tonan6e

2022-09-03

How to find the Laplace Transform of ${t}^{2}sin\left(t\right)$
Using the rule:
$\mathcal{L}\left({t}^{n}f\left(t\right)\right)=\left(-1{\right)}^{n}\frac{{d}^{n}}{d{s}^{n}}F\left(s\right)$
where in this case
$f\left(t\right)=\mathrm{sin}\left(t\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathcal{L}\left(\mathrm{sin}\left(t\right)\right)=F\left(s\right)=\frac{1}{{s}^{2}+1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n=2.$
Find the 2nd derivative of F(s):
$\frac{{d}^{2}}{d{s}^{2}}\left(\frac{1}{{s}^{2}+1}\right)=\frac{6{s}^{2}-2}{\left({s}^{2}+1{\right)}^{3}}$
The transform:
$\mathcal{L}\left({t}^{2}sin\left(t\right)\right)=\left(-1{\right)}^{2}\frac{6{s}^{2}-2}{\left({s}^{2}+1{\right)}^{3}}$

Xavier Jennings

$L\left[f\cdot g\right]\ne L\left[f\right]\cdot L\left[g\right]$
You should use :
$L\left[{t}^{n}f\left(t\right)\right]=\left(-1{\right)}^{n}{F}^{\left(n\right)}\left(s\right)$
Where $F\left(s\right)=L\left[f\right]$ and ${F}^{\left(n\right)}\left(s\right)$ is the nth derivative of F.