Question

Laplace Transform of t ^{2}sin(2t)?

Answer 1

Apply the same trick we just used twice.

Answer 2

I have this table where I know the laplace transform of f(t)=t\[ t ^{n}*f(t)\] is \[ (-1)^n *F^{n}(s)\] where n=1,2,3

I'm just confused because the solution said that I can use that with n=2 OR n=1 to solve for the laplace transform above . n=2 makes sense since the function has a t^2 but I don't see why you can use n=1??

Answer 3

To be clear, that's not the n-th power of F, that's the n-th derivative of F.

Answer 4

Yeah I realized that earlier but why can you use n=1 because isn't the function t^2?

Answer 5

because maybe you know the Laplace transform of t.sin(at)

Answer 6

Yeah I do but where does the other t go?

Answer 7

sorry I'm new to this stuff that's why all the questions

Answer 8

You would use the case n = 1, if you knew F where
F(s) = L[t.sin 2t].

Then the Laplace transform you want
L[t^2.sin2t] = - F'(s)
= (-1)^1 F^(1) (s)

Answer 9

But I'm guessing you don't; I don't off the top of my head. But it doesn't matter of course because I do know
G(s) = L[sin 2t]
and then it's easy to calculate
L[t^2.sin 2t] = G''(s)

Answer 10

Or put another way, G'(s) = -L[t.sin 2t] = -F(s), hence G''(s) = -F'(s)

Answer 11

Ohhh I see that make sense it just makes the calculation easier!! Yeah I do know that because it's in the table he provides

Answer 12

thank you!!