Explain $y''-y=0$












0












$begingroup$


I've solved this equation and got $c_0cosh x + c_1sinh x$. However, I've noticed that if the arbitrary constant $c_0$ doesn't equal $c_1$, this wouldn't work; you would have to convert the $cosh x $and $sinh x$ into solutions of $e^x$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?






sequences-and-series ordinary-differential-equations power-series differential







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the substitution $$y(x)=e^{lambda x}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 8 '18 at 14:55






  • 3




    $begingroup$
    Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
    $endgroup$
    – TonyK
    Dec 8 '18 at 15:00
















0












$begingroup$


I've solved this equation and got $c_0cosh x + c_1sinh x$. However, I've noticed that if the arbitrary constant $c_0$ doesn't equal $c_1$, this wouldn't work; you would have to convert the $cosh x $and $sinh x$ into solutions of $e^x$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?






sequences-and-series ordinary-differential-equations power-series differential







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Use the substitution $$y(x)=e^{lambda x}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 8 '18 at 14:55






  • 3




    $begingroup$
    Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
    $endgroup$
    – TonyK
    Dec 8 '18 at 15:00














0












0








0





$begingroup$


I've solved this equation and got $c_0cosh x + c_1sinh x$. However, I've noticed that if the arbitrary constant $c_0$ doesn't equal $c_1$, this wouldn't work; you would have to convert the $cosh x $and $sinh x$ into solutions of $e^x$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?






sequences-and-series ordinary-differential-equations power-series differential







share|cite|improve this question











$endgroup$




I've solved this equation and got $c_0cosh x + c_1sinh x$. However, I've noticed that if the arbitrary constant $c_0$ doesn't equal $c_1$, this wouldn't work; you would have to convert the $cosh x $and $sinh x$ into solutions of $e^x$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?






sequences-and-series ordinary-differential-equations power-series differential




sequences-and-series ordinary-differential-equations power-series differential






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 15:01









mrtaurho

4,24421234




4,24421234










asked Dec 8 '18 at 14:54









Fourth Fourth

514




514








  • 1




    $begingroup$
    Use the substitution $$y(x)=e^{lambda x}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 8 '18 at 14:55






  • 3




    $begingroup$
    Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
    $endgroup$
    – TonyK
    Dec 8 '18 at 15:00














  • 1




    $begingroup$
    Use the substitution $$y(x)=e^{lambda x}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 8 '18 at 14:55






  • 3




    $begingroup$
    Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
    $endgroup$
    – TonyK
    Dec 8 '18 at 15:00








1




1




$begingroup$
Use the substitution $$y(x)=e^{lambda x}$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:55




$begingroup$
Use the substitution $$y(x)=e^{lambda x}$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:55




3




3




$begingroup$
Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
$endgroup$
– TonyK
Dec 8 '18 at 15:00




$begingroup$
Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it?
$endgroup$
– TonyK
Dec 8 '18 at 15:00










3 Answers
3






active

oldest

votes


















2












$begingroup$

Why do you claim that it doesn't work? $$(c_0cosh +c_1sinh )''=c_0(sinh)'+c_1(cosh)'=c_0cosh+c_1sinh$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $cosh x$ and $sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.



    The hyperbolic functions are exponentials in disguise.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      It works for any $(c_0,c_1)$. Since $$(cosh(x))''=cosh(x)$$ and $$(sinh(x))''=sinh(x),$$
      if $$y=c_0 cosh(x)+c_1sinh(x),$$
      then
      $$y''=c_0cosh(x)+c_1sinh(x)=y,$$
      and so
      $$y''-y=0.$$



      It is also true that $cosh(x)$ and $sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because
      $$W(cosh(x),sinh(x))=left|begin{matrix}cosh(x)&sinh(x)\(cosh(x))'&(sinh(x))'\end{matrix}right|=$$
      $$=left|begin{matrix}cosh(x)&sinh(x)\sinh(x)&cosh(x)\end{matrix}right|=cosh^2(x)-sinh^2(x)=1,quad forall x.$$



