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HallsofIvy
#9
May10-09, 06:25 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,524
The point is that all three methods:
a)[tex]\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),[/tex]
where F(s) is the Laplace transform of t.

b)[tex]\displaystyle \mathcal{L}[tf(t)]= -F'(s)[/tex]
where F(s) is the Laplace transform of [itex]e^{3t}[/itex].

c)[tex]\displaystyle \mathcal{L}[te^{at}]= \int_0^\infty te^{(-s+3)t}dt[/tex]

will give the same result.

It would be a good exercise to try each method and see.