Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

The equation for logistic growth is $$ \frac{d y}{d t}=k y(L-y) $$ Show that this differential equation has the solution $$ y=\frac{L y_{0}}{y_{0}+\left(L-y_{0}\right) e^{-L k t}} $$ Hint \(: \frac{1}{y(L-y)}=\frac{1}{L y}+\frac{1}{L(L-y)} .\)

Step-by-Step Solution

Answer

The solution is verified as \( y = \frac{L y_0}{y_0 + (L-y_0)e^{-L k t}} \).

Step 1: Separate Variables

We start by separating the variables to prepare for integration. Rearrange the given differential equation: \[ \frac{d y}{y(L-y)} = k \cdot d t. \]

Step 2: Partial Fraction Decomposition

Use the hint provided to decompose the fraction: \( \frac{1}{y(L-y)} = \frac{1}{Ly} + \frac{1}{L(L-y)} \). Substitute this back into the equation: \[ d y \left( \frac{1}{Ly} + \frac{1}{L(L-y)} \right) = k \cdot d t. \]

Step 3: Integrate Both Sides

Integrate both sides of the equation. The left side integrates to: \[ \int \left( \frac{1}{Ly} + \frac{1}{L(L-y)} \right) \, dy = \frac{1}{L}\ln|y| + \frac{1}{L}\ln|L-y|. \] The right side integrates to: \[ \int k \, dt = kt + C, \] where \( C \) is a constant of integration.

Step 4: Solve for \( y \)

Combine the integrals: \[ \frac{1}{L} \ln \left| \frac{y}{L-y} \right| = kt + C. \] Exponentiate both sides to solve for \( y \): \[ \frac{y}{L-y} = e^{Lkt + LC} = Ae^{Lkt}, \] where \( A = e^{LC} \).

Step 5: Express \( y \) Explicitly

Rearrange to express \( y \): \[ y = \frac{ALe^{Lkt}}{1 + Ae^{Lkt}}. \] Substitute the initial condition \( y(0) = y_0 \) to find \( A \): \[ A = \frac{y_0}{L-y_0}. \] So the solution becomes \[ y = \frac{Ly_0}{y_0 + (L-y_0)e^{-Lkt}}. \]

Step 6: Verify the Initial Condition

Ensure that the solution satisfies the initial condition. At \( t = 0 \), \( y(0) = \frac{Ly_0}{y_0 + (L-y_0)e^{0}} = y_0. \) This confirms our solution is correct.

Key Concepts

Understanding Differential Equations

In mathematics, differential equations model how quantities change over time or space. They are equations involving derivatives, which represent rates of change. In our logistic growth equation,\[\frac{d y}{d t}=k y(L-y),\]it describes the rate at which the population \( y \) changes with respect to time \( t \). Here, \( k \) is the growth rate constant, and \( L \) is the carrying capacity, the maximum population size the environment can sustain. Such equations are fundamental in modeling biological growth, physics, and economics.
Differential equations can usually be classified into two types:

• Ordinary Differential Equations (ODEs) involve functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs) involve functions of multiple variables.

The logistic growth equation is an ODE, as it has one independent variable, \( t \). Solving this equation typically involves finding a function \( y(t) \) that satisfies the relation given. We achieve this by separating variables—grouping terms involving \( y \) and \( dy \) on one side and \( t \) and \( dt \) on the other—which sets the stage for integration.

The Role of Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex fractions, making them easier to integrate. When dealing with a fraction like \[ \frac{1}{y(L-y)}, \]it helps to break it down into simpler fractions.
The hint in the exercise gives the decomposition:\[ \frac{1}{y(L-y)} = \frac{1}{Ly} + \frac{1}{L(L-y)}. \]
This step transforms the original integrand into a sum of simpler fractions. Each term can now be integrated separately, which is a much simpler process.
Partial fraction decomposition is especially useful in solving integrals of rational functions, where the degree of the numerator is less than the degree of the denominator. Once decomposed, each part can be matched with standard forms that have known integration results.
Thus, using this technique assists us in proceeding with the integration step of the logistic growth equation solution.

Integration Techniques and Finding Solutions

Integration is the reverse process of differentiation and is essential for solving differential equations like the logistic growth model. After separating the variables and applying partial fraction decomposition, the next task is to integrate each side.
On the left side, we integrate:\[ \int \left( \frac{1}{Ly} + \frac{1}{L(L-y)} \right) \, dy. \]Each component can be integrated using the natural logarithm rule:\[ \int \frac{1}{u} \, du = \ln|u| + C, \]which gives us:\[ \frac{1}{L}\ln|y| + \frac{1}{L}\ln|L-y|. \]
On the right side,\[ \int k \, dt = kt + C. \]
Integration techniques like these, which involve recognizing patterns and standard integral results, are powerful tools in calculus. Once both sides are integrated, the constant \( C \) must be considered, often determined using initial conditions. With these steps, the solution can be found, verifying its validity by substituting it back into the original equation.