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First order linear ODE's.

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve this first order linear ODE

    x (dy/dx)-2y=6x5


    2. Relevant equations



    3. The attempt at a solution


    x (dy/dx)-2y=6x5

    divide through by x

    dy/dx -2y/x = 6x4

    I= e∫-2/x dx

    =e-ln(2) =-2

    ∴ -2(dy/dx) + 4y/x = -12x4

    d/dx(-2y) = -12x5/5 + C

    y= 6/5(x5) - C/2

    this is wrong but where did I go wrong?

    Thanks
     
  2. jcsd
  3. Sep 8, 2012 #2

    vela

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    You didn't calculate the integrating factor correctly. Recheck the integral.
     
  4. Sep 8, 2012 #3
    ohh it should be a half; thanks

    so it should be y = y=-6/5x5 - 2C

    the answer in my book has 2x5 + Ax2
     
  5. Sep 8, 2012 #4

    vela

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    No, it should have an x in it.

    EDIT: It refers to the integrating factor.
     
    Last edited: Sep 8, 2012
  6. Sep 8, 2012 #5

    LCKurtz

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    The book is correct. Show your steps and we can help you find where you went astray.
     
  7. Sep 9, 2012 #6
    This isn't true; take a look at log/exponential rules.
     
  8. Sep 9, 2012 #7
    okbthanks I figured it out now../.

    I=1/x2

    although I am alittle confused about this integral

    if you work out the integral of -2/x

    -2ln(x)

    so I thought e^-2ln(x)

    which is e^1/-2ln(x)


    should be -1/2x

    and not -1/x^2

    ??

    thanks.
     
  9. Sep 9, 2012 #8

    HallsofIvy

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    You seem to be making one algebra mistake after another. Yes, the integral of -2/x is -2ln(x) and the exponential of that is [itex]e^{-2\ln(x)}[/itex] but that is NOT [itex]e/(-2\ln(x))[/itex]. I can't imagine where you got that. It is, rather [itex]1/e^{2\ln(x)}[/itex].

    But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?
     
    Last edited by a moderator: Sep 9, 2012
  10. Sep 9, 2012 #9

    yeah I made a stupid mistake of course e-2ln(x) = 1/e2ln(x)

    its been a while so first I should of done

    -2ln(x) = -ln(x2)

    Thanks
     
    Last edited by a moderator: Sep 9, 2012
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