A rabbit population in a reserve starts at 800 and grows at a rate proportional to its size, with growth constant 0.12 per year. Solve the differential equation, find the population after 5 years, and find the doubling time.

Step-by-step solution

1. 1

   Separate the variables. The equation $\dfrac{dP}{dt}=0.12P$ is separable — collect $P$ on one side and $t$ on the other:

   $\frac{dP}{P}=0.12\,dt$

2. 2

   Integrate both sides and solve for P.

   $\ln|P|=0.12t+C\quad\Longrightarrow\quad P(t)=Ae^{0.12t}$

   Applying $P(0)=800$ gives $A=800$, so

   $P(t)=800e^{0.12t}$

3. 3

   Evaluate at t = 5.

   $P(5)=800e^{0.12(5)}=800e^{0.6}=800(1.82212)=1457.7$

   Rounding to whole rabbits,

   $P(5)\approx 1458\ \text{rabbits}$

4. 4

   Set up the doubling condition. Doubling means $P(t)=2P(0)$, i.e.

   $800e^{0.12t}=1600\quad\Longrightarrow\quad e^{0.12t}=2$

5. 5

   Solve for the doubling time.

   $0.12t=\ln 2\quad\Longrightarrow\quad t=\frac{\ln 2}{0.12}=\frac{0.69315}{0.12}=5.776\ \text{years}$

   $\boxed{P(5)\approx 1458\ \text{rabbits};\qquad t_{\text{double}}\approx 5.78\ \text{years}}$

6. 6

   Note that the doubling time is independent of the starting size. The $800$ cancels in $\tfrac{P(t)}{P(0)}=e^{kt}$, so any exponentially growing population with $k=0.12$ doubles every $5.78$ years. Cross-check: after $5.776$ years the model gives $800e^{0.12(5.776)}=800e^{0.6931}=1600$ ✓, and after $5$ years the ratio $\tfrac{1458}{800}=1.822$ is just under $2$, consistent with a doubling time slightly over $5$ years.