Question

Solve the logistic differential equation representing population growth with the given initial condition. Then use the solution to predict the population size at time \(t=3 .\) \(y^{\prime}=y(1-y), y(0)=0.5\)

Short answer

At \(t=3\), the population size is approximately 0.951.

Step-by-step solution

Understand the Problem

We are given a logistic differential equation \(y' = y(1-y)\) with the initial condition \(y(0) = 0.5\). This represents the growth of a population \(y(t)\) over time \(t\). Our goal is to solve this differential equation to find \(y(t)\) and then use this solution to find \(y(3)\).

Set Up the Logistic Differential Equation

The logistic differential equation can be rewritten as \(\frac{dy}{dt} = y(1-y)\). This is a separable differential equation, which means it can be rearranged to integrate each side separately.

Separate Variables

Rearrange the equation to separate variables: \(\frac{dy}{y(1-y)} = dt\). Now both sides of the equation have variables that can be integrated.

Integrate Both Sides

Integrate both sides: \(\int \frac{1}{y(1-y)} \, dy = \int \, dt\). Use partial fraction decomposition on the integrand on the left: \(\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}\). So you integrate to get: \(\ln |y| - \ln |1-y| = t + C\).

Solve for the General Solution

Combine the logarithms on the left: \(\ln \left| \frac{y}{1-y} \right| = t + C\). Exponentiate both sides to solve for \(y\): \(\frac{y}{1-y} = e^{t+C} = e^C e^t\). Let \(A = e^C\), so \(\frac{y}{1-y} = Ae^t\).

Solve for the Particular Solution Using Initial Condition

Use \(y(0) = 0.5\) to find \(A\). Plug \(t = 0\) and \(y = 0.5\) into \(\frac{y}{1-y} = A e^t\): \(\frac{0.5}{0.5} = A \cdot e^0\). This simplifies to \(A = 1\). Thus, \(\frac{y}{1-y} = e^t\) for the particular solution.

Solve for \(y(t)\)

From \(\frac{y}{1-y} = e^t\), solve for \(y\): \(y = \frac{e^t}{1+e^t}\). This is the particular solution to the differential equation.

Evaluate the Solution at \(t=3\)

To find \(y(3)\), substitute \(t = 3\) into the solution: \(y(3) = \frac{e^3}{1+e^3}\). Compute \(e^3\), then simplify to find \(y(3)\).

Final Step: Compute Approximate Value

Calculating \(y(3)\), we find that \(e^3 \approx 20.0855\). So \(y(3) = \frac{20.0855}{1+20.0855} = \frac{20.0855}{21.0855} \approx 0.951\).

Key concepts

Population Growth

Population growth modeled by a logistic differential equation is an interesting topic in mathematics and biology. In reality, population growth isn't always exponential. It tends to level off as resources become limited. This type of growth is captured by the logistic model, where the rate of increase is proportional both to the size of the population and the remaining capacity of the environment. - **Logistic Model:** The equation \(y' = y(1-y)\) reflects this idea. Here, \(y(t)\) represents the population size at time \(t\), and \(1-y\) represents available resources or space.- **Carrying Capacity:** As \(y\) approaches 1, the growth rate \(y'\) decreases. This is because \(1-y\) approaches zero, signaling that the carrying capacity is being reached.Logistic equations like this are valuable for understanding phenomena like population dynamics, spread of diseases, and even some economic models.

Separable Differential Equations

Separable differential equations are a class of differential equations that can be solved by separating the variables. This makes them accessible to solve analytically, meaning you can frequently find an exact solution.- **Separation of Variables:** The equation \(y' = y(1-y)\) can be rewritten as \(\frac{dy}{dt} = y(1-y)\), and separates into \(\frac{dy}{y(1-y)} = dt\). This means each side of the equation contains only one variable and its differential.- **Integration:** Once separated, each side can be integrated. The left side with respect to \(y\), and the right side with respect to \(t\).By finding antiderivatives for each side, we progress towards solving the equation. Recognizing when you have a separable equation is a powerful tool in differential equations.

Partial Fraction Decomposition

Partial fraction decomposition is a vital method in calculus used to simplify the integration of rational expressions. When dealing with fractions, such as the one we encounter in \(\frac{dy}{y(1-y)}\), it can make the integration process far more manageable.- **Decomposition:** In this case, \(\frac{1}{y(1-y)}\) can be broken into simpler terms: \(\frac{1}{y} + \frac{1}{1-y}\). - **Integration:** This decomposition allows us to integrate each component separately, turning a difficult integration into a series of simpler logarithmic integrals.Once decomposed, solving the integral leads to a solution for the differential equation, an essential step in the logistic equation process.

Initial Condition Problem Solving

Initial condition problem solving allows us to find a particular solution to a differential equation. This means tailoring the general solution to meet specific criteria—or conditions—at a particular moment.- **Given:** We had an initial condition \(y(0) = 0.5\). This condition means that at \(t = 0\), the population was half of its eventual carrying capacity.- **Application:** Plugging into our solved equation, \(\frac{y}{1-y} = A e^t\), setting \(t = 0\) and \(y = 0.5\), it simplifies to \(A = 1\).- **Particular Solution:** This particular value of \(A\) ensures the equation accurately represents not just general behavior, but the exact situation at \(t = 0\).Using initial conditions like this aligns mathematical models with real-world scenarios, providing precise predictions or projections.