Question

In Problems 3 - 24 fill in the blanks or answer true or false.

10.$\mathit{L}\left\{e^{-3t}sin 2t\right\}$
=……..

Short answer

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

Step-by-step solution

Consider the function

The function is,

$\begin{array}{l}L\left\{e^{-3t}\sin 2t\right\}\\ Thes\tan dardformulaisgivenby,\\ L\left\{\sin \left(at\right)\right\}=\frac{a}{s^2+a^2}\end{array}$

Compute the Laplace transform

consider the function,

$\begin{array}{l}L\left\{e^{-3t}\sin 2t\right\}\\ \mathrm{If} L\left\{f\left(t\right)\right\}=F\left(s\right) \mathrm{then} L\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right)\\ Here,a=-3\\ AndL\left\{\sin \left(2t\right)\right\}=\frac{2}{s^2+2^2}\end{array}$

$\begin{array}{l}Replaceswith\left(s+(-3)\right)\\ SubstitutethevalueandfindtheLaplacetransform\end{array}$

$\begin{array}{l}L\left\{e^{-3t}\sin \left(2t\right)\right\}=L\left\{\sin \left(2t\right)\right\}\left(s+3\right)\\ \Rightarrow L\left\{e^{-3t}\sin \left(2t\right)\right\}=\frac{2}{{\left(s+3\right)}^2+4}\end{array}$

$\begin{array}{l}Therefore,thelaplacetransformofL\left\{e^{-3t}\sin \left(2t\right)\right\}is,\\ \therefore L\left\{e^{-3t}\sin 2t\right\}=\frac{2}{{\left(s+3\right)}^2+4}\end{array}$