Problem 21 - Advanced Engineering Mathematics

Question

Determine a region of the \(x y\)-plane for which the given differential equation would have a unique solution whose graph passes through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ \left(4-y^{2}\right) y^{\prime}=x^{2} $$

Step-by-Step Solution

Answer

The unique region is where \(-2 < y < 2\) in the xy-plane.

Step 1: Rewrite the Differential Equation

The given differential equation is \((4-y^2) y' = x^2\). Let's rewrite it in the form that is more directly associated with an initial value problem. We can express the differential equation explicitly in terms of \(y'\): \ y' = \frac{x^2}{4-y^2}\.

Step 2: Identify Necessary Conditions for Uniqueness

To ensure a unique solution around a point \((x_0, y_0)\), the function \(f(x, y) = \frac{x^2}{4-y^2}\) and its partial derivative with respect to \(y\) should be continuous around \((x_0, y_0)\).

Step 3: Find Points of Discontinuity

The function \(f(x, y) = \frac{x^2}{4-y^2}\) becomes undefined and discontinuous when \(y^2 = 4\) because the denominator becomes zero. This implies that the lines \(y = 2\) and \(y = -2\) are points of discontinuity.

Step 4: Define Region for Uniqueness

To maintain continuity, the region should exclude these lines. The region of the \(xy\)-plane where the solution is unique is any region that does not include points where \(y = 2\) or \(y = -2\). Hence, the valid regions for uniqueness are \(y \in (-\infty, -2)\), \(y \in (-2, 2)\), and \(y \in (2, \infty)\).

Key Concepts

Unique Solution

When solving a differential equation, finding a unique solution is crucial. The solution of a differential equation is considered unique if, for a given point located within a specified region, there exists only one solution that satisfies both the equation and the initial conditions. In the context of our differential equation, \( y' = \frac{x^2}{4-y^2} \),we aim to determine specific settings under which a unique solution exists.To ascertain uniqueness, we utilize the **Picard-Lindelöf Theorem (or Picard's Theorem)**, which necessitates that the function\( f(x, y) = \frac{x^2}{4-y^2} \)be continuous in the region and that it possesses a continuous partial derivative, especially with respect to \( y \), in a neighborhood around the initial point \((x_0, y_0)\).
This guarantees a locally unique solution to the initial value problem.

Continuity

Continuity plays an essential role in solving differential equations as it ensures no abrupt changes in the value of a function within a given region. For our differential equation\( y' = \frac{x^2}{4-y^2} \),continuity requires that the function \( f(x, y) = \frac{x^2}{4-y^2} \) is defined without interruptions within the region being analyzed.Continuity depends significantly on the domain. For this function to remain continuous:

• The denominator must not be zero, implying \( 4 - y^2 eq 0 \).
• This occurs when \( y eq 2 \) and \( y eq -2 \).

Thus, as long as \( y \) stays outside \( 2 \) and \( -2 \), continuity—and consequently a unique solution—can be preserved.

Points of Discontinuity

Discontinuity signifies places where a function ceases to operate predictably or return real values. It generally indicates points or lines where the function becomes undefined and can't deliver solutions.In our specific problem, the function\( f(x, y) = \frac{x^2}{4-y^2} \)exhibits discontinuity by developing a zero denominator when \( y^2 = 4 \). As a result,

• The values \( y = 2 \) and \( y = -2 \) create vertical asymptotes or "walls," disrupting continuity.
• These values must be explicitly avoided when seeking solutions.

Understanding these breaking points allows us to accurately outline regions devoid of discontinuities, vital for ensuring the equation remains solvable with a unique solution.

Initial Value Problem

An initial value problem, in the realm of differential equations, involves equations paired with a predefined point, known as an initial condition. This enables us to solve the differential equation uniquely and tailor it specific to a boundary condition.In the case of\((4-y^2) y^{\prime}=x^2 \),we explore solutions of the type \( y(x) \) such that not only does the differential equation hold but also the conditions at the starting point \((x_0, y_0)\).
Given the initial condition, the problem morphs into a more directed task, where methods and theories about continuous functions and partial derivatives guide us to unique solutions specific to the entire equation from inception through result. Make sure to pick initial conditions outside the discontinuity points, particularly \( y = 2 \) and \( y = -2 \).