Solving the Equation for Bacterial Growth in Experimentation

[iamthegelo]

Homework Statement

The growth of bacteria is proportional to the number present but is being reduced at a constant rate for experimentation. Find the equation for N

Homework Equations

d(N)/dt = KN-R , N = number of bacteria, K = proportionality const., R = reduction rate.

The Attempt at a Solution

Above was my guess to what the differential equation should look like...it's a separable equation. My question: Is the above formula the right formula based on my understanding of the question? With that equation I can't seem to solve it.

 

[clamtrox]

Seems reasonable. Try doing the regular thing, which is to separate t and N to different sides of the equation like this

\frac{dN}{KN-R} = dt

and then integrate. You also need to assume that KN>R. What does this mean?

 

[iamthegelo]

I tried that but I just can't seem to get it right. I don't get the same solution as the book. By the way the question is No. 22 of Section 2 in Chapter 8 of Boas' Math Methods for those that have the book.

I integrate those and get,

(1/K)*ln(KN-R) = t + ln(C) , C is the initial number of bacteria...I put in ln(C) for simplification.

 

[HallsofIvy]

Okay, so you have ln(KN- R)= KT+ Kln(C) and from that KN-R= (Ce^K) e^{KT}. KN= R+ (Ce^K)e^{KT} and, finally, N= R/K+ (Ce^K/K)e^{KT}. What is the answer in the book?

 

[iamthegelo]

The answer in the book is,

N = Ce^{Kt} - (R/K)(e^{Kt}-1)

 

[clamtrox]

So, how do you solve the DE?

\int_{N(t_0)}^{N(t)} \frac{dN}{KN-R} = \int_{t_0}^t dt

where t_0 is arbitrary (choose zero)

 

[iamthegelo]

Thanks, I just got it. Can you tell me why I needed the limits? The other one before this didn't have the R (reduction) so the equation was just

N = N_0e^{Kt}

When I was solving that I didn't need limits.

 

[clamtrox]

The limits contain the constant of integration in a more transparent way; you don't need to be guessing the correct forms. You could have just as well derived the case R=0 with the limits, probably with less work too.