Given $y = Acos(kt) + Bsin(kt)$, where $A, B, k$ are constants, prove that $y^{(2)} + (k^2)y = 0$












0












$begingroup$


EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!



In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.



I have noted that $y^n$ refers to the nth derivative.



I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.



That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.



The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.



The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.



Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.



I am not sure if this is relevant in breaking down the proof into cases...



Thank you in advance for any insight.






calculus derivatives trigonometry proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
    $endgroup$
    – WW1
    Dec 3 '18 at 3:01








  • 2




    $begingroup$
    Ah! In that case, I just find the second derivative and sub in!
    $endgroup$
    – user424712
    Dec 3 '18 at 3:02
















0












$begingroup$


EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!



In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.



I have noted that $y^n$ refers to the nth derivative.



I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.



That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.



The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.



The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.



Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.



I am not sure if this is relevant in breaking down the proof into cases...



Thank you in advance for any insight.






calculus derivatives trigonometry proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
    $endgroup$
    – WW1
    Dec 3 '18 at 3:01








  • 2




    $begingroup$
    Ah! In that case, I just find the second derivative and sub in!
    $endgroup$
    – user424712
    Dec 3 '18 at 3:02














0












0








0





$begingroup$


EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!



In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.



I have noted that $y^n$ refers to the nth derivative.



I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.



That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.



The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.



The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.



Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.



I am not sure if this is relevant in breaking down the proof into cases...



Thank you in advance for any insight.






calculus derivatives trigonometry proof-writing







share|cite|improve this question











$endgroup$




EDIT: There was a silly typo, $y^n$ is actually $y''$... sorry everyone!



In that case just finding $y''$ and substituting this and $y$ into the left side will yield the answer 0.



I have noted that $y^n$ refers to the nth derivative.



I found that the 1st, 5th, 9th, ..., $(4m+1)$th derivatives have a sign pattern of - and +, respectively, in front of each trig ratio.



That is, $y^n = -Ak^nsin kt + Bk^ncos kt$.



The 2nd, 6th, 10th, ..., $(4m+2)$th derivates have a sign pattern of - and -.



The 3rd, 7th, 11th, ... $(4m+3)$th derivates have a sign pattern of + and -.



Finally, the 4th, 8th, 12th, ..., $(4m+4)$th derivatives have a sign pattern of + and +.



I am not sure if this is relevant in breaking down the proof into cases...



Thank you in advance for any insight.






calculus derivatives trigonometry proof-writing




calculus derivatives trigonometry proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 20:33









Paras Khosla

699213




699213










asked Dec 3 '18 at 2:58









user424712user424712

12




12








  • 2




    $begingroup$
    I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
    $endgroup$
    – WW1
    Dec 3 '18 at 3:01








  • 2




    $begingroup$
    Ah! In that case, I just find the second derivative and sub in!
    $endgroup$
    – user424712
    Dec 3 '18 at 3:02














  • 2




    $begingroup$
    I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
    $endgroup$
    – WW1
    Dec 3 '18 at 3:01








  • 2




    $begingroup$
    Ah! In that case, I just find the second derivative and sub in!
    $endgroup$
    – user424712
    Dec 3 '18 at 3:02








2




2




$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01






$begingroup$
I'm pretty sure that your are looking at a misprint, $y^n$ is probably supposed to be $y''$ ( second derivative)
$endgroup$
– WW1
Dec 3 '18 at 3:01






2




2




$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02




$begingroup$
Ah! In that case, I just find the second derivative and sub in!
$endgroup$
– user424712
Dec 3 '18 at 3:02










2 Answers
2






active

oldest

votes


















0












$begingroup$

$$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.



    $frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$



    Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$



    Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved






    share|cite|improve this answer











    $endgroup$













      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',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      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%2f3023555%2fgiven-y-a-coskt-b-sinkt-where-a-b-k-are-constants-prove-that-y%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
      You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
        You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
          You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.






          share|cite|improve this answer









          $endgroup$



          $$y^{''} =frac {d^2}{dt^2}y= frac {d}{dt}left({frac d{dt}y}right)= frac {d}{dt}(-Aksin(kt)+Bkcos(kt))=kfrac {d}{dt}(-Asin(kt)+Bcos(kt))=k^2(-Acos(kt)-Bsin(kt))=-k^2y$$
          You must have done something erroneous! may be the derivatives of $sin(x)$ and $cos(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 11:16









          Sameer BahetiSameer Baheti

          5168




          5168























              0












              $begingroup$

              We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.



              $frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$



              Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$



              Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.



                $frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$



                Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$



                Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.



                  $frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$



                  Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$



                  Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved






                  share|cite|improve this answer











                  $endgroup$



                  We have $y=A cos(kt)+B sin(kt)$ where ${ A, B, k }$ are constants. We are required to prove that $y^{(2)}+(k^2)y=0$ where $y^{(2)}$ denotes the second derivative of $y$ with respect to $t$.



                  $frac{dy}{dt}=-Ak sin(kt)+Bk cos(kt) implies frac{d^2y}{dt^2}=-Ak^2 cos(kt)+Bk^2 sin(kt) tag1$



                  Also, $k^2y=Ak^2 cos(kt)+Bk^2 sin(kt) tag2$



                  Adding equations $(1)$ and $(2)$ together we indeed get $0$ as the result. Hence, Proved







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 11 at 12:25

























                  answered Dec 3 '18 at 19:38









                  Paras KhoslaParas Khosla

                  699213




                  699213






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023555%2fgiven-y-a-coskt-b-sinkt-where-a-b-k-are-constants-prove-that-y%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                      Image
                      2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                      Read more

                      ComboBox Display Member on multiple fields

                      Image
                      2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
                      Read more

                      Is it possible to collect Nectar points via Trainline?

                      Image
                      2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
                      Read more