# use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=2+2(e^{-t}-1)u_1(t)

Question
Laplace transform
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$$f(t)=2+2(e^{-t}-1)u_1(t)$$

2020-12-29
$$\text{Step 1}$$
$$\text{Given, }f(t)=2+2(e^{-t}-1)u_1(t)$$
$$\text{To determine the Laplace Transform } L\left[f(t)\right]$$
$$\text{Step 2}$$
$$L\left[f(t)\right]=L\left[2+2(e^{-t}-1)u_1(t)\right]$$
$$=L\left[2\right]+L\left[2(e^{-t}-1)u_1(t)\right]$$
$$=2L\left[1\right]+2L\left[e^{-t}u_1(t)\right]-2L\left[u_1(t)\right] \text{ using linearity of Laplace Tranform, }$$
$$=2\frac{1}{s}+2e^{-s}L\left[e^{-t}\right]-2\frac{e^{-s}}{s}$$
$$=\frac{2}{s}+\frac{2e^{-s}}{s+1}-\frac{2e^{-s}}{s}$$

### Relevant Questions

use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $$f(t)=e^{3t}\cos5t-e^{-t}\sin2t$$
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$$f(t)=\frac{e^{-5t}}{\sqrt{t}}$$
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
$$f(t)=2(t-5)u_5(t)$$
use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $$f(t)=\int_0^t (t-w)\cos(2w)dw$$
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.
Use properties of the Laplace transform to answer the following
(a) If $$f(t)=(t+5)^2+t^2e^{5t}$$, find the Laplace transform,$$L[f(t)] = F(s)$$.
(b) If $$f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})$$, find the Laplace transform, $$L[f(t)] = F(s)$$. HINT:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$$
(c) If $$F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}$$ find the inverse Laplace transform, $$L^{-1}|F(s)| = f(t)$$
(d) If $$F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})$$ , find the inverse Laplace transform, $$L^{-1}[F(s)] = f(t)$$
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) - ?
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
a)$$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)$$
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$