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).


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

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

  • $\begingroup$ 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

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