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Math Help - inverse laplace

  1. #1
    Member i_zz_y_ill's Avatar
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    inverse laplace

    y'' + 2y' = x sub to y(0) = 1 and y'(0) = 0

    answer is y = \frac{9}{8} -\frac{x}{4} + \frac{x^2}{4} - \frac{exp(-2x)}{8} but cant see how they did it,,,some version of inverse laplace?
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    y'' + 2y' = x sub to y(0) = 1 and y'(0) = 0

    answer is y = \frac{9}{8} -\frac{x}{4} + \frac{x^2}{4} - \frac{exp(-2x)}{8} but cant see how they did it,,,some version of inverse laplace?
    y'' + 2y' = x,~~y(0) = 1,~y'(0) = 0.

    First, find the image of the equation by Laplace:

    \mathcal{L}\left\{y(x)\right\} = y(s),

    \mathcal{L}\left\{y'(x)\right\} = sy(s) - y(0) = sy(s) - 1,

    \mathcal{L}\left\{y''(x)\right\} = s^2y(s) - y(0)s - y'(0) = s^2y(s) - s.

    \mathcal{L}\left\{x\right\} = \frac{1}{s^2}.

    So, you have

    s^2y(s) - s + 2\left(sy(s) - 1\right) = \frac{1}{s^2};

    \left( s^2 + 2s\right)y(s) = \frac{1}{s^2} + s + 2;

    y(s) = \frac{s^3 + 2s^2 + 1}{s^3 (s + 2)} = \frac{9}{8s} + \frac{1}{2s^3} - \frac{1}{4s^2} - \frac{1}{8(s + 2)}.

    Finally you have

    y(x) = {\mathcal{L}^{-1}}\left\{y(s)\right\} = \mathcal{L}^{-1}\left\{\frac{9}{8s} + \frac{1}{2s^3} - \frac{1}{4s^2} - \frac{1}{8(s + 2)} \right\} = \frac{9}{8} - \frac{x}{4} + \frac{x^2}{4} - \frac{e^{-2x}}{8}.
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