Question #291400

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 dndn / dtdt = 1 / 20k20k 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]


1
Expert's answer
2022-01-28T14:46:37-0500

The given DE is;

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

a] By separation of variables we have;

1n(kn)dndt=120k\displaystyle \frac{1}{n(k-n)}\frac{dn}{dt}=\frac{1}{20k}\\

Integrating both sides wrt t yields;

1n(kn)dndt dt=120kdt1n(kn) dn=120k dt(1kn+1k(kn)) dn=120k dt1kn dn+1k(kn) dn=120k dtln(n)kln(kn)k=t20k+c, where c is an arbitrary constantln(n)ln(kn)=t20+ck,ln(nkn)=0.05t+cknkn=c1e0.05t, where c1=eckne0.05t=c1(kn)ne0.05t+nc1=c1kn=c1kc1+e0.05t=2001+1c1e0.05t=2001+c2e0.05t, where c2=1c1 and k=200\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 t=0, n=50\displaystyle t=0,\ n=50 thus;

50=2001+c21+c2=4c2=3.\displaystyle 50=\frac{200}{1+c_2}\Rightarrow1+c_2=4\Rightarrow c_2=3.

Thus, n=2001+3e0.05t\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;

n(2023)=2001+3e0.05×2023=200 bears\displaystyle n(2023)=\frac{200}{1+3e^{-0.05\times 2023}}=200 \text{ bears}


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