# To find: The Laplace transform of L[e^{-3t}t^4]

To find:
The Laplace transform of $L\left[{e}^{-3t}{t}^{4}\right]$
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Step 1
To find:
Laplace transform of $L\left[{e}^{-3t}{t}^{4}\right]$
Step 2
Use the Laplace transform:$L\left\{{t}^{k}f\left(t\right)\right\}=\left(-1{\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left\{f\left(t\right)\right\}\right)$

$L\left\{{t}^{4}{e}^{-3t}\right\}=\left(-1{\right)}^{4}\frac{{d}^{4}}{d{s}^{4}}\left(L\left\{{e}^{-3t}\right\}\right)$
$=\frac{{d}^{4}}{d{s}^{4}}\left(L\left\{{e}^{-3t}\right\}\right)$
Use the Laplace transform: $L\left\{{e}^{-at}\right\}=\frac{1}{s+a}$
$L\left\{{t}^{4}{e}^{-3t}\right\}=\frac{{d}^{4}}{d{s}^{4}}\left(L\left\{{e}^{-3t}\right\}\right)$
$=\frac{{d}^{4}}{d{s}^{4}}\left(\frac{1}{s+3}\right)$
$=\frac{{d}^{3}}{d{s}^{3}}\frac{d}{ds}\frac{1}{s+3}$
$=\frac{{d}^{3}}{d{s}^{3}}\left(\frac{-1}{\left(s+3{\right)}^{2}}\right)$
$=\frac{{d}^{2}}{d{s}^{2}}\frac{d}{ds}\left(\frac{-1}{\left(s+3{\right)}^{2}}\right)$
$=\frac{{d}^{2}}{d{s}^{2}}\left(\frac{1\cdot 2}{\left(s+3{\right)}^{3}}\right)$
$=\frac{d}{ds}\frac{d}{ds}\left(\left(1\cdot 2\right)/\left(s+3{\right)}^{3}\right)$
$=\frac{d}{ds}\left(-\frac{1\cdot 2\cdot 3}{\left(s+3{\right)}^{4}}\right)$
$=\left(\frac{1\cdot 2\cdot 3\cdot 4}{\left(s+3{\right)}^{5}}\right)$
$=\frac{\left(4!\right)}{\left(s+3{\right)}^{5}}$
Hence the Laplace transform $L\left[{e}^{-3t}{t}^{4}\right]=\frac{\left(4!\right)}{\left(s+3{\right)}^{5}}$