The Laplace transform of the function L{6e−3t−t2+2t−8}

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opsadnojD · Accepted Answer

Calculation:   Consider,  L{6e−3t−t2+2t−8}  The objective is to determine the Laplace Transform   Consider f, f1, and f2 be the functions whose Laplace Transform exists for s &amp;gt; α and c be a constant  Then, for s &amp;gt; α  L{f1+f2}=L{f1}+L{f2}  Also,  L{cf}=cL{f}  Using above properties:  L{6e−3t−t2+2t−8}=L{{6e−3t}−L{t2}+L{2t}−8L{1}  =6L{e−3t}−L{t2}−L{t2}+2L{t}−8L{1}  Since the Laplace Transform from the Laplace Transform table:  L{e0⋅t=1s,∀s&amp;gt;0  L{ea⋅t}=1s−a,∀s&amp;gt;a  L{tn}=n!sn+1,∀s&amp;gt;0,n=1,2,3,…  Using above formula:  6L{e−3t}−L{t2}+2L{t}−8L{1}=6s−(−3)−2!s2+1+2×1!s1+1−8×1s,  ∀s&amp;gt;0  =6s+3−2s3+2s2−8s,∀s&amp;gt;0  Laplace transforms into   F(s)=6s+3−2s2+2s2−8s,∀s&amp;gt;0  Conclusion:  The required Laplace transform for this function is   F(s)=6s+3−2s2+2s2−8s,∀s&amp;gt;0

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

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