Question

Find and classify each singular point as regular or irregular: (x+1)y" + (x-4)⁵/³/(x²+4x+3) y' + (x-5) y=0

Answer 1

The singular points of the given differential equation are x = -3, x = -1, and x = -1.

Step-by-step explanation:

A singular point of a differential equation is a point where the coefficients of the equation become infinite or where the equation is not defined. To find the singular points of the given equation, we need to determine the values of 'x' where the coefficients are infinite or where the denominator of the coefficient is zero.

For the given equation, (x+1)y'' + (x-4)^(5/3)/(x^2+4x+3) y' + (x-5) y=0, the singular points occur when:

• The coefficient (x+1) becomes zero, giving x = -1.
• The denominator (x^2+4x+3) becomes zero, giving x = -3 or x = -1.

Therefore, the singular points of the given equation are x = -3, x = -1, and x = -1.