Question

L^-1{1/s^2 - 48/s^5}

Answer 1

please try to make your question clear .tq

Answer 2

\[\mathcal{L}^{-1}\{\frac{1}{s^2} - \frac{48}{s^5}\} = \mathcal{L}^{-1}\{\frac{1}{s^2}\} - \mathcal{L}^{-1}\{\frac{48}{s^5}\} = t - 48 \left[\mathcal{L}^{-1}\{\frac{1}{s^5}\}\right]\]
which is
\[t - 2t^4\]