Answer to Question #268208 in Differential Equations for Hemanth

Question

The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If

the population of a colony of yeast bacteria triples in one hour, find the number of bacteria which will be

present at the end of five hours.

Answer 1

$\displaystyle \text{Let N be the no of bacteria and $\frac{dN}{dt}$ = rate of increase of bacteria}\\ \frac{dN}{dt}=kN\\ \implies \frac{dN}{N}= kdt\\ \ln N = kt + A\\ e^{\ln N} = e^{kt + A}\\ \implies N = ce^{kt}\\ \text{Let N = 3c, t = 1, then we have that}\\ \implies 3c = ce^{kt}\\ \therefore e^k =3\implies k = 1.0986\\ \text{Next, we find the no of bacteria present after 5 hours}\\ N = ce^{5k}\\ \implies nc = ce^{1.0986\cdot 5}\\ \implies n = e^{1.0986\cdot 5}= 243\\ \text{Therefore, the number of bacteria present after 5 hours is 243 times more than the }\\ \text{initial population of bacteria}$