Prove {F}{left({s}right)}=frac{1}{{{s}-{2}}} text{ then } f{{left({t}right)}} is a) {e}^{{{2}{t}}}{u}{left({t}right)} b) u(t+2) c) u(t-2) d) {e}^{{-{2}{t}}}{u}{left({t}right)}

Question
Laplace transform
Prove $${F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}\ \text{ then }\ f{{\left({t}\right)}}$$ is
a) $${e}^{{{2}{t}}}{u}{\left({t}\right)}$$
b) $$u(t+2)$$
c) $$u(t-2)$$
d) $${e}^{{-{2}{t}}}{u}{\left({t}\right)}$$

2021-01-28
Step 1
Given $${F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}$$
We have to find the f(t)
Use definition of inverse Laplace transform, which is given below
$$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}\ldots{\left({1}\right)}$$
Step 2
Take inverse Laplace transform of $${F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}$$
Hence, $${L}^{ -{{1}}}{\left\lbrace{F}\right\rbrace}{\left({s}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{2}}}\right\rbrace}…{\left({2}\right)}$$
From equation (1) and equation (2)
$$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{2}}}\right\rbrace}$$
$$={e}^{{{2}{t}}}{u}{\left({t}\right)}$$
Therefore, $$f{{\left({t}\right)}}={e}^{{{2}{t}}}{u}{\left({t}\right)}$$
Hence, option (a) is correct.

Relevant Questions

Given that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}$$
a) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ using the theorem for differentiation
c) Repeat a) and b) for the case that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
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) $${\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
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 $$L\left\{y_1(t)+y_2(t)+y_3(t)\right\}$$
$$a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
$$B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}$$
$$c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}$$
$$d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}$$
$$e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
Hypothetical potential energy curve for aparticle of mass m
If the particle is released from rest at position r0, its speed atposition 2r0, is most nearly
a) $$\displaystyle{\left({\frac{{{8}{U}{o}}}{{{m}}}}\right)}^{{1}}{\left\lbrace/{2}\right\rbrace}$$
b) $$\displaystyle{\left({\frac{{{6}{U}{o}}}{{{m}}}}\right)}^{{\frac{{1}}{{2}}}}$$
c) $$\displaystyle{\left({\frac{{{4}{U}{o}}}{{{m}}}}\right)}^{{\frac{{1}}{{2}}}}$$
d) $$\displaystyle{\left({\frac{{{2}{U}{o}}}{{{m}}}}\right)}^{{\frac{{1}}{{2}}}}$$
e) $$\displaystyle{\left({\frac{{{U}{o}}}{{{m}}}}\right)}^{{\frac{{1}}{{2}}}}$$
if the potential energy function is given by
$$\displaystyle{U}{\left({r}\right)}={b}{r}^{{P}}-\frac{{3}}{{2}}\rbrace+{c}$$
where b and c are constants
which of the following is an edxpression of the force on theparticle?
1) $$\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left({r}^{{-\frac{{5}}{{2}}}}\right)}$$
2) $$\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left\lbrace{3}{b}\right\rbrace}{\left\lbrace{2}\right\rbrace}{\left({r}^{{-\frac{{1}}{{2}}}}\right)}$$
3) $$\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left\lbrace{3}\right\rbrace}{\left\lbrace{2}\right\rbrace}{\left({r}^{{-\frac{{1}}{{2}}}}\right)}$$
4) $$\displaystyle{2}{b}{\left({r}^{{-\frac{{1}}{{2}}}}\right)}+{c}{r}$$
5) $$\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left\lbrace{2}{b}\right\rbrace}{\left\lbrace{5}\right\rbrace}{\left({r}^{{-\frac{{5}}{{2}}}}\right)}+{c}{r}$$
Find the Laplace transform of $$\displaystyle f{{\left({t}\right)}}={t}{e}^{{-{t}}} \sin{{\left({2}{t}\right)}}$$
Then you obtain $$\displaystyle{F}{\left({s}\right)}=\frac{{{4}{s}+{a}}}{{\left({\left({s}+{1}\right)}^{2}+{4}\right)}^{2}}$$
$$2y'+y=0 , y(0)=-3$$
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}}}}$$