Problem 7 - Advanced Engineering Mathematics

Question

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. $$ \left(x^{2}+x-6\right) y^{\prime \prime}+(x+3) y^{\prime}+(x-2) y=0 $$

Step-by-Step Solution

Answer

The singular points are \(x = -3\) and \(x = 2\), both are regular.

Step 1: Identify the coefficients of the differential equation

The given differential equation is \((x^2 + x - 6) y'' + (x + 3) y' + (x - 2) y = 0\). Identify the coefficients:\[P(x) = x^2 + x - 6, \quad Q(x) = x + 3, \quad R(x) = x - 2\]

Step 2: Find the roots of P(x)

Set \(P(x) = 0\) to find the singular points, which are the roots of \(x^2 + x - 6 = 0\).Factor the quadratic:\((x + 3)(x - 2) = 0\) leading to roots \(x = -3\) and \(x = 2\). These are the singular points.

Step 3: Check regularity of each singular point

A singular point \(x = x_0\) is regular if \(\frac{(x-x_0)Q(x)}{P(x)}\) and \(\frac{(x-x_0)^2R(x)}{P(x)}\) are analytic at \(x = x_0\). Otherwise, it is irregular. - For \(x = -3\): * \(P(x) = (x+3)(x-2)\), \(Q(x) = x+3\), and \(R(x) = x-2\). * \(\frac{(x+3)(x+3)}{(x+3)(x-2)} = \frac{x+3}{x-2}, \quad \frac{(x+3)^2(x-2)}{(x+3)(x-2)} = x+3\). * Both are analytic at \(x = -3\), thus \(x = -3\) is a regular singular point.- For \(x = 2\): * \(P(x) = (x+3)(x-2)\), \(Q(x) = x+3\), and \(R(x) = x-2\). * \(\frac{(x-2)(x+3)}{(x+3)(x-2)} = 1, \quad \frac{(x-2)^2(x-2)}{(x+3)(x-2)} = \frac{x-2}{x+3}\). * Both are analytic at \(x = 2\), thus \(x = 2\) is a regular singular point.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that involve functions and their derivatives. These equations help us understand systems and phenomena where change is a notable factor. For example, they can describe how populations grow, heat transfers, or objects fall under gravity. In simpler terms, they are equations that relate a function and its rates of change.
For a given differential equation, identifying the order and degree is essential. The order is determined by the highest derivative present, while the degree is the power of that derivative. In our exercise, we deal with a second-order differential equation because the highest derivative involved is the second derivative, denoted by \( y'' \). Understanding these basics lays the foundation for analyzing singular points within the equation.

Singular Points Classification

When solving differential equations, singular points play a crucial role. These are specific values of \( x \) where the coefficients of the equation or its solutions behave differently. Let's classify them to understand their nature better.

• Singular Points: These are points where the differential equation's coefficients become undefined or infinite. For instance, roots of the coefficient \( P(x) \) in our exercise become singular points.
• Regular vs. Irregular Singular Points: Classification depends on how the differential equation behaves near these points. Whether a singular point is deemed regular or irregular influences the type of solutions expected.

To detect singular points, one sets the coefficient of the highest derivative (\( P(x) \)) to zero and solves for \( x \). Identifying these points is a pivotal step before determining their regularity.

Regular and Irregular Singular Points

Regular and irregular singular points refer to how "well-behaved" a differential equation is at specific singular points. A regular singular point is where solutions may be problematic but are still manageable using certain techniques, while irregular singular points reflect more complex behavior.
Determining regularity involves specific checks:

• Regular Singular Point: At a singular point \( x = x_0 \), it's considered regular if the expressions \( \frac{(x-x_0)Q(x)}{P(x)} \) and \( \frac{(x-x_0)^2R(x)}{P(x)} \) are analytic. Analytic simply means they can be expressed as a convergent power series around the point \( x_0 \).
• Irregular Singular Point: If the conditions for a regular singular point aren't met, it's classified as irregular. This often leads to solutions that are harder to handle and might require advanced techniques for solving.

In our exercise, checking these conditions showed that both \( x = -3 \) and \( x = 2 \) are regular singular points, meaning the solutions, though singular at these points, are more predictable and can be managed analytically.