Q7Advanced Engineering Mathematics I

Let $F(s)=\frac{s+2}{(s-2{)}^3}$

We rewrite the numerator in terms of $(s-2)$: $s+2=(s-2)+4$

So, $F(s)=\frac{(s-2)+4}{(s-2{)}^3}=\frac{1}{(s-2{)}^2}+\frac{4}{(s-2{)}^3}$

Using the First Shifting Theorem $L^{-1}\{F(s-a)\}=e^{at}f(t)$: $L^{-1}\{F(s)\}=e^{2t}{L}^{-1}\left\{\frac{1}{s^2}\right\}+4e^{2t}{L}^{-1}\left\{\frac{1}{s^3}\right\}$

Since $L^{-1}\{1/s^n\}=\frac{t^{n-1}}{(n-1)!}$: $f(t)=e^{2t}(t)+4e^{2t}\left(\frac{t^2}{2!}\right)=e^{2t}(t+2t^2)$