1
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

We should add the term for the removal at a constant rate $K_2>0$ that is

$$\dfrac{dN(t)}{dt} = KN(t)-K_2 $$

$\endgroup$
  • $\begingroup$ 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 $N = \frac{N_0e^{Kt}+R}{K}$ $\endgroup$ – David Scott Nov 26 '18 at 18:23
  • $\begingroup$ You also need to adjust for the initial conditions $N(0)=N_0$, check that! $\endgroup$ – gimusi Nov 26 '18 at 18:26
  • $\begingroup$ Your solution is not compatible with the initial condition. If you show your derivation I’ll take a look to it. $\endgroup$ – gimusi Nov 26 '18 at 18:42
  • $\begingroup$ It does work for this boundary condition, THANK YOU. $\endgroup$ – David Scott Nov 26 '18 at 21:07
  • $\begingroup$ @DavidScott You are welcome! Well done, Bye. $\endgroup$ – gimusi Nov 26 '18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.