The first part is a simple question, but I cant find a clear answer.
Does:
$$\mathcal{L}(ax''(t)) = \mathcal{L}(a)\times\mathcal{L}(x''(t))$$
$a$ is a constant
$x(t)$ is a variable that changes with t
so
$$\mathcal{L}(a) = \frac{a}{s}$$ $$\mathcal{L}(x''(t))= s^2X(s)-sx(0)-x'(0)$$ $$\mathcal{L}(ax''(t)) = asX(s)-ax(0)-\frac{ax'(0)}{s}$$
Is this correct?
The alternative is that:
$$\mathcal{L}(ax''(t)) = a\times\mathcal{L}(x''(t))$$
so
$$\mathcal{L}(ax''(t)) = as^2X(s)-asx(0)-ax'(0)$$
If the latter is correct, why does the $a$ not get "laplaced"? The laplace transform of a constant number is that number divided by $s$, so why would the $a$ not get "laplaced" when it is multiplied by the variable $x(t)$?
Well, formally by definition: $\displaystyle\mathcal{L}_t\left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt $, then (since $a$ is constant) $$\displaystyle\mathcal{L}_t\left\{a \ f(t)\right\}=\int_0^{\infty} e^{-st} (a \ f(t)) \,dt= a\int_0^{\infty} e^{-st}f(t) \,dt=a \ \displaystyle\mathcal{L}_t\left\{f(t)\right\}.$$