Answer to Question #232318 in Math for abi

Question

L^-1{ e^-2s / s³}

Answer 1

$$L^{-1}\{\dfrac{e^{-2s}}{s^3}\}=H(t-2)L^{-1}\{\dfrac{1}{s^3}\}(t-2)$$

$$=H(t-2)\dfrac{(t-2)^2}{2}=\dfrac{H(t-2)\cdot(t-2)^2}{2}$$