# 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
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 $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$

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.
$$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
$${L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}={s}{L}{\left( f{{\left({t}\right)}}\right)}- f{{\left({0}\right)}}$$
$$={s}{\left({L}{\left({4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}$$
$$={4}{s}{\left({L}{\left({e}^{{-{3}{t}}}{e}^{{{12}}}{u}{\left({t}-{4}\right)}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}$$
$$={4}{s}{e}^{{{12}}}{L}{\left({e}^{{-{3}{t}}}{u}{\left({t}-{4}\right)}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}$$
$$={4}{s}{e}^{{{12}}}{e}^{{-{4}{s}}}{\left(\frac{1}{{{s}+{3}}}\right)}-{4}{e}^{{{12}}}{u}{\left(-{4}\right)}$$
$${L}{\left(\frac{{{d}{f}}}{{{\left.{d}{t}\right.}}}\right)}=\frac{{{4}{e}^{{{12}}}}}{{{s}+{3}}}{\left[{s}{e}^{{-{4}{s}}}-{\left({s}+{3}\right)}{u}{\left(-{4}\right)}\right]}$$

### Relevant Questions

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) $$z(t)=e^{-2t}\left[t \cos (3t)\right]$$
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f) $$z(t)=t^4e^{-2t}$$
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$$F(s)=\arctan \frac{23}{s}$$
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:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
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y(t) - ?
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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) $$\displaystyle{\frac{{{3}{b}}}{{{2}}}}{\left\lbrace{2}{b}\right\rbrace}{\left\lbrace{5}\right\rbrace}{\left({r}^{{-\frac{{5}}{{2}}}}\right)}+{c}{r}$$
Solve by using Laplace Transform and explain the steps in brief
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$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
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) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$