use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$f(t)={e}^{3t}\mathrm{cos}5t-{e}^{-t}\mathrm{sin}2t$

Emily-Jane Bray
2021-03-04
Answered

use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$f(t)={e}^{3t}\mathrm{cos}5t-{e}^{-t}\mathrm{sin}2t$

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Bella

Answered 2021-03-05
Author has **81** answers

Step 1
Given function is $f(t)={e}^{3t}\mathrm{cos}5t-{e}^{-t}\mathrm{sin}2t$
Step 2
Obtain the Laplace transform.
$L\{{e}^{at}\mathrm{cos}bt\}=\frac{s-a}{(s-a{)}^{2}+{b}^{2}}$

$L\{{e}^{at}\mathrm{sin}bt\}=\frac{b}{(s-a{)}^{2}+{b}^{2}}$

$L\{f(t)\}=L\{{e}^{3t}\mathrm{cos}5t-{e}^{-t}\mathrm{sin}2t\}$

$=L\{{e}^{3t}\mathrm{cos}5t\}-L\{{e}^{-t}\mathrm{sin}2t\}$

$=\frac{s-3}{(s-3{)}^{2}+{5}^{2}}-\frac{2}{(s-(-1){)}^{2}+{2}^{2}}$

$=\frac{s-3}{(s-3{)}^{2}+25}-\frac{2}{(s+1{)}^{2}+4}$

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