Question

Suppose that you are told that \(x\) and \(x^{2}\) are solutions of a differential equation \(P(x) y^{\prime \prime}+\) \(Q(x) y^{\prime}+R(x) y=0 .\) Can you say whether the point \(x=0\) is an ordinary point or a singular point? Hint: Use Theorem \(3.2 .1,\) and note the values of \(x\) and \(x^{2}\) at \(x=0 .\)

Short answer

Answer: x=0 is an ordinary point.

Step-by-step solution

Recall the definition of analytic functions

A function is analytic at a point if it has a convergent power series expansion around that point.

Evaluate x and x^2 at x=0

At \(x=0\), we have: 1. \(x = 0\) 2. \(x^2 = 0^2 = 0\)

Determine analyticity of x and x^2 at x=0

Both \(x=0\) and \(x^2=0\) are polynomial functions. Since polynomial functions are analytic everywhere, both functions are analytic at \(x=0\).

Apply Theorem 3.2.1

Since both \(x\) and \(x^2\) are analytic at \(x=0\), and they are solutions to the given differential equation, by Theorem 3.2.1 we can conclude that \(x=0\) is an ordinary point of the differential equation \(\displaystyle P(x) y^{\prime \prime}+\) \(Q(x) y^{\prime}+R(x) y=0\).

Key concepts

Singular Point

In the world of differential equations, understanding the concept of a singular point is crucial. A singular point, unlike an ordinary point, is a point where a differential equation's coefficients do not behave well. This typically happens for second-order differential equations of the form \( P(x) y^{\prime \prime} + Q(x) y^{\prime} + R(x) y = 0 \). At a singular point, the coefficients \( P(x) \), \( Q(x) \), and \( R(x) \) may not be well-defined, or their behaviors might not allow for certain types of solutions, like power series solutions.

An ordinary point, on the other hand, is where the coefficients are analytic (or can be expressed as a convergent power series). This makes the equation easier to solve.

Consider the hint from our original exercise. Knowing that \( x \) and \( x^2 \) are solutions implies \( x = 0 \) is a well-behaved point, thus an ordinary point. This means the coefficients of the differential equation at \( x = 0 \) do not disrupt its calculus solutions.

Analytic Functions

Analytic functions play a key role in determining ordinary and singular points. For a function to be analytic at a point means it can be represented by a power series around that point, and this series must converge.

Therefore, these are crucial characteristics:

• They have derivatives of all orders.
• The derivatives near the point form a series that is valid within a specific radius of convergence.
• Being analytic ensures the function behaves nicely in its convergence zone.

Polynomial functions, for example, are inherently analytic functions everywhere on the real line because they satisfy these properties.

This property is utilized in solving differential equations, as we see in our example problem. Both \( x \) and \( x^2 \) are analytic at \( x=0 \) because they are polynomial functions.

Polynomial Functions

Polynomial functions are a fundamental part of mathematics, defined as expressions involving variables raised to whole number exponents with real number coefficients. A polynomial function can be simple like \( x \) or \( x^2 \), or more complex like \( 4x^3 - 3x + 6 \).

Key features of polynomial functions include:

• They are continuous and differentiable everywhere in their domain.
• Polynomials have derivatives of all orders, making them useful in approximating other functions.
• The roots of polynomials are finite in number, providing a solid point of analysis for equations where these functions appear.

Because polynomial functions are always analytic, they qualify as being usable in establishing points like \( x = 0 \) as ordinary rather than singular, aiding in the solvability of differential equations. This is why in the original exercise, since our solutions involve \( x \) and \( x^2 \), the problem is more approachable.