To find: The Laplace transform of the function L{t4−t2−t+sin⁡2t}

Question

To find: The Laplace transform of the function  L{t4−t2−t+sin⁡2t}

Fatema Sutton · Accepted Answer

Approach:  With the help of Laplace Transform table and linearity property of Laplace Transformation calculation is carried out as follows:  Calculation:  L{t4−t2−t+sin⁡2t}  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{t4−t2−t+sin⁡2t}=L{t4}+L{−t2}+L{−t}+L{sin⁡2t}  =L{t4}−L{t2}−L{t}+L{sin⁡2t}  Since the Laplace Transform from the Laplace Transform table:  L{tn}=n!sn+1,∀s&amp;gt;0,n=1,2,3,…  L{sin⁡at}=as2+a2,∀s&amp;gt;0  Using above  formulae:  L{t4}−L{t2}−L{t}+L{sin⁡2t}=4!s4+1−2!s2+1−1!s1+1+2s2+(2)2,∀s&amp;gt;0  =24s5−2s3−1s2+/(s2+2),∀s&amp;gt;0  Gives, the Laplace Transform is  F(s)=24s5−2s3−1s2+/(s2+2),∀s&amp;gt;0  Conclusion:  Therefore, the required Laplace Transform for the given function is  F(s)=24s5−2s3−1s2+/(s2+2),∀s&amp;gt;0

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

Answered question

Answer & Explanation

New Questions in Integral Calculus