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Math Help - D Operator

  1. #1
    Newbie khanim's Avatar
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    D Operator

    Need to find the Particular Integral

    D^{2}y+2Dy+2y=65e^{x}sin 2x

    Last edited by khanim; September 15th 2009 at 05:13 AM.
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  2. #2
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    Quote Originally Posted by khanim View Post
    Need to find the Particular Integral

    D^{2}y+2Dy+2=65e^{x}sin 2x

    What you have written would be more simply written as D^2y+ 2Dy= 65e^x sin(2x)- 2. Was that what you meant?

    Or did you mean D^2y+ 2Dy+ 2y= 65e^x sin(2x)?

    Or D^2y+ 2Dy+ y= 65e^x sin(2x)?
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  3. #3
    Newbie khanim's Avatar
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    D^{2}y+2Dy+2y=65e^{x}sin 2x

    Sorry about that, I have edited the question, I was missing the y after the two
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  4. #4
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    y=\frac1{D^2+2D+2}65\mathrm e^x\sin 2x=65\mathrm e^x\frac1{(D+1)^2+2(D+1)+2}\sin 2x =65\mathrm e^x\frac1{D^2+4D+5}\sin 2x

    {}=65\mathrm e^x\frac1{-2^2+4D+5}\sin 2x=65\mathrm e^x\frac1{4D+1}\sin 2x=65\mathrm e^x\frac{4D-1}{16D^2-1}\sin 2x

    {}=65\mathrm e^x\frac{4D-1}{16(-2^2)-1}\sin 2x=\mathrm e^x(1-4D)\sin 2x=\mathrm e^x(\sin 2x-8\cos 2x).
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