1. The problem is to compute the expression $l[e^{-2t}\sinh 4t]$. 2. Recall that the Laplace transform of $\sinh(at)$ is $\frac{a}{s^2 - a^2}$ for $s > |a|$. 3. Using the linearity and time-shifting properties of the Laplace transform, write $e^{-2t}\sinh(4t)$. 4. We can write the Laplace transform as $\mathcal{L}\{e^{-2t}\sinh(4t)\} = \int_0^\infty e^{-st} e^{-2t} \sinh(4t) dt = \int_0^\infty e^{-(s+2)t} \sinh(4t) dt$. 5. This is the Laplace transform of $\sinh(4t)$ evaluated at $s+2$. 6. So, $l[e^{-2t}\sinh(4t)] = \frac{4}{(s+2)^2 - 16}$. 7. Simplify the denominator: $(s+2)^2 - 16 = s^2 + 4s + 4 - 16 = s^2 + 4s - 12$. 8. Final answer: $$l[e^{-2t}\sinh(4t)] = \frac{4}{s^2 + 4s - 12}.$$