 # Let y(t)=int_0^tf(t)dt If the Laplace transform of y(t) is given Y(s)=frac{19}{(s^2+25)} , find f(t) a) f(t)=19 sin(5t) b) none c) f(t)=6 sin(2t) d) f(t)=20 cos(6t) e) f(t)=19 cos(5t) Armorikam 2021-01-15 Answered
Let $y\left(t\right)={\int }_{0}^{t}f\left(t\right)dt$ If the Laplace transform of y(t) is given $Y\left(s\right)=\frac{19}{\left({s}^{2}+25\right)}$ , find f(t)
a) $f\left(t\right)=19\mathrm{sin}\left(5t\right)$
b) none
c) $f\left(t\right)=6\mathrm{sin}\left(2t\right)$
d) $f\left(t\right)=20\mathrm{cos}\left(6t\right)$
e) $f\left(t\right)=19\mathrm{cos}\left(5t\right)$
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Step 1
The Laplace transform of y(t) is given $Y\left(s\right)=\frac{19}{\left({s}^{2}+25\right)}$
The Laplace transform is written as,
$L\left\{f\left(t\right)\right\}=F\left(s\right)$
$L\left[{\int }_{0}^{t}f\left(t\right)dt\right]=\frac{1}{s}F\left(s\right)$
It is given that , $y\left(t\right)={\int }_{0}^{t}f\left(t\right)dt$
Take Laplace on both sides,
$L\left\{y\left(t\right)\right\}=L\left\{{\int }_{0}^{t}f\left(t\right)dt\right\}$
Step 2
Substitute Laplace of y(t) ,
$\frac{1}{s}L\left\{f\left(t\right)\right\}=\frac{19}{\left({s}^{2}+25\right)}$
$Lf\left(t\right)=\frac{19s}{\left({s}^{2}+25\right)}$
Take inverse on both sides,
$f\left(t\right)={L}^{-1}\left[\frac{19}{\left({s}^{2}+25\right)}\right]$
$=19{L}^{-1}\left[\frac{s}{\left({s}^{2}+25\right)}\right]$
The inverse Laplace transform of $\mathrm{cos}\left(at\right)$ is ${L}^{-1}\left[\frac{s}{\left({s}^{2}+{a}^{2}\right)}\right]$
Step 3
Therefore ,
$f\left(t\right)={L}^{-1}\left[\frac{19s}{\left({s}^{2}+25\right)}\right]$
$=19{L}^{-1}\left[\frac{s}{\left({s}^{2}+25\right)}\right]$
$=19\mathrm{cos}5t$
The value of $f\left(t\right)=19\mathrm{cos}\left(5t\right)$ and the correct option is E.