[tex]f(s)\Longrightarrow L^{-1}=\{\frac{-4s-9}{s^2+25-8}\}[/tex]
First dismantle,
[tex]L^{-1}=\{-\frac{4s}{s^2+25-8}-\frac{9}{s^2+25-8}\}[/tex]
Now use the linearity property of Inverse Laplace Transform which states,
For functions [tex]f(s),g(s)[/tex] and constants [tex]a, b[/tex] rule applies,
[tex]L^{-1}=\{a\cdot f(s)+b\cdot g(s)\}=aL^{-1}\{f(s)\}+bL^{-1}\{f(s)\}[/tex]
Hence,
[tex]-4L^{-1}\{\frac{s}{s^2+25-8}\}-9L^{-1}\{\frac{1}{s^2+25-8}\}[/tex]
The first part simplifies to,
[tex]
L^{-1}\{\frac{s}{s^2+25-8}\} \\
\frac{d}{dt}(\frac{1}{\sqrt{17}}\sin(t\sqrt{17})) \\
\cos(t\sqrt{17})
[/tex]
The second part simplifies to,
[tex]
L^{-1}\{\frac{1}{s^2+25-8}\} \\
\frac{1}{\sqrt{17}}\sin(t\sqrt{17})
[/tex]
And we result with,
[tex]\boxed{-4\cos(t\sqrt{17})-\frac{9}{\sqrt{17}}\sin(t\sqrt{17})}[/tex]
Hope this helps.
If you have any additional questions please ask. I made process of solving as quick as possible therefore you might be left over with some uncertainty.
Hope this helps.
r3t40
