Question

Do the laplace transform of L\left\{t-e^{3t}\right\}

Laplace transform
ANSWERED
asked 2021-09-19

Do the laplace transform of \(L\left\{t-e^{3t}\right\}\)

Expert Answers (1)

2021-09-20

Step 1
\(L\left\{t-e^{3t}\right\}\)
We know that the Laplace transformation of
\(\displaystyle{L}{\left({t}\right)}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}\cdot{t}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}^{{2}}}}}\)
\(\displaystyle{L}{\left({e}^{{{a}{t}}}\right)}={\int_{{0}}^{{\infty}}}{e}^{{-{s}{t}}}\cdot{e}^{{{a}{t}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{s}-{a}}}},{s}{>}{a}\)
So \(\displaystyle\rightarrow{L}{\left({t}-{e}^{{{3}{t}}}\right)}\)
\(\displaystyle\Rightarrow{L}{\left({t}\right)}-{L}{\left({e}^{{{3}{t}}}\right)}\) by the defination of laplace transformation
\(\displaystyle\Rightarrow{\frac{{{1}}}{{{s}^{{2}}}}}-{\frac{{{1}}}{{{s}-{3}}}}\)
\(\displaystyle\Rightarrow{\frac{{{\left({s}-{3}\right)}-{s}^{{2}}}}{{{s}^{{2}}{\left({s}-{3}\right)}}}}\)
\(\displaystyle\Rightarrow{L}{\left({t}-{e}^{{{3}{t}}}\right)}={\frac{{{\left({s}-{3}\right)}-{s}^{{2}}}}{{{s}^{{2}}{\left({s}-{3}\right)}}}}\)

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