Step 1

The following formulae are used in solving the given problem.

\(L^{-1}{f(x)+g(x)}=L^{-1}(f(x))+L^{-1}(g(x))\)

\(L^{-1}(a \cdot f(x))=a L^{-1}(f(x))\)

\(L^{-1}\left(\frac{1}{s+a}\right)=e^{-at}\)

Step 2

Evaluate the inverse Laplace transform of the given function as follows.

\(L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}=L^{-1}\left\{-\frac{2}{11}(s-3)+\frac{13}{11}(s+8)\right\} (\text{ by partial fractions })\)

\(L^{-1}\left\{-\frac{2}{11(s-3)}\right\}+L^{-1}\left\{\frac{13}{11(s+8)}\right\}\)

\(-\frac{2}{11}L^{-1}\left\{1(s-3)\right\}+\frac{13}{11}L^{-1}\left\{\frac{1}{s+8}\right\}\)

\(-\frac{2}{11}e^{3t}+\frac{13}{11}e^{-8t}\)

Therefore, the inverse Laplace transform of the given function is \(-\frac{2}{11}e^{3t}+\frac{13}{11}e^{-8t}\)

The following formulae are used in solving the given problem.

\(L^{-1}{f(x)+g(x)}=L^{-1}(f(x))+L^{-1}(g(x))\)

\(L^{-1}(a \cdot f(x))=a L^{-1}(f(x))\)

\(L^{-1}\left(\frac{1}{s+a}\right)=e^{-at}\)

Step 2

Evaluate the inverse Laplace transform of the given function as follows.

\(L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}=L^{-1}\left\{-\frac{2}{11}(s-3)+\frac{13}{11}(s+8)\right\} (\text{ by partial fractions })\)

\(L^{-1}\left\{-\frac{2}{11(s-3)}\right\}+L^{-1}\left\{\frac{13}{11(s+8)}\right\}\)

\(-\frac{2}{11}L^{-1}\left\{1(s-3)\right\}+\frac{13}{11}L^{-1}\left\{\frac{1}{s+8}\right\}\)

\(-\frac{2}{11}e^{3t}+\frac{13}{11}e^{-8t}\)

Therefore, the inverse Laplace transform of the given function is \(-\frac{2}{11}e^{3t}+\frac{13}{11}e^{-8t}\)