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Why do:

  1. $L(5t)=\frac{5}{s^2}$ and not $\frac{5}{s}$?

  2. $L(te^{-2t})=?$

I can work with laplace transformation table

2 Answers 2

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Since you say you can work with transform tables, rather than deriving from the definition I will use two common results from the tables.

Using $\mathcal{L}\{y^\prime\}=s\mathcal{L}\{y\}-y(0)$ let

  1. $y_1=5$
  2. $y_2=5t$

$\mathcal{L}\{y_1\prime\}=s\mathcal{L}\{y_1\}-y_1(0)$ therefore

$\mathcal{L}\{0\}=s\mathcal{L}\{5\}-5$ so $\mathcal{L}\{5\}=\dfrac{5}{s}$

$\mathcal{L}\{y_2^\prime\}=s\mathcal{L}\{y_2\}-y_2(0)$ therefore

$\mathcal{L}\{5\}=s\mathcal{L}\{5t\}-5(0)$. Thus

$\dfrac{5}{s}=s\mathcal{L}\{5t\}$ so

$\mathcal{L}\{5t\}=\dfrac{5}{s^2}$.

Using $\mathcal{L}\{ty\}=-\dfrac{d}{dy}\mathcal{L}\{y\}$

$\mathcal{L}\{te^{-2t}\}=-\dfrac{d}{dy}\left(\dfrac{1}{s+2}\right)=\dfrac{1}{(s+2)^2}$

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$$\mathcal{L}(5t)(s) = \int_0^{+\infty} 5t\ e^{-st}\ \text{d}t = \frac{5}{s^2}$$

$$\mathcal{L}(te^{-2t})(s) = \int_0^{+\infty} t\ e^{-2t}\ e^{-st}\ \text{d}t = \frac{1}{(s+2)^2}$$

Integrals are trivial. If you reached this level of theory, you can compute them without problems.

Important LT properties here

http://lpsa.swarthmore.edu/LaplaceZTable/LaplacePropTable.html