对于
\[\frac{d^2u(z)}{dz^2} + p(z)\frac{du(z)}{dz} + q(z)u(z) = 0
\]
\(z_0\) — 方程的正则奇点
进行洛朗级数展开：\(p(z) = \sum_{m=-1}^{\infty} p_m(z - z_0)^m, \quad q(z) = \sum_{m=-2}^{\infty} q_m(z - z_0)^m\)
设 \(u(z)\) 展开的最低次幂为 \(s\) 次 \(u(z) = \sum_{k=0}^{\infty} a_k(z - z_0)^{s+k}\) 代入方程
\[\sum_{k=0}^{\infty} (s+k)(s+k-1)a_k(z-z_0)^{s+k-2} + \sum_{k=0}^{\infty} (s+k)a_k(z-z_0)^{s+k-1} \sum_{m=-1}^{\infty} p_m(z-z_0)^m
\]
\[+ \sum_{k=0}^{\infty} a_k(z-z_0)^{s+k} \sum_{m=-2}^{\infty} q_m(z-z_0)^m = 0
\]
最低幂次项系数 \(a_0[s(s - 1) + sp_{-1} + q_{-2}] = 0\)
\(a_0 \neq 0\)，指标方程 \(s^2 + (p_{-1} - 1)s + q_{-2} = 0\) \(\rightarrow s_1 \geq s_2\)
取 \(s = s_1\) \(u_1(z) = \sum_{k=0}^{\infty} a_k(z - z_0)^{s_1+k}\)
\[\sum_{k=0}^{\infty} (s_1+k)(s_1+k-1)a_k(z-z_0)^{s_1+k-2} + \sum_{k=0}^{\infty} (s_1+k)a_k(z-z_0)^{s_1+k-1} \sum_{m=-1}^{\infty} p_m(z-z_0)^m
\]
\[+ \sum_{k=0}^{\infty} a_k(z-z_0)^{s_1+k} \sum_{m=-2}^{\infty} q_m(z-z_0)^m = 0
\]
\((z - z_0)^{s_1-1}\) 项前面的系数
\(a_1 [s_1(s_1 + 1) + (s_1 + 1)p_{-1} + q_{-2}] + a_0 [s_1p_0 + q_{-1}] = 0\)
\[a_1 = -\frac{s_1p_0 + q_{-1}}{s_1(s_1 + 1) + (s_1 + 1)p_{-1} + q_{-2}} a_0
\]
\((z - z_0)^{s_1}\) 项前面的系数
\(a_2 [(s_1 + 1)(s_1 + 2) + (s_1 + 2)p_{-1} + q_{-2}]+a_1 [(s_1 + 1)p_0 + q_{-1}]+a_0 [s_1p_1 + q_0] = 0\)
\[a_2 = -\frac{(s_1 + 1)p_0 + q_{-1}}{(s_1 + 1)(s_1 + 2) + (s_1 + 2)p_{-1} + q_{-2}} a_1-\frac{s_1p_1 + q_0}{(s_1 + 1)(s_1 + 2) + (s_1 + 2)p_{-1} + q_{-2}} a_0
\]
\[\vdots
\]
\[a_k = -\frac{(s_1 + k - 1)p_0 + q_{-1}}{(s_1 + k - 1)(s_1 + k) + (s_1 + k)p_{-1} + q_{-2}} a_{k-1}-\cdots-\frac{s_1p_{k-1} + q_{k-2}}{(s_1 + k - 1)(s_1 + k) + (s_1 + k)p_{-1} + q_{-2}} a_0
\]
\[u_1(z) = \sum_{k=0}^{\infty} a_k(z - z_0)^{s_1+k} \qquad \text{— 第一类型解}
\]
对于 \(s = s_2\) \(u_2(z) = \sum_{k=0}^{\infty} b_k(z - z_0)^{s_2+k}\)
\[\sum_{k=0}^{\infty} (s_2+k)(s_2+k-1)a_k(z-z_0)^{s_2+k-2} + \sum_{k=0}^{\infty} (s_2+k)a_k(z-z_0)^{s_2+k-1} \sum_{m=-1}^{\infty} p_m(z-z_0)^m
\]
\[+ \sum_{k=0}^{\infty} a_k(z-z_0)^{s_2+k} \sum_{m=-2}^{\infty} q_m(z-z_0)^m = 0
\]
当 \(s_1 - s_2\) 不等于整数时，与前相同的分析过程必然可得
\[b_k = -\frac{(s_2 + k - 1)p_0 + q_{-1}}{(s_2 + k - 1)(s_2 + k) + (s_2 + k)p_{-1} + q_{-2}} b_{k-1}-\cdots-\frac{s_2p_{k-1} + q_{k-2}}{(s_2 + k - 1)(s_2 + k) + (s_2 + k)p_{-1} + q_{-2}} b_0
\]
当 \(s_1 - s_2 = m\) 为整数时
（在\(k=m\)时）\((s_2 + m - 1)(s_2 + m) + (s_2 + m)p_{-1} + q_{-2} = (s_1 - 1)s_1 + s_1p_{-1} + q_{-2} \equiv 0\)
\(b_m = ?\)
需要另一种方法，得到
第二类型解:
\[u_2(z) = Au_1(z) \ln(z - z_0) + \sum_{k=0}^{\infty} d_k(z - z_0)^{s_2+k}
\]
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