91Ó°ÊÓ

The differential equation is given as \( x^3(x^2-25)(x-2)^2 y'' + 3x(x-2) y' + 7(x+5)y = 0 \). We can compare this with the standard form \( P(x) y'' + Q(x) y' + R(x) y = 0 \). Here, \( P(x) = x^3(x^2-25)(x-2)^2 \), \( Q(x) = 3x(x-2) \), and \( R(x) = 7(x+5) \). We will use these to find singular points.

The singular points of the differential equation occur where \( P(x) = 0 \). Solving \( x^3(x^2-25)(x-2)^2 = 0 \), we find the singular points as \( x = 0, x = 5, x = -5, \) and \( x = 2 \).

For a singular point \( x = x_0 \), if \( (x-x_0)\frac{Q(x)}{P(x)} \) and \( (x-x_0)^2\frac{R(x)}{P(x)} \) are analytic (meaning they can be expressed as a power series) at \( x = x_0 \), it is a regular singular point; otherwise, it is irregular. Let's check each singular point:1. **For \( x = 0 \):** - \( \frac{Q(x)}{P(x)} = \frac{3(x-2)}{x^2(x^2-25)(x-2)^2} \), so \( x \frac{Q(x)}{P(x)} \) is not analytic at \( x = 0 \). - \( \frac{R(x)}{P(x)} = \frac{7(x+5)}{x^3(x^2-25)(x-2)^2} \), so \( x^2 \frac{R(x)}{P(x)} \) is not analytic at \( x = 0 \). - Thus, \( x = 0 \) is an irregular singular point.2. **For \( x = 5 \):** - \( \frac{Q(x)}{P(x)} = \frac{3(x-2)}{x^3(x^2-25)(x-2)^2} \), hence \( (x-5) \frac{Q(x)}{P(x)} \) is analytic at \( x = 5 \). - \( \frac{R(x)}{P(x)} = \frac{7(x+5)}{x^3(x^2-25)(x-2)^2} \), so \( (x-5)^2 \frac{R(x)}{P(x)} \) is analytic at \( x = 5 \). - Therefore, \( x = 5 \) is a regular singular point.3. **For \( x = -5 \):** - \( (x+5) \) factors cancel out analytically, following similar steps as above, verifies that \( x = -5 \) is also a regular singular point.4. **For \( x = 2 \):** - Similar calculations show that \( x = 2 \) is a regular singular point.