# The Laplace transform of tu(t-2)=e^{-2s}left[frac{s+2s^2}{s^3}right] True or False?

The Laplace transform of $tu\left(t-2\right)={e}^{-2s}\left[\frac{s+2{s}^{2}}{{s}^{3}}\right]$
True or False?
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Step 1
The given function is $tu\left(t-2\right)$
To find the Laplace transform proceed as follows
$L\left\{tu\left(t-2\right)\right\}$ . The function $u\left(t-2\right)$ is unit step function .
Find the Laplace of $u\left(t-2\right)$
$L\left\{u\left(t-2\right)\right\}=\frac{{e}^{-2s}}{s}$
The function $u\left(t-2\right)$ is multiplied by t, use the formula for the multiplication by t^n use the formula
here n =1
therefore, $L\left\{tu\left(t-2\right)\right\}=\left(-1\right)\frac{d}{ds}\frac{{e}^{-2s}}{s}$
Step 2
Solve $L\left\{tu\left(t-2\right)\right\}=\left(-1\right)\frac{d}{ds}\frac{{e}^{-2s}}{s}$
$L\left\{tu\left(t-2\right)\right\}=\left(-1\right)\frac{\frac{sd}{ds}{e}^{-2s}-{e}^{-2s}\frac{d}{ds}s}{{s}^{2}}$
$=\left(-1\right)\left(\frac{s\left(-2\right){e}^{-2s}-{e}^{-2s}\left(1\right)}{{s}^{2}}\right)$
$=\left(\frac{2s{e}^{-2s}+{e}^{-2s}}{{s}^{2}}\right)$
$={e}^{-2s}\left(\frac{2s+1}{{s}^{2}}\right)\cdot \frac{s}{s}$
$={e}^{-2s}\left(\frac{2{s}^{2}+s}{{s}^{3}}\right)$
The Laplace transform of the given function is ${e}^{-2s}\left(\frac{2{s}^{2}+s}{{s}^{3}}\right)$
The answer is TRUE