Find all singular points of xy"+(1-x)y'+xy=0 and determine whether each one is regular or irregular.

Answer: x=0, regular

I know that x=0 but I need to take the limit as x approaches to 0 but what function should I take the limit?

Find all singular points of xy"+(1-x)y'+xy=0 and determine whether each one is regular or irregular.

Answer: x=0, regular

I know that x=0 but I need to take the limit as x approaches to 0 but what function should I take the limit?

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Marked as Best Answer

Hello Sun,

Recall that a singular point x = x_{0} of an equation such as

P(x)y'' + Q(x)y' + R(x)y = 0

is called a regular singular point if both of the following are analytic at x_{0}:

(x - x_{0})( Q(x)/P(x) ) and (x - x_{0})^{2}( R(x)/P(x) ).

Since your equation has P(x) = x, Q(x) = 1-x, and R(x) = x, and these are all polynomials, checking that the above quotients are analytic at x_{0} amounts to seeing whether their limits are finite as you x → x_{0}. That is, compute the limits

lim_{x → 0} (x(1-x)/x) and lim_{x → 0} (x(x)/x)

and see if they are finite (they are). If so, the point x = 0 is a regular singular point.

Hope this clears things up.

Regards,

Hassan H.

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