Someone can explain step by step how i can manage this P(s) in order to apply the inverse laplace transform ?
$ P(s) = \frac{s^3+5s^2+3}{s^2(s^2-3s-18)}$
I tried this way of fractioning but it doesnt work:
$\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+3} +\frac{D}{s-6} = \frac{s^3+5s^2+3}{s^2(s^2-3s-18)}$
Why do you say it doesn't work? It does. Since\begin{multline}\frac As+\frac B{s^2}+\frac C{s+3}+\frac D{s-6}=\\=\frac{A s^3-3 A s^2-18 A s+B s^2-3 B s-18 B+C s^3-6 C s^2+D s^3+3 D s^2}{s^2(s^2-3s-18)},\end{multline}you should solve the system$$\left\{\begin{array}{l}-18 B=3\\-18 A - 3 B=0\\-3 A + B - 6 C + 3 D=5\\A + C + D=1.\end{array}\right.$$You will get that $A=\frac1{36}$, that $B=-\frac16$, that $C=-\frac7{27}$, and that $D=\frac{133}{108}$.