How to write this as a differential equation?
I was solving differential equations problems from Mary Boas book, and there was a problem that I couldn't write it as a differential equation
Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant
rate by the removal of bacteria for experimental purposes.
How can I add the removal rate to my differential equation (I write the increase rate as $$dfrac{dN}{dt} = KN(t)$$where $K$ is a constant, and $N(t)$ is the number of bacteria at any time).
differential-equations
asked Nov 26 at 18:17
David Scott
82
add a comment |
I was solving differential equations problems from Mary Boas book, and there was a problem that I couldn't write it as a differential equation
Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant
rate by the removal of bacteria for experimental purposes.
How can I add the removal rate to my differential equation (I write the increase rate as $$dfrac{dN}{dt} = KN(t)$$where $K$ is a constant, and $N(t)$ is the number of bacteria at any time).
differential-equations
asked Nov 26 at 18:17
David Scott
82
add a comment |
I was solving differential equations problems from Mary Boas book, and there was a problem that I couldn't write it as a differential equation
Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant
rate by the removal of bacteria for experimental purposes.
How can I add the removal rate to my differential equation (I write the increase rate as $$dfrac{dN}{dt} = KN(t)$$where $K$ is a constant, and $N(t)$ is the number of bacteria at any time).
differential-equations
asked Nov 26 at 18:17
David Scott
82
I was solving differential equations problems from Mary Boas book, and there was a problem that I couldn't write it as a differential equation
Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant
rate by the removal of bacteria for experimental purposes.
How can I add the removal rate to my differential equation (I write the increase rate as $$dfrac{dN}{dt} = KN(t)$$where $K$ is a constant, and $N(t)$ is the number of bacteria at any time).
differential-equations
differential-equations
asked Nov 26 at 18:17
David Scott
82
asked Nov 26 at 18:17
David Scott
82
asked Nov 26 at 18:17
David Scott
82
asked Nov 26 at 18:17
David Scott
82
asked Nov 26 at 18:17
David Scott
82
82
add a comment |
add a comment |
1 Answer
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We should add the term for the removal at a constant rate $K_2>0$ that is
$$dfrac{dN(t)}{dt} = KN(t)-K_2 $$
answered Nov 26 at 18:19
gimusi
1
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We should add the term for the removal at a constant rate $K_2>0$ that is
$$dfrac{dN(t)}{dt} = KN(t)-K_2 $$
answered Nov 26 at 18:19
gimusi
1
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
add a comment |
We should add the term for the removal at a constant rate $K_2>0$ that is
$$dfrac{dN(t)}{dt} = KN(t)-K_2 $$
answered Nov 26 at 18:19
gimusi
1
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
add a comment |
We should add the term for the removal at a constant rate $K_2>0$ that is
$$dfrac{dN(t)}{dt} = KN(t)-K_2 $$
answered Nov 26 at 18:19
gimusi
1
We should add the term for the removal at a constant rate $K_2>0$ that is
$$dfrac{dN(t)}{dt} = KN(t)-K_2 $$
answered Nov 26 at 18:19
gimusi
1
answered Nov 26 at 18:19
gimusi
1
answered Nov 26 at 18:19
gimusi
1
answered Nov 26 at 18:19
gimusi
1
1
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
add a comment |
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
At the end of the book, the final answer is $N = N_0e^{Kt}-(R/K)(e^{Kt}-1)$, however, if I just add it like that I get $N = frac{N_0e^{Kt}+R}{K}$
– David Scott
Nov 26 at 18:23
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
You also need to adjust for the initial conditions $N(0)=N_0$, check that!
– gimusi
Nov 26 at 18:26
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it.
– gimusi
Nov 26 at 18:42
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
It does work for this boundary condition, THANK YOU.
– David Scott
Nov 26 at 21:07
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
@DavidScott You are welcome! Well done, Bye.
– gimusi
Nov 26 at 21:30
add a comment |
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