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Y''+xy'-y=0 differential equation

  1. Dec 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Given [itex]y_1=x[/itex] is a solution, solve the differential equation


    2. Relevant equations

    y''+xy'-y=0

    3. The attempt at a solution

    Since I am given [itex]y_1=x[/itex] (is there a hotkey for adding TeX tags so I don't have to manually type these tags over and over? So tedious.) then I wish to vary a parameter A that y is multiplied by, A(x)...

    [itex]y_2=Ax[/itex]
    [itex]y_2'=A'x+A[/itex]
    [itex]y_2''=A''x+2A'[/itex]
    Plugging into the original equation, I end up with

    [itex]A''x+(x^2+2)A'=0[/itex]

    This is another differential equation, and I do not know where to go from here.
     
  2. jcsd
  3. Dec 2, 2013 #2

    FeDeX_LaTeX

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    You could reduce the order of your new DE with a simple substitution.
     
  4. Dec 2, 2013 #3

    vela

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  5. Dec 2, 2013 #4

    LCKurtz

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    Code (Text):

     Use ##tex here## instead of [itex] tags for inline and
    $$tex here$$ instead of [tex] tags.
     
     
  6. Dec 2, 2013 #5
    Regarding the # # tags, I was using those for a while, but, a few posts ago, they completely stopped working for me. Maybe it was a script error in the browser. I've noticed that I have to refresh quite a few times to get the equations to show, no matter which browser or computer I'm using.

    Anyway, I reduced the order of the equation, by substituting P = A', then I get a separable DE.

    I get a solution of the form:


    ##\text{Ce}^{-\frac{x^3}{3}-2 x}##

    I do not think this is right. I simply replaced every instance of A' with P. I do not know how to integrate from this point, because I have a nasty integral that evades substitution methods. This just feels wrong.
     
  7. Dec 2, 2013 #6
    (can't delete this)
     
  8. Dec 2, 2013 #7

    vela

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    Using Mathematica, I find the solution is ##P = Ce^{-x^2/2}/x^2##. Not sure that makes it any easier for you though.
     
  9. Dec 2, 2013 #8
    I've been using Mathematica, as well, which is why I figured that my answer was wrong.
     
  10. Dec 2, 2013 #9

    FeDeX_LaTeX

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    Where are you getting the ##\frac{x^3}{3}## term from?
     
  11. Dec 2, 2013 #10

    vela

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    You seem to have forgotten the factor of ##x## multiplying ##P' = A''##.
     
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