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Old 07-01-2005, 02:27 PM   #11
Azred
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Well, I didn't substitute exactly in the manner you suggest, but that isn't important.

What is important is that my answer might be wrong and yours might be right. My "big plans" of looking into the problem when I got home fell through and then I got really lazy on my day off. [img]tongue.gif[/img] [img]graemlins/petard.gif[/img]


hmmm....I found this interesting site which gives the result of cos(sqrt(3)*t) + cosh (sqrt(3) * t). This is going to require a little more investigation....


[ 07-01-2005, 02:38 PM: Message edited by: Azred ]
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Old 07-02-2005, 06:35 PM   #12
Azred
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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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Old 07-02-2005, 07:20 PM   #13
ZFR
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I'm afraid it dont work that way

take for e.g

1/s^2 can be broken into (1/s) * (1/s)

the inverse of 1/s = 1

but the invesre of 1/s^2 is not 1*1
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Old 07-03-2005, 02:04 AM   #14
Azred
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Well, I worked it out by hand with my handy differential equations book by my side, but...

...I'm not going to quibble about it. [img]graemlins/petard.gif[/img]
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