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2020-11-29

Find the Laplace transform of $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)$
Then you obtain $F\left(s\right)=\frac{4s+a}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}$
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davonliefI

Step 1
Given that $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)$
We have to Laplace transform of
Use Laplace transform formula
$L\left\{{t}^{k}f\left(t\right)\right\}={\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left\{f\left(t\right)\right\}\right)$
For, $t{e}^{-t}\mathrm{sin}\left(2t\right),f\left(t\right)={e}^{-t}\mathrm{sin}\left(2t\right)$
$L\left\{t{e}^{-t}\mathrm{sin}\left(2t\right)\right\}=-\frac{d}{dx}\left(L\left\{{e}^{-t}\mathrm{sin}\left(2t\right)\right\}\right)\dots \left(1\right)$
Now, use formula $L\left\{{e}^{bt}\mathrm{sin}\left(at\right)\right\}=\frac{a}{{\left(s-b\right)}^{2}+{a}^{2}}$
Now, $L\left\{{e}^{-t}\mathrm{sin}\left(2t\right)\right\}=\frac{2}{{\left(s+1\right)}^{2}+4}$
Substitute $L\left\{{e}^{-t}\mathrm{sin}\left(2t\right)\right\}=\frac{2}{{\left(s+1\right)}^{2}+4}$ in equation (1)
$L\left\{t{e}^{-t}\mathrm{sin}\left(2t\right)\right\}=-\frac{d}{dx}\left(\frac{2}{{\left(s+1\right)}^{2}+4}\right)$
Step 2
Now,
$L\left\{t{e}^{-t}\mathrm{sin}\left(2t\right)\right\}=-\frac{d}{dx}\left(\frac{2}{{\left(s+1\right)}^{2}+4}\right)$
$=-\left(-\frac{4\left(s+1\right)}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}\right)$
$=\frac{4s+4}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}$
Therefore, $F\left(s\right)=\frac{4s+4}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}$
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