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Math Help - Annihilator Operators and higher than second order results for aux equation

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    Annihilator Operators and higher than second order results for aux equation

    Hello,

    I am solving a DE by the Annihilator method.

    I know that if you have a solution for the auxillary repeated then you must add an (x) before the next constant.

    i.e. (D-1)^2 would be c1e^x + xc2e^x.

    However, what if I had D^3? or (D+1)^3? Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

    So would it be  c1e^-x + xc2e^-x + ((x^2)/2)c3e^-x?


    What do you think?
    Last edited by eingeist; February 26th 2010 at 02:19 PM.
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  2. #2
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    Quote Originally Posted by eingeist View Post
    Hello,

    I am solving a DE by the Annihilator method.

    I know that if you have a solution for the auxillary repeated then you must add an (x) before the next constant.

    i.e. (D-1)^2 would be c1e^x + xc2e^x.

    However, what if I had D^3? or (D+1)^3? Judging by what happens in a square, I am guessing I am integrating? But I am unsure.

    So would it be  c1e^-x + xc2e^-x + )(x^2)/2)c3e^-x?


    What do you think?
    Actually for (D+1)^3 you would get y=c_1e^{-t}+c_2te^{-t}+c_3t^2e^{-t}

    One way to derive this is to use reduction of order.

    In general If you have (D+1)^n=c_1e^{-t}+c_2te^{-t}+...c_nt^{n-1}e^{-t}

    By taking the Wronskian you can see that these are linearly independant.
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  3. #3
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    read  csub1 , csub2, etc. where I have c1 and c2.. I'm just learning the code...

    edit: well, still learning. can't figure out subscripts
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  4. #4
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    The original equation is:

     y^{II} + 5y^I + 6y = 2-3x+4e^{-x}

    Shouldn't I be able to use  D^2+5D+6 , Factor:

     (D+2)(D+3) and use annihilators on the right side and add them to the left:

     (D-1) D^2 D

    So I would have (D-1)D^2D(D+2)(D+3) = 0

    See my problem? I have repeated D three times. So

    <br />
y=c_1e^{-3x} + c_2e^{-2x} + c_3 + xc_4 + ?
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