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) `40-300""e^(-t//2)`

Options

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

Answer

C

Explanation

We have,
`(d)/(dt)(f(t))=(1)/(2)p(t)-200`
`rArr" "(d)/(dt)(p(t))+(-(1)/(2))p(t)=-200" …(i)"`
This is a linear differential equation with I.F. `=e^(int-(1)/(2)dt)=e^(-(1)/(2))`
Multiplying both sides of (i) by I.F. `=e^(-t//2)`, we obtain
`e^(-t//2)(d)/(dt)(p(t))+(-(1)/(2))p(t)e^(-t//2)=-200e^(-t//2)`
Integrating both sides with respect to t, we get
`P(t)e^(-t//2)=400e^(-t//2)+C" ...(ii)"`
Putting t = 0 and p(0) = 100, we get
`100=400+CrArr C=-300`
Putting C =` -300`, we get
`p(t)e^(-t//2)=400e^(-t//2)-300`
`rArr" "p(t)=400-300e^(t//2)`