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  • The population P=P(t) at time 't' of acertain species follows the differential equation

The population P=P(t) at time 't' of acertain species follows the differential equation \frac{dp}{dt}=0.5P-450.\;I\! \! f \; P(0)=850. Then the time at which population become zero:
Option: 1 \log_{e}18
 
Option: 2 \frac{1}{2}\log_{e}18
Option: 3 \log_{e}9  
Option: 4 2\log_{e}18

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\begin{aligned} &\frac{\mathrm{dP}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{P}(\mathrm{t})-900}{2}\\ &\int_{0}^{t} \frac{\mathrm{d} \mathrm{P}(\mathrm{t})}{\mathrm{P}(\mathrm{t})-900}=\int_{0}^{\mathrm{t}} \frac{\mathrm{dt}}{2}\\ &\{\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|\}_{0}^{\mathrm{t}}=\left\{\frac{\mathrm{t}}{2}\right\}_{0}^{\mathrm{t}}\\ &\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ell \mathrm{n}|\mathrm{P}(0)-900|=\frac{\mathrm{t}}{2}\\ &\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}}{2}\\ &\text { Let at } t=t_{1}, P(t)=0 \text { hence }\\ &\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ln 50=\frac{\mathrm{t}_{1}}{2}\\ &\mathrm{t}_{1}=2 \ell \mathrm{n} 18 \end{aligned}

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Suraj Bhandari

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