Laplace Transforms of $(t2)u(t2)$ and $t u(t2)2u(t2)$?
Your solution is right. Note that the Laplace transform of $tu(t2)$ is not $\dfrac{e^{2s}}{s^2}$. Rather $$\mathcal{L}(t u(t2)) = \int_{2}^{\infty} t e^{st}dt = \frac{2s+1}{s^2}e^{2s}. \tag{1}$$ Using $(1)$, you can easily verify that $$\mathcal{L}((t2)u(t2)) = \mathcal{L}(tu(t2)2u(t2))=\frac{e^{2s}}{s^2}.\tag{2}$$
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most venerable sir
Updated on August 01, 2022Comments

most venerable sir over 1 year
Let $u(t)$ be the unitstep function; i.e., it is equal to $0$ for $t<0$, and $1$ for $t>0$. My question is why are not the Laplace transforms of $(t2)u(t2)$ and $t u(t2)  2u(t2)$ equal to each other?
I first calculated the Laplace transform of $(t2)u(t2)$. I used the formula for the Laplace transform of $f(ta)u(ta)$ is equal to $e^{as}F(s)$. Therefore, for $a=2$, and $f(t)=tu(t)$, whose $F(s)=\frac{1}{s^2}$, I get the Laplace transform of $(t2)u(t2)$ to be $\frac{e^{2s}}{s^2}$.
But my professor split it into $tu(t2)2u(t2)$ and got its Laplace tranform to be $e^{2s}(\frac{1}{s^2}\frac{2}{s})$.
I don't know who is wrong.