find the inverse Laplace transform of the given function. {F}{left({s}right)}=frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}

Question
Laplace transform
find the inverse Laplace transform of the given function.
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$

2021-02-17
Step 1
Given function is:
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$
Since the Laplace transform of:
$${L}{\left( \sinh{{a}}{t}\right)}=\frac{a}{{{s}^{2}-{a}^{2}}}$$
And
$${L}{\left( \cosh{{a}}{t}\right)}=\frac{s}{{{s}^{2}-{a}^{2}}}$$
Step 2
Now,
Using the above transform:
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$
$$=\frac{{{2}{s}}}{{{s}^{2}-{4}}}-\frac{3}{{{s}^{2}-{4}}}$$
$$={2}\cdot\frac{s}{{{s}^{2}-{2}^{2}}}-\frac{3}{{2}}\cdot\frac{2}{{{s}^{2}-{2}^{2}}}$$
$$={2}{L}{\left( \cosh{{2}}{t}\right)}-\frac{3}{{2}}{L}{\left( \sinh{{2}}{t}\right)}$$
$$={L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}$$
inverse Laplace transform of $${L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}$$:
$${L}^{ -{{1}}}{\left({L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}\right)}={2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}$$
THEREFORE Inverse Laplace transform of given function is $${2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}$$

Relevant Questions

Explain why the function is discontinuous at the given number a. Sketch the graph of the function. $$f(x) = \left\{\frac{1}{x+2}\right\}$$ if $$x \neq -2$$ a= -2
1 if x = -2
Find the inverse Laplace transform of the given function by using the convolution theorem. $${F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
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 the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$
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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
determine the inverse Laplace transform of the function.
$$\displaystyle{Q}{\left({s}\right)}=\frac{s}{{{s}^{2}+{64}}}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
find the inverse Laplace transform of the given function
$$F(s)=\frac{e^{-2}+e^{-2s}-e^{-3s}-e^{-4s}}{s}$$
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 $$F(s)=\frac{(s+4)}{(s^2+9)}$$
a)$$\cos(t)+\frac{4}{3}\sin(t)$$
b)non of the above
c) $$\cos(3t)+\sin(3t)$$
d) $$\cos(3t)+\frac{4}{3} \sin(3t)$$
e)$$\cos(3t)+\frac{2}{3} \sin(3t)$$
f) $$\cos(t)+4\sin(t)$$
...