Laplace transform of a product

Question

I tried to solve the product below:

$$3t\sin(6t)$$

but it seems that getting the transform of each and multiply the result is not leading to a correct answer:

$$\frac{3}{s^2}\frac{6}{s^2+36}$$

How does one solve such transforms?

Answer 1

We know that if $ L(f(t))=F(s)$ so $ L(t.f(t))=-F’(s)$ in which $F’(s)=\frac{dF}{ds}$. Here you need just to derivative the second part of the last formula above with respect to $s$ and then multiply the result by $3$.

Answer 2

I think you can use the following formula. $$\mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds} F(s)$$