Trent Carpenter

2021-09-09

First using a trigonometric
identity, find $L\left\{f\left(t\right)\right\}$
$f\left(t\right)=\mathrm{sin}2t\mathrm{cos}2t$

Arham Warner

Linearity Property: suppose ${f}_{1}\left(p\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{2}\left(p\right)$ are Laplace transformations of ${F}_{1}\left(t\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{F}_{2}\left(t\right)$ respectively. Then:
$L\left\{{c}_{1}{F}_{1}\left(t\right)+{c}_{2}{F}_{2}\left(t\right)\right\}={c}_{1}L\left\{{F}_{1}\left(t\right)\right\}+{c}_{2}L\left\{{F}_{2}\left(t\right)\right\}={c}_{1}{f}_{1}\left(p\right)+{c}_{2}{f}_{2}\left(p\right)$
Also, $L\left\{\mathrm{sin}at\right\}=\frac{a}{{p}^{2}+{a}^{2}}$
Given, $f\left(t\right)=\mathrm{sin}2t\mathrm{cos}2t$
We have to find $L\left\{f\left(t\right)\right\}$ i.e. the Laplace Transformation of giveb $f\left(t\right)$.
$f\left(t\right)=\mathrm{sin}2t\mathrm{cos}2t=$
$=\frac{1}{2}\left(2\mathrm{sin}2t\mathrm{cos}2t\right)=$
$=\frac{1}{2}\mathrm{sin}4t$
Taking Laplace transformation on both sides,
$L\left\{f\left(t\right)\right\}=L\left\{\frac{1}{2}\mathrm{sin}4t\right\}=$
$=\frac{1}{2}L\left\{\mathrm{sin}4t\right\}=$
$=\frac{1}{2}\frac{4}{{p}^{2}+{4}^{2}}=$
$=\frac{2}{{p}^{2}+{4}^{2}}$
The required solution $L\left\{f\left(t\right)\right\}=\frac{2}{{p}^{2}+{4}^{2}}$

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