Question

Solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{\mathrm{o}}\) the initial population. $$r=0.2, K=300, P_{0}=50$$

Short answer

Answer: The general equation for the population function, P(t), is given by: \(P(t) = \frac{300 \cdot 50e^{60t}}{50e^{60t} + 300 - 50}\).

Step-by-step solution

Write down the logistic growth equation with given values

We have the logistic growth equation: $$\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{300}\right)$$

Solve the logistic growth differential equation

To solve this separable differential equation, we can rewrite it as: $$\frac{dP}{P(1 - P/300)} = 0.2dt$$ Integrate both sides: $$\int \frac{dP}{P(1 - P/300)} = \int 0.2 dt$$

Use partial fraction decomposition to integrate the left side of the equation

Using partial fraction decomposition to integrate the left side of the equation, we have: $$\frac{1}{P(1 - P/300)} = \frac{A}{P} + \frac{B}{1 - P/300}$$ Then: $$A(1 - P/300) + BP = 1$$ Solving for \(A\) and \(B\), we get \(A = \frac{1}{300}\) and \(B = -\frac{1}{300}\). So the integral becomes: $$\int \frac{1}{300P} dP - \int \frac{1}{300(1 - P/300)} dP$$

Integrate both terms and apply the constant to one side

Integrate both terms: $$\frac{1}{300} \int \frac{1}{P} dP - \frac{1}{300} \int \frac{1}{1 - P/300} dP = 0.2 \int dt$$ Which gives: $$\frac{1}{300}\ln |P| - \frac{1}{300} \ln |1 - P/300| = 0.2t + C$$

Solve for the population \(P(t)\)

Take the exponential of both sides and solve for \(P(t)\): $$\frac{|P|}{|1 - P/300|} = e^{300(0.2t + C)}$$ Since \(t\) and \(P\) are both positive, we can remove the absolute value signs: $$\frac{P}{1 - P/300} = Ce^{60t}$$ We know that at \(t = 0\), \(P(0) = P_{0} = 50\): $$\frac{50}{1 - 50/300} = C$$ Solving for \(C\), we get \(C = 50\). Now, we can write the general equation for \(P(t)\). $$\frac{P}{1 - P/300} = 50e^{60t}$$ Rearrange to isolate \(P\): $$P(t) = \frac{300 \cdot 50e^{60t}}{50e^{60t} + 300 - 50}$$

Graph the solution

To graph the solution, you can plot \(P(t)\) on the vertical axis and \(t\) on the horizontal axis. The function \(P(t) = \frac{300 \cdot 50e^{60t}}{50e^{60t} + 300 - 50}\) shows a sigmoid shape, with a rapid increase in population followed by a leveling off towards the carrying capacity, \(K = 300\).

Key concepts

Initial Value Problem

When solving differential equations, an initial value problem requires finding a function that not only satisfies the differential equation but also meets specific initial conditions. In our logistic growth exercise, the differential equation represents the rate of population growth and is influenced by the initial population, symbolized by P₀, along with the growth rate and carrying capacity.

By incorporating the given initial conditions (in this case, when t = 0, P = P₀ = 50), we can determine the constant of integration which is essential for pinpointing the exact solution out of the family of potential solutions. This tailored solution provides a quantitative prediction of population dynamics over time, describing how the initial population evolves in a specific environmental context.

Carrying Capacity

The concept of carrying capacity, denoted as K, is central to the logistic growth model. It represents the maximum population size that the environment can sustain indefinitely, given the food, habitat, water, and other essentials available in the environment.

The carrying capacity is a limiting factor that affects the rate of population growth; as the population P approaches K, the growth rate slows down and eventually stops. In our logistic growth model, K is valued at 300, implying that when the population reaches this size, it will stabilize. The carrying capacity is a vital parameter for understanding the sustainability of ecosystems and for managing natural resources.

Natural Growth Rate

The natural growth rate, expressed as r in the logistic growth equation, influences how rapidly the population can increase when not limited by resources. It is essentially the intrinsic rate at which the population grows in optimal conditions.

In our exercise, the natural growth rate is 0.2, which can be interpreted as the potential growth rate per unit of time if there were no constraints. However, as the population size P gets closer to the carrying capacity K, the effective growth rate diminishes. This rate isn't constant but depends on the interaction between the population and its environment, indicative of real-world population dynamics where multiple factors influence growth.

Partial Fraction Decomposition

Partial fraction decomposition is a mathematical technique used to simplify the integration process when dealing with complex rational functions. It's especially useful when the denominator can be factored into simpler parts, as in the differential equation of logistic growth.

In this case, partial fraction decomposition allows us to express the complicated fraction 1 / (P(1 - P/K)) as the sum of simpler fractions, which can be integrated individually. Through this method, we found the values for A and B to be 1/300 and -1/300, which facilitated the calculation of the integral and ultimately helped us solve for the population function P(t). This illustrates how partial fraction decomposition is a powerful tool for solving differential equations analytically.