Question

The Laplace transform of the function {left({2}{t}-{3}right)}{e}^{{frac{{{t}+{2}}}{{3}}}} is equal to: a) {e}^{{frac{2}{{3}}}}{left(frac{2}{{left({s}-

Laplace transform
ANSWERED
asked 2021-01-15
The Laplace transform of the function \({\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\) is equal to:
a) \({e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right)}\)
b) \({e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\cdot\frac{3}{{{s}-\frac{1}{{3}}}}\right)}\)
c) \({e}^{{\frac{2}{{3}}}}{\left(\frac{6}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\right)}\)
d) \({\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}\)

Answers (1)

2021-01-16
Step 1
Given that
\({\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\)
To find the Laplace transformation of given function
Step 2
Given
\({\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\)
\(={2}{t}{e}^{{\frac{{{t}+{2}}}{{3}}}}-{3}{e}^{{\frac{{{t}+{2}}}{{3}}}}\)
Apply Laplace transform on both side
\({L}{\left\lbrace{\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\right\rbrace}={L}{\left\lbrace{2}{t}{e}^{{\frac{{{t}+{2}}}{{3}}}}-{3}{e}^{{\frac{{{t}+{2}}}{{3}}}}\right\rbrace}\)
\(={L}{\left\lbrace{2}{t}{e}^{{\frac{{{t}+{2}}}{{3}}}}\right\rbrace}-{L}{\left\lbrace{3}{e}^{{\frac{{{t}+{2}}}{{3}}}}\right\rbrace}\)
\(={2}{L}{\left\lbrace{t}{e}^{{\frac{t}{{3}}}}\cdot{e}^{{\frac{2}{{3}}}}\right\rbrace}-{L}{\left\lbrace{3}{e}^{{\frac{t}{{3}}}}\cdot{e}^{{\frac{2}{{3}}}}\right\rbrace}\)
\(={2}{e}^{{\frac{2}{{3}}}}{L}{\left\lbrace{t}{e}^{{\frac{t}{{3}}}}\right\rbrace}-{3}{e}^{{\frac{2}{{3}}}}{L}{\left\lbrace{e}^{{\frac{t}{{3}}}}\right\rbrace}\)
\(={2}{e}^{{\frac{2}{{3}}}}\frac{1}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-{3}{e}^{{\frac{2}{{3}}}}\frac{1}{{{s}-\frac{1}{{3}}}}\)
\({\left[\text{because }\ {L}{\left\lbrace{t}{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{\left({s}-{a}\right)}^{2}}{L}{\left\lbrace{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{{s}-{a}}}\right]}\)
\(={e}^{{\frac{2}{{3}}}}{\left[\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right]}\)
\({L}{\left\lbrace{\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}\right\rbrace}={e}^{{\frac{2}{{3}}}}{\left[\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right]}\)
Answer A
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