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Annihilator Operators in Differential Equations

Discussion in 'The Okie Corral' started by eingeist, Feb 26, 2010.

  1. eingeist

    eingeist

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    Solving for an auxiliary equation, if you come up with an exponent higher than 2, how do you write the third coeffecient?

    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. [math](D-1)^2[/math] would be [math]c1e^x + xc2e^x.[/math]

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

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


    What do you think?


    edit: ooooh, is there a way to write math in the code like that?
     
    Last edited: Feb 26, 2010
  2. KevinFACE

    KevinFACE

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    Brain overload!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!