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Math Help - Differential equation

  1. #1
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    Differential equation

    Use a suitable trial function to find the particular solution of

    \frac{dy}{dx}+6y=3sin(4x)

    y=Acos(4x)+Bsin(4x)

    \frac{dy}{dx}=-4Asin(4x)+4Bcos(4x)

    From the equation, -4Asin(4x)+4Bcos(4x)+6[Acos(4x)+Bsin(4x)]=3sin(4x)

    Now I don't understand the last bit where you have to pick the two parts which correspond to 0 and 3. Could someone explain it for me?
    Last edited by Haris; February 11th 2009 at 12:31 PM.
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  2. #2
    MHF Contributor
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    Hi

    -4Asin(4x)+4Bcos(4x)+6[Acos(4x)+Bsin(4x)]=3sin(4x)

    Separate cos and sin
    (-4A+6B-3)sin(4x)+(4B+6A)cos(4x)=0

    Therefore
    -4A+6B-3=0 and 4B+6A=0
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