Exponential growth/decay formula: what happened to the other constant of integration?Differential Inequality...

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Exponential growth/decay formula: what happened to the other constant of integration?

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Exponential growth/decay formula: what happened to the other constant of integration?


Differential Inequality Conditions to Determine Exponential Growth/DecayIntegrating two equations that equal, what happens to the constant on one of the sides?Simple integration for absolute energy formulationIdentifying the constant of integrationIntegration by Parts and the Constant of Integrationfinding the half life of an Exponential decay problem.Find minimization of the difference between integrals of exponential decay, and step exponential decay functionsWhy does the constant of integration move?Why do we ignore constant produced by integration of derivative when we derive integration by parts formula?Explanation of proof nedeed: Why is $y'=c cdot y$ always a exponential growth/decay function?













3












$begingroup$


The standard equation for exponential growth and decay starts starts like this:



$$ {dPover dt}=kP$$



$$ {dPover P}=kdt$$



$$ int{dPover P}=int kdt$$



$$ color{red}{ln |P|}=kt+C$$



I don't understand the left hand side at this point, isn't $int{1over x}dx = ln |x| +C$? Where did the constant go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
    $endgroup$
    – JoseSquare
    2 hours ago
















3












$begingroup$


The standard equation for exponential growth and decay starts starts like this:



$$ {dPover dt}=kP$$



$$ {dPover P}=kdt$$



$$ int{dPover P}=int kdt$$



$$ color{red}{ln |P|}=kt+C$$



I don't understand the left hand side at this point, isn't $int{1over x}dx = ln |x| +C$? Where did the constant go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
    $endgroup$
    – JoseSquare
    2 hours ago














3












3








3





$begingroup$


The standard equation for exponential growth and decay starts starts like this:



$$ {dPover dt}=kP$$



$$ {dPover P}=kdt$$



$$ int{dPover P}=int kdt$$



$$ color{red}{ln |P|}=kt+C$$



I don't understand the left hand side at this point, isn't $int{1over x}dx = ln |x| +C$? Where did the constant go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$




The standard equation for exponential growth and decay starts starts like this:



$$ {dPover dt}=kP$$



$$ {dPover P}=kdt$$



$$ int{dPover P}=int kdt$$



$$ color{red}{ln |P|}=kt+C$$



I don't understand the left hand side at this point, isn't $int{1over x}dx = ln |x| +C$? Where did the constant go?






calculus integration ordinary-differential-equations




calculus integration ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Eevee Trainer

7,13321337




7,13321337










asked 2 hours ago









agbltagblt

315114




315114








  • 3




    $begingroup$
    Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
    $endgroup$
    – JoseSquare
    2 hours ago














  • 3




    $begingroup$
    Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
    $endgroup$
    – JoseSquare
    2 hours ago








3




3




$begingroup$
Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
$endgroup$
– JoseSquare
2 hours ago




$begingroup$
Well if you mean that it should be $ln |P| + K_1 = kt + K_2$, then call $C=K_2-k_1$ and that is
$endgroup$
– JoseSquare
2 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

When you integrate both sides, each has a constant - you'd get, for constants $A,B$:



$$ int{dPover P}=int kdt implies ln|P|+A = kt+B$$



Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then



$$ln|P|+A = kt+B implies ln|P|=kt+B-A=kt+C$$



This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Well, notice that:



    $$lnleft|text{P}left(tright)right|+text{C}_1=text{k}cdot t+text{C}_2tag1$$



    Getting $text{C}_1$ on the other side gives:



    $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}_2-text{C}_1tag2$$



    But $text{C}_2-text{C}_1$ is another constant, so:
    $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}tag3$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You can write $$|P|=e^{kt}cdot e^{C}$$ so $$P=C_1e^{kt}$$ defining $$C_1=e^{C}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
        $endgroup$
        – Eevee Trainer
        2 hours ago










      • $begingroup$
        Exactly this have i written!
        $endgroup$
        – Dr. Sonnhard Graubner
        2 hours ago










      • $begingroup$
        Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
        $endgroup$
        – Eevee Trainer
        2 hours ago











      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












      $begingroup$

      When you integrate both sides, each has a constant - you'd get, for constants $A,B$:



      $$ int{dPover P}=int kdt implies ln|P|+A = kt+B$$



      Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then



      $$ln|P|+A = kt+B implies ln|P|=kt+B-A=kt+C$$



      This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        When you integrate both sides, each has a constant - you'd get, for constants $A,B$:



        $$ int{dPover P}=int kdt implies ln|P|+A = kt+B$$



        Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then



        $$ln|P|+A = kt+B implies ln|P|=kt+B-A=kt+C$$



        This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          When you integrate both sides, each has a constant - you'd get, for constants $A,B$:



          $$ int{dPover P}=int kdt implies ln|P|+A = kt+B$$



          Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then



          $$ln|P|+A = kt+B implies ln|P|=kt+B-A=kt+C$$



          This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.






