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Solve ODE

  1. Oct 5, 2006 #1
    Hi i need to use Variation of Parameters to solve this ODE

    [tex] x^2 y'' - 2xy' + 2y = x^(9/2)[/tex]

    So far I was thinking to use Euler's Equation and I really don't know if it will work please help me out with a hint. THanks.
     
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 5, 2006 #2

    arildno

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    Well, you will have two distinct contributions:
    1. The general solution to the associated homogenous problem (i.e, with the right-hand side equal to zero)

    2. A particular solution of your diff.eq. Here, try a simple power solution as your trial function, i.e [tex]y_{p}=Ax^{9}[/tex]
     
  4. Oct 5, 2006 #3
    Ok. so this is what I did...

    [tex] y_p = Ax^9 y'_p = 9Ax^8 y''_p = 72Ax^7 [/tex]

    Then i plug in into the original equation

    [tex] 72Ax^9 - 18Ax^9 + 2Ax^9 = x^9 [/tex]

    after that I get [tex] 1/52 [/tex] as an answer.

    Sorry i dont get your point.
     
  5. Oct 5, 2006 #4

    arildno

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    That would give you the particular solution, yes (A=1/52)
    What have you done about the other contribution?
     
  6. Oct 5, 2006 #5

    arildno

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    If you don't get my point, then you have no business doing math, since you evidently haven't bothered to read your textbook or go to class. Goodbye.
     
  7. Oct 5, 2006 #6
    [tex] y_p = Ax^(9/2) y'_p = 9/2 Ax^(7/2) y''_p = 63/4Ax^(95/2) [/tex]

    then i get [tex] A = 34/5 [/tex]
     
  8. Oct 5, 2006 #7
    im sorry i haved worked on these problems for too long and my english is not that good maybe I word my sentence wrong. Would you reconsider.
     
  9. Oct 5, 2006 #8
    You need a general solution and a particular solution. The sum of the two is the general solution you are after.
     
  10. Oct 5, 2006 #9

    arildno

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    Whatever nonsense is this???

    You are to solve the homogenous equation:
    [tex]x^{2}y_{h}''-2xy_{h}'+2y_{h}=0[/tex]

    Use the Euler trial solution: [tex]y_{h}=Cx^{n}[/tex]
    in order to determine allowable values for n.
     
  11. Oct 5, 2006 #10
    I know but whats going on is that I need to use Variation of parameters and for that I need y_1 and y_2 so im stuck. One of my friends told me they have used Euler's method to find y_1 and y_2
     
  12. Oct 5, 2006 #11

    arildno

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    Well, why didn't you say at the beginning you were required to do this with variation of parameters, then?
     
  13. Oct 6, 2006 #12
    I did. Oh and i was trying to solve the homogeneous eq and I dont know if I should take the x's into my quadratic formula or just the constants.??
     
  14. Oct 6, 2006 #13

    arildno

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    Asking that shows you haven't really tried it:
    Let's see:
    [tex]y_{h}''=n*(n-1)Cx^{n-2}, y_{h}=nCx^{(n-1}}[/tex]
    Inserting this into the homog. equation yields:
    [tex]x^{2}n*(n-1)Cx^{n-2}-2xnCx^{n-1}+2Cx^{n}=0\to{n(n-1)Cx^{n}-2nCx^{n}+2Cx^{n}=0\to(n(n-1)-2n+2)Cx^{n}=0[/tex]
    Now, what must you require so that the last left hand side is zero for all choices of x, but so that in general, [itex]y_{h}[/tex] is distinct from the zero function?
     
  15. Oct 6, 2006 #14

    arildno

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    This would be correct particular solution for your EDITED ODE, something that occurred AFTER my first replies.
     
  16. Oct 6, 2006 #15
    Sorry to bug you too much but I am confused on the post before the last one.
     
  17. Oct 6, 2006 #16

    arildno

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    What's your problem with it? You INSERT a trial solution into your diff.eq and see what happens!
    Please say EXACTLY where your problem with that lies!
     
  18. Oct 6, 2006 #17
    I got it! LOL thank you so much for your time and im sorry to bother you Ill be prepared next time.
     
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