Laplace transform please check my answer

In summary, the Laplace transform of (t+2)sinh2t is (4s/(s^2 + 4)^2 ) + (4/(s^2 - 4)), and to check if it is simplified, one could put it into a common denominator or perform an inverse Laplace transform.
  • #1
cabellos
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1
I have to find the laplace transform of (t+2)sinh2t

my answer is (4s/(s^2 + 4)^2 ) + (4/(s^2 - 4))

is this simplified as much as possible...?
 
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  • #2
Depends on what you regard as "simplification".
You might throw everything ont a common denominator, but I'm not too sure that is a simplification.
 
  • #3
first off sorry for double posting...

secondly, is my answer correct...Id really aprreciate it if someone could check.

Thanks
 
  • #4
using maple:

laplace:
[tex] (t+2)\sinh 2t [/tex]

equals:
[tex] \frac{4(s^2+s-4)}{(s+2)^2(s-2)^2} [/tex]

You can put that in the form you need to check. Or if really lazy (as I'm being) plug in s values to get an idea if you are correct, don't use this as proof that they are the same.

Also, why don't you just perform an inverse laplace on your expression to check?
 

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