Problem 02 | Second Shifting Property of Laplace Transform

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$
 

$F(s) = \mathcal{L} (t^3)$   and   $a = 2$

$F(s) = \dfrac{3!}{s^{3 + 1}}$

$F(s) = \dfrac{6}{s^4}$
 

Thus,
$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \left( \dfrac{6}{s^4} \right)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \dfrac{6e^{-2s}}{s^4}$           answer

$g(t) = \begin{cases} f(t - 2)^3 & t \gt 2 \\ 0 & t \lt 2 \end{cases}$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-st} g(t) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^2 e^{-st} (0) \, dt + \int_2^\infty e^{-st} (t - 2)^3 \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_2^\infty e^{-st} (t - 2)^3 \, dt$
 

$z = t - 2$

$t = z + 2$

$dt = dz$
 

when   $t = 2, \, z = 0$

when   $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-s(z + 2)} z^3 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz - 2s} z^3 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz} e^{-2s} z^3 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \int_0^\infty e^{-sz} z^3 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \mathcal{L} (z^3)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \left( \dfrac{3!}{s^{3 + 1}} \right)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-2s} \left( \dfrac{6}{s^4} \right)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \dfrac{6e^{-2s}}{s^4}$       okay