Question

Evaluate L^-1 {(s^2 - 29s + 5) / (s-4)^ (s^2 +3)} inverse laplace transform

Answer 1

Is this what you want to find the inverse Lalplace Transform of
\[
\frac{s^2-29 s+5}{(s-4)^2 \left(s^2+3\right)}
\]
I will suppose that that is it, since your are not online

Answer 2

First use partial fraction
\[
\frac{s^2-29 s+5}{(s-4)^2
\left(s^2+3\right)}=\frac{2-s}{s^2+3}+\frac{1}{s-4}-\frac{5}{(s-4)^2}=\\
\frac{2}{s^2+3}- \frac{s}{s^2+3}+\frac{1}{s-4}-\frac{5}{(s-4)^2}
\]

Answer 3

\[
\mathcal{L}_s^{-1}\left[\frac{2}{s^2+3}\right](t)=\frac{2 \sin
\left(\sqrt{3} t\right)}{\sqrt{3}}
\]

Answer 4

\[
\mathcal{L}_s^{-1}\left[-\frac{s}{s^2+3}\right](t)=-\cos \left(\sqrt{3}
t\right)
\]

Answer 5

\[
\mathcal{L}_s^{-1}\left[\frac{1}{s-4}\right](t)=e^{4 t}
\]

Answer 6

\[
\mathcal{L}_s^{-1}\left[-\frac{5}{(s-4)^2}\right](t)=-5 e^{4 t} t
\]

Answer 7

good work elias, but where's nackulen?

Answer 8

yeap that's the one..(referring to your first reply.)

Answer 9

Did you understand the details?

Answer 10

yes i do now.
thank you so much.

Answer 11

YW