Differential Equations · Question of the Day

Differential Equations Question of the Day

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Thursday, September 17, 2026

The differential equation $y'' + P(x)y' + Q(x)y = 0$ has an ordinary point at $x = x_0$ if both $P(x)$ and $Q(x)$ are analytic at $x_0$ . For the equation $(x^2 - 4)y'' + xy' + (x+1)y = 0$ , which statement about the point $x = 1$ is correct?

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$x = 1$ is an ordinary point since the original coefficients are polynomials at this point
$x = 1$ is a regular singular point because the equation has polynomial coefficients
$x = 1$ is an ordinary point because $\frac{x}{x^2-4}$ and $\frac{x+1}{x^2-4}$ are both analytic at $x = 1$ (correct answer)
$x = 1$ is an irregular singular point due to the behavior of $Q(x)$ near $x = 1$

Explanation: To determine the nature of $x = 1$ , we must write the equation in standard form: $y'' + \frac{x}{x^2-4}y' + \frac{x+1}{x^2-4}y = 0$ . Since $x^2-4 = (x-2)(x+2)$ , at $x = 1$ we have $x^2-4 = -3 \neq 0$ . Therefore both $P(x) = \frac{x}{x^2-4}$ and $Q(x) = \frac{x+1}{x^2-4}$ are analytic at $x = 1$ , making it an ordinary point. Choice A incorrectly focuses on the original form; Choice B misclassifies the point type; Choice D incorrectly identifies it as irregular singular.