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Solution by variation of parameters

  1. Nov 13, 2005 #1
    i'm confused about that method:

    1) when proving that method works, why do you have to make u and v satisfy the 2nd condition u`y1+v`y2=0

    2) when you're integrating to find yp, why do you leave out the constant that results from the integration?
     
  2. jcsd
  3. Nov 13, 2005 #2

    saltydog

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    When you let:

    [tex]y=A(x)y_1(x)+B(x)y_2(x)[/tex]

    where [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] are solutions to the homogeneous equation, and you substitute that into:

    [tex]y^{''}+p(x)y^{'}+q(x)y=R(x)[/tex]

    Letting:

    [tex]A^{'}y_1+B^{'}y_2=0[/tex]

    reduces the complexity of the resulting equation to the simplified form:

    [tex]A^{'}y_1^{'}+B^{'}y_2^{'}=R(x)[/tex]

    Technically you could let it be any non-zero function but zero makes the math easier.

    As far as ignoring the constants of integration, note you're looking only for a particular solution to the original DE right: homogeneous part+particular part=solution. And a partcular solution does not contain any arbitrary constants.
     
  4. Nov 14, 2005 #3
    thank you very much for clearing those things up!!! :)
     
  5. Nov 14, 2005 #4

    HallsofIvy

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    Remember that there are, in general, an infinite number of functions
    A(x), B(x) such that [tex]y=A(x)y_1(x)+B(x)y_2(x)[/tex]
    Requiring that [tex]A'y_1+ B'y_2= 0[/tex]
    just "narrows the field" and guaranties that there will not be any second derivatives of A or B in the final equations.
     
    Last edited: Nov 14, 2005
  6. Nov 14, 2005 #5
    thank you very much!!! :)
     
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