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Math Help - Annihilators/ Differential Operator Question

  1. #1
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    Annihilators/ Differential Operator Question

    I want to use A(D)= (D+2)^2 to annihilate F(X)=5xe^(-2x).

    However, I also need to compute A(D) operating on y''+4y'+4y in order to find Yp.


    Thus, my question is this.

    How do I compute A(D) operating on y''+4y'+4y?

    I know it's simple but I'm stuck.
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  2. #2
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    Quote Originally Posted by divinelogos View Post
    I want to use A(D)= (D+2)^2 to annihilate F(X)=5xe^(-2x).

    However, I also need to compute A(D) operating on y''+4y'+4y in order to find Yp.


    Thus, my question is this.

    How do I compute A(D) operating on y''+4y'+4y?

    I know it's simple but I'm stuck.
    The equation that we want to solve is

    y''+4y'+4y=5xe^{-2x}

    If we write this in terms of the operator D we get

    D^2y+4Dy+4y=5xe^{-2x} \iff (D^2+4D+4)y=5e^{-2x} \iff (D+2)^2y=5xe^{-2x}

    From here we see that the complimentary solution is

    y_c=c_1e^{-2x}+c_2xe^{-2x}

    Now find the particular solution we need to annihilate the right hand side.
    Using the Annihilator that you have above gives

    (D+2)^4y=(D+2)^2(5xe^{-2x})=0

    This "new" equation have solution

    y_p=ae^{-2x}+bxe^{-2x}+mx^2e^{-2x}+nx^3e^{-2x}

    Now we need to remove any terms that are in the complimentary solution. If we relabel the constants we get


    y_p=Ax^2e^{-2x}+Bx^3e^{-2x}

    From here it is just finding A and B.
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  3. #3
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    Quote Originally Posted by divinelogos View Post
    I want to use A(D)= (D+2)^2 to annihilate F(X)=5xe^(-2x).

    However, I also need to compute A(D) operating on y''+4y'+4y in order to find Yp.


    Thus, my question is this.

    How do I compute A(D) operating on y''+4y'+4y?
    If that is what you meant to say, you don't! y''+ 4y'+ 4y is already (D+ 2)^2y.
    (D+ 2)^2y means (D+ 2)(D+ 2)y= (D+ 2)(Dy+ 2y)= (D+ 2)(y'+ 2y)= D(y')+ D(2y)+ 2(y'+ 2y)= y''+ 2y'+ 2y'+ 4y= y''+ 4y'+ 4y.

    I know it's simple but I'm stuck.
    In order to find yp you need to find a function y(x) such that (D+ 2)^2y= 5xe^{-2x}
    I recommend trying something of the form y= (Ax^3+ Bx^2)e^{-2x}. Then Dy= [(Ax^3+ Bx^2)e^{-2x}]'= (3Ax^2+ 2Bx)e^{-2x}- 2(Ax^3+ Bx^2)e^{-2x}= (-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x} and D^2y= [(-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x}]'= (-6Ax^2+ (6A- 4B)x+ 2B)e^{-2x}- 2(-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x}= (4Ax^3+ (-12A+ 4B)x^2+ (-6A- 4B)x+ 2B)e^{-2x}. Put those into (D^2+ 4D+ 4)y= 5xe^{-2x} and try to find A and B.
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