Find Inverse Laplace of $\frac{e^{-2s}}{(s^{3})}$ and evaluate it $f(3)$.

Question

I was able to solve it by using second shift theorem which led my answer to be $y(t)=u(t-4)-(t-4)^2/2$ but how would I evaluate it for $f(3)$?

I am unsure on what to do with u.

I tried to use unit step function but I am still unclear on how to solve it.

The answer is supposed to be a numerical value but u in the equation is throwing me off.

Answer 1

$y(3) = u(-1) - (1/2)(-1)^2 = 0 - 1/2 = - \frac{1}{2}$

Edit, it seems like the real solution (as someone posted earlier was t-2 instead of t-4).

Therefore, the solution is: $y(3) = u(1) - (1/2)(1)^2 = 1 - 1/2 = \frac{1}{2}$

Answer 2

I get $\frac {e^{-2s}}{s^3} = \int_2^{\infty} \frac 12(t-2)^2e^{-st}\ dt$

$f(t) = \mathcal L^{-1}\{\frac {e^{-2s}}{s^3}\} = \frac 12u(t-2)(t-2)^2$

$f(3) = \frac 12$