Question

Find the inverse Laplace Transform of {1/s^3(s^2+1)}

Answer 1

We shall find the inverse laplace transform analytically of

$\displaystyle\frac{1}{s^{3}(s^{2}+1)}$

We shall first resolve the expression by partial fraction

$\displaystyle\frac{1}{s^{3}(s^{2}+1)}=\frac{A}{s}+\frac{B}{s^{2}}+\frac{C}{s^{3}}+\frac{Ds+E}{s^{2}+1}$

$1\equiv As^{2}(s^{2}+1)+Bs(s^{2}+1)+C(s^{2}+1)+Ds^{4}+Es^{3}$

It follows that

$[s^{4}]$ $A+D=0$

$[s^{3}]$ $B+E=0$

$[s^{2}]$ $A+C=0$

$[s]$ $B=0 \implies E=0$

We have that $C=1\implies A=-1, D=1$

NOW, we have

$\displaystyle\frac{1}{s^{3}(s^{2}+1)}=\frac{-1}{s}+\frac{1}{s^{3}}+\frac{s}{s^{2}+1}$

$\mathcal{L}^{-1}\left\{\displaystyle\frac{1}{s^{3}(s^{2}+1}\right\}=\mathcal{L}^{-1}\left\{\frac{-1}{s}+\frac{1}{s^{3}}+\frac{s}{s^{2}+1}\right\}$

$=\mathcal{L}^{-1}\left\{\frac{-1}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{1}{s^{3}}\right\}+\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+1}\right\}$

$=-1+\frac{t^{2}}{2}+cos(t)$

$=cos(t)+\frac{t^{2}}{2}-1$

which is the inverse laplace transform of $\displaystyle\frac{1}{s^{3}(s^{2}+1)}$