Find the Laplace transform F of the following functions and give the domain of F. f(x)=cos(2x)

kuCAu 2021-09-23 Answered
The Laplace transform of a continuous function over the interval \(\displaystyle{\left[{0},∞\right)}{\left[{0},∞\right)}\) is defined by \(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}.\)
This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of F is the set of all real numbers s such that the improper integral converges.
Find the Laplace transform F of the following functions and give the domain of F. \(\displaystyle{f{{\left({x}\right)}}}={\cos{{\left({2}{x}\right)}}}\)

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Expert Answer

yunitsiL
Answered 2021-09-24 Author has 12381 answers
Since the Laplace transform of a continious function over the interval \(\displaystyle{\left[{0},∞\right)}\) is defined by
\(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{x}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\infty}}{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\lim_{{{t}\to\infty}}{\int_{{0}}^{{t}}}{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}\)
Using integration by parts we get
\(\displaystyle\int{\cos{{\left({2}{x}\right)}}}{e}^{{{s}{x}}}{\left.{d}{x}\right.}=\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{r}}}{\left({2}{\sin{{\left({2}{x}\right)}}}-{s}{\cos{{2}}}{x}\right)}\)
Hence
\(\displaystyle{F}{\left({s}\right)}=\lim_{{{t}\to\infty}}{\left(\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{r}}}{\left({2}{\sin{{\left({2}{x}\right)}}}-{s}{\cos{{2}}}{x}\right)}\right)}\)
\(\displaystyle=\lim_{{{t}\to\infty}}{\left(\frac{{s}}{{{s}^{{2}}+{4}}}-\frac{{1}}{{{s}^{{2}}+{4}}}{e}^{{-{s}{t}}}{\left({2}{\sin{{\left({2}{t}\right)}}}-{t}{\cos{{\left({2}{t}\right)}}}\right)}\right)}\)
\(\displaystyle=\frac{{s}}{{{s}^{{2}}+{4}}}\)
Domain of F is \(\displaystyle{\left[{0},∞\right)}\)
Answer
\(\displaystyle{F}{\left({s}\right)}=\frac{{1}}{{s}}\)
Domain of F is \(\displaystyle{\left[{0},∞\right)}\)
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