Tobias Ali

2020-11-06

Find Laplace transform of the given function
$t{e}^{-4t}\mathrm{sin}3t$

Expert

Step 1
Concept Used:
$L\left\{{t}^{k}f\left(t\right)\right\}=\left(-1\right)k\left({d}^{k}\right)\frac{L\left\{f\left(t\right)\right\}}{d}{s}^{k}$
also,
$L\left\{\mathrm{sin}\left(at\right)\right\}=\frac{a}{\left({s}^{2}+{a}^{2}\right)}$
$L\left\{{e}^{-bt}\mathrm{sin}\left(at\right)\right\}=\frac{a}{\left(s+b{\right)}^{2}+{a}^{2}}$
Step 2
Given: Given function is $f\left(t\right)=t{e}^{-4t}\mathrm{sin}\left(3t\right)$
We want to calculate Laplace transform
Calculation:
Given function is $f\left(t\right)=t{e}^{-4t}\mathrm{sin}\left(3t\right)$
Therefore,
$L\left\{f\left(t\right)\right\}=L\left\{t{e}^{-4t}\mathrm{sin}\left(3t\right)\right\}$
$=\left(-1\right)1d\frac{\left[L\left\{{e}^{-4t}\mathrm{sin}\left(3t\right)\right\}\right]}{ds}$
$=-\frac{d}{ds}\left[\frac{3}{\left(s+4{\right)}^{2}+9}\right]$
$=-\frac{d}{ds}\left[\frac{3}{{s}^{2}+8s+16+9}\right]$
$=-\frac{d}{ds}\left[\frac{3}{{s}^{2}+8s+25}\right]$
$=-3\frac{d\left({s}^{2}+8s+25{\right)}^{-1}}{ds}$
$=-3\left(-1\right)\left({s}^{2}+8s+25{\right)}^{-2}\frac{d\left({s}^{2}+8s+25\right)}{ds}$

$=3\left({s}^{2}+8s+25{\right)}^{-}2\left(2s+8\right)$
$L\left\{f\left(t\right)\right\}=\frac{6\left(s+4\right)}{\left({s}^{2}+8s+25{\right)}^{2}}$
Step 3
Answer: The Laplace transform of $f\left(t\right)=t{e}^{-4t}\mathrm{sin}\left(3t\right)$ is $L\left\{f\left(t\right)\right\}=\frac{6\left(s+4\right)}{\left({s}^{2}+8s+25{\right)}^{2}}$

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