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Math Help - homogenous differencial equation

  1. #1
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    homogenous differencial equation

    write out a homogeneous linear D.E for which 5e^(2x)+2e^(-x)-3xe^(-x)+cos 3x is a solution
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  2. #2
    Member Glaysher's Avatar
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    y'' - 2y' + y = 0

    gives y = Ae^(-x) + Bxe^(-x)

    Let y = 5e^(2x)

    y' = 10e^(2x)

    y'' = 20e^(2x)

    So y'' - 2y' + y = 20e^(2x) - 20e^(2x) + 5e^(2x) = 5e^(2x)

    Let y = cos 3x

    y' = -3 sin 3x

    y'' = -9 cos 3x

    y'' - 2y' + y = -9 cos 3x + 6 sin 3x + cos 3x = 6 sin 3x - 8 cos 3x

    So have y'' - 2y' + y = 6 sin 3x - 8 cos 3x + 5e^(2x)

    Need boundary condition for the A and B above as need A = 2, B =-3

    y = 5e^(2x) + Ae^(-x) + Bxe^(-x) + cos 3x

    x = 0, y = 6 + A

    Want A = 2 so y = 8

    y' = 10e^(2x) - 2e^(-x) + Be^(-x) - Bxe^(-x) - 3 sin 3x

    x = 0, y' = 8 + B

    Want B = -3 so y' = -11

    So differential equation y'' - 2y' + y = 6 sin 3x - 8 cos 3x + 5e^(2x) subject to boundary conditions y' = -11 and y = 8 when x = 0

    EDIT: I forget what homogeneous means so if this isn't homogeneous you will need to adapt this method to make one that is
    Last edited by Glaysher; April 7th 2007 at 11:49 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Glaysher View Post

    EDIT: I forget what homogeneous means so if this isn't homogeneous you will need to adapt this method to make one that s
    homogenious basically means equal to zero. so for instance, if you wanted to find a second order D.E you would have to find one of the form:

    a(x)y'' + b(x)y' + c(x)y = 0

    where y = 5e^(2x)+2e^(-x)-3xe^(-x)+cos 3x

    and a(x), b(x), and c(x) are linear funcions of x
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