Problem 37 - Advanced Engineering Mathematics

Question

Is \(x=0\) an ordinary or a singular point of the differential equation \(x y^{\prime \prime}+(\sin x) y=0 ?\) Defend your answer with sound mathematics.

Step-by-Step Solution

Answer

The point \(x=0\) is an ordinary point for the differential equation.

Step 1: Identify the standard form

The given differential equation is \(x y^{\prime \prime} + (\sin x) y = 0\). To understand the nature of the point \(x = 0\), we start by writing the equation in the standard form for a second-order linear differential equation, \(y^{\prime \prime} + P(x)y^{\prime} + Q(x)y = 0\).

Step 2: Rearrange into standard form

Initially, divide the entire equation by \(x\) to isolate \(y^{\prime \prime}\): \[ y^{\prime \prime} + \frac{(\sin x)}{x} y = 0. \] Here, \(P(x) = 0\) and \(Q(x) = \frac{\sin x}{x}\).

Step 3: Analyze the coefficients at x=0

We now need to analyze whether \(x = 0\) is an ordinary or singular point by checking the nature of the coefficients at \(x = 0\). A point is an ordinary point if both \(P(x)\) and \(Q(x)\) are analytic at that point.

Step 4: Check analyticity of P(x) and Q(x)

\(P(x) = 0\) is clearly analytical everywhere. However, \(Q(x) = \frac{\sin x}{x}\) presents a challenge at \(x = 0\). We expand \(\sin x\) in a Taylor series to see the behavior, \[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \] Thus, \(Q(x) = \frac{x - \frac{x^3}{6} + \cdots}{x} \) simplifies to \[ Q(x) = 1 - \frac{x^2}{6} + \cdots \] which is analytic at \(x = 0\).

Step 5: Conclusion on the nature of x=0

Since both \(P(x)\) (which is 0) and \(Q(x)\) (which is \(1 - \frac{x^2}{6} + \cdots\)) are analytic at \(x=0\), the point \(x = 0\) is an ordinary point for the differential equation.

Key Concepts

Ordinary Points

In the realm of differential equations, an ordinary point is a location where the coefficients in the equation behave nicely. This means the coefficients don't have any breaks, jumps, or indeterminate forms there.
An ordinary point is mathematically defined as a point where all the coefficient functions involved in the differential equation are analytic.
Analytic basically means that these functions can be expressed as a power series—a fancy set of sums involving powers of the variable.

Finding an ordinary point often requires rewriting a differential equation into its standard form. For a second-order linear differential equation, the standard form is usually given as:

• \(y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0\)

Here, both \(P(x)\) and \(Q(x)\) need to be checked for analyticity at the point in question. If they are both analytic, then the point is considered ordinary.

In the given exercise, by transforming the differential equation and examining \(P(x)\) and \(Q(x)\), we determined that both are analytic at \(x = 0\), confirming it as an ordinary point.

Singular Points

Contrasting ordinary points, singular points occur where the differential equation takes on unusual properties, making it more challenging to solve. A singular point is where one or more coefficient functions become non-analytic. For example, they might have a discontinuity or an undefined value at the particular point.

Identifying a singular point involves analyzing the coefficient functions once the differential equation is in standard form:

• Again, \(y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0\)

If either \(P(x)\) or \(Q(x)\) is not analytic at some point, then that point is singular.

In practical terms, singular points may be classified further into regular or irregular types, based on how the coefficients "misbehave"—regular points allow some series solutions, whereas irregular points do not.
In this exercise, we found \(x = 0\) to be an ordinary point, showing that even challenging-looking terms can sometimes simplify and resolve analytic at the point in question.

Second-Order Linear Differential Equation

Second-order linear differential equations are essential in modeling various physical phenomena like vibrations, electrical circuits, and fluid flow.
These equations involve a second derivative of the function and typically have the form:

• \(y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0\)

Where \(P(x)\) and \(Q(x)\) are known coefficient functions. The equation is linear because the unknown function \(y\) and its derivatives appear only to the first power.

Finding solutions to these equations depends on the nature of their coefficients especially at specific points like ordinary or singular points. The general process involves determining these points via the coeficients' behavior.

In our particular problem, transforming the equation and assessing the coefficients around \(x = 0\) helps decide on the point's nature and ultimately guides the further process for finding solutions if required. This kind of analysis lays the foundation for understanding behaviors and particular solutions for a wide variety of real-world systems.