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Math Help - help with annihilator method

  1. #1
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    help with annihilator method

    alright. Heres the problem:
    1. y''' + y'' = 8x^2

    Heres what I do:

    m^3 + m^2 = 0

    m = 0, m = -1

    D^3(D^3+D^2)y = D^38x^2

    m^3(m^3 + m^2) = 0      auxillary equation

    m^5(m+1) = 0

    m = 0,0,0,0,0, -1

    Yc = C1 + C2x + C3x^2 +C4+x^3 +C5x^4 + C6e^-^x

    The particular solution has the form:

    Yp = Ax^2 + Bx + C

    Y'p = 2Ax + B

    Y''p = 2A

    Y'''p = 0

    Plugging in the particular solution into the DE gives:

    2A = 8x^2

    The book says the answer is:

    y = C1 + C2x +C3e^-^x + \frac{2}{3}x^4 - \frac{8}{3}x^3 + 8x^2

    What im lost on(yes its algebra) on how they get 8/3, 2/3, and 8. And how they knew where to put them.


    Anyone willing to help me out? Thanks!!!
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  2. #2
    Behold, the power of SARDINES!
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    The homogeneous equation is

    y'''+y''=0 \iff D^3y+D^2y=0 \iff D^2(D+1)y=0

    So the solution to the homogeneous solution is

    y_c=c_1+c_2x+c_3e^{-x}

    Now to find the particular solution we

    D^2(D+1)y=8x^2 \iff D^5(D+1)y=0

    So the solution to this new equation are

    A+Bx+Cx^2+Dx^3+Ex^4+Fe^{-x}

    Now we need to removed any terms that are in the complimentary solution this gives

    y_p=Cx^2+Dx^3+Ex^4

    Now just plug this in to the ODE and this gives

    (6D+24Ex)+(2C+6Dx+12Ex^2)=8x^2

    12Ex^2+(24E+6D)x+(6D+2C)=8x^2

    This gives the system of equation

    12E=8 \quad 24E+6D=0 \quad 6D+2C=0

    Then just plug these back into the particular solution.
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  3. #3
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    Thank you. You are the man!
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