 # Use properties of the Laplace transform to answer the following (a) If f(t)=(t+5)^2+t^2e^{5t}, find the Laplace transform,L[f(t)] = F(s). (b) If f(t) fortdefruitI 2021-02-25 Answered
Use properties of the Laplace transform to answer the following
(a) If $f\left(t\right)=\left(t+5{\right)}^{2}+{t}^{2}{e}^{5t}$, find the Laplace transform,$L\left[f\left(t\right)\right]=F\left(s\right)$.
(b) If $f\left(t\right)=2{e}^{-t}\mathrm{cos}\left(3t+\frac{\pi }{4}\right)$, find the Laplace transform, $L\left[f\left(t\right)\right]=F\left(s\right)$. HINT:
$\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)-\mathrm{sin}\left(\alpha \right)\mathrm{sin}\left(\beta \right)$
(c) If $F\left(s\right)=\frac{7{s}^{2}-37s+64}{s\left({s}^{2}-8s+16\right)}$ find the inverse Laplace transform, ${L}^{-1}|F\left(s\right)|=f\left(t\right)$
(d) If $F\left(s\right)={e}^{-7s}\left(\frac{1}{s}+\frac{s}{{s}^{2}+1}\right)$ , find the inverse Laplace transform, ${L}^{-1}\left[F\left(s\right)\right]=f\left(t\right)$
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Step 1
We use the results from table of Laplace transform.
$a\right)f\left(t\right)=\left(t+5{\right)}^{2}+{t}^{2}{e}^{5t}={t}^{2}+10t+25+{t}^{2}{e}^{5t}$
$⇒L\left\{f\left(t\right)\right\}=L\left\{{t}^{2}+10t+25+{t}^{2}{e}^{5t}\right\}=L\left\{{t}^{2}\right\}+10L\left\{t\right\}+L\left\{25\right\}+L\left\{{t}^{2}{e}^{5t}\right\}$
$F\left(s\right)=\frac{2}{{s}^{3}}+\frac{10}{{s}^{2}}+\frac{25}{s}+\frac{2}{\left(s-5{\right)}^{3}}$
$b\right)f\left(t\right)=2{e}^{-t}\mathrm{cos}\left(3t+\frac{\pi }{4}\right)⇒2{e}^{-t}\left[\mathrm{cos}\left(3t\right)\cdot \mathrm{cos}\left(\frac{\pi }{4}\right)-\mathrm{sin}\left(3t\right)\cdot \mathrm{sin}\left(\frac{\pi }{4}\right)\right]$
$⇒2{e}^{-t}\mathrm{cos}\left(3t\right)\cdot \mathrm{cos}\left(\frac{\pi }{4}\right)-2{e}^{-t}\mathrm{sin}\left(3t\right)\cdot \mathrm{sin}\left(\frac{\pi }{4}\right)$
$⇒L\left\{f\left(t\right)\right\}=2\mathrm{cos}\left(\frac{\pi }{4}\right)L\left\{{e}^{-t}\mathrm{cos}\left(3t\right)\right\}-2\mathrm{sin}\left(\frac{\pi }{4}\right)L\left\{{e}^{-t}\mathrm{sin}\left(t\right)\right\}$
we take constant term Laplace
$⇒F\left(s\right)=\frac{\sqrt{2}s}{\left(s+1{\right)}^{2}+9}-\frac{2}{\left(s+1{\right)}^{2}+9}$
$c\right)F\left(\frac{7{s}^{2}-37s+64}{s\left({s}^{2}-8s+16\right)}\right)=\frac{4}{s}+\frac{3}{\left(s-4\right)}+\frac{7}{\left(s-4{\right)}^{2}}$
(by partial fraction)
Then ${L}^{-1}\left\{F\left(s\right)\right\}={L}^{-1}\left\{\frac{4}{s}\right\}+{L}^{-1}\left\{\frac{3}{\left(s-4\right)}\right\}+{L}^{-1}\left\{\frac{7}{\left(s-4{\right)}^{2}}\right\}$
$f\left(t\right)=4+3{e}^{4t}+7t{e}^{4t}$
$d\right)F\left(s\right)={e}^{-7s}\left(\frac{1}{s}+\frac{s}{\left({s}^{2}+1\right)}\right)=\frac{{e}^{-7s}}{s}+\frac{s{e}^{-7s}}{\left({s}^{2}+7\right)}$
$⇒{L}^{-1}\left\{F\left(s\right)\right\}={L}^{-1}\left\{\frac{{e}^{-7s}}{s}\right\}+{L}^{-1}\left\{\frac{s{e}^{-7s}}{\left({s}^{2}+7\right)}\right\}$
$=u\left(t-7\right)+u\left(t-7\right)\mathrm{cos}\left(t-7\right)$
All the results teple of Laplace transform