Question

The population $p\left(t\right)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{dp\left(t\right)}{dt}=0.5p\left(t\right)-450$. If $p\left(0\right)=850$, then the time at which the population becomes zero is:

• A. $2\ln 18$
• B. $\ln 9$
• C. $\frac{1}{2}\ln 18$
• D. $\ln 18$

Answer 1

The correct option is A $2\ln 18$

$\frac{dp\left(t\right)}{dt}=\frac{1}{2}p\left(t\right)-450\Rightarrow \frac{dp\left(t\right)}{dt}-\frac{1}{2}p\left(t\right)=-450$

$\therefore I.F.=e^{\int (-1/2)dt}=e^{-t/2}$

Hence, general solution is
$p\left(t\right)\cdot e^{-t/2}=\int (-450\times e^{-t/2})dt+c$
$\Rightarrow p\left(t\right)\cdot e^{-t/2}=900p\left(t\right)\cdot e^{-t/2}+c$
$\Rightarrow p\left(t\right)=900+c{e}^{t/2}$

Given, $p\left(0\right)=850$
$\Rightarrow c=-50$

$\therefore p\left(t\right)=900-50e^{t/2}$

Let, at time $t_1$ the population becomes zero, then
$\Rightarrow 0=900-50e^{t_1/2}$
$\Rightarrow t_1=2\ln 18$