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)}

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)}

Question
Laplace transform
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 \(f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\)

Answers (1)

2020-11-09
Step 1
According to the given information, the given function is:
\(f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}\)
Step 2
For part (a) it is required to find \({L}\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\) by differentiating f(t) and then using Laplace transform.
So,
\(\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{\left(-{3}\right)}=-{12}{e}^{{-{3}{\left({t}-{4}\right)}}}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={L}{\left(-{12}{e}^{{-{3}{\left({t}-{4}\right)}}}\right)}={L}{\left(-{12}{e}^{{-{3}{t}}}{e}^{{{12}}}\right)}\)
\(=-{12}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}\right)}{\left[{L}{\left({e}^{{-{a}{t}}}\right)}=\frac{1}{{{s}+{a}}}\right]}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}=-{12}{e}^{{{12}}}{\left(\frac{1}{{{s}+{3}}}\right)}\)
Step 3
Now, it is required to find \({L}{\left({d}\frac{ f{{\left({t}\right)}}}{{\left.{d}{t}\right.}}\right)}\) using theorem for differentiation.
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={s}{L}{\left( f{{\left({t}\right)}}\right)}- f{{\left({0}\right)}}\)
\(={s}{\left({L}{\left({4}{e}^{{-{3}{\left({t}-{4}\right)}}}\right)}\right)}-{4}{e}^{{-{3}{\left({0}-{4}\right)}}}\)
\(={s}{\left({L}{\left({4}{e}^{{-{3}{t}}}{e}^{{{12}}}\right)}\right)}-{4}{e}^{{{12}}}\)
\(={4}{s}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}\right)}-{4}{e}^{{{12}}}\)
\(={4}{s}{e}^{{{12}}}{\left(\frac{1}{{{s}+{3}}}\right)}-{4}{e}^{{{12}}}\)
\(={4}{e}^{{{12}}}{\left(\frac{s}{{{s}+{3}}}-{1}\right)}\)
\(={4}{e}^{{{12}}}{\left(\frac{{{s}-{\left({s}+{3}\right)}}}{{{s}+{3}}}\right)}\)
\(={4}{e}^{{{12}}}{\left(-\frac{3}{{{s}+{3}}}\right)}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}=\frac{{-{12}{e}^{{{12}}}}}{{{s}+{3}}}\)
Step 4
Now, for part (c) it is required to repeat part (a) and (b) for the function:
\(f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\)
Step 5
Now, first find the derivative and then find the laplace transform.
\(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}={4}{\left[{e}^{{-{3}{\left({t}-{4}\right)}}}{u}'{\left({t}-{4}\right)}+{u}{\left({t}-{4}\right)}{e}^{{-{3}{\left({t}-{4}\right)}}}{\left(-{3}\right)}\right]}\)
\(={4}{e}^{{-{3}{t}}}{e}^{{{12}}}{\left[{u}'{\left({t}-{4}\right)}-{3}{u}{\left({t}-{4}\right)}\right]}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={4}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}{\left({u}'{\left({t}-{4}\right)}-{3}{u}{\left({t}-{4}\right)}\right)}\right)}\)
now , first find
\({L}{\left({u}'{\left({t}-{4}\right)}-{3}{u}{\left({t}-{4}\right)}\right)}={L}{\left({u}'{\left({t}-{4}\right)}\right)}-{3}{L}{\left({u}{\left({t}-{4}\right)}\right)}\)
\(={s}{L}{\left({u}{\left({t}-{4}\right)}\right)}-{u}{\left({0}-{4}\right)}-{3}{L}{\left({u}{\left({t}-{4}\right)}\right)}\)
\(={\left({s}-{3}\right)}{L}{\left({u}{\left({t}-{4}\right)}\right)}-{u}{\left(-{4}\right)}\)
but , \({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={4}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}{\left({u}'{\left({t}-{4}\right)}-{3}{u}{\left({t}-{4}\right)}\right)}\right)}\)
to obtain \({L}{\left({e}^{{-{3}{t}}}{\left({u}'{\left({t}-{4}\right)}-{3}{u}{\left({t}-{4}\right)}\right)}\right)}\)
use shifting theorem and put s+3 in place of s in
\(L(u'(t-4)-3u(t-4))\)
Step 6
Solving further to get: \({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={4}{e}^{{{12}}}{\left[{\left({s}+{3}-{3}\right)}{L}{\left({u}{\left({t}-{4}\right)}\right)}-{u}{\left(-{4}\right)}\right]}={4}{e}^{{{12}}}\)
\({\left[{s}{\left(\frac{{{e}^{{-{4}{s}}}}}{{{s}+{3}}}\right)}-{u}{\left(-{4}\right)}\right]}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}=\frac{{{4}{e}^{{{12}}}}}{{{s}+{3}}}{\left[{s}{e}^{{-{4}{s}}}-{\left({s}+{3}\right)}{u}{\left(-{4}\right)}\right]}\)
Step 7
Now, find using theorem for differentiation.
\(f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={s}{L}{\left( f{{\left({t}\right)}}\right)}- f{{\left({0}\right)}}\)
\(={s}{\left({L}{\left({4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}\)
\(={4}{s}{\left({L}{\left({e}^{{-{3}{t}}}{e}^{{{12}}}{u}{\left({t}-{4}\right)}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}\)
\(={4}{s}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}{u}{\left({t}-{4}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}\)
\(={4}{s}{e}^{{{12}}}{e}^{{-{4}{s}}}{\left(\frac{1}{{{s}+{3}}}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}\)
\({L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}=\frac{{{4}{e}^{{{12}}}}}{{{s}+{3}}}{\left[{s}{e}^{{-{4}{s}}}-{\left({s}+{3}\right)}{u}{\left(-{4}\right)}\right]}\)
0

Relevant Questions

asked 2021-06-06
Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.
a) \(x(t)=\cos(3t)\)
b)\(y(t)=t \cos(3t)\)
c) \(z(t)=e^{-2t}\left[t \cos (3t)\right]\)
d) \(x(t)=3 \cos(2t)+5 \sin(8t)\)
e) \(y(t)=t^3+3t^2\)
f) \(z(t)=t^4e^{-2t}\)
asked 2021-05-16
Find the Laplace transform of the function \(L\left\{f^{(9)}(t)\right\}\)
asked 2021-05-12
One property of Laplace transform can be expressed in terms of the inverse Laplace transform as \(L^{-1}\left\{\frac{d^nF}{ds^n}\right\}(t)=(-t)^n f(t)\) where \(f=L^{-1}\left\{F\right\}\). Use this equation to compute \(L^{-1}\left\{F\right\}\)
\(F(s)=\arctan \frac{23}{s}\)
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) - ?
asked 2021-05-01
Use Theorem 7.4.3 to find the Laplace transform F(s) of the given periodic function.
F(s)=?
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 2020-12-17
Solve by using Laplace Transform and explain the steps in brief
\(\displaystyle\frac{{{d}^{2}{x}}}{{{\left.{d}{t}\right.}^{2}}}+{2}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}+{x}={3}{t}{e}^{{-{t}}}\)
Given \(x(0)=4, x'(0)=2\)
asked 2021-02-21
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)\(f(t)=1+2t\) b)\(f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}\)
c)\(f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)\)
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 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) \(\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}\)
...