Trent Carpenter

2021-09-09

First using a trigonometric

identity, find$L\left\{f\left(t\right)\right\}$

$f\left(t\right)=\mathrm{sin}2t\mathrm{cos}2t$

identity, find

Arham Warner

Skilled2021-09-10Added 102 answers

Linearity Property: suppose ${f}_{1}\left(p\right){\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{f}_{2}\left(p\right)$ are Laplace transformations of ${F}_{1}\left(t\right){\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{F}_{2}\left(t\right)$ respectively. Then:

$L\{{c}_{1}{F}_{1}\left(t\right)+{c}_{2}{F}_{2}\left(t\right)\}={c}_{1}L\left\{{F}_{1}\left(t\right)\right\}+{c}_{2}L\left\{{F}_{2}\left(t\right)\right\}={c}_{1}{f}_{1}\left(p\right)+{c}_{2}{f}_{2}\left(p\right)$

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$L\left\{f\left(t\right)\right\}=\frac{2}{{p}^{2}+{4}^{2}}$

Also,

Given,

We have to find

Taking Laplace transformation on both sides,

The required solution

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