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

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

2021-02-04
Step 1
The given function is $$tu(t-2)$$
To find the Laplace transform proceed as follows
$$L\left\{tu(t-2)\right\}$$ . The function $$u(t-2)$$ is unit step function .
Find the Laplace of $$u(t-2)$$
$$L\left\{u(t-2)\right\}=\frac{e^{-2s}}{s}$$
The function $$u(t-2)$$ is multiplied by t, use the formula for the multiplication by t^n use the formula $$L\left\{f(t)\right\}= L\left\{f(t)\right\}=(\bar f)(s) \text{ then } L\left\{t^nf(t)\right\}=(-1)^n \frac{d^n}{ds^n}(\bar f)(s)$$
here n =1
therefore, $$L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}$$
Step 2
Solve $$L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}$$
$$L\left\{tu(t-2)\right\}=(-1)\frac{\frac{sd}{ds}e^{-2s}-e^{-2s}\frac{d}{ds}s}{s^2}$$
$$=(-1)\left(\frac{s(-2)e^{-2s}-e^{-2s}(1)}{s^2}\right)$$
$$=\left(\frac{2se^{-2s}+e^{-2s}}{s^2}\right)$$
$$=e^{-2s}\left(\frac{2s+1}{s^2}\right) \cdot \frac{s}{s}$$
$$=e^{-2s}\left(\frac{2s^2+s}{s^3}\right)$$
The Laplace transform of the given function is $$e^{-2s}\left(\frac{2s^2+s}{s^3}\right)$$