Solution

For

y^{''} + P (x) y^{'} + Q (x) y = 0,

x_{0}

is an ordinary point when

P

and

Q

are analytic there, and a singular point otherwise. A singular point is regular singular when

(x - x_{0}) P (x)

and

(x - x_{0})^{2} Q (x)

are analytic there. These are the ordinary-point and regular-singular criteria.

y = \sum_{m = 0}^{\infty} c_{m} x^{m} .

Equating the coefficient of

x^{m}

gives

(m + 1) (m + b) c_{m + 1} = (m + a) c_{m} .

Taking

c_{0} = 1

c_{m} (a, b) = \frac{( a ) _{m}}{( b ) _{m} m !},

where

(a)_{m} = a (a + 1) \dots (a + m - 1)

is the rising factorial. Thus

y_{1} = M (x, a, b) = \sum_{m = 0}^{\infty} \frac{( a ) _{m}}{( b ) _{m} m !} x^{m} .

Putting

y = x^{1 - b} u

and simplifying gives

x u^{''} + (2 - b - x) u^{'} - (a - b + 1) u = 0.

Therefore

y_{2} = x^{1 - b} M (x, a - b + 1, 2 - b) .

For nonintegral

b

, the powers

x^{0}

and

x^{1 - b}

at zero are distinct, so these solutions are linearly independent.

When

b \to 1

, both solutions tend to

M (x, a, 1)

. Differentiate their difference with respect to

b

. Writing

M_{a}, M_{b}

for derivatives with respect to the second and third arguments,

b \to 1 lim \frac{y _{2} - y _{1}}{b - 1} = - M (x, a, 1) lo g x - M_{a} (x, a, 1) - 2 M_{b} (x, a, 1) .

Hence at

b = 1

one may take

M (x, a, 1), - M (x, a, 1) lo g x - M_{a} (x, a, 1) - 2 M_{b} (x, a, 1)

as two linearly independent solutions. Their independence follows from the logarithmic term in the second solution.

Solved by gpt-5.6-sol high.

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