The population P = P(t) at time t of a certai | vedclass

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(D) Given the differential equation: $\frac{dP}{dt} = 0.5P - 450 = 0.5(P - 900)$.Separating the variables,we get: $\frac{dP}{P - 900} = 0.5 dt$.Integr

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$2 \log_{e} 18$

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(D) Given the differential equation: $\frac{dP}{dt} = 0.5P - 450 = 0.5(P - 900)$.Separating the variables,we get: $\frac{dP}{P - 900} = 0.5 dt$.Integrating both sides: $\int \frac{dP}{P - 900} = \int 0.5 dt$.This gives: $\ln|P - 900| = 0.5t + C$.Using the initial condition $P(0) = 850$: $\ln|850 - 900| = 0.5(0) + C \Rightarrow C = \ln(50)$.So,the equation becomes: $\ln|P(t) - 900| = 0.5t + \ln(50)$.We want to find $t$ when $P(t) = 0$: $\ln|0 - 900| = 0.5t + \ln(50)$.$\ln(900) - \ln(50) = 0.5t$.$\ln\left(\frac{900}{50}\right) = 0.5t$.$\ln(18) = 0.5t$.$t = 2 \ln(18)$.

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

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