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I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform \(F(s)\) and \(G(s)\), then \(\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}\)
Given that \(\dfrac{1}{s^2} = \dfrac{1!}{s^2}\) and \(-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}\)
So you have \(\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}\)
From Table of Laplace Transform, you have \(\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}\) and hence \(\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n\)
So you have \(\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}\).
Hope this helps...
