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Math Help - Undetermined Coefficient question

  1. #1
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    Undetermined Coefficient question

    (d-1)^2(d+1)y = 10cos2x

    Work I've done:

    yc:
    1. (r-1)^2(r+1) = 0, r=1,1,-1
    2. yc=(c1 +c2x)e^x + c3e^-x
    3. Then expand the original: r^3 - r^2 - r + 1 = 10cos2x
    4. So y^(3) - y" - y' +y = 10cos2x

    How do I go about finding yp? yp=Acos2x + Bsin2x, and then find to the third derivative? Thanks,

    JN
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jim Newt View Post
    (d-1)^2(d+1)y = 10cos2x

    Work I've done:

    yc:
    1. (r-1)^2(r+1) = 0, r=1,1,-1
    2. yc=(c1 +c2x)e^x + c3e^-x
    3. Then expand the original: r^3 - r^2 - r + 1 = 10cos2x
    4. So y^(3) - y" - y' +y = 10cos2x

    How do I go about finding yp? yp=Acos2x + Bsin2x, and then find to the third derivative? Thanks,

    JN
    y_p=A\cos(2x)+B\sin(2x)
    y_p'=-2A\sin(2x)+2B\cos(2x)
    y_p''=-4A\cos(2x)-4B\sin(2x)
    y_p^{(3)}=8A\sin(2x)-8B\cos(2x)

    Plug these into the equation to solve for A and B.

    I hope this helps.
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  3. #3
    Junior Member
    Joined
    Feb 2008
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    Quote Originally Posted by TheEmptySet View Post
    y_p=A\cos(2x)+B\sin(2x)
    y_p'=-2A\sin(2x)+2B\cos(2x)
    y_p''=-4A\cos(2x)-4B\sin(2x)
    y_p^{(3)}=8A\sin(2x)-8B\cos(2x)

    Plug these into the equation to solve for A and B.

    I hope this helps.
    Thanks Empty Set, I originally came up with what you did. I reworked it and discovered another simple mistake in the arithmetic that screwed me up. Thanks,

    JN
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