Laplace transform of $t^2 y''$

Question

So I need to take the Laplace transform of the following function:

$$ Eqn. 1: \\\frac{1}{2}\mathcal{L}{\{t^2 \ddot y\}} $$

I've found that

$$ Eqn. 2:\\ \mathcal{L}{\{t^{n} f(t)\}} = (-1)^{n}F^{n}(s) \,\,\, \text{for}\, n\, =\, 1,\, 2,\, 3, ... $$

So is Laplace associative? i.e.

$$ Eqn. 3:\\ \mathcal{L}{\{f_{1}(t)f_{2}(t)\}} = \mathcal{L}{\{f_{1}(t)\}} \cdot \mathcal{L}{\{f_{2}(t)\}} = F_{1}(s) \cdot F_{2}(s) $$

or is there already a transform for the thing I'm trying to transform?

Edit: I'm transforming Eqn. 1.

Eqn 2 and 3 are the way I think I would proceed to solve the problem. Sorry, original post wasn't quite as clear as it could have been.

Answer 1

$$ \mathcal{L}(t f(t)) = \int_0^{\infty} t e^{-st}f(t) dt = -\frac{d}{ds}\int_0^{\infty}e^{-st}f(t) dt = (-1)\frac{d}{ds}F(s) $$

so

$$ \mathcal{L}(t^n f(t)) = (-1)^n\frac{d^n}{ds^n}F(s) $$

then

$$ \mathcal{L}(t^n f''(t)) = (-1)^n\frac{d^n}{ds^n}\left(s^2F(s)-s f'(0)-f(0)\right) $$

Answer 2

No, just as the integral of $f(x)g(x)$ is not the product of the integrals of $f(x)$ and $g(x)$.

The correct form is a convoluton $$\mathcal{L}{\{f_{1}(t)f_{2}(t)\}} = \frac 1{2\pi i}\lim_{T \to \infty}\int_{c-iT}^{c+iT}F_1(\sigma)(F_2(s-\sigma)d\sigma$$