Problem 49 - Calculus Early Transcendentals

Question

Assume that \(y=y(t)\) satisfies the logistic equation with \(y_{0}=y(0)\) the initial value of \(y .\) (a) Use separation of variables to derive the solution $$ y=\frac{y_{0} L}{y_{0}+\left(L-y_{0}\right) e^{-k t}} $$ (b) Use part (a) to show that \(\lim _{t \rightarrow+\infty} y(t)=L\)

Step-by-Step Solution

Answer

(a) Derivation matches the given solution. (b) \\( \lim_{t \rightarrow +\infty} y(t) = L \\) using the algebraic simplification.

Step 1: Separate the Variables

The logistic equation is given by \( \frac{dy}{dt} = ky \left(1 - \frac{y}{L}\right) \). To solve using separation of variables, we'll rewrite the equation such that all terms involving \( y \) are on one side and terms involving \( t \) are on the other. This results in: \[\frac{1}{y(1 - \frac{y}{L})} \, dy = k \, dt\]

Step 2: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \). This gives:\[\int \frac{1}{y(1 - \frac{y}{L})} \, dy = \int k \, dt\]Use partial fraction decomposition for the left side:\[\frac{1}{y(1 - \frac{y}{L})} = \frac{1}{y} + \frac{1}{L-y}\]Therefore,\[\int \left(\frac{1}{y} + \frac{1}{L-y}\right) \, dy = \int k \, dt\]This results in \[\ln |y| - \ln |L-y| = kt + C\] where \( C \) is the constant of integration.

Step 3: Solve for y in terms of t

Exponentiate both sides to eliminate the logarithm:\[|y| (y) = C_1 e^{kt} (\frac{L-y}{y})\]Let \( C_1 = e^C \). Solving for \( y \) in terms of \( t \) yields:\[y = \frac{L}{1 + Ce^{-kt}}\]where \( C \) is a constant. Using initial condition \( y(0) = y_0 \), \[y_0 = \frac{L}{1+C} \Rightarrow C = \frac{L-y_0}{y_0}\]Substituting \( C \) back into \( y \), we obtain:\[y = \frac{y_0 L}{y_0 + (L-y_0) e^{-kt}}\]This matches the form given in the question.

Step 4: Show Limit as t approaches Infinity

Analyze the expression:\[y = \frac{y_0 L}{y_0 + (L-y_0)e^{-kt}}\]As \( t \rightarrow \infty \), \( e^{-kt} \rightarrow 0 \). Thus, the expression simplifies to:\[y = \frac{y_0 L}{y_0 + 0} = L\]Therefore, \( \lim\limits_{t \to \infty} y(t) = L \).

Key Concepts

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations by isolating the variables on each side of the equation. This method becomes particularly useful with the logistic equation.
The logistic equation can often be written as a differential equation in the form \[ \frac{dy}{dt} = ky \left(1 - \frac{y}{L}\right) \]. This is structured in a way where each side of the equation can be manipulated to isolate each variable. By moving all terms with \( y \) to one side and all terms with \( t \) to the other, we achieve: \[ \frac{1}{y \left(1 - \frac{y}{L}\right)} \, dy = k \, dt \].
Once separated, each side of the equation can be integrated, allowing us to solve for a solution using known integration techniques. This step often leads to a logarithmic form that can be simplified by exponentiating both sides, paving the way to solve for \( y \) as a function of \( t \). This transformation plays a critical role in converting complex relationships in differential equations into more manageable forms.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a complex fraction as a sum of simpler fractions. This technique is significant in simplifying the integration process involved in separation of variables.
For the logistic equation, we face an integral of the form \( \int \frac{1}{y(1 - \frac{y}{L})} \, dy \). By applying partial fraction decomposition, we rewrite this integral into a sum of two easier integrals:\[ \frac{1}{y(1 - \frac{y}{L})} = \frac{1}{y} + \frac{1}{L-y} \].
This decomposition simplifies the integration into two distinct, more manageable logarithmic integrals: \( \int \frac{1}{y} \, dy \) and \( \int \frac{1}{L-y} \, dy \). The result translates into \( \ln |y| - \ln |L-y| \), making the subsequent steps and application of integration constants much more straightforward. By reducing the complexity of the fraction through partial fraction decomposition, we significantly ease the process of finding the explicit solution to the differential equation.

Limit at Infinity

Understanding the limit at infinity is crucial when working with logistic equations because they often model growth that stabilizes over time. Examining the behavior of the function \( y(t) \) as \( t \to \infty \) gives meaningful insights into the model's asymptotic behavior.
For the logistic equation \( y = \frac{y_0 L}{y_0 + (L-y_0)e^{-kt}} \), observe that as \( t \to \infty \), the exponential term \( e^{-kt} \to 0 \). This dramatically simplifies the equation to \( y = \frac{y_0 L}{y_0 + 0} = L \), indicating that \( y \) approaches a limit of \( L \).
This limit represents the upper bound or carrying capacity of the system that the logistic model is describing. It shows that despite initially rapid growth, the function eventually stabilizes, reflecting realistic constraints such as resources or space. This analysis highlights a defining feature of logistic growth models, showing how they tend to approach a stable equilibrium at infinity, providing robust predictions about long-term behavior.