Question

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. Logistic population growth Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right)\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-r t}},\) where \(C\) is an arbitrary constant. b. Find the value of \(C\) that corresponds to the initial condition \(P(0)=50\) c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\) d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).

Short answer

#Answer# C = 5, P(t) = 300/(1 + 5e^(-0.1t)), Limit as t → ∞ = 300

Step-by-step solution

Verify the general solution of the logistic equation.

To verify the general solution of the logistic equation given as \(P(t) = \frac{K}{1 + Ce^{-rt}}\), we'll substitute \(P(t)\) into the given differential equation, \(P'(t) = rP\left(1−\frac{P}{K}\right)\). Taking the derivative of \(P(t)\) with respect to \(t\), we get: \(P'(t) = -\frac{K(-r)(Ce^{-rt})}{(1 + Ce^{-rt})^2}\). We can simplify it to: \(P'(t) = \frac{rKP}{K^2P^2}Ce^{-rt}\). Now, we'll replace \(P'(t)\) and \(P\) in the differential equation: \(\frac{rKP}{K^2P^2}Ce^{-rt} = rP\left(1 - \frac{P}{K}\right)\). By simplification, we get: \(1 - \frac{P}{K} = Ce^{-rt}\). Thus, we have verified the general solution of the logistic equation.

Find the value of C based on the initial condition.

Given the initial condition \(P(0) = 50\), we can substitute this into the general solution to find the value of C: \(50 = \frac{K}{1 + Ce^0}\) Now, we need to isolate C: \(C = \frac{K}{50} - 1\).

Graph the solution for the given initial condition and parameters.

Using the given parameters \(P(0) = 50, r = 0.1,\) and \(K = 300\), substitute these values into the general solution: \(P(t) = \frac{300}{1 + 5e^{-0.1t}}\) Plot this function on a graph for various t-values to visualize the population growth over time.

Calculate the limit as time tends to infinity and examine the consistency with the graph.

We want to find the limit as \( t\rightarrow \infty\): \(\lim_{t\to\infty} P(t) = \lim_{t\to\infty} \frac{300}{1 + 5e^{-0.1t}}\) As \(t \to \infty\), the term \(5e^{-0.1t} \to 0\). Therefore, \(\lim_{t \to \infty} P(t) = \frac{300}{1} = 300\) The result of the limit is consistent with the graph obtained in step 3 since the population approaches \(K=300\) over time. This confirms the graph and verifies that the solution is correct.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the context of logistic population growth, the equation takes the form \(P'(t) = rP\left(1 - \frac{P}{K}\right)\), where \(P(t)\) represents the population at time \(t\), \(r\) is the intrinsic growth rate, and \(K\) is the carrying capacity of the environment. This particular type of differential equation models how a population grows rapidly at first but slows down as it approaches the carrying capacity.

Understanding how to solve differential equations is crucial for students studying calculus or biology because it allows us to predict and understand real-world phenomena, like population dynamics, using mathematical models.

Solution Verification

Solution verification is the process of ensuring that a proposed solution to a differential equation actually solves the equation. It requires substituting the solution back into the original equation and confirming that all terms satisfy the equation. For instance, after substituting the general solution \(P(t) = \frac{K}{1 + Ce^{-rt}}\) into the logistic differential equation and simplifying, we should obtain an identity.

This step is a critical aspect of problem-solving in mathematics, as it proves the correctness of the solution beyond a mere speculative match. It's the mathematical equivalent of checking your work in an exam—it ensures the solution is not only plausible but also entirely accurate within the equation's constraints.

Initial Conditions

In the realm of differential equations, initial conditions are specific values assigned to the function, and possibly its derivatives, at a certain point, which in turn determine the particular solution to the differential equation. For the logistic model \(P(0) = 50\), we use this initial condition to calculate the constant \(C\) by plugging it into the formula for \(P(t)\) when \(t=0\).

Initial conditions allow us to tailor general solutions to specific scenarios, making our mathematical models more applicable to real-world situations. This concept is not limited to mathematics but is also integral in sciences like physics and engineering, where describing a system's state at a given moment is vital for predicting future behavior.

Limits of Functions

The study of the limits of functions, especially as time goes to infinity, is essential in understanding long-term behavior within mathematical models, such as logistic population growth. In our example, the limit \(\lim_{t \to \infty} P(t)\) shows us what happens to the population size as time passes indefinitely.

In the logistic model, finding this limit involves examining the exponential decay of \(Ce^{-rt}\) as \(t\) increases. Since the exponential function decays to zero, the population levels off at the carrying capacity \(K\), confirming the model's prediction that a population cannot exceed its environment's resources indefinitely. Understanding limits is critical for students as it provides a window into the behavior of functions beyond the scope of finite computation or observation.