Question

The Laplace transform of the function {L}{leftlbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}rightrbrace}

Transformation properties
ANSWERED
asked 2020-10-21
The Laplace transform of the function \({L}{\left\lbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}\right\rbrace}\)

Answers (1)

2020-10-22
Calculation:
Consider,
\({L}{\left\lbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}\right\rbrace}\)
The objective is to determine the Laplace Transform
Consider \(f,\ f_{1},\ \text{and}\ f_{2}\ \text{be the functions whose Laplace Transform exists for}\ s\ >\ \alpha\) and c be a constant
Then, for \(s\ >\ \alpha\)
\({L}{\left\lbrace{{f}_{{1}}+}{f}_{{2}}\right\rbrace}={L}{\left\lbrace{f}_{{1}}\right\rbrace}+{L}{\left\lbrace{f}_{{2}}\right\rbrace}\)
Also,
\({L}{\left\lbrace{c}{f}\right\rbrace}={c}{L}{\left\lbrace{f}\right\rbrace}\)
Using above properties:
\({L}{\left\lbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}\right\rbrace}={L}{\left\lbrace{\left\lbrace{6}{e}^{{-{3}{t}}}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}+{L}{\left\lbrace{2}{t}\right\rbrace}-{8}{L}{\left\lbrace{1}\right\rbrace}\right.}\)
\(={6}{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}+{2}{L}{\left\lbrace{t}\right\rbrace}-{8}{L}{\left\lbrace{1}\right\rbrace}\)
Since the Laplace Transform from the Laplace Transform table:
\({L}{\left\lbrace{e}^{{{0}\cdot{t}}}=\frac{1}{{s}},\forall{s}>{0}\right.}\)
\({L}{\left\lbrace{e}^{a}\cdot{t}\right\rbrace}=\frac{1}{{{s}-{a}}},\forall{s}>{a}\)
\({L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}},\forall{s}>{0},{n}={1},{2},{3},\ldots\)
Using above formula:
\({6}{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}-{L}{\left\lbrace{t}^{2}\right\rbrace}+{2}{L}{\left\lbrace{t}\right\rbrace}-{8}{L}{\left\lbrace{1}\right\rbrace}=\frac{6}{{{s}-{\left(-{3}\right)}}}-\frac{{{2}!}}{{{s}^{{{2}+{1}}}}}+{2}\times\frac{{{1}!}}{{{s}^{{{1}+{1}}}}}-{8}\times\frac{1}{{s}},\)
\(\forall{s}>{0}\)
\(=\frac{6}{{{s}+{3}}}-\frac{2}{{s}^{3}}+\frac{2}{{s}^{2}}-\frac{8}{{s}},\forall{s}>{0}\)
Laplace transforms into
\({F}{\left({s}\right)}=\frac{6}{{{s}+{3}}}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{{s}},\forall{s}>{0}\)
Conclusion:
The required Laplace transform for this function is
\({F}{\left({s}\right)}=\frac{6}{{{s}+{3}}}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{{s}},\forall{s}>{0}\)
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