The population $\mathrm{P}=\mathrm{P}(\mathrm{t})$ at time $\mathrm{t}$ of certain species follows the…

The population $\mathrm{P}=\mathrm{P}(\mathrm{t})$ at time $\mathrm{t}$ of certain species follows the differential equation $\frac{\mathrm{dP}}{\mathrm{dt}}=0.5 \mathrm{P}-450$. If $\mathrm{P}(0)=850$, then the time at which population becomes zero is
  1. $2 \log 18$
  2. $\log 9$
  3. $\frac{1}{2} \log 18$
  4. $\log 18$

Solution

Given differential equation is $\begin{aligned} & \quad \frac{\mathrm{dP}}{\mathrm{dt}}=0.5 \mathrm{P}-450 \\ & \quad=\frac{\mathrm{P}}{2}-\frac{900}{2} \\ & \therefore \quad \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{P}-900}{2} \\ & \therefore \quad \frac{2 \mathrm{dP}}{\mathrm{P}-900}=\mathrm{dt} \\ & \quad \text { Integrating on both sides, we get } \\ & 2 \log |\mathrm{P}-900|=\mathrm{t}+\mathrm{c} \\ & \mathrm{P}(0)=850 \text { i.e., } \mathrm{P}=850 \text { when } \mathrm{t}=0 \\ & \mathrm{c}=2 \log 50 \\ & 2 \log |\mathrm{P}-900|=\mathrm{t}+2 \log 50 \\ & \mathrm{When} \mathrm{P}=0, \\ & 2 \log 900=\mathrm{t}+2 \log 50 \\ & \Rightarrow \mathrm{t}=2(\log 900-\log 50) \\ & \quad=2 \log \frac{900}{50}=2 \log 18 \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 1)

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