# To find: The Laplace transform of the function {L}{leftlbrace{t}^{4}-{t}^{2}-{t}+ sin{sqrt{{{2}{t}}}}rightrbrace}

To find: The Laplace transform of the function
$L\left\{{t}^{4}-{t}^{2}-t+\mathrm{sin}\sqrt{2t}\right\}$
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Fatema Sutton
Approach:
With the help of Laplace Transform table and linearity property of Laplace Transformation calculation is carried out as follows:
Calculation:
$L\left\{{t}^{4}-{t}^{2}-t+\mathrm{sin}\sqrt{2t}\right\}$
Consider and c be a constant.
Then, for
$L\left\{{f}_{1}+{f}_{2}\right\}=L\left\{{f}_{1}\right\}+L\left\{{f}_{2}\right\}$
Also,
$L\left\{cf\right\}=cL\left\{f\right\}$
Using above properties:
$L\left\{{t}^{4}-{t}^{2}-t+\mathrm{sin}\sqrt{2t}\right\}=L\left\{{t}^{4}\right\}+L\left\{-{t}^{2}\right\}+L\left\{-t\right\}+L\left\{\mathrm{sin}\sqrt{2t}\right\}$
$=L\left\{{t}^{4}\right\}-L\left\{{t}^{2}\right\}-L\left\{t\right\}+L\left\{\mathrm{sin}\sqrt{2t}\right\}$
Since the Laplace Transform from the Laplace Transform table:
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},\mathrm{\forall }s>0,n=1,2,3,\dots$
$L\left\{\mathrm{sin}at\right\}=\frac{a}{{s}^{2}+{a}^{2}},\mathrm{\forall }s>0$
Using above formulae:
$L\left\{{t}^{4}\right\}-L\left\{{t}^{2}\right\}-L\left\{t\right\}+L\left\{\mathrm{sin}\sqrt{2t}\right\}=\frac{4!}{{s}^{4+1}}-\frac{2!}{{s}^{2+1}}-\frac{1!}{{s}^{1+1}}+\frac{\sqrt{2}}{{s}^{2}+{\left(\sqrt{2}\right)}^{2}},\mathrm{\forall }s>0$
$=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}\left({s}^{2}+2\right),\mathrm{\forall }s>0$
Gives, the Laplace Transform is
$F\left(s\right)=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}\left({s}^{2}+2\right),\mathrm{\forall }s>0$
Conclusion:
Therefore, the required Laplace Transform for the given function is
$F\left(s\right)=\frac{24}{{s}^{5}}-\frac{2}{{s}^{3}}-\frac{1}{{s}^{2}}+\sqrt{/}\left({s}^{2}+2\right),\mathrm{\forall }s>0$