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Math Help - private solutions

  1. #1
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    private solutions

    y''-4y+4y=xe^2x

    what is the general private solution

    usuallu
    its
    y_p=(ax+b)e^2x
    but here we have
    a double root of 2
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by transgalactic View Post
    y''-4y+4y=xe^2x

    what is the general private solution

    usuallu
    its
    y_p=(ax+b)e^2x
    but here we have
    a double root of 2
    I think you mean particular solution.

    From just looking at the right hand side we would "guess" that

    y_p=(ax+b)^{2x}

    This is correct, but since the complimentary solution is c_1e^{2x}+c_2xe^{2x} as you said a double root we have to multiply our guess by x^2 This is justified by using reduction of order. so we get

    y_p=(ax^3+bx^2)e^{-2x}

    Now just plug in to find a and b.
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