# Find inverse laplace find y(t) for the following displaystyle{Y}{left({s}right)}=frac{8}{{s}}+frac{1}{{{s}-{3}}}cdot{e}^{{-{2}{s}}}

Question
Laplace transform
Find inverse laplace
find y(t) for the following
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}$$

2020-12-01
Step 1
Consider the given function,
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+{e}^{{-{2}{s}}}\cdot\frac{1}{{{s}-{3}}}$$
use the inverse transform rule which states that,
$$\displaystyle\text{if }\ {L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}= f{{\left({t}\right)}}\ \text{ then }\ {L}^{ -{{1}}}{\left\lbrace{e}^{{-{a}{s}}}{Y}{\left({s}\right)}\right\rbrace}={U}{\left({t}-{a}\right)} f{{\left({t}-{a}\right)}}$$
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{1}{{s}}\right\rbrace}={1}$$
\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{a}}}\right\rbrace}={e}^{{-{a}{t}}}\)
Step 2
by using the above rules the inverse Laplace transform of the function will be,
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{8}{{s}}+{e}^{{-{2}{s}}}\frac{1}{{{s}-{3}}}\right\rbrace}$$
$$\displaystyle f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace\frac{8}{{s}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace{e}^{{-{2}{s}}}\frac{1}{{{s}-{3}}}\right\rbrace}$$
$$\displaystyle={8}+{U}{\left({t}-{2}\right)}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{3}}}\right\rbrace}{\left({t}-{2}\right)}$$
$$\displaystyle={8}+{U}{\left({t}-{2}\right)}{e}^{{{2}{\left({t}-{2}\right)}}}$$
hence the inverse Laplace transform is $$\displaystyle f{{\left({t}\right)}}={8}+{U}{\left({t}-{2}\right)}{e}^{{{2}{\left({t}-{2}\right)}}}$$

### 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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$

The following quadratic function in general form, $$\displaystyle{S}{\left({t}\right)}={5.8}{t}^{2}—{81.2}{t}+{1200}$$ models the number of luxury home sales, S(t), in a major Canadian urban area, according to statistical data gathered over a 12 year period. Luxury home sales are defined in this market as sales of properties worth over \$3 Million (inflation adjusted). In this case, $$\displaystyle{\left\lbrace{t}\right\rbrace}={\left\lbrace{0}\right\rbrace}\ \text{represents}\ {2000}{\quad\text{and}\quad}{\left\lbrace{t}\right\rbrace}={\left\lbrace{11}\right\rbrace}$$represents 2011. Use a calculator to find the year when the smallest number of luxury home sales occurred. Without sketching the function, interpret the meaning of this function, on the given practical domain, in one well-expressed sentence.

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.
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) - ?
If the Laplace Transforms of fimetions $$y_1(t)=\int_0^\infty e^{-st}t^3dt , y_2(t)=\int_0^\infty e^{-st} \sin 2tdt , y_3(t)=\int_0^\infty e^{-st}e^t t^2dt$$ exist , then Which of the following is the value of $$L\left\{y_1(t)+y_2(t)+y_3(t)\right\}$$
$$a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
$$B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}$$
$$c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}$$
$$d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}$$
$$e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
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 = ?
$$f(t)=tu_2(t)$$
Ans. $$F(s)=\left(\frac{1}{s^2}+\frac{2}{s}\right)e^{-2s}$$
$$\displaystyle{F}{\left({s}\right)}=\frac{10}{{{s}{\left({s}^{2}+{9}\right)}}}$$
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$
$$\displaystyle f{{\left({t}\right)}}={2}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}$$