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November 2nd, 2014, 03:57 AM   #1
szz
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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.
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November 2nd, 2014, 04:18 AM   #2
szz
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Posts: 224
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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 04:27 AM.
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