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

Answer 1

Answer: (B)

$2\ln 18$

We have,

$\frac{dp\left(t\right)}{dt}=0.5p\left(t\right)-450$

$\Rightarrow \int \frac{dp\left(t\right)}{0.5p\left(t\right)-450}=\int dt$

$\Rightarrow \frac{1}{0.5}In\left(0.5p\left(t\right)-450\right)=t+c$

$Sincep\left(0\right)=850$

$\frac{1}{0.5}In\left(\frac{0.5}{10}\times 850-450\right)=0+c\frac{1}{0.5}In\left(-25\right)+cc=In{\left(-25\right)}^2=In\left(625\right)$

$So,\frac{1}{0.5}In\left(0.5p\left(t\right)-450\right)=t+In\left(625\right)$

$Populationbecomeszero,i.ep\left(t\right)=0\frac{1}{0.5}In\left(0.5\times 0-450\right)=t+In\left(625\right)$

$t=In\left\{{\left(-450\right)}^2\right\}-In\left(625\right)$

$t=\ln \left(324\right)$

$\therefore t=2In18$

Hence, this is the answer.