Problem 20 - Advanced Engineering Mathematics

Question

In Problems 19 and 20, verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution \(y=\phi(x)\) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval \(I\) of the definition of each solution \(\phi\). $$ 2 x y d x+\left(x^{2}-y\right) d y=0 ; \quad-2 x^{2} y+y^{2}=1 $$

Step-by-Step Solution

Answer

The explicit solution is \(y = x^2 + \sqrt{x^4 + 1}\) with interval \(I = (-\infty, \infty)\).

Step 1: Analyze the Given Problem

In this problem, we need to verify that the given expression \(-2x^2 y + y^2 = 1\) is an implicit solution of the differential equation: \(2xy\,dx + (x^2 - y)\,dy = 0\). We will also find the explicit solution \(y = \phi(x)\) and determine the interval \(I\) on which this solution is valid.

Step 2: Implicit Differentiation

Differentiate \(-2x^2 y + y^2 = 1\) with respect to \(x\) to check if it's a solution of the differential equation. Using implicit differentiation, we find:\[\frac{d}{dx}(-2x^2 y + y^2) = \frac{d}{dx}(1)\]Applying the product rule and chain rule, we get:\[-4xy - 2x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0\]

Step 3: Simplify and Compare

Rearrange the differentiated expression:\[-4xy = (2x^2 - 2y) \frac{dy}{dx}\]Simplify it further:\[\frac{dy}{dx} = \frac{-4xy}{2x^2 - 2y} = \frac{2xy}{y - x^2}\]This matches the differential equation, confirming \(-2x^2 y + y^2 = 1\) as an implicit solution.

Step 4: Solve for Explicit Solution

Let's solve the implicit equation \(-2x^2 y + y^2 = 1\) for \(y\):Substituting back, we simplify to find:\[y^2 - 2x^2 y - 1 = 0\]This is a quadratic equation in \(y\). We can use the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = -2x^2\), and \(c = -1\). Solving, we find:\[y = x^2 \pm \sqrt{x^4 + 1}\]

Step 5: Select an Explicit Solution and Determine the Interval

We select one solution: \(y = x^2 + \sqrt{x^4 + 1}\). The square root ensures non-negativity, so \(x\) can be any real number such that \(x^4 + 1 \geq 0\), which is valid for all \(x \in \mathbb{R}\). Thus, the interval \(I\) is all real numbers: \(I = (-\infty, \infty)\).

Step 6: Graph the Explicit Solution

Using a graphing utility, plot the function \(y = x^2 + \sqrt{x^4 + 1}\). The graph should show a smooth curve, illustrating a part of a parabola offset by the square root term. It reflects symmetry about the y-axis.

Key Concepts

Implicit Solution

In differential equations, an implicit solution is a type of solution where the dependent variable, typically denoted as \( y \), is not isolated or distinctly expressed in terms of the independent variable \( x \). Instead, both variables appear intermingled within a single equation. This kind of solution is useful when it's challenging or impossible to express \( y \) explicitly.

Take for instance our differential equation. We were given an expression \(-2x^2 y + y^2 = 1\) to verify as an implicit solution. Notice how \( y \) and \( x \) are intertwined within a single equation rather than presenting \( y \) on one side alone. By differentiating this expression in context with the provided differential equation, we managed to confirm it serves as an implicit solution.

In practical terms, the implicit solution offers a complete representation of the relationship between \( x \) and \( y \) and serves as a crucial step in the process of finding explicit solutions.

Explicit Solution

An explicit solution of a differential equation refines the problem further by isolating \( y \) and expressing it exclusively using \( x \). This makes it easier to graph and analyze the behavior of solutions over particular intervals.

In this exercise, simplifying and solving the implicit equation \(-2x^2 y + y^2 = 1\) using the quadratic formula allowed us to extract two possible explicit solutions:

• \( y = x^2 + \sqrt{x^4 + 1} \)
• \( y = x^2 - \sqrt{x^4 + 1} \)

The explicit solution provides a clearer understanding of how \( y \) behaves as \( x \) changes. It offers a straightforward function that we can easily plot to visualize the solution's behavior on a coordinate plane.

First-Order Differential Equations

First-order differential equations are a fundamental class of differential equations characterized by the highest derivative being the first derivative, meaning they involve \( \frac{dy}{dx} \). Such equations are widely applicable in fields like physics, biology, and economics due to their simplicity and the systems they can model.

In our specific example, the given first-order differential equation is \( 2xy\,dx + (x^2 - y)\,dy = 0 \). This equation derived from a simple physical or theoretical model features terms involving the first derivative of \( y \) with respect to \( x \). The goal often with these equations is not only to verify implicit solutions but also to find explicit solutions, analyzing them further.

First-order differential equations provide a vital groundwork that helps us build more complex models by first understanding basic relationships.

Interval of Definition

The interval of definition refers to the range or domain of values for \( x \) over which the solution, whether implicit or explicit, is valid and meaningful. It's crucial to identify this interval to ensure the solution is accurately interpreted across applicable scenarios.

For the explicit solution \( y = x^2 + \sqrt{x^4 + 1} \), the interval of definition was determined to be all real numbers, \( I = (-\infty, \infty) \). This is because the expression under the square root is always non-negative regardless of \( x \)'s value, guaranteeing the function remains valid across the entire real number line.

Understanding this concept ensures that we don't make incorrect assumptions about the applicability of solutions outside their valid range, which can be critical in real-world applications where the context dictates specific constraints.