Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform L\{t^4+t^3-t^2-t+\sin \sqrt2t \}

glasskerfu 2021-09-15 Answered

Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform
\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}\)

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Expert Answer

Macsen Nixon
Answered 2021-09-16 Author has 23272 answers

The given Laplace transform is
\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}\)
By using the linearity property of Laplace transform
For Function \(f(t), g(t)\) and constant a,b:
\(L\left\{a f(t)+b g(t)\right\}=aL\left\{f(t)\right\}+bL\left\{g(H)\right\}\)
So from eqn (1)
\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}=L\left\{t^4\right\}+\left\{t^3\right\}-\left\{t^2\right\}-\left\{t\right\}+\left\{\sin \sqrt2t\right\}\) (2)
By using \(L\left\{t^n\right\}=\frac{n!}{s^{n+1}}\) and
\(L\left\{\sin(at)\right\}=\frac{a}{s^2+a^2}\) we will get from eqn (2)
\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}=\frac{4!}{s^5}+\frac{3!}{s^4}-\frac{2!}{s^3}-\frac{1}{s^2}+\frac{\sqrt2}{s^2+(\sqrt2)^2}\)
\(L\left\{t^4+t^3-t^2-t+\sin \sqrt2t\right\}=\frac{4!}{s^5}+\frac{3!}{s^4}-\frac{2!}{s^3}-\frac{1}{s^2}+\frac{\sqrt2}{s^2+(\sqrt2)^2}\)

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