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}

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asked 2020-12-25

\(L\left\{t-e^{-3t}\right\}\]
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asked 2021-02-16

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asked 2021-02-02

Obtain the Laplace Transform of
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asked 2021-02-08

Use the Laplace transform to solve the following initial value problem:
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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 \(L\left\{y_1(t)+y_2(t)+y_3(t)\right\}\)
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\(B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}\)
\(c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}\)
\(d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}\)
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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:
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a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
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y(t) - ?

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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.

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Answer 1

\(\text{Step 1}\)
\(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 }
\(L\left\{e^{3t}\sin(t)-t^{4}+e^{t}\right\}=\frac{4}{(s-3)^{2}+16}-\frac{24}{s^{5}}+\frac{1}{s-1}\)

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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}

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