Problem 39 - Advanced Engineering Mathematics

Question

Every autonomous first-order equation \(d y / d x=f(y)\) is separable. Find explicit solutions \(y_{1}(x), y_{2}(x), y_{3}(x)\), and \(y_{4}(x)\) of the differential equation \(d y / d x=y-y^{3}\) that satisfy, in turn, the initial conditions \(y_{1}(0)=2, y_{2}(0)=\frac{1}{2}, y_{3}(0)=-\frac{1}{2}\), and \(y_{4}(0)=-2\). Use a graphing utility to plot the graphs of each solution. Compare these graphs with those predicted in Problem 19 of Exercises \(2.1\). Give the exact interval of definition for each solution.

Step-by-Step Solution

Answer

Solutions are separated, partial fractions applied, and integrated. Initial conditions find constants, yielding specific intervals for each defined solution. Graphing confirms theoretical behaviors.

Step 1: Separate Variables

First, we express the differential equation \( \frac{dy}{dx} = y - y^3 \) in a separable form. This involves rearranging terms to express it in the form \( \int g(y) \, dy = \int f(x) \, dx \). For this equation, we write it as \( \frac{dy}{y - y^3} = dx \).

Step 2: Partial Fraction Decomposition

The expression \( \frac{1}{y - y^3} \) must be decomposed using partial fractions to integrate it. Factorize the denominator: \( y - y^3 = y(1-y)(1+y) \). The partial fraction decomposition becomes: \( \frac{A}{y} + \frac{B}{1-y} + \frac{C}{1+y} = \frac{1}{y(1-y)(1+y)} \). Solving yields \( A = 1, B = 1/2, C = -1/2 \).

Step 3: Integrate Each Term

Integrate both sides. For the left side: \( \int \left( \frac{1}{y} + \frac{1/2}{1-y} - \frac{1/2}{1+y} \right) dy \). The right side is \( \int dx \). This gives: \( \ln|y| - \frac{1}{2}\ln|1-y| - \frac{1}{2}\ln|1+y| = x + C \).

Step 4: Simplify and Solve for y

Combine logs: \( \ln|y| - \frac{1}{2}\ln|(1-y)(1+y)| = x + C \). Simplifying gives \( \ln\left| \frac{y}{\sqrt{1-y^2}} \right| = x + C \), which leads to \( |y| = e^{x+C} \sqrt{1-y^2} \). Solve this for \( y \) explicitly.

Step 5: Apply Initial Conditions

Solve for the constant \( C \) using each initial condition. For example, if \( y_1(0) = 2 \), substitute into the explicit form and solve for \( C \). Do the same for \( y_2(0) = \frac{1}{2} \), \( y_3(0) = -\frac{1}{2} \), and \( y_4(0) = -2 \).

Step 6: Determine Intervals of Definition

Examine the domain of the solution for different initial conditions to find the interval of definition for each \( y \). These intervals depend on where the expression inside the square root, \( 1-y^2 \), remains non-negative. Check endpoints where the function becomes undefined due to division by zero or other constraints.

Step 7: Graph Solutions and Compare

Use a graphing utility to plot each solution \( y_i(x) \). Compare these graphs to those from Problem 19 of Exercises 2.1 to identify any distinct differences or confirmations. Ensure the graphs are consistent with theoretical predictions about behavior near critical points.

Key Concepts

Separable Differential Equations

Differential equations are a type of equation wherein the unknown element is a function. In the context of separable differential equations, we focus on a special kind. These equations can be expressed as a product of functions, one of which depends solely on the independent variable (e.g., x) and the other solely on the dependent variable (e.g., y).

To solve them, begin by manipulating the equation so that all terms containing the dependent variable are on one side and all terms containing the independent variable are on the other. This means transforming the equation into an integral form:

• Integrate both sides separately.
• Result in an equation that implicitly or explicitly describes the function.

For example, the given equation \(\frac{dy}{dx} = y - y^3\) can be rewritten as \(\frac{dy}{y - y^3} = dx\). Once placed in this form, you can integrate both sides to find the solution.

Partial Fraction Decomposition

When dealing with complex rational expressions, like the one in our differential equation \(\frac{1}{y - y^3}\), direct integration is not possible. That's where partial fraction decomposition comes in.

This technique allows you to express a complex rational expression as a sum of simpler fractions that are easier to integrate. The first step is to factor the denominator completely. In our example:

• Factor \(y - y^3\) into \(y(1-y)(1+y)\).

Next, express \(\frac{1}{y - y^3}\) using partial fractions:

• Find constants A, B, and C such that \(\frac{A}{y} + \frac{B}{1-y} + \frac{C}{1+y} = \frac{1}{y(1-y)(1+y)}\).

Solving for these constants allows us to then integrate each term separately, thus simplifying the overall integration process.

Initial Value Problems

Initial value problems (IVPs) involve finding a specific solution to a differential equation that satisfies a given initial condition. These initial conditions impose a value for the function at a specific point.

In our exercise, initial conditions are given to determine explicit solutions for \(y\) at different starting points, namely:

• \(y_1(0) = 2\)
• \(y_2(0) = \frac{1}{2}\)
• \(y_3(0) = -\frac{1}{2}\)
• \(y_4(0) = -2\)

Each condition allows you to solve for the constant \(C\) after integrating the differential equation. Matching the solution to the initial condition ensures the solution is unique to that specific starting point.

Interval of Definition

The interval of definition specifies the domain over which a solution to a differential equation is valid. Certain conditions, such as division by zero or undefined expressions, determine these intervals.

For the equation \(\frac{dy}{dx} = y - y^3\), consider where the expression inside the square root, \(1-y^2\), remains non-negative. This forms part of determining the interval of definition:

• The function is defined wherever \(1-y^2\geq 0\), implying that \(-1 \leq y \leq 1\).
• Evaluate this range in conjunction with other elements in the equation that may impose further restrictions.

Analyzing these conditions carefully helps us find intervals where the solutions are valid and behave as expected. The intervals for each specific initial condition will depend on where the functions remain continuous and differentiable.