Problem 39 - Advanced Engineering Mathematics

Question

(a) Separating variables we have \(2 y d y=(2 x+1) d x .\) Integrating gives \(y^{2}=x^{2}+x+c .\) When \(y(-2)=-1\) we find \(c=-1,\) so \(y^{2}=x^{2}+x-1\) and \(y=-\sqrt{x^{2}+x-1} .\) The negative square root is chosen because of the initial condition. (b) From the figure, the largest interval of definition appears to be approximately \((-\infty,-1.65)\). (c) Solving \(x^{2}+x-1=0\) we get \(x=-\frac{1}{2} \pm \frac{1}{2} \sqrt{5},\) so the largest interval of definition is \(\left(-\infty,-\frac{1}{2}-\frac{1}{2} \sqrt{5}\right)\) The right-hand endpoint of the interval is excluded because \(y=-\sqrt{x^{2}+x-1}\) is not differentiable at this point.

Step-by-Step Solution

Answer

The largest interval of definition is \( \left(-\infty, -\frac{1}{2} - \frac{1}{2}\sqrt{5}\right) \).

Step 1: Separate Variables

Start with the differential equation: \( 2y \, dy = (2x+1) \, dx \). This equation is already in a separated form, which allows us to integrate both sides independently.

Step 2: Integrate Both Sides

Integrate both sides to find the relationship between \(y\) and \(x\). Integrating the left side gives \( \int 2y \, dy = y^2 + C_1 \) and integrating the right side gives \( \int (2x+1) \, dx = x^2 + x + C_2 \). Combine constants to get \( y^2 = x^2 + x + C \).

Step 3: Apply Initial Condition

Use the initial condition \( y(-2) = -1 \) to find \( C \). Substitute \( x = -2 \), \( y = -1 \) into the integrated equation: \((-1)^2 = (-2)^2 + (-2) + C\). Solving gives \( C = -1 \). Thus, the equation becomes \( y^2 = x^2 + x - 1 \).

Step 4: Solve for y

Solve the equation \( y^2 = x^2 + x - 1 \) for \( y \). Since \( y(-2) = -1 \) and we need \( y < 0 \), take the negative square root: \( y = -\sqrt{x^2 + x - 1} \).

Step 5: Determine the Interval of Definition

Identify where \( y = -\sqrt{x^2 + x - 1} \) is defined and differentiable. This requires determining when \( x^2 + x - 1 \geq 0 \), which involves solving the quadratic inequality. The roots of \( x^2 + x - 1 = 0 \) are found using the quadratic formula: \( x = -\frac{1}{2} \pm \frac{1}{2} \sqrt{5} \). These roots are real, so the solution interval is determined by testing intervals. The function is defined on \( \left(-\infty, -\frac{1}{2} - \frac{1}{2}\sqrt{5}\right) \), because it is not differentiable at the endpoints.

Key Concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations, where the variables can be rearranged so that each variable is on a different side of the equation. This setup allows independent integration of each side.
For example, consider the differential equation given in the exercise: \( 2y \, dy = (2x+1) \, dx \). Both sides of the equation contain differentials (\( dy \) and \( dx \)), and each side has either a function of \( y \) or \( x \). This separation enables us to integrate both sides individually, making the equation manageable.
Simply integrate both sides: The left side becomes \( \int 2y \, dy = y^2 + C_1 \), while the right side transforms to \( \int (2x+1) \, dx = x^2 + x + C_2 \). After integration, the constants \( C_1 \) and \( C_2 \) are usually combined into a single constant \( C \). This yields the expression \( y^2 = x^2 + x + C \). This equation can now be analyzed to understand how \( y \) changes with \( x \).
Separation of variables works well when you can isolate each variable on one side of the equation, paving the way for straightforward integration.

Initial Value Problem

An initial value problem is a differential equation complemented by a specific condition, known as an initial condition. This condition specifies the value of the unknown function at a particular point. It is key to determining a unique solution.
In our example, the initial condition given is \( y(-2) = -1 \). This means that when \( x = -2 \), the function \( y \) has the value \(-1\). After integrating the separated variables, the resulting expression \( y^2 = x^2 + x + C \) includes an arbitrary constant \( C \).
To find \( C \), plug the initial condition into this equation. Substitute \( x = -2 \) and \( y = -1 \) into \((y)^2 = x^2 + x + C\), which results in the equation \((-1)^2 = (-2)^2 + (-2) + C\). Solving this will give you \( C = -1 \).
This value for \( C \) allows you to refine your solution to \( y^2 = x^2 + x - 1 \), tailoring the general solution to meet the initial value condition and ensuring that the solution is specific to the situation defined in the problem.

Quadratic Inequality

A quadratic inequality involves an expression of the form \( ax^2 + bx + c \) compared to zero, instead of being set equal to it. Solving these inequalities helps determine where a function is defined or behaves in a particular way.
For the exercise's function \( y = -\sqrt{x^2 + x - 1} \), we must find where the expression under the square root, \( x^2 + x - 1 \), is non-negative.
First, solve the quadratic equation \( x^2 + x - 1 = 0 \) using the quadratic formula: \( x = -\frac{1}{2} \pm \frac{1}{2} \sqrt{5} \). These roots split the number line into intervals. Determine where \( x^2 + x - 1 \leq 0 \) by testing these intervals with points, such as \( x = -2 \).
The expression represents a parabolic shape, which dips below the x-axis between the roots. Consequently, the function is well-defined on \( \left(-\infty, -\frac{1}{2} - \frac{1}{2}\sqrt{5}\right) \).
This allows us to state where the function is valid and ensures that its behavior is consistent within this particular range.

Integrating Factors

Integrating factors is a method used typically for solving linear first-order differential equations but are not used directly in this exercise. They come into play when an equation cannot be easily manipulated by separation of variables.
For a linear equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), multiplying through by an integrating factor \( \mu(x) \) enables the left side to become a single derivative \( \frac{d}{dx}[\mu(x)y] \). This allows us to integrate both sides easily.
To determine the integrating factor,

• Calculate \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the entire differential equation by \( \mu(x) \).

Such a method could be applied in situations where direct separation is not possible. Integrating factors make it possible to transform complex equations into solvable forms, opening a path to clear solutions.