A population \( P \) obeys the logistic model. It satisfies the equation \( \frac{d P}{d t}=\frac{5}{900} P(9-P) \) for \( P>0 \). (c) Assume that \( P(0)=4 \). Find \( P(84) \). \( P(84)= \)

To solve the logistic growth problem, we start with the differential equation: \[ \frac{dP}{dt} = \frac{5}{900} P(9 - P) \] This can be rewritten as: \[ \frac{dP}{P(9 - P)} = \frac{5}{900} dt \] Next, we can use partial fractions on the left side: \[ \frac{1}{P(9 - P)} = \frac{A}{P} + \frac{B}{9 - P} \] Solving for \( A \) and \( B \), we find \( A = \frac{1}{9} \) and \( B = \frac{1}{9} \). Therefore, we can express our equation as: \[ \frac{1}{9} \left( \frac{1}{P} + \frac{1}{9 - P} \right) dP = \frac{5}{900} dt \] Integrating both sides, we get: \[ \frac{1}{9} \ln |P| - \frac{1}{9} \ln |9 - P| = \frac{5}{900} t + C \] This simplifies to: \[ \ln \left| \frac{P}{9 - P} \right| = \frac{5}{100} t + C' \] Exponentiating gives: \[ \frac{P}{9 - P} = e^{C'} e^{\frac{5}{100} t} = K e^{\frac{5}{100} t} \] At \( t = 0 \), we can solve for \( K \) using the initial condition \( P(0) = 4 \): \[ \frac{4}{9 - 4} = K \implies K = \frac{4}{5} \] Thus, the equation becomes: \[ \frac{P}{9 - P} = \frac{4}{5} e^{\frac{5}{100} t} \] Now we need to find \( P(84) \): Substituting \( t = 84 \): \[ \frac{P(84)}{9 - P(84)} = \frac{4}{5} e^{\frac{5}{100} \cdot 84} \] Calculating \( e^{4.2} \): \[ \frac{P(84)}{9 - P(84)} = \frac{4}{5} e^{4.2} \approx \frac{4}{5} \cdot 66.686 \] \[ \approx 53.349 \] Let \( P(84) = x \), then: \[ \frac{x}{9 - x} = 53.349 \] Cross multiplying gives: \[ x = 53.349(9 - x) \] \[ x + 53.349x = 480.141 \] \[ 54.349x = 480.141 \] \[ x \approx \frac{480.141}{54.349} \approx 8.826 \] Thus, \[ P(84) \approx 8.83 \]