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 November 2nd, 2014, 02:57 AM #1 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus Relatioship between Laplace and Inverse Laplace Transform from tables Hi all, Yesteday I proposed my self this exercice: \displaystyle \begin{aligned} & 6 {\mathrm{d}^3y \over \mathrm{d}x^3} = 0 \\ & 6 ( s^3 \mathcal{L} [y] - s^2y(0) - sy'(0) - y''(0) ) = \\ & 6 ( s^3 \mathcal{L} [y] - 2s^2 - 5s - 3 ) = \\ & 6 \mathcal{L} [y]s^3 -12s^2 -30s -18 = \\ & \mathcal{L} [y]= {18 + 30s + 12 s^2 \over 6s^3} = {a \over s} + {b \over s^2} + {c \over s^3} = \\ & a = 2 \,\,, b = 5 \,\,, c = 3 \\ & \mathcal{L} [y] = {2 \over s} + {5 \over s^2} + {3 \over s^3} \\ & y = \mathcal{L}^{-1} [y] = 2 + 5\mathrm{e}^t + {3 \over 2}\mathrm{e}^{t^2} \end{aligned} What it's not clear to me is the relationship between: $\displaystyle {3 \over s^3}$ and $\displaystyle {3 \over 2}\mathrm{e}^{t^2}$ retrieving the result from the tables. I my opinion the standard tables (for Laplace transform as integration) are often incomplete and this is why I do not like to depend on them. I know that: $\displaystyle \int_0^{\infty} {3 \over 2}\mathrm{e}^{t^2} \mathrm{e}^{-st} \,\mathrm{d}t = {3 \over s^3}$ and $\displaystyle {3 \over 2}\mathrm{e}^{t^2} = \mathcal{L}^{-1} \left [ {3 \over s^3} \right ]$ can be proofed with Inverse Laplace transform from principles but we did not studied it yet. So, the standard tables report: $\displaystyle t^n = {n ! \over s^{n + 1}}$ It's OK that: $\displaystyle {3 \over 2}\mathrm{e}^{t^2} \Rightarrow {n ! \over s^{n + 1}} \Rightarrow {3 \over s^3}$ But for the reverse process, I am quiet confused. I will appreciate if you make me understand better. Thank you in advance. szz Last edited by szz; November 2nd, 2014 at 03:10 AM.
 November 2nd, 2014, 03:18 AM #2 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus I deduce the true inverse relationship from Laplace domain to time domain from the tables is: $\displaystyle \mathcal{L}[y(t)] = {a \over s^n} \Rightarrow \mathcal{L}^{-1}[y(t)] = y(t) = {a \mathrm{e}^{t^{(n-1)}} \over (n - 1)!}$ Last edited by szz; November 2nd, 2014 at 03:27 AM.

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