Question

The population $p\left(t\right)$ a time $t$ of a certain mouse species satisfies the differential equation $\frac{dp\left(t\right)}{dt}=\frac{1}{2}p\left(t\right)-450$. If $p\left(0\right)=850$, then the time at which the population becomes zero is:

• A. $2ln18$
• B. $ln9$
• C. $\frac{1}{2}ln18$
• D. $ln18$

Answer 1

The correct option is A $2ln18$

$A.2\log 18$

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

$\frac{d\left(p\right(t\left)\right)}{dt}=\frac{p\left(t\right)-900}{2}$

$2\int \frac{d\left(p\right(t\left)\right)}{p\left(t\right)-900}=\int dt$

$2\ln |p\left(t\right)-900|=t+c$

$t=0$

$\Rightarrow 2\ln 50=0+c$

$\therefore 2\ln |p\left(t\right)-900|=t+2\ln 50$

$p\left(t\right)=0$

$\Rightarrow 2\ln 900=t+2\ln 50$

$t=2(\ln 900-\ln 50)$

$=2\ln \left(\frac{900}{50}\right)$

$=2\ln 18$