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  • Under natural circumstances, the population of a mice would increase at a rate that is proportional to the number of mice present at any time, provided the island has not cats at all !.
  • There were no cats from the beginning of $1970$ to the beginning of $1980$, during these time, the mouse population doubled,reaching all time high of $100,000$ at the beginning of $1980$.
  • People on the island alarmed by the increasing threat of mice important cats to kill them !. If the indicated increase of natural rate of mice is offset by the cats which kill $1000$ mice in one month.

What would be the mice remain at the beginning of $1981$ ?.


My attempt !:

Setting up a differential equation. let x denote the number of mice and t denote the year!

$$\frac{dx}{dt}∝x$$

k-constant of proportionality!

$$\frac{dx}{dt}=kx$$

Integrating we have

$$x=ce^{kt}$$

At 1970 we have $$x_0=ce^{1970k}$$ At 1980 we have $$2x_0=ce^{1980k}$$

Using ration of similar quantities, we then have

$$[2]^{\frac{1}{10}}=e^k$$ At 1980 we know can now determine c $$100,000=c[2]^{198}$$ $$c=\frac{100,000}{2^{198}}$$

The new differential equation is then denoted by. $$\frac{dx}{dt}=kx-12000$$ Solving we have $$kx-12000=ce^{kt}$$

Is this right way of doing it?

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  • $\begingroup$ how did you get 2^1/10 = e^k $\endgroup$ – Saketh Malyala Jun 15 '17 at 6:02
  • $\begingroup$ Ratio of the two equation then factor! $\endgroup$ – Crazy Jun 15 '17 at 6:03
  • $\begingroup$ Sorry I missed the second part where the mice also decrease. I deleted the answer. Please feel free to downvote it. $\endgroup$ – user1952500 Jun 15 '17 at 6:19
  • $\begingroup$ It is okay. No problem. The way you tackle the question is the same as me. Lol. $\endgroup$ – Crazy Jun 15 '17 at 6:22
  • $\begingroup$ Ah ok, I get it now, thanks for bearing with the comments. $\endgroup$ – user1952500 Jun 15 '17 at 6:39
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• The natural mouse population doubles in $10$ years.

• Beginning of $1980$, the population of mice is $100,000$.

• Cats begin to kill mice at $12000$ mice per year in $1980$.

That is the information we have.

So we have $\displaystyle \frac{dx}{dt}=\frac{\ln2}{10}(x)-12000$.

This is because normally, the exponential growth $\displaystyle ce^{kt}$ has differential equation $\displaystyle \frac{dx}{dt}=kx$.

We add the $-12000$ because those are the mouse deaths from cats over the years, assuming they live forever.

Solving this equation gives $\displaystyle x = C\left(2^\frac{t}{10}\right)+\frac{120000}{\ln(2)}$ .

Therefore, assuming $t=0$ corresponds to the year $1980$, we have $\displaystyle C=\left(100000-\frac{120000}{\ln 2}\right)$.

We then have $\displaystyle x = \left(100000-\frac{120000}{\ln 2}\right)2^{\frac{t}{10}}+\frac{120000}{\ln 2}$.

Evaluating at $t=1$ corresponding to the year $1981$ gives about $94751.6800374$ which is about $\boxed{94752}$ mice.

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