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Math Help - Second order ODE problem

  1. #1
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    Second order ODE problem

    y'' - 3y' + 2y = 5e^{2x}

    Homogenous solution is:
    Ae^{2x} + Be^x

    Particular solution guess:
    y = axe^{2x}
    y' = \alpha e^{2x}(2x + 1)
    y'' = 4\alpha e^{2x}(x + 1)

    Subbing into ODE:
    4\alpha e^{2x}(x + 1) - 3\alpha e^{2x}(2x + 1) + 2\alpha xe^{2x} = 5e^{2x}

    My text says this simplifies to:
    4\alpha e^{2x} - 3\alpha e^{2x} = 5e^{2x}

    How is this so? I understand that you can make this true by making x = 0, but wouldn't this make the RHS = 5?

    I'm sure I'm confusing myself, any help would be great.
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  2. #2
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    You need to expand and collect like terms...

    \displaystyle 4\alpha e^{2x}(x + 1) - 3\alpha e^{2x}(2x + 1) + 2\alpha x\,e^{2x} = 5e^{2x}

    \displaystyle 4\alpha x\,e^{2x} + 4\alpha e^{2x} - 6\alpha x\,e^{2x} - 3\alpha e^{2x} + 2\alpha x\,e^{2x} = 5e^{2x}.


    The \displaystyle x\,e^{2x} terms all cancel.
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  3. #3
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    Ahh right. I thought they factored it because there was some convenient way to solve it. I guess not! Thanks.
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