The population p(t) at time t of a certain mouse species satisfies the differential equation

$\frac{d}{dt}(p(t))= 0.5p(t)- 450$

If $p(0) = 850 ,$ then the time at which the population becomes zero is

$2 \, ln \, 18$

The correct answer is option (1) : $2 \, ln \, 18$

We have,

$\frac{d}{dt}(p(t))= 0.5p(t) - 450$

$⇒\frac{2}{p(t) - 900}d(p(t))= dt $

$⇒\frac{2}{900-p(t)}d(p(t))= -dt $

On integrating, we obtain

$-2\, ln (900- p(t)) = - t + C$ ...........(i)

Putting $t=0$ and $p(0)= 850$, we obtain : $-2\, ln \, 50 = C$

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

$⇒ln\, (900- p(t))^2 = t + ln \, 50^2 $

$⇒t = ln \begin{Bmatrix}\frac{900-p(t)}{50}\end{Bmatrix}^2$

Suppose $p(t) = 0$ when $t=t_1$. Then,

$t_1= ln\left(\frac{900}{50}\right)^2 = ln\, 18^2 = 2\, ln \, 18 $