Problem 41 - Advanced Engineering Mathematics

Question

(a) Find an explicit solution of the initial-value problem $$ \frac{d y}{d x}=\frac{2 x+1}{2 y}, \quad y(-2)=-1 $$ (b) Use a graphing utility to plot the graph of the solution in part (a). Use the graph to estimate the interval \(I\) of definition of the solution. (c) Determine the exact interval \(I\) of definition by analytical methods.

Step-by-Step Solution

Answer

(a) \( y = \pm \sqrt{x^2 + x - 1} \). (b) and (c) Interval: \( x \leq \frac{-1-\sqrt{5}}{2} \) or \( x \geq \frac{-1+\sqrt{5}}{2} \).

Step 1: Separate Variables

Start with the given differential equation \( \frac{dy}{dx} = \frac{2x+1}{2y} \). Separate the variables by multiplying both sides by \(2y\;dx\):\[ 2y \; dy = (2x+1) \; dx \]

Step 2: Integrate Both Sides

Integrate both sides separately:\[ \int 2y \, dy = \int (2x+1) \, dx \]The integration gives:\[ y^2 = x^2 + x + C \]

Step 3: Apply Initial Condition

Use the initial condition \( y(-2) = -1 \) to solve for \( C \). Substitute \( x = -2 \) and \( y = -1 \) into the equation:\[ (-1)^2 = (-2)^2 + (-2) + C \]\[ 1 = 4 - 2 + C \]\[ C = -1 \]

Step 4: Express the Explicit Solution

The explicit solution with the constant \( C \) is:\[ y^2 = x^2 + x - 1 \]Solving for \( y \), we get:\[ y = \pm \sqrt{x^2 + x - 1} \]

Step 5: Graph the Solution

Use a graphing utility to plot \( y = \pm \sqrt{x^2 + x - 1} \). Note that the graph only includes real numbers when the expression under the square root is non-negative: \( x^2 + x - 1 \geq 0 \).

Step 6: Estimate Interval of Definition from Graph

The graph indicates that the solution is defined for values of \( x \) where \( x^2 + x - 1 \geq 0 \). This visually appears to be outside the roots of \( x^2 + x - 1 = 0 \).

Step 7: Solve for Interval of Definition Analytically

Find the roots of the quadratic equation \( x^2 + x - 1 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{where } a=1, b=1, c=-1 \]\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]The roots are \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \). Thus, the interval of definition where \( x^2 + x - 1 \geq 0 \) is \( x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2} \).

Key Concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations. It involves rearranging the equation to isolate all terms with the dependent variable on one side and all terms with the independent variable on the other. For example, consider the differential equation \( \frac{dy}{dx} = \frac{2x+1}{2y} \). By multiplying both sides by \(2y\) and \(dx\), we get the separated form:

• \(2y \, dy = (2x+1) \, dx\)

This format allows each side to be integrated separately, simplifying the equation into a form that is more easily solvable. The integration results provide an implicit solution. Applying the initial conditions then helps find an explicit solution by determining any constant of integration.

Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable, typically expressed in the form \(ax^2 + bx + c = 0\). Solving such equations often requires the quadratic formula:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides solutions for \(x\) based on the coefficients \(a\), \(b\), and \(c\). For our example, we have the equation \(x^2 + x - 1 = 0\) related to the definition interval of the explicit solution. Using the quadratic formula:

• \[x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\]

These roots are crucial for determining where the expression under a radical is valid, ensuring real solutions are graphically represented.

Interval of Definition

The interval of definition is the set of all \(x\)-values for which the function is defined. For differential equations, this interval is determined by understanding where all components under operations such as square roots provide valid, real-number results. For example, given the solution \(y = \pm \sqrt{x^2 + x - 1}\), we examine the inequality:

• \(x^2 + x - 1 \geq 0\)

To solve this, refer to the roots found from the quadratic equation. The function is defined where \(x^2 + x - 1\) is non-negative, which is:

• \(x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2}\)

Thus, these values determine the valid range of \(x\) for the solution.

Graphing Solutions

Graphing solutions of differential equations provides a visual representation of their behavior over an interval. Using tools like graphing utilities can help pinpoint intervals where solutions are valid. For the function \(y = \pm \sqrt{x^2 + x - 1}\), you would only graph points where \(x^2 + x - 1 \geq 0\). Below this region in the graph, the solution is undefined as the expression under the square root becomes negative. Looking at the graph, it becomes evident that the solution includes open intervals outside the roots, confirming:

• \(x \leq \frac{-1 - \sqrt{5}}{2} \cup x \geq \frac{-1 + \sqrt{5}}{2}\)

Utilizing graphing utilities confirms analytical findings, sharpening understanding of how the solution behaves over different intervals.