The original question was to find $\mathcal{L}{g(t)}$ from
$$g(t) = \begin{cases} t, 0 \le t < 1 \\ 2-t , 1 \le t \le 2 \\ 0 , t>2 \end{cases} $$
I formed a Heaviside unit step function to solve it,
Here it is -
$g(t) = t [ g(t) - g(t-1) ] + (2-t) [g(t-1) - g(t-2)] + 0 [ g(t-2) - g(t- \infty) ] $
I ended up getting -
$g(t) = t g(t) - 2(t-1)g(t-1) - (2-t)g(t-2) $
Now I have to take Laplace transform of the entire thing, I have to use the formula of $\mathcal{L}{U(t-a)f(t-a)} = e^{-as} F(s)$ to solve the 2nd and 3rd one.
But what about $\mathcal{L}{tg(t)} $ I am not too sure. Can I bring out the t ? For example - $t \mathcal {L}{g(t)} $ I think I $t$ is a variable that I need to transform and it’s not a number (when I can do that).
I searched online and found this formula -
$U(t)f(t) = \mathcal{L}{f(t)} $
Do I have to use this then ? Is this the correct formula that I need to use ?
so... $\mathcal {L}{tg(t)} = \mathcal{L}{t} = \frac {1}{S^2} $ ?
There are various ways to write the Laplace transform you need. It is most commonly written
$$ \mathcal{L}\{f(t-a)U(t-a)\}=\mathcal{L}\{f(t)\}e^{-as}\tag{1} $$
An alternate version is
$$ \mathcal{L}\{f(t)U(t-a)\}=\mathcal{L}\{f(t+a)\}e^{-as}\tag{2} $$
Using the second rule, for example one has
\begin{eqnarray} \mathcal{L}\{(t-2)U(t-2)\}&=&\mathcal{L}\{(t+2)-2\}e^{-2s}\\ &=&\mathcal{L}\{t\}e^{-2s}\\ &=&\frac{e^{-2s}}{s^2} \end{eqnarray}