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Math Help - [SOLVED] How to find differential operators/annihilators?

  1. #1
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    Unhappy [SOLVED] How to find differential operators/annihilators?

    I don't understand how to find differential operators that annihilate functions... my professor just does them in lecture like they are nothing, but none of my friends (nor I) know how these work.

    Some examples...

    y=x^2 exp(2x) \rightarrow(D-2)^3
    y=x^3 exp(-x) \rightarrow(D+1)^4
    y=exp(-x) cos(3x)\rightarrow(D+1)^2+3^2
    y=x exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^2
    y=x^3 exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^4

    But how would I, for example, annihilate:
    x+cos(x)

    Thanks in advance!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by horan View Post
    I don't understand how to find differential operators that annihilate functions... my professor just does them in lecture like they are nothing, but none of my friends (nor I) know how these work.

    Some examples...

    y=x^2 exp(2x) \rightarrow(D-2)^3
    y=x^3 exp(-x) \rightarrow(D+1)^4
    y=exp(-x) cos(3x)\rightarrow(D+1)^2+3^2
    y=x exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^2
    y=x^3 exp(-x) cos(3x)\rightarrow[(D+1)^2+3^2 ]^4

    But how would I, for example, annihilate:
    x+cos(x)

    Thanks in advance!
    I've never done it quite this way, but probably the easiest method would be to find a homogeneous linear differential equation, such that y = x + cos(x) is a solution.

    So let
    y = x + cos(x)

    Then
    y^{\prime} = 1 - sin(x)
    and
    y^{\prime \prime} = -cos(x)

    Note that
    y^{\prime \prime} + y = x

    We aren't done yet because this is not a homogeneous equation. But
    y^{(3)} + y^{\prime} = 1

    y^{(4)} + y^{\prime \prime} = 0

    So your annihilator would be (D^4 + D^2).

    -Dan
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  3. #3
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    Thank you so much!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by horan View Post
    Thank you so much!
    You are welcome. I should add that the method I used in my solution is only guaranteed to find an annihilator. It will not necessarily find the simplest annihilator.

    -Dan
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