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Question
Laplace transform
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer

2020-12-28
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 $$-\frac{2}{11}e^{3t}+\frac{13}{11}e^{-8t}$$

### Relevant Questions

The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
d) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$
Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
Use properties of the Laplace transform to answer the following
(a) If $$f(t)=(t+5)^2+t^2e^{5t}$$, find the Laplace transform,$$L[f(t)] = F(s)$$.
(b) If $$f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})$$, find the Laplace transform, $$L[f(t)] = F(s)$$. HINT:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$$
(c) If $$F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}$$ find the inverse Laplace transform, $$L^{-1}|F(s)| = f(t)$$
(d) If $$F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})$$ , find the inverse Laplace transform, $$L^{-1}[F(s)] = f(t)$$
The Laplace inverse of $$L^{-1}\left[\frac{s}{s^2+5^2}\right]$$ is
$$a) \cos(5t)$$
$$b) \sin h(5t)$$
$$c) \sin(5t)$$
$$d) \cos h(5t)$$
Find the Laplace transform of $$\displaystyle f{{\left({t}\right)}}={t}{e}^{{-{t}}} \sin{{\left({2}{t}\right)}}$$
Then you obtain $$\displaystyle{F}{\left({s}\right)}=\frac{{{4}{s}+{a}}}{{\left({\left({s}+{1}\right)}^{2}+{4}\right)}^{2}}$$
Please type in a = ?
Find the inverse Laplace transform of the given function by using the convolution theorem. $${F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
$$L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1s}\right\}$$
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$