]
     
  6. Nov 12, 2006 #5

    Ah yes, interesting. So then I get it in the form

    [tex] 4z^3 dz / (4 - z^2) [/tex]

    I see you can simplify the [tex]4 - z^2 to:
    (2-z)(2+z)[/tex], but then I tried to do partial fractions, and that also went nowhere, as there is no existing way to do a partial fraction for that that I could find?

    and again I am stumped.. sorry, I'm just not getting this. :frown:
     
  7. Nov 12, 2006 #6

    Hurkyl

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    Cyclovenom: Why not just z = y^0.5? (or z^2 = y)

    This is a routine application of partial fractions. Maybe you forgot about a step? (the way I learned it... the first step)
     
  8. Nov 12, 2006 #7
    [tex]A/(2-x) + B/(2+x) = 4x^3/(4-x^2)[/tex]
    [tex]A(2+x) + B(2-x) = 4x^3[/tex]
    [tex](2A+2B) + x(A-B) = 4x^3 + 0x + 0[/tex]
    2A+2B = 0, so A=-B
    x(A-B)=0x, so A=B

    Therefore, no solution exists by the method I know.
     
  9. Nov 12, 2006 #8

    Hurkyl

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    There's something you're supposed to do before this...

    (if you can't find it in your book, the answer is: you're supposed to divide first)
     
  10. Nov 12, 2006 #9

    Pyrrhus

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    Yes, it's easier, i guess i am gettin' rusty.
     
  11. Nov 12, 2006 #10
    :frown: I'm sorry I can't find it.

    It's because I'm stupid.

    edit: Long division.. hmm..

    would it be the same as:
    [tex] -4x + 16x / (4-x^2) [/tex]

    I checked and looks fine.. integrating now



    so final answer of [tex] -2z^2 - 8ln(4-z^2) [/tex]
    Then just sub in root y for z^2
     
    Last edited: Nov 12, 2006
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