Question

Let the population of rabbits surviving at a time t be governed by the differential equation $dp(t)/dt=(1/2)p(t)-200.$ If $p(0)=100,$ then p(t) equals

• A. $600-500e^{t/2}$
• B. $400-300e^{-t/2}$
• C. $400-300e^{t/2}$
• D. $300-200e^{-t/2.}$

Answer 1

Answer: (C)

Since $\frac{dp}{dt}-\frac{1}{2}p\left(t\right)=-200$ is lmear in $y$
$\therefore I.F.=e^{\int \frac{1}{2}dt}=e^{-\frac{t}{2}}$
$\therefore$ role is $p\cdot e^{\frac{-t}{2}}=\int -200\cdot \left(e^{-\frac{t}{2}}\right)dt+C$
$=-200\cdot \frac{e^{-t/2}}{-1/2}+C$
$=400e^{-t/2}+C$
Since $p\left(0\right)=100$
$\therefore 100e^0=400e^0+C$
$\Rightarrow 100=400+C$
$\Rightarrow C=-300$
$\therefore p\left(t\right)=400-300e^{t/2}$