Question

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem $$P^{\prime}(t)=r P\left(1-\frac{P}{K}\right), \quad P(0)=P_{0}$$is \(P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-n}+1}\)

Short answer

Question: Determine the particular solution of the given logistic equation \(P'(t) = rP(t)\left(1 - \frac{P(t)}{K}\right)\) with an initial condition \(P(0) = P_0\). Answer: The particular solution of the given logistic equation with the given initial condition is \(P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-rt}+1}\).

Step-by-step solution

Separate variables

First, we rewrite the given equation and separate the variables: $$\frac{P'}{rP\left(1-\frac{P}{K}\right)} = 1$$ Separating the variables gives: $$\frac{dP}{P\left(1-\frac{P}{K}\right)} = r \, dt$$

Integrate both sides

Now we integrate both sides of the equation: $$\int \frac{dP}{P\left(1-\frac{P}{K}\right)} = \int r \, dt$$ Let's find the integral on the left side using the substitution method: Let \(u = 1 - \frac{P}{K}\), then \(\frac{-dP}{K} = du\). We rewrite the integral on the left side in terms of \(u\): $$\int \frac{-K \, du}{(K - KP)u} = \int \frac{-K \, du}{K(1 - u)u}$$ Now the integral becomes: $$-\int \frac{du}{u(1-u)}$$ Now, use partial fraction decomposition to separate the integral: $$-\frac{A}{u}+\frac{B}{(1-u)} = \frac{-1}{u(1-u)}$$ After solving for A and B, we have: $$A = 1, \quad B = 1$$ Thus, the left-hand integral becomes: $$-\int \left(\frac{1}{u} + \frac{1}{(1-u)}\right) du$$ Now, integrate both sides: $$-\int \left(\frac{1}{u} + \frac{1}{(1-u)}\right) du = r \int dt$$ Integrate: $$-\ln|u|+\ln|(1-u)|= rt+C$$

Substituting back and simplifying

Now substitute back the original variable P and simplify: $$-\ln\left|1-\frac{P}{K}\right|+\ln\left|\frac{P}{K}\right| = rt+C$$ Taking the exponential of both sides and rearranging gives: $$\frac{P}{K-P} = Ce^{rt}$$

Apply the initial condition

Now, apply the initial condition \(P(0) = P_0\): $$\frac{P_0}{K-P_0} = C e^0$$ $$C = \frac{P_0}{K- P_0}$$ Substitute this value of C back into the general solution: $$\frac{P}{K-P} = \frac{P_0}{K-P_0} e^{rt}$$

Solve for P(t)

To solve for P(t), we can first solve for \(\frac{1}{P}\) and then invert the result: $$\frac{1}{P} = \frac{1}{K\left(1+\frac{P_0}{K-P_0}e^{-rt}\right)}$$ After some algebraic simplification, we get the final solution: $$P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-rt}+1}$$ This is the solution of the logistic equation with the given initial condition.

Key concepts

Separation of Variables

Separation of variables is a powerful method commonly used to solve ordinary differential equations, where the equation is manipulated to isolate the variables on opposite sides of the equation. In this method, factors involving one variable are moved to one side of the equation, while factors involving the other variable are shifted to the other side.

For the logistic equation in the exercise, \(P'(t) = rP(1-\frac{P}{K})\), separation of variables allows us to rewrite the equation into a form that isolates \(dP\) (the differential of \(P\)) on one side and \(dt\) (the differential of \(t\)) on the other. This is achieved by dividing both sides by \(P(1-\frac{P}{K})\) and multiplying by \(dt\), leading to an integrable form involving the separate variables \(P\) and \(t\).

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe the rate of change of a quantity and are central in engineering, physics, biology, economics, and many other fields. The logistic equation is a first-order non-linear differential equation that models phenomena such as population growth and the spread of diseases.

Understanding the nature of differential equations, particularly how to express and handle them analytically, is crucial for solving the problems they represent. They can often be solved by techniques like separation of variables, integration, or by using more advanced methods depending on their complexity.

Initial Value Problem

In the context of differential equations, an initial value problem (IVP) involves finding a function that not only satisfies a differential equation but also meets initial conditions given at the start of the problem. The initial conditions provide specific values for the function or its derivatives at particular points, which help to determine the unique solution to a differential equation.

In the exercise, \(P(0) = P_0\) gives us the initial population size, which is essential for finding the particular solution to the logistic equation. Without this initial value, we would have a general solution with an arbitrary constant, but with it, we can solve for the exact behavior of the population over time.

Integration Techniques

Integration is the inverse operation of differentiation, and it plays a major role in solving differential equations. Various integration techniques are employed to evaluate integrals that arise from separating variables.

Some common techniques include substitution, which involves changing the variable of integration to simplify the integral, and integration by parts, which is based on the product rule for differentiation. In this problem, we use substitution to transform the integral into a more manageable form, then apply partial fraction decomposition to further simplify it before integrating.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique used to break down complex rational expressions into simpler ones that are easier to integrate. It is particularly useful when dealing with integrals of rational functions.

To utilize this method, we express the rational function as a sum of fractions with simpler denominators. The coefficients of these simpler fractions are determined through a process of equating coefficients or by evaluating the original expression at strategic values. In the logistic equation's solution, we decompose the integrand, allowing us to integrate it term by term to find a solution.