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

Burhan Hopper 2021-02-12 Answered
To find: The Laplace transform of the function
L{t4t2t+sin2t}
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Expert Answer

Fatema Sutton
Answered 2021-02-13 Author has 88 answers
Approach:
With the help of Laplace Transform table and linearity property of Laplace Transformation calculation is carried out as follows:
Calculation:
L{t4t2t+sin2t}
Consider f, f1 and f2 be the functions whose Laplace Transform exists for s > α and c be a constant.
Then, for s > α
L{f1+f2}=L{f1}+L{f2}
Also,
L{cf}=cL{f}
Using above properties:
L{t4t2t+sin2t}=L{t4}+L{t2}+L{t}+L{sin2t}
=L{t4}L{t2}L{t}+L{sin2t}
Since the Laplace Transform from the Laplace Transform table:
L{tn}=n!sn+1,s>0,n=1,2,3,
L{sinat}=as2+a2,s>0
Using above formulae:
L{t4}L{t2}L{t}+L{sin2t}=4!s4+12!s2+11!s1+1+2s2+(2)2,s>0
=24s52s31s2+/(s2+2),s>0
Gives, the Laplace Transform is
F(s)=24s52s31s2+/(s2+2),s>0
Conclusion:
Therefore, the required Laplace Transform for the given function is
F(s)=24s52s31s2+/(s2+2),s>0
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