The population $\mathrm{p}(\mathrm{t})$ at time $t$ of a certain mouse species satisfies the differential…

The population $\mathrm{p}(\mathrm{t})$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=0.5 \mathrm{~p}(\mathrm{t})$ $-450$. If $p(0)=850$, then the time at which the population becomes zero is
  1. $2 \ln 18$
  2. $\ln 9$
  3. $\frac{1}{2} \ln 18$
  4. $\ln 18$

Solution

$\frac{d(p(t))}{d t}=\frac{1}{2} p(t)-450$ $\frac{d(p(t))}{d t}=\frac{p(t)-900}{2}$ $2 \int \frac{d(p(t))}{p(t)-900}=\int d t$ $2 \ln |p(t)-900|=t+c$ $t=0 \quad \Rightarrow 2 \ln 50=0+c \quad \Rightarrow c=2 \ln 50$ $\therefore 2 \ln |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+2 \ln 50$ $P(\mathrm{t})=0 \quad \Rightarrow 2 \ln 900=\mathrm{t}+2 \ln 50$ $t=2(\ln 900-\ln 50)=2 \ln \left(\frac{900}{50}\right)=2 \ln 18$.

Asked in: JEE Main 2012 (Offline)

Practice more Differential Equations questions on Aicharya