The population P(t) of a certain mouse specie | vedclass

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(C) Given the differential equation: $\frac{dP(t)}{dt} = 0.5 P(t) - 450$. This can be written as: $\frac{dP(t)}{dt} = \frac{1}{2} P(t) - 450 = \frac{P

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$2 \log 18$

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(C) Given the differential equation: $\frac{dP(t)}{dt} = 0.5 P(t) - 450$. This can be written as: $\frac{dP(t)}{dt} = \frac{1}{2} P(t) - 450 = \frac{P(t) - 900}{2}$. Separating the variables,we get: $\int \frac{dP(t)}{P(t) - 900} = \int \frac{1}{2} dt$. Integrating both sides: $\log |P(t) - 900| = \frac{1}{2} t + C$. Multiplying by $2$: $2 \log |P(t) - 900| = t + C'$. Given $P(0) = 850$,we substitute these values: $2 \log |850 - 900| = 0 + C' \Rightarrow C' = 2 \log 50$. Thus,the equation is: $2 \log |P(t) - 900| = t + 2 \log 50$. To find the time $t$ when the population becomes zero,set $P(t) = 0$: $2 \log |0 - 900| = t + 2 \log 50$. $t = 2 \log 900 - 2 \log 50 = 2 \log \left( \frac{900}{50} \right) = 2 \log 18$.

The population $P(t)$ of a certain mouse species at time $t$ satisfies the differential equation $\frac{dP(t)}{dt} = 0.5 P(t) - 450$. If $P(0) = 850$,then the time at which the population becomes zero is

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