two simple laplace transformation with simple functions

Question

this is a simple laplace transformation can someone please help me solve this I am stuck

Answer 1

Let's remind the definition of the Laplace transform: $$ \mathcal L(f)(s) = \int_0^{\infty} f(t) e^{-st} \, dt $$ So for example, we have with $f(t) = e^{-t} \cos(2t)$ that $$ \mathcal L(f)(s) = \int_0^{\infty} \cos(2t) e^{-(s+1)t} \, dt = \mathcal L(g)(s+1)) $$ with $g(t) = \cos(2t)$. You can either compute it yourself or lookup a table, but $$ \mathcal L(t \mapsto \cos(at))(s) = \frac s{s^2+a^2} $$ so $\mathcal L(g)(s) = \frac{s}{s^2+4}$. It follows that $$ \mathcal L(f)(s) = \mathcal L(g)(s+1) = \frac{s+1}{(s+1)^2 + 4} = \frac {s+1}{s^2 + 2s + 5}. $$ Similarly, you expand $(4e^{2t} - 2)^3$ as a polynomial in $e^t$ and rely on tables to compute the Laplace transform of the individual terms, and then rely on the linearity of the Laplace transform.

Hope that helps,