Laplace transform of piecewise function with t*exp(2t) for t≥1

[Philip Wong]

Homework Statement

f(t) is a piecewise function:
{0 0<= t< 1
{t*exp(2t) t = >1

Homework Equations

F(s)= L{f(t)}

The Attempt at a Solution

F(s)= L{t*exp(2t)}

for this problem I just took the Laplace Transformer directly from the table which is: n!/ (s-a)^(n+1)

and after plucking in the relevant number I got:
1/(s-2)^2

Firstly I want to ask, is this correct? And have I missed any steps when setting up this problem?

Secondly if this is correct and I don't want to use the table to get the formula directly, can someone please show me other methods to come to the same answer.

thanks,
Phil

 

[HallsofIvy]

What you have is incorrect. That is the Laplace transform of $f(t)= te^{2t}$ which is NOT the same as the function given here.

Do not know the definition of the Laplace transform? It is
$$F(f)= \int_0^\infty e^{-st}f(t)dt$$

Integrate that.
The Laplace transform of this function is
$$\int_1^\infty e^{-st}te^{2t}dt= \int_1^\infty te^{(2- s)t}dt$$