$$\mathcal{L}(tf(t)) = -\dfrac {dF(s)}{ds}$$ If $y'' - 4y = 4H(t-2)$ then $$\mathcal {L}(4tH(t-2)) =-\dfrac {d(\mathcal {L}(y''- 4y))}{ds}$$
If there is a fault in this line of reasoning how could one solve this problem. Given that $y(0) = 0$ and $y''(0) = \sinh(-2)$