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 \(\displaystyle P(x) y^{\prime \prime}+\) \(Q(x) y^{\prime}+R(x) y=0\).