use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=2(t-5)u_5(t)

Question
Laplace transform
asked 2021-01-13
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
\(f(t)=2(t-5)u_5(t)\)

Answers (1)

2021-01-14
\(\text{Step 1}\)
\(\text{Given: }\)
\(f(t)=2(t-5)u_5(t)\)
\(L\left[f(t)\right]-?\)
\(\text{Step 2}\)
\(\text{Formula : } L\left[f(t-c)u_c(t)\right]=e^{-cs}L\left[f(t)\right]\)
\(\text{Step 3}\)
\(\text{Solution : } f(t)=2(t-5)u_5(t)\)
\(\text{Take Laplace Transform to both sides: }\)
\(L\left[f(t)\right]=L\left[2(t-5)u_5(t)\right]\)
\(L\left[f(t)\right]=2L\left[(t-5)u_5(t)\right]\)
\(L\left[f(t)\right]=2e^{-5s}L(t) \Rightarrow \left[L\left[f(t-c)u_c(t)\right]=e^{-cs}L\left[f(t)\right]\right]\)
\(L\left[f(t)\right]=2e^{-5s}\bigg(\frac{1!}{s^{1+1}}\bigg) \Rightarrow \left[L(t^n)=\frac{n!}{s^{n+1}}\right]\)
\(L\left[f(t)\right]=2e^{-5s}\bigg(\frac{1}{s^2}\bigg)\)
\(\begin{array}{|c|c|} \hline L\left[f(t)\right]=2\frac{e^{-5s}}{s^2} \\ \hline \end{array}\)
\(\text{Step 4}\)
\(\text{Answer : } L\left[f(t)\right]=2\frac{e^{-5s}}{s^2}\)
0

Relevant Questions

asked 2020-12-28
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
\(f(t)=2+2(e^{-t}-1)u_1(t)\)
asked 2021-03-04
use properties of the Laplace transform and the table of Laplace transforms to determine L[f] \(f(t)=e^{3t}\cos5t-e^{-t}\sin2t\)
asked 2021-03-02
use properties of the Laplace transform and the table of Laplace transforms to determine L[f] \(f(t)=\int_0^t (t-w)\cos(2w)dw\)
asked 2020-12-30
use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
\(f(t)=\frac{e^{-5t}}{\sqrt{t}}\)
asked 2020-11-01
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.
asked 2021-02-25
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, \(L^{-1}[F(s)] = f(t)\)
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 2020-12-12
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform
\(L\left\{3e^{-4t}-t^{2}+6t-9\right\}\)
asked 2021-03-07
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. \(L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}\)
asked 2021-03-02
\(\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }\)
\(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:}
\(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}\)
\(\text{ Verify the following Laplace transforms, where u is a real number. }\)
\(f(t)=t \rightarrow F(s)=\frac{1}{s^2}\)
...