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)}

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
asked 2021-01-27
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) \({e}^{{-{2}{t}}}{u}{\left({t}\right)}\)

Answers (1)

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.
0

Relevant Questions

asked 2020-11-08
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 \(f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\)
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) \(\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}\)
asked 2021-03-20
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) \(\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}\)
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) \({\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}\)
asked 2020-11-02
Find the Laplace transform \(L\left\{u_3(t)(t^2-5t+6)\right\}\)
\(a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)\)
\(b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)\)
\(c) F(s)=e^{-3s}\frac{2+s}{s^4}\)
\(d) F(s)=e^{-3s}\frac{2+s}{s^3}\)
\(e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}\)
asked 2021-01-27
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 \(L\left\{y_1(t)+y_2(t)+y_3(t)\right\}\)
\(a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)
\(B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}\)
\(c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}\)
\(d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}\)
\(e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)
asked 2020-12-01
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) \(\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}\)
asked 2021-05-10
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) \(\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left\lbrace{2}{b}\right\rbrace}{\left\lbrace{5}\right\rbrace}{\left({r}^{{-\frac{{5}}{{2}}}}\right)}+{c}{r}\)
asked 2020-11-29
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 = ?
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) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}\)
...