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Math Help - Stuck solving for a particular solution

  1. #1
    Newbie staevobr's Avatar
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    Stuck solving for a particular solution

    y'' +4y=6xcos(2x)

    I am using the method of undetermined coefficients. I had no trouble solving for the homogeneous solution, but the particular solution has me baffled. I used  y_p=Axcos(2x)+Bxsin(2x). After finding the derivatives of this and putting them back into the original equation, I end up with a system of equations that I cannot solve. Have I chosen this function poorly?
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  2. #2
    Newbie driegert's Avatar
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    Hey,

    Because of the x out front of the cos you have to choose something along the lines of:

    y_{p}=A\cos{2x} + Bx\cos{2x} + C\sin{2x} + Dx\sin{2x}

    In order to take care of the derivatives where you will have mixed x's on the sines and cosines.
    I didn't try to solve it, but I believe this is correct. Give it a shot .

    Hope it helps!
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  3. #3
    Newbie staevobr's Avatar
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    I tried that and it didn't work.
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  4. #4
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    Quote Originally Posted by staevobr View Post
    I tried that and it didn't work.
    Then please clearly post all the working for what you tried so that it can be reviewed.
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  5. #5
    Newbie staevobr's Avatar
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    y_p=Acos(2x)+Bxsin(2x)+Csin(2x)+Dxcos(2x)
    y'_p=-2Asin(2x)+Bsin(2x)+2Bxcos(2x)+2Ccos(2x)+Dcos(2x)-2Dxsin(2x)
    y''_p=-4Acos(2x)+2Bcos(2x)+2Bcos(2x)-4Bxsin(2x)-4Csin(2x)
     \> -2Dsin(2x)-2Dsin(2x)-4Dxcos(2x)

    y''_p + 4y = -4Acos(2x)+2Bcos(2x)+2Bcos(2x)-4Bxsin(2x)-4Csin(2x)-2Dsin(2x)
    -2Dsin(2x)-4Dxcos(2x) + 4Acos(2x) + 4Bxsin(2x) +4Csin(2x) + 4Dxcos(2x)


    4Bcos(2x)-4Dsin(2x)
    This makes B=3x/2, which is not a constant.
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  6. #6
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    Quote Originally Posted by staevobr View Post
    y'' +4y=6xcos(2x)

    I am using the method of undetermined coefficients. I had no trouble solving for the homogeneous solution, but the particular solution has me baffled. I used  y_p=Axcos(2x)+Bxsin(2x). After finding the derivatives of this and putting them back into the original equation, I end up with a system of equations that I cannot solve. Have I chosen this function poorly?
    Since the homogenous solution (which I was hoping you would include in your working because it can be important in deciding what particular solution to try - like in this question) is y_h = (Ax + B) (\sin (2x) + \cos (2x)), the 'particular solutions' mentioned so far in this thread will not work (why?).

    The particular solution you need to use is (why?):

    y_p = (a_1 x + a_2 x^2) (b_1 \cos (2x) + b_2 \sin (2x))
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  7. #7
    Newbie staevobr's Avatar
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    The homogeneous solution I actually got is

    Characteristic equation: r^2+4=0
    r=  $\pm$2i
    y_h=C_1e^{0x}cos(2x)+C_2e^{0x}sin(2x)
    y_h = C_1cos(2x) + C_2sin(2x)
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  8. #8
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    Quote Originally Posted by staevobr View Post
    The homogeneous solution I actually got is

    Characteristic equation: r^2+4=0
    r= $\pm$2i
    y_h=C_1e^{0x}cos(2x)+C_2e^{0x}sin(2x)
    y_h = C_1cos(2x) + C_2sin(2x)
    Yes, you are of course correct (an aberation on my part). Nevertheless, I think the particular solution I proposed is what you need to use.
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