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Find all singular points of the given differential equation, then classify as Irregular Singular Points (ISP) or Regular Singular Points (RSP):

(x² - 4) y'' + (1/(x-2)) y'

User Datenshi
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Final answer:

The given differential equation has two singular points: x = 2 (RSP) and x = -2 (ISP).

Step-by-step explanation:

To find the singular points of the given differential equation, let's first determine where the coefficients become zero. The coefficient (x² - 4) becomes zero when x = 2 and x = -2. The coefficient 1/(x-2) becomes zero at x = 2.

Therefore, we have two singular points: x = 2 and x = -2.

Since (x-2) is a factor in the coefficient 1/(x-2), the singular point x = 2 is a Regular Singular Point (RSP). On the other hand, since (x+2) is not a factor in any of the coefficients, the singular point x = -2 is an Irregular Singular Point (ISP).

User Pranav MS
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