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Laplace transform

 
 
Reply Tue 30 Mar, 2004 12:14 pm
Hi,
I have tried to solve this differential equation with Laplace transform:
x''+2x'- 8x= -256t^3, x(0) = 15, x'(0) = 36
this is Laplace transform:
s^2*X-15s - 36 +2sX-15-8X = -256*6/s^4
after some operations I've got this:
X = 15/s+36/s^2+48/s^3+192/s^4-(41/2)/(S+4) + 14/(s-2)
If I transform this back I get this:
x(t) = -41/2*e^(-4t) + 14e^(2t) + 32t^3 + 24t^2 + 36t + 15
but x(0) is not 15!
Does anyone know where did I go wrong?
Thank you,
Niko
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Type: Discussion • Score: 1 • Views: 730 • Replies: 2
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tideman76
 
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Reply Thu 1 Apr, 2004 08:49 am
I think I found an initial mistake.


In the transform step, on the transform of 2x', it should transform to

2*[sX - 15] which equals 2sX-30.

On your transform line, you do not have the -30, instead you have -15.

This is as far as I have gotten on the problem so far.

Hope this helps.
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niko2000
 
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Reply Thu 1 Apr, 2004 02:10 pm
Thank you
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