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}

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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}

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asked 2020-11-01

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.

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asked 2020-12-27

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

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asked 2021-02-09

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative \(\frac{dy}{dt}\) also appears. Consider the following initial value problem, defined for t > 0:
\(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}\)
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
\({Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?\)
b) Obtain the solution y(t).
y(t) - ?

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asked 2021-02-16

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

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asked 2021-01-13

Find the inverse Laplace transform of the given function by using the convolution theorem. \({F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}\)

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asked 2021-02-25

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, \(L^{-1}[F(s)] = f(t)\)

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asked 2021-01-05

Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
\({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\)

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asked 2020-10-25

Please provide steps
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) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)

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asked 2021-01-15

The Laplace transform of the function \({\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\) is equal to:
a) \({e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right)}\)
b) \({e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\cdot\frac{3}{{{s}-\frac{1}{{3}}}}\right)}\)
c) \({e}^{{\frac{2}{{3}}}}{\left(\frac{6}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\right)}\)
d) \({\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}\)

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asked 2021-01-30

Use Laplace transform to find the solution of the IVP
\(2y'+y=0 , y(0)=-3\)
a) \(f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}\)
b)\(f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}\)
c)\(f{{\left({t}\right)}}={6}{e}^{{{2}{t}}}
d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}\)

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Answer 1

Step 1 It is given that, \(L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}\) Step 2 Obtain the inverse Laplace transform as follows: \(L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}=L^{-1}\left\{\frac{1}{(s-1)^2}\right\}-L^{-1}\left\{\frac{120}{(s+3)^6}\right\}\)
\(=e^{t}t-e^{-3t}t^5(\text{If } L^{-1}\left\{F(s)\right\}=f(t) \text{ then } L^{-1}\left\{F(s-a)\right\}=e^{at}f(t))\)

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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}

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}

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