Let the population of rabbits surviving at a time $$t$$ be governed by the differential equation $$\frac{dp(t)}{dt} = \frac{1}{2}\{p(t) - 400\}$$. If $$p(0) = 100$$, then $$p(t)$$ equals:

$$\frac{dp}{p - 400} = \frac{1}{2} dt$$

$$\int \frac{dp}{p - 400} = \int \frac{1}{2} dt$$

$$\ln|p - 400| = \frac{t}{2} + C$$

$$p - 400 = e^{t/2 + C} = e^C \cdot e^{t/2}$$

$$p(t) = 400 + K e^{t/2} \quad (\text{where } K = e^C)$$

Using $$p(0) = 100$$: $$100 = 400 + K e^0 \implies K = -300$$

$$p(t) = 400 - 300 e^{t/2}$$

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