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





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      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


















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