- #1

- 101

- 25

$$

xy'' + (b-x)y' - ay = 0

$$

By considering the series solution ##y=\sum c_mx^m##, I have obtained two solutions of the following form:

$$

\begin{aligned}

y_1 &= M(x, a, b) \\

y_2 &= x^{1-b}M(x, a-b+1, 2-b) \\

\end{aligned}

$$

Where ##M## is a function after solving the recurrence coefficients, please feel free to check if my result is correct:

$$

M(x, a, b) = \sum_{m=0}^\infty \frac{\prod_{i=0}^{m-1}(a+i)}{m!\prod_{i=0}^{m-1}(b+i)}x^m

$$

The question is if ##b=1##, we get repeated root and ##y_1 = y_2## hence linearly dependent. How to obtain a second independent solution? The problem has a hint of considering the following limit:

$$

\lim_{b \to 1} \frac{y_2 - y_1}{b - 1}

$$