Inverse Laplace transform (first shifting theorem)

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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 ?




calculus laplace-transform




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












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 ?




calculus laplace-transform




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 ?




calculus laplace-transform




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 ?





calculus laplace-transform





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

1466




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
1






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






















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885442%2finverse-laplace-transform-first-shifting-theorem%23new-answer', 'question_page');

    );

    Post as a guest

















    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






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885442%2finverse-laplace-transform-first-shifting-theorem%23new-answer', 'question_page');

            );

            Post as a guest
































































            -

            這個網誌中的熱門文章

            Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
            閱讀完整內容

            Why am i infinitely getting the same tweet with the Twitter Search API?

            圖片
            up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
            閱讀完整內容

            Carbon dioxide

            圖片
            "CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
            閱讀完整內容