      This implies that all the solutions are of the form
      $$y=c_0 cosh(x)+c_1sinh(x).$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $cosh'$ is $sinh$, not $cosh$.
        $endgroup$
        – Saucy O'Path
        Dec 8 '18 at 15:05










      • $begingroup$
        Thanks, I forgot another prime on each expression.
        $endgroup$
        – Alejandro Nasif Salum
        Dec 8 '18 at 15:10











      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%2f3031200%2fexplain-y-y-0%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Why do you claim that it doesn't work? $$(c_0cosh +c_1sinh )''=c_0(sinh)'+c_1(cosh)'=c_0cosh+c_1sinh$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Why do you claim that it doesn't work? $$(c_0cosh +c_1sinh )''=c_0(sinh)'+c_1(cosh)'=c_0cosh+c_1sinh$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Why do you claim that it doesn't work? $$(c_0cosh +c_1sinh )''=c_0(sinh)'+c_1(cosh)'=c_0cosh+c_1sinh$$






          share|cite|improve this answer









          $endgroup$



          Why do you claim that it doesn't work? $$(c_0cosh +c_1sinh )''=c_0(sinh)'+c_1(cosh)'=c_0cosh+c_1sinh$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 15:01









          Saucy O'PathSaucy O'Path

          5,9191626




          5,9191626























              2












              $begingroup$

              $cosh x$ and $sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.



              The hyperbolic functions are exponentials in disguise.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $cosh x$ and $sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.



                The hyperbolic functions are exponentials in disguise.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $cosh x$ and $sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.



                  The hyperbolic functions are exponentials in disguise.






                  share|cite|improve this answer









                  $endgroup$



                  $cosh x$ and $sinh x$ are two independent linear combinations of $e^x$ and $e^{-x}$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.



                  The hyperbolic functions are exponentials in disguise.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 15:03









                  Yves DaoustYves Daoust

                  126k672225




                  126k672225























                      1












                      $begingroup$

                      It works for any $(c_0,c_1)$. Since $$(cosh(x))''=cosh(x)$$ and $$(sinh(x))''=sinh(x),$$
                      if $$y=c_0 cosh(x)+c_1sinh(x),$$
                      then
                      $$y''=c_0cosh(x)+c_1sinh(x)=y,$$
                      and so
                      $$y''-y=0.$$



                      It is also true that $cosh(x)$ and $sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because
                      $$W(cosh(x),sinh(x))=left|begin{matrix}cosh(x)&sinh(x)\(cosh(x))'&(sinh(x))'\end{matrix}right|=$$
                      $$=left|begin{matrix}cosh(x)&sinh(x)\sinh(x)&cosh(x)\end{matrix}right|=cosh^2(x)-sinh^2(x)=1,quad forall x.$$



                      This implies that all the solutions are of the form
                      $$y=c_0 cosh(x)+c_1sinh(x).$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $cosh'$ is $sinh$, not $cosh$.
                        $endgroup$
                        – Saucy O'Path
                        Dec 8 '18 at 15:05










                      • $begingroup$
                        Thanks, I forgot another prime on each expression.
                        $endgroup$
                        – Alejandro Nasif Salum
                        Dec 8 '18 at 15:10
















                      1












                      $begingroup$

                      It works for any $(c_0,c_1)$. Since $$(cosh(x))''=cosh(x)$$ and $$(sinh(x))''=sinh(x),$$
                      if $$y=c_0 cosh(x)+c_1sinh(x),$$
                      then
                      $$y''=c_0cosh(x)+c_1sinh(x)=y,$$
                      and so
                      $$y''-y=0.$$



                      It is also true that $cosh(x)$ and $sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because
                      $$W(cosh(x),sinh(x))=left|begin{matrix}cosh(x)&sinh(x)\(cosh(x))'&(sinh(x))'\end{matrix}right|=$$
                      $$=left|begin{matrix}cosh(x)&sinh(x)\sinh(x)&cosh(x)\end{matrix}right|=cosh^2(x)-sinh^2(x)=1,quad forall x.$$



