# Find the Laplace transform L{u_3(t)(t^2-5t+6)} a) F(s)=e^(-3s)(2/s^4-5/s^3+6/s^2)

Find the Laplace transform $L\left\{{u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right\}$
$a\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{4}}-\frac{5}{{s}^{3}}+\frac{6}{{s}^{2}}\right)$
$b\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{3}}-\frac{5}{{s}^{2}}+\frac{6}{s}\right)$
$c\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{4}}$
$d\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{3}}$
$e\right)F\left(s\right)={e}^{-3s}\frac{2-11s+30{s}^{2}}{{s}^{3}}$

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Step 1
According to the given information, it is required to determine the laplace transform of the given function.
$L\left\{{u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right\}$
Step 2
Use the below formula to find the laplace transform.
$L\left({u}_{c}\left(t\right)g\left(t\right)\right)={e}^{-cs}L\left(g\left(t+c\right)\right)$
$L\left({u}_{c}\left(t\right)\right)=\frac{{e}^{-cs}}{s}$
Step 3
Solving further to get:
$L\left({u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right)=L\left({u}_{3}\left(t\right){t}^{2}\right)-5L\left({u}_{3}\left(t\right)t\right)+6L\left({u}_{3}\left(t\right)\right)$
$={e}^{-3s}L\left(\left(t+3{\right)}^{2}\right)-5{e}^{-3s}L\left(t+3\right)+\frac{6{e}^{-3s}}{s}$
$={e}^{-3s}\left[\frac{2}{{s}^{3}}+\frac{6}{{s}^{2}}+\frac{9}{s}\right]-5{e}^{-3s}\left[\frac{1}{{s}^{2}}+\frac{3}{s}\right]+\frac{6{e}^{-3s}}{s}$
$={e}^{-3s}\left[\frac{2}{{s}^{3}}+\frac{6}{{s}^{2}}+\frac{9}{s}-\frac{5}{{s}^{2}}-\frac{15}{s}+\frac{6}{s}\right]$
$={e}^{-3s}\left[\frac{2}{{s}^{3}}+\frac{1}{{s}^{2}}\right]$
$L\left({u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right)={e}^{-3s}\left[\frac{2+s}{{s}^{3}}\right]$
Step 4
Therefore, the correct option is (D).