Differential Equations Find the Laplace transform of t*e^(4t)*sin(9t)

The question is:

Find the Laplace transform of t*e^(4t)*sin(9t)


I know from the webwork answers that the answer is (18*(s-4))/((s-4)^2-81)^2. However, I have no idea how this answer was obtained. Could someone please walk me through it?

You first need to look at a Laplace transform table and observe that L{(-t)^n * f(t)} = F^n(s) where n F^n(s) is the nth derivative of the Laplace transform of f(t). In this case, if we re-wrote our function as
-(-te^(4t)*sin(9t)) and applied the Laplace transform:

L{-(-te^(4t)*sin(9t)} = -L{-te^(4t)*sin(9t)}

Comparing -te^(4t)*sin(9t) to (-t)^n*f(t), you'll see that n = 1 and f(t) = e^(4t)*sin(9t). So therefore F(s) = 9/((s - 4)^2 + 81) as per the rule that L{e^(at)*sin(bt)} = b/((s - a)^2 + b^2).

Now we just need to take the derivative of 9/((s - 4)^2 + 81) once to get F '(s) and we should be good.

So F '(s) = -18(s - 4)/((s - 4)^2 + 81)^2

Therefore, the Laplace transform of te^(4t)*sin(9t) is just -F '(s) = (18(s - 4))/((s - 4)^2 + 81)^2(That negative came from the negative we took out of the Laplace before.)

I hope this makes sense!

By the way, in your answer it should be ((s - 4)^2 + 81)^2 on the bottom instead of ((s - 4)^2 - 81)^2.