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Reduction order does not reduce

  1. Jun 20, 2005 #1
    Here is one more I am having a problem with:

    [tex]4x^2y'' + 4xy' + (4x^2 - 1)y = 0, x > 0 [/tex]

    one solution is
    y1(x) = sin(x)* x-1/2

    but when I try to work it all out with another linearly indep. solution

    y2(x) = u(x) sin(x) * x-1/2

    the order does not reduce and the y'' is huge... so I was wondering if I am missing some trick or is it really that painfully tedious :redface:

    ps: couldn't figure out how to put x ^(-1/2) in latex

    EDIT: I think I got it, but I did have to go through pages of writing out these derivatives.
     
    Last edited: Jun 20, 2005
  2. jcsd
  3. Jun 21, 2005 #2

    saltydog

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    You mind posting the reduced-order ODE?
     
  4. Jun 21, 2005 #3
    result

    Yeah, sure. It took me only like an hour to do the algebra :biggrin: so I will not overload my post with the whole thing, although if you are interested I can post :surprised
    In the end u term got cancelled so here's what I ended up with:

    u''(4x3/2sin(x)) + u'(8x3/2cos(x)) = 0

    ln(u') = ln|sin(x)-2|;

    u = -cot(x) + C;

    so y2(x) = -x-1/2cos(x);

    I hope this is correct, gonna find out in class today :confused:

    EDIT: yep, it's correct only i forgot about the coefficients: take A = -1 and C = 0

    => y2(x) = x-1/2cos(x);
     
    Last edited: Jun 21, 2005
  5. Jun 21, 2005 #4

    saltydog

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    Hello EvLer. Thanks for posting that. I have a question: How did you obtain the first solution? Normally, this type of problem is approached via power series, indical equation, logarithms maybe, messy expressions, the lot.
     
  6. Jun 21, 2005 #5
    sorry to disappoint you, but first solution was given as a condition, i.e. the problem consisted of finding second linear indep. solution given *that* solution.

    EDIT:hmmmm, may be a good question to my instructor :bugeye:
     
    Last edited: Jun 21, 2005
  7. Jun 22, 2005 #6

    HallsofIvy

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    I haven't calculated the whole thing but, in general any term involving "u" rather than u' or u" MUST be multiplied by the correct derivate of the functions multipling u. If that function satisfied the equation, then your final equation will involve u" and u' but NOT u. You certainly should be able to reduce the order by letting v= u'.

    Post your work or at least the final equation you got.
     
  8. Jun 22, 2005 #7

    GCT

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    Did you try the Wronskian approach (I'm actually on this chapter myself, test tommorrow)?
     
  9. Jun 22, 2005 #8

    GCT

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    Never mind it seems that you've found the solution yourself, didn't notice all of the editing.
     
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