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(A) Given the differential equation: $\frac{dP}{dt} = 0.05 P$. Separating the variables,we get: $\frac{dP}{P} = 0.05 dt$. Integrating both sides: $\in

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$20 \ln 2$ years

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(A) Given the differential equation: $\frac{dP}{dt} = 0.05 P$. Separating the variables,we get: $\frac{dP}{P} = 0.05 dt$. Integrating both sides: $\int \frac{dP}{P} = \int 0.05 dt$. This gives: $\ln P = 0.05 t + C$. At $t = 0$,let the initial population be $P_0$,so $\ln P_0 = C$. Substituting $C$ back into the equation: $\ln P = 0.05 t + \ln P_0$,which simplifies to $\ln(\frac{P}{P_0}) = 0.05 t$. We want to find $t$ when the population doubles,i.e.,$P = 2P_0$. Substituting this: $\ln(\frac{2P_0}{P_0}) = 0.05 t$. $\ln 2 = 0.05 t$. Since $0.05 = \frac{1}{20}$,we have $\ln 2 = \frac{t}{20}$. Therefore,$t = 20 \ln 2$ years.

$A$ population $P$ grows at the rate given by the equation $\frac{dP}{dt} = 0.05 P$. In how many years will the population double?

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