Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform. Lleft{e^{3t}sin(4t)-t^{4}+e^{t}right}

Question
Laplace transform
asked 2021-02-08
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
\(L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\}\)

Answers (1)

2021-02-09
\(\text{Step 1}\)
\(L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\}
\(=L\left\{e^{3t}\sin(4t)\right\}-L\left\{t^{4}\right\}+L\left\{e^{t}\right\}\ \ \ \ \ \ \text{1)}
\(\text{we know that } L\left\{t^{n}\right\}=\frac{n!}{s^{n+1}}
\(L\left\{e^{at}\right\}=\frac{1}{s-a}
\(L\left\{\sin(at)\right\}=\frac{a}{s^{2}+a^{2}}
\(L\left\{\sin(4t)\right\}=\frac{4}{s^{2}+16}:
\(\text{by first shifting } L\left\{e^{at}\sin(at)\right\}=\frac{b}{(s-a)^{2}+b^{2}}
\(L\left\{e^{3t}\sin(4t)\right\}=\frac{4}{(s-3)^{2}+16} ,
\(L\left\{t^{4}\right\}=\frac{4!}{s^{4+1}}=\frac{24}{s^{5}}
\(L\left\{e^{t}\right\}=\frac{1}{s-1}
\(\text{Step 2}
\(\text{Then 1) will becomes as }
\(L\left\{e^{3t}\sin(t)-t^{4}+e^{t}\right\}=\frac{4}{(s-3)^{2}+16}-\frac{24}{s^{5}}+\frac{1}{s-1}\)
0

Relevant Questions

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 2021-02-25
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 2021-03-04
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 2020-12-25
\(L\left\{t-e^{-3t}\right\}\]
which of the laplace transform is
\(1.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s-3}\)
\(2.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s-3}\)
\(3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}\)
\(4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}\)
asked 2021-02-16
\(\text{Find the Laplace transform }\ F(s)=L\left\{f(t)\right\}\ \text{of the function }\ f(t)=6+\sin(3t) \ \text{defined on the interval }\ t\geq0\)
asked 2021-02-02
Obtain the Laplace Transform of
\(L\left\{t^3-t^2+4t\right\}\)
\(L\left\{3e^{4t}-e^{-2t}\right\}\)
asked 2021-02-08
Use the Laplace transform to solve the following initial value problem:
\(2y"+4y'+17y=3\cos(2t)\)
\(y(0)=y'(0)=0\)
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for \(L\left\{y(t)\right\}\) b) Express the solution \(y(t)\) in terms of a convolution integral
asked 2021-01-27
If the Laplace Transforms of fimetions \(y_1(t)=\int_0^\infty e^{-st}t^3dt , y_2(t)=\int_0^\infty e^{-st} \sin 2tdt , y_3(t)=\int_0^\infty e^{-st}e^t t^2dt\) exist , then Which of the following is the value of \(L\left\{y_1(t)+y_2(t)+y_3(t)\right\}\)
\(a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)
\(B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}\)
\(c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}\)
\(d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}\)
\(e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}\)
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 2020-11-01
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.
...