# Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform. Lleft{e^{3t}sin(4t)-t^{4}+e^{t}right}

Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
$L\left\{{e}^{3t}\mathrm{sin}\left(4t\right)-{t}^{4}+{e}^{t}\right\}$
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$\text{Step 1}$
$L\left\{{e}^{3t}\mathrm{sin}\left(4t\right)-{t}^{4}+{e}^{t}\right\}$

$L\left\{{e}^{at}\right\}=\frac{1}{s-a}$
$L\left\{\mathrm{sin}\left(at\right)\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
$L\left\{\mathrm{sin}\left(4t\right)\right\}=\frac{4}{{s}^{2}+16}:$

$L\left\{{e}^{3t}\mathrm{sin}\left(4t\right)\right\}=\frac{4}{\left(s-3{\right)}^{2}+16},$
$L\left\{{t}^{4}\right\}=\frac{4!}{{s}^{4+1}}=\frac{24}{{s}^{5}}$
$L\left\{{e}^{t}\right\}=\frac{1}{s-1}$
$\text{Step 2}$

$L\left\{{e}^{3t}\mathrm{sin}\left(t\right)-{t}^{4}+{e}^{t}\right\}=\frac{4}{\left(s-3{\right)}^{2}+16}-\frac{24}{{s}^{5}}+\frac{1}{s-1}$

Jeffrey Jordon