Respuesta :

[tex]\mathcal L^{-1}_t\left\{\dfrac1{s^2}-\dfrac{48}{s^5}\right\}=\mathcal L^{-1}_t\left\{\dfrac{1!}{s^2}\right\}-2\mathcal L^{-1}_t\left\{\dfrac{4!}{s^5}\right\}=t-2t^4[/tex]