                      This implies that all the solutions are of the form
                      $$y=c_0 cosh(x)+c_1sinh(x).$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $cosh'$ is $sinh$, not $cosh$.
                        $endgroup$
                        – Saucy O'Path
                        Dec 8 '18 at 15:05










                      • $begingroup$
                        Thanks, I forgot another prime on each expression.
                        $endgroup$
                        – Alejandro Nasif Salum
                        Dec 8 '18 at 15:10














                      1












                      1








                      1





                      $begingroup$

                      It works for any $(c_0,c_1)$. Since $$(cosh(x))''=cosh(x)$$ and $$(sinh(x))''=sinh(x),$$
                      if $$y=c_0 cosh(x)+c_1sinh(x),$$
                      then
                      $$y''=c_0cosh(x)+c_1sinh(x)=y,$$
                      and so
                      $$y''-y=0.$$



                      It is also true that $cosh(x)$ and $sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because
                      $$W(cosh(x),sinh(x))=left|begin{matrix}cosh(x)&sinh(x)\(cosh(x))'&(sinh(x))'\end{matrix}right|=$$
                      $$=left|begin{matrix}cosh(x)&sinh(x)\sinh(x)&cosh(x)\end{matrix}right|=cosh^2(x)-sinh^2(x)=1,quad forall x.$$



                      This implies that all the solutions are of the form
                      $$y=c_0 cosh(x)+c_1sinh(x).$$






                      share|cite|improve this answer











                      $endgroup$



                      It works for any $(c_0,c_1)$. Since $$(cosh(x))''=cosh(x)$$ and $$(sinh(x))''=sinh(x),$$
                      if $$y=c_0 cosh(x)+c_1sinh(x),$$
                      then
                      $$y''=c_0cosh(x)+c_1sinh(x)=y,$$
                      and so
                      $$y''-y=0.$$



                      It is also true that $cosh(x)$ and $sinh(x)$ are L.I. functions, since the wronskian determinant is not zero, because
                      $$W(cosh(x),sinh(x))=left|begin{matrix}cosh(x)&sinh(x)\(cosh(x))'&(sinh(x))'\end{matrix}right|=$$
                      $$=left|begin{matrix}cosh(x)&sinh(x)\sinh(x)&cosh(x)\end{matrix}right|=cosh^2(x)-sinh^2(x)=1,quad forall x.$$



                      This implies that all the solutions are of the form
                      $$y=c_0 cosh(x)+c_1sinh(x).$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 8 '18 at 15:09

























                      answered Dec 8 '18 at 15:03









                      Alejandro Nasif SalumAlejandro Nasif Salum

                      4,765118




                      4,765118












                      • $begingroup$
                        $cosh'$ is $sinh$, not $cosh$.
                        $endgroup$
                        – Saucy O'Path
                        Dec 8 '18 at 15:05










                      • $begingroup$
                        Thanks, I forgot another prime on each expression.
                        $endgroup$
                        – Alejandro Nasif Salum
                        Dec 8 '18 at 15:10


















                      • $begingroup$
                        $cosh'$ is $sinh$, not $cosh$.
                        $endgroup$
                        – Saucy O'Path
                        Dec 8 '18 at 15:05










                      • $begingroup$
                        Thanks, I forgot another prime on each expression.
                        $endgroup$
                        – Alejandro Nasif Salum
                        Dec 8 '18 at 15:10
















                      $begingroup$
                      $cosh'$ is $sinh$, not $cosh$.
                      $endgroup$
                      – Saucy O'Path
                      Dec 8 '18 at 15:05




                      $begingroup$
                      $cosh'$ is $sinh$, not $cosh$.
                      $endgroup$
                      – Saucy O'Path
                      Dec 8 '18 at 15:05












                      $begingroup$
                      Thanks, I forgot another prime on each expression.
                      $endgroup$
                      – Alejandro Nasif Salum
                      Dec 8 '18 at 15:10




                      $begingroup$
                      Thanks, I forgot another prime on each expression.
                      $endgroup$
                      – Alejandro Nasif Salum
                      Dec 8 '18 at 15:10


















                      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%2f3031200%2fexplain-y-y-0%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more