Find the laplace transform of the following f(t)=tu_2(t) Ans. F(s)=left(frac{1}{s^2}+frac{2}{s}right)e^{-2s}

Find the laplace transform of the following
$f\left(t\right)=t{u}_{2}\left(t\right)$
Ans. $F\left(s\right)=\left(\frac{1}{{s}^{2}}+\frac{2}{s}\right){e}^{-2s}$
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Step 1
first calculate Laplace transform of ${u}_{2}\left(t\right)$
${F}_{u}\left(s\right)={\int }_{-}{\mathrm{\infty }}^{\mathrm{\infty }}{u}_{2}\left(t\right){e}^{-st}dt$
$={\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt$
$=\frac{{e}^{-2s}}{s}$
Step 2
therefore, Laplace transform of $t{u}_{2}\left(t\right)$ is :
$-\frac{d}{ds}{F}_{u}\left(s\right)=-\frac{d}{ds}\cdot \frac{{e}^{-2s}}{s}$
$=-\frac{-2s{e}^{-2s}-{e}^{-2s}}{{s}^{2}}$
$=\left(\frac{1}{{s}^{2}}+\frac{2}{s}\right){e}^{-2s}$
Therefore, Laplace transform of $t{u}_{2}\left(t\right)$ is
$=\left(\frac{1}{{s}^{2}}+\frac{2}{s}\right){e}^{-2s\right)}$