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November 2nd, 2014, 03:57 AM  #1 
Senior Member Joined: Oct 2014 From: I don't know... Posts: 220 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 04:10 AM. 
November 2nd, 2014, 04:18 AM  #2 
Senior Member Joined: Oct 2014 From: I don't know... Posts: 220 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^{(n1)}} \over (n  1)!} $ Last edited by szz; November 2nd, 2014 at 04:27 AM. 

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inverse, laplace, relatioship, tables, transform 
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