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 $2\ln 18$

Given differential equation is
$\frac{dp\left(t\right)}{dt}=0.5p\left(t\right)-450$
$\Rightarrow \frac{dp\left(t\right)}{dt}=\frac{1}{2}p\left(t\right)-450=\frac{p\left(t\right)-900}{2}$
$\Rightarrow 2\frac{dp\left(t\right)}{dt}=-[900-p(t\left)\right]\Rightarrow 2\frac{dp\left(t\right)}{900-p\left(t\right)}=-dt$
Integrating both sides, we get
$-2\int \frac{dp\left(t\right)}{900-p\left(t\right)}=\int dt$
$\Rightarrow 2\ln [900-p(t\left)\right]=t+c$
when $t=0,p\left(0\right)=850$
$2\ln 50=c$
$\Rightarrow \left[\ln \left(\frac{900-p\left(t\right)}{50}\right)\right]=\frac{t}{2}\Rightarrow 900-p\left(t\right)=50e^{\frac{t}{2}}$
$\Rightarrow p\left(t\right)=900-50e^{\frac{t}{2}}$
Let $p\left(t_1\right)=0$
So, $0=900-50e^{\frac{t_1}{2}}$
$\therefore t_1=2\ln 18$