Let the population of rabbits surviving at a time t be governed by the differential equation `(d p(t)/(dt)=1/2p(t)-200.` If `p(0)""=""100` , then p(t) equals (1) `400-300""e^(t//2)` (2) `300-200""e^(-t//2)` (3) `600-500""e^(t//2)` (4) `400-300""e^(-t//2)`

Options

A. `40-300e^(t//2)`
B. `200-200e^(-t//2)`
C. `600-500e^(t//2)`
D. `400-300e^(-t//2)`

Answer

A

Explanation

`(dp)/(dt) = (p-400)/(2)`
`rArr (dp)/(p-400)=1/2dt`
Integrating, we get
`"ln "|p-400|=1/2t+c`
When `t=0, p=100`, we have ln 300=c
`therefore "ln"|(p-400)/(300)|=t/2`
`rArr |p-400|=300e^(t//2)`
`rArr 400-p=300e^(t//2)` (as `p lt 400)`
`rArr p=400-300e^(t//2)`