# The Laplace transform of tu(t-2)=e^{-2s}left[frac{s+2s^2}{s^3}right] True or False?

Question
Laplace transform
The Laplace transform of $$tu(t-2)=e^{-2s}\left[\frac{s+2s^2}{s^3}\right]$$
True or False?

2021-02-04
Step 1
The given function is $$tu(t-2)$$
To find the Laplace transform proceed as follows
$$L\left\{tu(t-2)\right\}$$ . The function $$u(t-2)$$ is unit step function .
Find the Laplace of $$u(t-2)$$
$$L\left\{u(t-2)\right\}=\frac{e^{-2s}}{s}$$
The function $$u(t-2)$$ is multiplied by t, use the formula for the multiplication by t^n use the formula $$L\left\{f(t)\right\}= L\left\{f(t)\right\}=(\bar f)(s) \text{ then } L\left\{t^nf(t)\right\}=(-1)^n \frac{d^n}{ds^n}(\bar f)(s)$$
here n =1
therefore, $$L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}$$
Step 2
Solve $$L\left\{tu(t-2)\right\}=(-1)\frac{d}{ds}\frac{e^{-2s}}{s}$$
$$L\left\{tu(t-2)\right\}=(-1)\frac{\frac{sd}{ds}e^{-2s}-e^{-2s}\frac{d}{ds}s}{s^2}$$
$$=(-1)\left(\frac{s(-2)e^{-2s}-e^{-2s}(1)}{s^2}\right)$$
$$=\left(\frac{2se^{-2s}+e^{-2s}}{s^2}\right)$$
$$=e^{-2s}\left(\frac{2s+1}{s^2}\right) \cdot \frac{s}{s}$$
$$=e^{-2s}\left(\frac{2s^2+s}{s^3}\right)$$
The Laplace transform of the given function is $$e^{-2s}\left(\frac{2s^2+s}{s^3}\right)$$

### Relevant Questions

Find the laplace transform of the following
$$f(t)=tu_2(t)$$
Ans. $$F(s)=\left(\frac{1}{s^2}+\frac{2}{s}\right)e^{-2s}$$
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) $${\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}$$
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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
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) - ?
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 = ?
Find the laplace transform of the following:
Change of Scale
$$\text{If } L\left\{f(t)\right\}=\frac{s^2-s+1}{(2s+1)^2(s-2)} \text{ , find } L\left\{f(2t)\right\}$$
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.
$$2y'+y=0 , y(0)=-3$$
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}}}}$$
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$