91Ó°ÊÓ

Using the logistic equation and the given parameter values, we can write the equation as: $$\frac{dP}{dt} = 0.2P \left(1-\frac{P}{300}\right)$$

To solve this differential equation, we need to solve for \(P(t)\) given the initial value \(P_{0}=50\). Let's rewrite the given ODE in the following form $$\frac{dP}{t} = \frac{0.2P(300-P)}{300}$$ We have the pattern of a separable equation: $$\frac{dP}{P(300-P)} = \frac{0.2 dt}{300}$$ Now integrate both sides: $$\int\left(\frac{1}{P}+\frac{1}{300-P}\right)dP = \int\frac{0.2}{300}dt$$ Use partial fraction decomposition on the left-hand side: $$\int\left(\frac{A}{P}+\frac{B}{300-P}\right)dP = A \int\frac{dP}{P} +B \int\frac{dP}{300-P}$$ where \(A\) and \(B\) are constants. By matching coefficients, we find \(A=1\) and \(B=1\). Therefore, the equation becomes: $$\int\frac{dP}{P} + \int\frac{dP}{300-P} = \int\frac{0.2}{300}dt$$ Now integrate: $$\ln |P|-\ln |300-P| = \frac{0.2t}{300} + C$$ Apply exponentiation to eliminate the logarithm: $$\frac{P}{300-P} = e^{\frac{0.2t}{300}+C}$$

Solve for the constant \(C\) using the initial condition \(P_{0}=50\) at \(t=0\): $$\frac{50}{300-50} = e^{C}$$ $$C = \ln\left(\frac{50}{250}\right)$$

Rewriting the previous solution we get $$P(t) = \frac{300e^{\frac{0.2t}{300}+\ln(\frac{50}{250})}}{1+e^{\frac{0.2t}{300}+\ln(\frac{50}{250})}}$$ Simplifying this expression gives us the final solution: $$P(t) = \frac{300(50)}{250(1-e^{-\frac{0.2t}{300}})+50}$$

To graph the solution, plot the function \(P(t)\) over a suitable time interval, say \(0 \leq t \leq 100\). Note that the population will approach the carrying capacity \(K=300\) as time goes on, and observing the graph will help visualize this.