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constant evaluation when using differential equations.


About the Legendre differential equationInhomogeneous 2nd-order linear differential equationDetermining $y' = 1 - y^2$ generally excludes $y = pm 1$Solving a set of coupled first order differential equationsSolution of a Modified Bessel Differential Equation with Complex CoefficientFirst order nonlinear differential equation 2Should I include constants of integration while solving for the particular integral or not?“Trivial” differential equations solving, in Hilbert spaces






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


This is regards to constant evaluation when using differential equations.



  • A solution is given to be:
    $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2x$$

  • A simplified solution in an answer book is given as:
    $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1 ) e^x+(c_2 ) e^2x$$

There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.



Sincerely,
Mary A. Marion










share|cite|improve this question











$endgroup$




















    2












    $begingroup$


    This is regards to constant evaluation when using differential equations.



    • A solution is given to be:
      $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2x$$

    • A simplified solution in an answer book is given as:
      $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1 ) e^x+(c_2 ) e^2x$$

    There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.



    Sincerely,
    Mary A. Marion










    share|cite|improve this question











    $endgroup$
















      2












      2








      2





      $begingroup$


      This is regards to constant evaluation when using differential equations.



      • A solution is given to be:
        $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2x$$

      • A simplified solution in an answer book is given as:
        $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1 ) e^x+(c_2 ) e^2x$$

      There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.



      Sincerely,
      Mary A. Marion










      share|cite|improve this question











      $endgroup$




      This is regards to constant evaluation when using differential equations.



      • A solution is given to be:
        $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2x$$

      • A simplified solution in an answer book is given as:
        $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1 ) e^x+(c_2 ) e^2x$$

      There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.



      Sincerely,
      Mary A. Marion







      ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      LutzL

      67.6k4 gold badges22 silver badges61 bronze badges




      67.6k4 gold badges22 silver badges61 bronze badges










      asked 8 hours ago









      Mary A. MarionMary A. Marion

      136 bronze badges




      136 bronze badges























          3 Answers
          3






          active

          oldest

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          3












          $begingroup$

          As $c_1$ and $c_2$ are constants, we could define two other constants $tildec_1=-(c_1+1)$ and $tildec_2=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.



          The answer key shows that



          $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2xtag1$$



          can be simplified to



          $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1) e^x+(c_2 ) e^2xtag2$$



          where $(2)$ could be rewritten as



          $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(tildec_1 ) e^x+(tildec_2 ) e^2xtag3$$



          provided that



          $$tildec_1=-(c_1+1)$$
          $$tildec_2=c_2-1$$



          The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.






          share|cite|improve this answer











          $endgroup$






















            2












            $begingroup$

            Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$



            Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.






            share|cite|improve this answer









            $endgroup$






















              2












              $begingroup$

              Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=fracepsilon2,$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.



              Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.






              share|cite|improve this answer









              $endgroup$

















                Your Answer








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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                As $c_1$ and $c_2$ are constants, we could define two other constants $tildec_1=-(c_1+1)$ and $tildec_2=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.



                The answer key shows that



                $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2xtag1$$



                can be simplified to



                $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1) e^x+(c_2 ) e^2xtag2$$



                where $(2)$ could be rewritten as



                $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(tildec_1 ) e^x+(tildec_2 ) e^2xtag3$$



                provided that



                $$tildec_1=-(c_1+1)$$
                $$tildec_2=c_2-1$$



                The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.






                share|cite|improve this answer











                $endgroup$



















                  3












                  $begingroup$

                  As $c_1$ and $c_2$ are constants, we could define two other constants $tildec_1=-(c_1+1)$ and $tildec_2=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.



                  The answer key shows that



                  $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2xtag1$$



                  can be simplified to



                  $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1) e^x+(c_2 ) e^2xtag2$$



                  where $(2)$ could be rewritten as



                  $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(tildec_1 ) e^x+(tildec_2 ) e^2xtag3$$



                  provided that



                  $$tildec_1=-(c_1+1)$$
                  $$tildec_2=c_2-1$$



                  The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.






                  share|cite|improve this answer











                  $endgroup$

















                    3












                    3








                    3





                    $begingroup$

                    As $c_1$ and $c_2$ are constants, we could define two other constants $tildec_1=-(c_1+1)$ and $tildec_2=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.



                    The answer key shows that



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2xtag1$$



                    can be simplified to



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1) e^x+(c_2 ) e^2xtag2$$



                    where $(2)$ could be rewritten as



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(tildec_1 ) e^x+(tildec_2 ) e^2xtag3$$



                    provided that



                    $$tildec_1=-(c_1+1)$$
                    $$tildec_2=c_2-1$$



                    The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.






                    share|cite|improve this answer











                    $endgroup$



                    As $c_1$ and $c_2$ are constants, we could define two other constants $tildec_1=-(c_1+1)$ and $tildec_2=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.



                    The answer key shows that



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )-(c_1+1) e^x+(c_2-1) e^2xtag1$$



                    can be simplified to



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(c_1) e^x+(c_2 ) e^2xtag2$$



                    where $(2)$ could be rewritten as



                    $$y=(e^2x+e^x ) ln⁡(1+e^-x )+(tildec_1 ) e^x+(tildec_2 ) e^2xtag3$$



                    provided that



                    $$tildec_1=-(c_1+1)$$
                    $$tildec_2=c_2-1$$



                    The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 6 hours ago

























                    answered 7 hours ago









                    Axion004Axion004

                    1,2906 silver badges17 bronze badges




                    1,2906 silver badges17 bronze badges


























                        2












                        $begingroup$

                        Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$



                        Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.






                        share|cite|improve this answer









                        $endgroup$



















                          2












                          $begingroup$

                          Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$



                          Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.






                          share|cite|improve this answer









                          $endgroup$

















                            2












                            2








                            2





                            $begingroup$

                            Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$



                            Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.






                            share|cite|improve this answer









                            $endgroup$



                            Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$



                            Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 7 hours ago









                            Mohammad Riazi-KermaniMohammad Riazi-Kermani

                            51.1k4 gold badges27 silver badges72 bronze badges




                            51.1k4 gold badges27 silver badges72 bronze badges
























                                2












                                $begingroup$

                                Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=fracepsilon2,$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.



                                Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.






                                share|cite|improve this answer









                                $endgroup$



















                                  2












                                  $begingroup$

                                  Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=fracepsilon2,$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.



                                  Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.






                                  share|cite|improve this answer









                                  $endgroup$

















                                    2












                                    2








                                    2





                                    $begingroup$

                                    Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=fracepsilon2,$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.



                                    Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=fracepsilon2,$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.



                                    Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 7 hours ago









                                    AllawonderAllawonder

                                    3,7478 silver badges18 bronze badges




                                    3,7478 silver badges18 bronze badges






























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