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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 [tex]F(s)[/tex] and [tex]G(s)[/tex], then [tex]\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}[/tex]
Given that [tex]\dfrac{1}{s^2} = \dfrac{1!}{s^2}[/tex] and [tex]-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}[/tex]
So you have [tex]\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\}[/tex]
From Table of Laplace Transform, you have [tex]\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}[/tex] and hence [tex]\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n[/tex]
So you have [tex]\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}[/tex].
Hope this helps...
