Inverse Laplace transform (first shifting theorem)

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This is my textbook from what I am studying and the “green” highlighted part is where I am questioning myself.



Let’s pull out the expression-



$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$



From here, I distribute it to become



$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $



Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?







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  • the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
    – Alla Tarighati
    Aug 17 at 7:08















up vote
0
down vote

favorite












enter image description here



This is my textbook from what I am studying and the “green” highlighted part is where I am questioning myself.



Let’s pull out the expression-



$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$



From here, I distribute it to become



$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $



Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?







share|cite|improve this question






















  • the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
    – Alla Tarighati
    Aug 17 at 7:08













up vote
0
down vote

favorite









up vote
0
down vote

favorite











enter image description here



This is my textbook from what I am studying and the “green” highlighted part is where I am questioning myself.



Let’s pull out the expression-



$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$



From here, I distribute it to become



$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $



Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?







share|cite|improve this question














enter image description here



This is my textbook from what I am studying and the “green” highlighted part is where I am questioning myself.



Let’s pull out the expression-



$$mathcalL ( frac frac1925 (s) + frac1425 (s+2)^2 + 3^2 )$$



From here, I distribute it to become



$ frac1925 mathcalL ( fracs (s+2)^2 + 3^2 )+ frac1425 mathcalL (frac1 (s+2)^2 + 3^2 ) $



Where does the $ mathcal L frac23 frac3 (s+2)^2 + 3^2 $ (green highlighted portion) come from ?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 17 at 7:04

























asked Aug 17 at 5:49









user185692

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1466











  • the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
    – Alla Tarighati
    Aug 17 at 7:08

















  • the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
    – Alla Tarighati
    Aug 17 at 7:08
















the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
– Alla Tarighati
Aug 17 at 7:08





the green part is just added to cancel the added part in orange part (that +2/denom). If you add green and orange parts, you get s/denom.
– Alla Tarighati
Aug 17 at 7:08











1 Answer
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Let us go back one line, then we have three parts:



$$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$



Now let us focus on the second part (containing orange and green parts):



$$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$



And adding a zero to the nomerator will not change anything:



$$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$



Or:



$$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Let us go back one line, then we have three parts:



    $$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$



    Now let us focus on the second part (containing orange and green parts):



    $$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$



    And adding a zero to the nomerator will not change anything:



    $$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$



    Or:



    $$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$






    share|cite|improve this answer


























      up vote
      1
      down vote



      accepted










      Let us go back one line, then we have three parts:



      $$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$



      Now let us focus on the second part (containing orange and green parts):



      $$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$



      And adding a zero to the nomerator will not change anything:



      $$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$



      Or:



      $$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Let us go back one line, then we have three parts:



        $$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$



        Now let us focus on the second part (containing orange and green parts):



        $$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$



        And adding a zero to the nomerator will not change anything:



        $$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$



        Or:



        $$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$






        share|cite|improve this answer














        Let us go back one line, then we have three parts:



        $$leftfrac625frac1s-2right + leftfracfrac19s25(s+2)^2+3^2right + leftfracfrac1425(s+2)^2+3^2right $$



        Now let us focus on the second part (containing orange and green parts):



        $$leftfracfrac19s25(s+2)^2+3^2right = frac1925fracs(s+2)^2+3^2 $$



        And adding a zero to the nomerator will not change anything:



        $$frac1925fracs+(2-2)(s+2)^2+3^2 = frac1925fracs+2(s+2)^2+3^2+frac1925frac-2(s+2)^2+3^2$$



        Or:



        $$frac1925leftfracs+2(s+2)^2+3^2+frac23frac-3(s+2)^2+3^2right$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 20 at 6:18

























        answered Aug 17 at 7:18









        Alla Tarighati

        2353




        2353






















             

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