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

$$\frac{dp(t)}{dt}=\frac{1}{2}{p(t)-400}$$dtdp(t)​=21​p(t)−400

$$p(0)=100$$p(0)=100

$$\frac{2dp(t)}{p(t)-400}=dt$$p(t)−4002dp(t)​=dt

Integrating,

$$\Rightarrow {\int }_{0}^{p(t)}\frac{\mathrm{dp}(t)}{p(t)-400}={\int }_{0}^{t}\frac{1}{2}\mathrm{dt}\Rightarrow {(\mathrm{ln}|p(t)-400|)}_{0}^{p(t)}=\frac{t}{2}$$⇒∫0p(t)​p(t)−400dp(t)​=∫0t​21​dt⇒(ln∣p(t)−400∣)0p(t)​=2t​

$$\Rightarrow \mathrm{ln}|\frac{p(t)-400}{300}|=\frac{t}{2}$$⇒ln∣300p(t)−400​∣=2t​

$$\Rightarrow |p(t)-400|=300{e}^{\frac{t}{2}}$$⇒∣p(t)−400∣=300e2t​

$$\Rightarrow 400-p(t)=300{e}^{\frac{t}{2}} (\therefore p(t)<400)$$⇒400−p(t)=300e2t​(∴p(t)<400)

$$\Rightarrow p(t)=400-300{e}^{\frac{t}{2}}$$⇒p(t)=400−300e2t​