Problem 23 - Thomas Calculus

Question

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y-\) window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b] .\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {0 \leq x \leq 4, \quad 0 \leq y \leq 3 ; \quad b=3}\end{array} $$

Step-by-Step Solution

Answer

The exact solution is \(y = \frac{2}{1 + 3e^{-2x}}\), and Euler's error decreases with more intervals.

Step 1: Slope Field Plotting

To plot the slope field for the differential equation \(y' = y(2-y)\), we use a CAS (Computer Algebra System). Set the \(xy\)-window to cover the range \(0 \leq x \leq 4\) and \(0 \leq y \leq 3\). The slope of the field at any point \((x, y)\) is given by the expression \(y(2-y)\). Observe the patterns in the slope field which suggests the behavior of solution curves.

Step 2: Finding the General Solution

Using a CAS DE solver, solve the differential equation \(y' = y(2-y)\). The general solution can be found by separating variables and integrating: \(\frac{dy}{y(2-y)} = dx\). Integrating both sides gives: \(-\ln|2-y| + \ln|y| = 2x + C\). Solving for \(y\), we find the general solution: \(y = \frac{2}{1 + Ce^{-2x}}\).

Step 3: Graphing Solutions for C Values

Using the general solution \(y = \frac{2}{1 + Ce^{-2x}}\), plot graphs for \(C = -2, -1, 0, 1, 2\) on the same slope field. This demonstrates how solutions behave with different starting conditions and illustrates potential equilibria.

Step 4: Solution with Initial Condition

Substitute the initial condition \(y(0) = \frac{1}{2}\) into the general solution: \(\frac{1}{2} = \frac{2}{1 + C}\) to solve for \(C\). This gives \(C = 3\). Substitute \(C=3\) back into the solution to get \(y = \frac{2}{1 + 3e^{-2x}}\). Plot this specific solution over \(0 \leq x \leq 3\).

Step 5: Euler's Method with 4 Intervals

Approximate the solution using Euler's Method over 4 intervals in the range \([0,3]\). Choose step size \(h = 0.75\). The initial value is \((0, 0.5)\). Iterate the Euler formula: \(y_{n+1} = y_{n} + h\cdot y_{n}(2-y_{n})\). Plot this approximation and compare with the exact solution.

Step 6: Euler's Method with More Intervals

Repeat the Euler method for 8, 16, and 32 subintervals. Decrease the step size: \(h = 0.375\), \(h = 0.1875\), and \(h = 0.09375\) respectively. Compute and plot each Euler approximation.

Step 7: Error Calculation and Discussion

Compute the error at \(x=3\) for each Euler approximation by subtracting the approximate \(y\) value from the exact solution \(y(3)\). Discuss improvements in precision as the number of intervals increases, showing a reduction in percentage error as the step size decreases.

Key Concepts

Slope Field

A slope field, also known as a direction field, provides a visual representation of differential equations. For the differential equation \( y' = y(2-y) \), a slope field helps illustrate the behavior of solution curves without solving the equation analytically. It is essentially a graph plotted in an \( xy \)-plane containing short line segments, each with a slope that corresponds to the differential equation's value at that point.

By examining the pattern formed by these line segments, you can interpret how solutions are likely to behave. For instance, in our logistic equation, the slope nature indicates where solutions might increase, decrease, or stabilize. This is why slope fields are particularly powerful; they provide insights into the qualitative behavior of solutions, especially useful when a direct solution is not possible.

Euler's Method

Euler's Method is a numerical method used for approximating solutions to initial value problems when an analytic solution is difficult to derive. This method is straightforward and provides an iterative approach to solve differential equations.

To apply Euler's method for the equation \( y' = y(2-y) \) with an initial condition \( y(0) = \frac{1}{2} \), you'll begin at the known point and calculate subsequent points using the formula:

• \( y_{n+1} = y_n + h \, f(x_n, y_n) \)
• where \( h \) is the step size and \( f(x_n, y_n) = y(2-y) \) in this scenario.

In our exercise, Euler's method allows for solutions of increasing accuracy with smaller step sizes. By refining the step size, you can minimize errors between the numerical approximation and the actual solution.

Logistic Equation

The logistic equation is a common model for population dynamics and is expressed as \( y' = y(2-y) \). This type of differential equation describes how a population changes over time and tends to stabilize, making it particularly useful for biological contexts.

In this specific form, the logistic equation involves a growth rate that decreases as the population \( y \) approaches a certain carrying capacity, here simplified to \( y = 2 \). As a result, the solutions exhibit a phase of rapid growth followed by stabilization, reflecting real-world constraints on growth due to factors like resource limitation. This feature makes logistic equations insightful for modeling sustainable population sizes.

Initial Value Problem

An initial value problem involves solving a differential equation with a given starting point, typically presented in the form \( y(x_0) = y_0 \). This starting condition allows for determining a unique solution from a family of potential solutions.

For the exercise problem, substituting the initial condition \( y(0) = \frac{1}{2} \) into the logistic equation sources a specific trajectory among all possible ones suggested by the general solution \( y = \frac{2}{1 + Ce^{-2x}} \). Identifying the initial value enables you to determine the constant \( C \), resulting in a particular solution; in this case, it is necessary for plotting the precise behavior over your specified interval.

Initial value problems are crucial for scenarios where the exact starting conditions can dramatically influence the path of a solution, highlighting their importance in both theoretical and applied differential analysis.