Question

Inverse laplace of [(1/s^2)-(48/s^5)]

Answer 1

`\mathcal L^(-1)_t\left\{\frac1{s^2}-(48)/(s^5)\right\}=\mathcal L^(-1)_t\left\{(1!)/(s^2)\right\}-2\mathcal L^(-1)_t\left\{(4!)/(s^5)\right\}=t-2t^4`