 My Math Forum Relatioship between Laplace and Inverse Laplace Transform from tables
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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. Tags inverse, laplace, relatioship, tables, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Calculus 1 November 1st, 2014 03:14 PM Rydog21 Calculus 2 October 15th, 2013 09:35 AM Deiota Calculus 1 April 28th, 2013 10:28 AM defunktlemon Real Analysis 1 April 14th, 2012 02:19 PM timh Applied Math 1 June 7th, 2010 06:44 PM

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