Laplace transform using 2nd shifting Theorem

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$$mathcalL(U(t-5)e^t-5 )$$



I’ve been taught this method and im not sure of and whether it will be correct, so I came here to ask.



$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) $



a = 5,
Therefore, $f(t-5) = e^t-5 $



since $mathcalL (f(t)) = F(s) $



$f(t) = e^t $



and $mathcalL (f(t)) = frac1s-1 $



Therefore the answer is - $e^-5s frac1s-1 $



Am I correct ?




laplace-transform




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    up vote
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    down vote

    favorite












    $$mathcalL(U(t-5)e^t-5 )$$



    I’ve been taught this method and im not sure of and whether it will be correct, so I came here to ask.



    $mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) $



    a = 5,
    Therefore, $f(t-5) = e^t-5 $



    since $mathcalL (f(t)) = F(s) $



    $f(t) = e^t $



    and $mathcalL (f(t)) = frac1s-1 $



    Therefore the answer is - $e^-5s frac1s-1 $



    Am I correct ?




    laplace-transform




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $$mathcalL(U(t-5)e^t-5 )$$



      I’ve been taught this method and im not sure of and whether it will be correct, so I came here to ask.



      $mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) $



      a = 5,
      Therefore, $f(t-5) = e^t-5 $



      since $mathcalL (f(t)) = F(s) $



      $f(t) = e^t $



      and $mathcalL (f(t)) = frac1s-1 $



      Therefore the answer is - $e^-5s frac1s-1 $



      Am I correct ?




      laplace-transform




      share|cite|improve this question











      $$mathcalL(U(t-5)e^t-5 )$$



      I’ve been taught this method and im not sure of and whether it will be correct, so I came here to ask.



      $mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) $



      a = 5,
      Therefore, $f(t-5) = e^t-5 $



      since $mathcalL (f(t)) = F(s) $



      $f(t) = e^t $



      and $mathcalL (f(t)) = frac1s-1 $



      Therefore the answer is - $e^-5s frac1s-1 $



      Am I correct ?





      laplace-transform





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 22 at 9:37









      user185692

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          Yes, you have applied $$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) = e^-5s frac1s-1$$



          correctly to get your answer $e^-5s frac1s-1$






          share|cite|improve this answer





















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            up vote
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            down vote



            accepted










            Yes, you have applied $$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) = e^-5s frac1s-1$$



            correctly to get your answer $e^-5s frac1s-1$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Yes, you have applied $$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) = e^-5s frac1s-1$$



              correctly to get your answer $e^-5s frac1s-1$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Yes, you have applied $$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) = e^-5s frac1s-1$$



                correctly to get your answer $e^-5s frac1s-1$






                share|cite|improve this answer













                Yes, you have applied $$mathcalL(U(t-5)e^t-5 = mathcalL (U(t-a) f(t-a) ) = e^-asF(s) = e^-5s frac1s-1$$



                correctly to get your answer $e^-5s frac1s-1$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 22 at 9:59









                Mohammad Riazi-Kermani

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