Question

For the population$p\left(t\right)$ at the time $t$ of a certain mouse species, 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. $2\log 18$
• B. $\log 9$
• C. $\frac{1}{2}\log 18$
• D. $\log 18$

Answer 1

The correct option is A

$2\log 18$

The explanation for the correct options:

Step1. The given differential equation:

Given that

The differential equation is $\frac{d\left(p\right(t\left)\right)}{dt}=0.5p\left(t\right)-450$

$$\begin{gathered} \Rightarrow \frac{d\left(p\right(t\left)\right)}{dt}=\frac{1}{2}p\left(t\right)-450 \\ \Rightarrow \frac{d\left(p\right(t\left)\right)}{dt}=\frac{p\left(t\right)-900}{2} \end{gathered}$$

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

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

Put $t=0$

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

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

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

Step2. Finding the time at which the population becomes zero:

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

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

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

$\Rightarrow$ $t=2\log \left(\frac{900}{50}\right)$

$\Rightarrow$ $t=2\log 18$

Hence, Option(A) is the correct answer.