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Math Help - urgent, impossible differential equation

  1. #1
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    urgent, impossible differential equation

    ive been given a differential equation:

    (d2y/dx2)-3(dy/dx)+2y=xe^2x

    and cannot work out the particular solution. i have tried
    y=Ax^{2} e^{x}

    getting my y' as 2Cxe^{x} + Cx^{2}e^{x}

    and y'' as Cx^{2}e^{x} + 4Cxe^{x} + 2Ce^{x}
    but then end up getting 2C=0

    is this question impossible or have i used the wrong particular solution and if so how would i go about it?

    HELP!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by james30 View Post
    ive been given a differential equation:

    (d2y/dx2)-3(dy/dx)+2y=xe^2x

    and cannot work out the particular solution. i have tried
    y=Ax^{2} e^{x}

    getting my y' as 2Cxe^{x} + Cx^{2}e^{x}

    and y'' as Cx^{2}e^{x} + 4Cxe^{x} + 2Ce^{x}
    but then end up getting 2C=0

    is this question impossible or have i used the wrong particular solution and if so how would i go about it?

    HELP!!
    \frac{d^2y}{dx^2} -3 \frac{dy}{dx} + 2y = xe^{2x}

    The problem is that if we try only one term we get multiple conditions on it. So typically we'd try y_p(x) = Cxe^{2x} + De^{2x}, but the last term is already a part of the homogeneous solution. So try instead:
    y_p(x) = Cx^2e^{2x} + Dxe^{2x}

    y_p^{\prime}(x) = 2Cx^2e^{2x} + (2C + 2D)xe^{2x} + De^{2x}

    y_p^{\prime \prime}(x) = 4Cx^2e^{2x} + (8C + 4D)xe^{2x} + (2C + 4D)e^{2x}

    So
    (4Cx^2e^{2x} + (8C + 4D)xe^{2x} + (2C + 4D)e^{2x}) -3(2Cx^2e^{2x} + (2C + 2D)xe^{2x}  + De^{2x}) + 2(Cx^2e^{2x} + Dxe^{2x}) = xe^{2x}

    Simplifying:
    2Cxe^{2x} + (2C + D)e^{2x} = xe^{2x}

    So
    2C = 1
    and
    2C + D = 0

    Thus
    C = \frac{1}{2} and D = -1.

    Thus the particular solution is
    y_p(x) = \frac{1}{2}x^2e^{2x} - xe^{2x}

    -Dan
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  3. #3
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    So
    2C = 1

    how does this work? i dont see how you get the 2C=1
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    I'd try y_p = (Ax + B)e^{2x}
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  5. #5
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    i just didnt understand

    Simplifying:


    So

    and


    Thus
    and .

    how the value of 2C=1 was calculated from the

    what do i set x as to make it all work because i know you set x as 0 to get 2C+D=0
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by james30 View Post
    i just didnt understand

    Simplifying:


    So

    and


    Thus
    and .

    how the value of 2C=1 was calculated from the

    what do i set x as to make it all work because i know you set x as 0 to get 2C+D=0
    I'm setting the coefficients of like terms on each side of the equation to be equal to each other.

    There is an xe^{2x} on each side so the coefficients must be the same. Thus 2C = 1.

    Similarly, there is no e^{2x} on the RHS, so the coefficient 2C + D = 0.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    I'd try y_p = (Ax + B)e^{2x}
    This won't work as Be^{2x} is a solution to the homogeneous equation. We simply won't have enough independent conditions left at the end to solve the system.

    -Dan
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