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












4












$begingroup$


The standard equation for exponential growth and decay starts and is derived 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 of integration from the left integral go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    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
    5 hours ago
















4












$begingroup$


The standard equation for exponential growth and decay starts and is derived 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 of integration from the left integral go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    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
    5 hours ago














4












4








4





$begingroup$


The standard equation for exponential growth and decay starts and is derived 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 of integration from the left integral go?






calculus integration ordinary-differential-equations







share|cite|improve this question











$endgroup$




The standard equation for exponential growth and decay starts and is derived 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 of integration from the left integral 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 1 hour ago









Eevee Trainer

7,17321337




7,17321337










asked 5 hours ago









agbltagblt

320114




320114








  • 4




    $begingroup$
    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
    5 hours ago














  • 4




    $begingroup$
    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
    5 hours ago








4




4




$begingroup$
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
5 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
5 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

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









$endgroup$





















    4












    $begingroup$

    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












      $begingroup$

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






      share|cite|improve this answer











      $endgroup$









      • 1




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










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






      • 1




        $begingroup$
        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
        4 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      6












      $begingroup$

      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









      $endgroup$


















        6












        $begingroup$

        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









        $endgroup$
















          6












          6








          6





          $begingroup$

          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 5 hours ago









          Eevee TrainerEevee Trainer

          7,17321337




          7,17321337























              4












              $begingroup$

              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









              $endgroup$


















                4












                $begingroup$

                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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 5 hours ago









                  JanJan

                  22k31340




                  22k31340























                      -3












                      $begingroup$

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






                      share|cite|improve this answer











                      $endgroup$









                      • 1




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










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






                      • 1




                        $begingroup$
                        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
                        4 hours ago
















                      -3












                      $begingroup$

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






                      share|cite|improve this answer











                      $endgroup$









                      • 1




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










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






                      • 1




                        $begingroup$
                        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
                        4 hours ago














                      -3












                      -3








                      -3





                      $begingroup$

                      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 4 hours ago

























                      answered 5 hours ago









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      76.8k42866




                      76.8k42866








                      • 1




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










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






                      • 1




                        $begingroup$
                        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
                        4 hours ago














                      • 1




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










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






                      • 1




                        $begingroup$
                        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
                        4 hours ago








                      1




                      1




                      $begingroup$
                      OP's concern is not with trying to find $P$, it's about what happened to the constants of integration.
                      $endgroup$
                      – Eevee Trainer
                      5 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
                      5 hours ago












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




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




                      1




                      1




                      $begingroup$
                      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
                      4 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
                      4 hours ago


















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