Answer to Question #291400 in Differential Equations for chickenapple

Question

A research is being done on a particular species of bear on an animal reserve territory. Initially, there are 50 bears on the reserve. After t years the number of bears, n, satisfies the differential equation $dn$ / $dt$ = 1 / $20k$ n(k-n), where k are constant.

a.) By using saperable viarable, show that the general solution of the differential equation is n = 200 / 1 +3e^-0.05t given that k = 200 [14 marks]

b.) In 2012, the population of the bears in the territory are 200. Estimate the population of the bears that will be able to survive after 2022. [3 marks]

Answer 1

The given DE is;

$\displaystyle \frac{dn}{dt}=\frac{n(k-n)}{20k}$

a] By separation of variables we have;

$\displaystyle \frac{1}{n(k-n)}\frac{dn}{dt}=\frac{1}{20k}\\$

Integrating both sides wrt t yields;

$\displaystyle \int \frac{1}{n(k-n)}\frac{dn}{dt}\ dt=\int\frac{1}{20k}dt\\ \Rightarrow\int\frac{1}{n(k-n)}\ dn=\frac{1}{20k}\int\ dt\\ \Rightarrow\int\left(\frac{1}{kn}+\frac{1}{k(k-n)}\right)\ dn=\frac{1}{20k}\int\ dt\\ \Rightarrow \int\frac{1}{kn}\ dn+\int\frac{1}{k(k-n)}\ dn=\frac{1}{20k}\int\ dt\\ \Rightarrow\frac{ln(n)}{k}-\frac{\ln(k-n)}{k}=\frac{t}{20k}+c,\ \text{where c is an arbitrary constant}\\ \Rightarrow \ln(n)-\ln(k-n)=\frac{t}{20}+ck,\\ \Rightarrow \ln\left(\frac{n}{k-n}\right)=0.05t+ck\\ \Rightarrow \frac{n}{k-n}=c_1e^{0.05t},\ \text{where } c_1=e^{ck}\\ \Rightarrow ne^{-0.05t}=c_1(k-n)\\ \Rightarrow ne^{-0.05t}+nc_1=c_1k\\ \Rightarrow n=\frac{c_1k}{c_1+e^{-0.05t}}=\frac{200}{1+\frac{1}{c_1}e^{-0.05t}}=\frac{200}{1+c_2e^{-0.05t}},\ \text{where }c_2=\frac{1}{c_1}\text{ and }k=200\\$

Now, at $\displaystyle t=0,\ n=50$ thus;

$\displaystyle 50=\frac{200}{1+c_2}\Rightarrow1+c_2=4\Rightarrow c_2=3.$

Thus, $\displaystyle n=\frac{200}{1+3e^{-0.05t}}$

b] Since the question says after 2022, then the population of bears that will be able to survive after 2022 is;

$\displaystyle n(2023)=\frac{200}{1+3e^{-0.05\times 2023}}=200 \text{ bears}$