# 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}

Question
Laplace transform
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$

2021-03-08
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: $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}=L^{-1}\left\{\frac{1}{(s-1)^2}\right\}-L^{-1}\left\{\frac{120}{(s+3)^6}\right\}$$
$$=e^{t}t-e^{-3t}t^5(\text{If } L^{-1}\left\{F(s)\right\}=f(t) \text{ then } L^{-1}\left\{F(s-a)\right\}=e^{at}f(t))$$

### Relevant Questions

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.
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer
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) - ?
find the inverse Laplace transform of the given function.
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$
Find the inverse Laplace transform of the given function by using the convolution theorem. $${F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}$$
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, $$L^{-1}[F(s)] = f(t)$$
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.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
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) $${\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}$$
$$2y'+y=0 , y(0)=-3$$
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}}}}$$