how do you derive the laplace transform for t^2 sin(3t)?
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$\displaystyle \mathcal L (t^2 sin(3t)) = \int_0^{\infty} t^2sin(3t)e^{-st} dt $ I'm afraid it's a rather long and boring integration by parts.
Last edited by Mush; Dec 17th 2008 at 01:13 AM.
Originally Posted by AmyZheng how do you derive the laplace transform for t^2 sin(3t)? Use the following operational theorem: $\displaystyle LT[t^2 \, f(t)] = F''(s)$ where $\displaystyle LT[f(t)] = F(s)$. In your case $\displaystyle f(t) = \sin (3t) \Rightarrow F(s) = \frac{3}{s^2 + 3^2}$.
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