Thread: Inverse Laplace
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Old 07-02-2005, 06:35 PM   #12
Azred
Drow Priestess
 

Join Date: March 13, 2001
Location: a hidden sanctorum high above the metroplex
Age: 55
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And now for the definitive answer....

Begin with the original equation s^2/(s^4 - 9), which breaks down into (s/(s^2 + 3))*(s/(s^2 - 3)). The inverse Laplace of s/(s^2 + 3) is cos (3t) and the inverse Laplace of s/(s^2 - 3) is cosh (3t), therefore...

...the inverse Laplace transform of the original equation is cos(3t) * cosh (3t), taking advantage of the convolution theorem. Of course, s > abs(sqrt(3)).

That should end the disussion. [img]graemlins/petard.gif[/img]
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