Let the population of rabbits surviving at a time $t$ be governed by the differential equation $\frac{d}{d t}(p(t))=\frac{1}{2} p(t)-200$. If $p(0)=100$, then $p(t)$ equals

$400-300 e^{t / 2}$

We have,

$\frac{d}{d t}(p(t))=\frac{1}{2} p(t)-200$

$\Rightarrow \frac{d}{d t}(p(t))+\left(-\frac{1}{2}\right) p(t)=-200$           ....(i)

This is a linear differential equation with integrating factor $=e^{\int-\frac{1}{2} d t}=e^{-\frac{t}{2}}$

Multiplying both sides of (i) by I.F. $=e^{-t / 2}$, we obtain

$e^{-t / 2} \frac{d}{d t}(p(t))+\left(-\frac{1}{2}\right) p(t) e^{-t / 2}=-200 e^{-t / 2}$

Integrating both sides with respect to $t$, we get

$p(t) e^{-t / 2}=400 e^{-t / 2}+C$          ......(ii)

Putting $t=0$ and $p(0)=100$, we get

$100=400+C \Rightarrow C=-300$

Putting $C=-300$, we get

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

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