          share|cite|improve this answer









          $endgroup$



          When you integrate both sides, each has a constant - you'd get, for constants $A,B$:



          $$ int{dPover P}=int kdt implies ln|P|+A = kt+B$$



          Well, we can subtract $A$ from both sides and define a constant $C = B-A$; then



          $$ln|P|+A = kt+B implies ln|P|=kt+B-A=kt+C$$



          This combination of constants is often implicit in solving differential equations - you'll integrate on two sides and then just combine the constants on whichever side of the equation is more convenient.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Eevee TrainerEevee Trainer

          7,13321337




          7,13321337























              3












              $begingroup$

              Well, notice that:



              $$lnleft|text{P}left(tright)right|+text{C}_1=text{k}cdot t+text{C}_2tag1$$



              Getting $text{C}_1$ on the other side gives:



              $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}_2-text{C}_1tag2$$



              But $text{C}_2-text{C}_1$ is another constant, so:
              $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}tag3$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Well, notice that:



                $$lnleft|text{P}left(tright)right|+text{C}_1=text{k}cdot t+text{C}_2tag1$$



                Getting $text{C}_1$ on the other side gives:



                $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}_2-text{C}_1tag2$$



                But $text{C}_2-text{C}_1$ is another constant, so:
                $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}tag3$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Well, notice that:



                  $$lnleft|text{P}left(tright)right|+text{C}_1=text{k}cdot t+text{C}_2tag1$$



                  Getting $text{C}_1$ on the other side gives:



                  $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}_2-text{C}_1tag2$$



                  But $text{C}_2-text{C}_1$ is another constant, so:
                  $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}tag3$$






                  share|cite|improve this answer









                  $endgroup$



                  Well, notice that:



                  $$lnleft|text{P}left(tright)right|+text{C}_1=text{k}cdot t+text{C}_2tag1$$



                  Getting $text{C}_1$ on the other side gives:



                  $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}_2-text{C}_1tag2$$



                  But $text{C}_2-text{C}_1$ is another constant, so:
                  $$lnleft|text{P}left(tright)right|=text{k}cdot t+text{C}tag3$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  JanJan

                  22k31340




                  22k31340























                      0












                      $begingroup$

                      You can write $$|P|=e^{kt}cdot e^{C}$$ so $$P=C_1e^{kt}$$ defining $$C_1=e^{C}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago










                      • $begingroup$
                        Exactly this have i written!
                        $endgroup$
                        – Dr. Sonnhard Graubner
                        2 hours ago










                      • $begingroup$
                        Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago
















                      0












                      $begingroup$

                      You can write $$|P|=e^{kt}cdot e^{C}$$ so $$P=C_1e^{kt}$$ defining $$C_1=e^{C}$$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago










                      • $begingroup$
                        Exactly this have i written!
                        $endgroup$
                        – Dr. Sonnhard Graubner
                        2 hours ago










                      • $begingroup$
                        Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago














                      0












                      0








                      0





                      $begingroup$

                      You can write $$|P|=e^{kt}cdot e^{C}$$ so $$P=C_1e^{kt}$$ defining $$C_1=e^{C}$$






                      share|cite|improve this answer











                      $endgroup$



                      You can write $$|P|=e^{kt}cdot e^{C}$$ so $$P=C_1e^{kt}$$ defining $$C_1=e^{C}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 hours ago

























                      answered 2 hours ago









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      76.8k42866




                      76.8k42866












                      • $begingroup$
                        OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago










                      • $begingroup$
                        Exactly this have i written!
                        $endgroup$
                        – Dr. Sonnhard Graubner
                        2 hours ago










                      • $begingroup$
                        Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago


















                      • $begingroup$
                        OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago










                      • $begingroup$
                        Exactly this have i written!
                        $endgroup$
                        – Dr. Sonnhard Graubner
                        2 hours ago










                      • $begingroup$
                        Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                        $endgroup$
                        – Eevee Trainer
                        2 hours ago
















                      $begingroup$
                      OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                      $endgroup$
                      – Eevee Trainer
                      2 hours ago




                      $begingroup$
                      OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                      $endgroup$
                      – Eevee Trainer
                      2 hours ago












                      $begingroup$
                      Exactly this have i written!
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      2 hours ago




                      $begingroup$
                      Exactly this have i written!
                      $endgroup$
                      – Dr. Sonnhard Graubner
                      2 hours ago












                      $begingroup$
                      Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                      $endgroup$
                      – Eevee Trainer
                      2 hours ago




                      $begingroup$
                      Still not addressing OP's concern. OP's concern is what happened to the constants after doing the integration $$int frac{dP}{P} = int k dt$$ Each side should generate a constant of integration, but in the work OP has seen, only the right side has that constant of integration. OP's concern is why there isn't one for the left side - it has nothing to do with how to find $P$ or that $e^C$ is another constant or any of that.
                      $endgroup$
                      – Eevee Trainer
                      2 hours ago


















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