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

Why do people break up the derivative notation when setting $u$ and $v$ for integration by parts?Constant of Integration of the Integrating FactorIntegrating two equations that equal, what happens to the constant on one of the sides?The constant of integration during integration by partsIdentifying the constant of integrationSkipping integration constantsIntegration by Parts and the Constant of IntegrationFind 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?

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

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The constant on the right hand side was integrated into the constant on the left hand side. Or is it the other way around? I can never differentiate left from right.
  $endgroup$
  – Servaes
  2 hours ago
  

  
  
  
  

add a comment | 

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

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The constant on the right hand side was integrated into the constant on the left hand side. Or is it the other way around? I can never differentiate left from right.
  $endgroup$
  – Servaes
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

$endgroup$

share|cite|improve this question

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

share|cite|improve this question

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

edited 5 hours ago

Eevee Trainer

7,23821337

edited 5 hours ago

Eevee Trainer

7,23821337

asked 8 hours ago

agbltagblt

320114

asked 8 hours ago

agbltagblt

320114

3 Answers
3

active

oldest

votes

7

$begingroup$

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

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

$endgroup$

add a comment | 

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

answered 8 hours ago

JanJan

22k31340

$endgroup$

add a comment | 

-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

edited 8 hours ago

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

• 
  
  
  

  
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

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

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

$endgroup$

add a comment | 

7

$begingroup$

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

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

$endgroup$

add a comment | 

$begingroup$

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

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

answered 8 hours ago

Eevee TrainerEevee Trainer

7,23821337

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

answered 8 hours ago

JanJan

22k31340

$endgroup$

add a comment | 

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

answered 8 hours ago

JanJan

22k31340

$endgroup$

add a comment | 

$begingroup$

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

answered 8 hours ago

JanJan

22k31340

$endgroup$

share|cite|improve this answer

answered 8 hours ago

JanJan

22k31340

answered 8 hours ago

JanJan

22k31340

answered 8 hours ago

JanJan

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

edited 8 hours ago

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

• 
  
  
  

  
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment | 

-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

edited 8 hours ago

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

• 
  
  
  

  
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866