The population p(t) at time t of a certain mo | vedclass

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(A) Given the differential equation: $\frac{dp(t)}{dt} = 0.5p(t) - 450 = \frac{p(t) - 900}{2}$.Separating the variables,we get: $\int \frac{dp(t)}{p(t

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$2 \ln 18$

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(A) Given the differential equation: $\frac{dp(t)}{dt} = 0.5p(t) - 450 = \frac{p(t) - 900}{2}$.Separating the variables,we get: $\int \frac{dp(t)}{p(t) - 900} = \int \frac{1}{2} dt$.Integrating both sides: $\ln |p(t) - 900| = \frac{1}{2} t + C$.Using the initial condition $p(0) = 850$: $\ln |850 - 900| = \frac{1}{2}(0) + C \implies C = \ln 50$.Thus,the equation is: $\ln |p(t) - 900| = \frac{1}{2} t + \ln 50$.We want to find $t$ when $p(t) = 0$: $\ln |0 - 900| = \frac{1}{2} t + \ln 50$.$\ln 900 - \ln 50 = \frac{1}{2} t$.$\ln \left( \frac{900}{50} \right) = \frac{1}{2} t$.$\ln 18 = \frac{1}{2} t$.$t = 2 \ln 18$.

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

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