inverse laplace

**i_zz_y_ill** · January 16th 2010, 01:17 AM

$y'' + 2y' = x$ 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?

**DeMath** · January 16th 2010, 02:25 AM

Originally Posted by i_zz_y_ill

$y'' + 2y' = x$ 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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