anonymous
  • anonymous
Find the Laplace Transform of the function: L{(t-4)sin2t} Fan and medal, please help. Posted below is what I have so far. My confusion is how did they get the answer..
Mathematics
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katieb
  • katieb
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anonymous
  • anonymous
\[L[(t-4)\sin2t] = L(tsin2t) - 4L(\sin2t) = L(tsin2t) - 4(\frac{ 1 }{ s^{2}+4 })\] I don't know how to handle L(tsin2t) I can find it using the formal definition given to me by the solution is: \[\frac{ 4 }{ (s^{2}+4)^{2} } - \frac{ 8 }{ s^{2}+4}\]
anonymous
  • anonymous
*** I can find the answer using the formal definition of laplace but there's another way and it's not coming to me...
mathmale
  • mathmale
I think you ought to look up "Laplace Transform Tables" on the 'Net. Next, look for the form \[t^n*f(t)\]

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mathmale
  • mathmale
and the transform of this configuration. Laplace transform tables range from awful to wonderful, so you may need to look at more than one such table to find on that has a transform pair applicable to your t*sin 2t. It does look like your'e off to a good start.
IrishBoy123
  • IrishBoy123
i don't know if this will help but a slightly off beat way to get this is to cut through the formal definition: As \(\mathcal{L} \{ \sin 2t \}= \int\limits_0^{\infty} \; e^{-st} \, \sin 2t \; dt = F(s)\) then \(\dfrac{dF}{ds} = \int\limits_0^{\infty} \; \dfrac{\partial}{\partial s} \left[ e^{-st} \, \sin 2t \right)] \; dt\) \(= \int\limits_0^{\infty} \; -t e^{-st} \sin 2t \; dt= -\mathcal{L} \{ t \, \sin 2t \}\) so \(\mathcal{L} \{ t \, \sin 2t \} = - \dfrac{dF}{ds} = - \dfrac{d}{ds} \dfrac{2}{s^2 + 4}\) \(=\dfrac{4s}{(s^2 + 4)^2}\) of course, you should be able to find this result in generalised form in any good table of LP transforms as it applies to all functions that have a Laplace Transform...which is the smart way to do it.
anonymous
  • anonymous
Okay that's good to know then. I'll be given a table during the exam but just wanted to be sure I wasn't missing anything obvious.

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