Did I Inverse Laplace correctly?

Question

$$L^{-1}\frac{4s}{(s-6)^{3}}$$ $$4L^{-1}\frac{s}{s^{3}}|s=s-6$$ $$4L^{-1}\frac{1}{s^{2}}|s=s-6$$ $$4L^{-1}\frac{1!}{s^{1+1}}|s=s-6$$ $$4te^{6t}$$

Is this correct? symbolab and Wolfram are giving me different answers...

Answer 1

If you don't know residues, you can rewrite your function as $$ \frac{4s}{(s-6)^3} = \frac{4(s-6)}{(s-6)^3} + \frac{24}{(s-6)^3} = \frac{4}{(s-6)^2} + \frac{24}{(s-6)^3} $$ and use the "translation rule" to invert the two terms separately.

Answer 2

The ILT may be computed via the residue theorem as follows:

$$\frac{4}{2!} \frac{d^2}{ds^2} \left [s e^{s t} \right ]_{s=6} = 2 \frac{d}{ds} \left [(1+s t) e^{s t} \right ]_{s=6} = 2 \left [(t (1+s t) + t) e^{s t} \right ]_{s=6} = (4 t+12 t^2) e^{6 t}$$

You seem to be missing the $t^2$ term.