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

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