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

Wierzycaz 2020-10-21 Answered
The Laplace transform of the function L{6e3tt2+2t8}
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Expert Answer

opsadnojD
Answered 2020-10-22 Author has 95 answers
Calculation:
Consider,
L{6e3tt2+2t8}
The objective is to determine the Laplace Transform
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{6e3tt2+2t8}=L{{6e3t}L{t2}+L{2t}8L{1}
=6L{e3t}L{t2}L{t2}+2L{t}8L{1}
Since the Laplace Transform from the Laplace Transform table:
L{e0t=1s,s>0
L{eat}=1sa,s>a
L{tn}=n!sn+1,s>0,n=1,2,3,
Using above formula:
6L{e3t}L{t2}+2L{t}8L{1}=6s(3)2!s2+1+2×1!s1+18×1s,
s>0
=6s+32s3+2s28s,s>0
Laplace transforms into
F(s)=6s+32s2+2s28s,s>0
Conclusion:
The required Laplace transform for this function is
F(s)=6s+32s2+2s28s,s>0
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