Consider the following second order differential equation
$$ z^2 \psi''(z)+(z a+b) \psi'(z)+(b z+c) \psi(z) = 0 $$
where $a,b$ and $c$ are arbitrary real numbers. The equation has two irregular singular points at $z = 0$ and $z = \infty$. Mathematica is unable to solve this problem. Of course if $b = 0$ then the solution is trivially given by $z$ to some power. But otherwise, I'm not sure how to attack this equation. Any help is greatly appreciated!
Motivated by this, you can build an asymptotic expansion with $b \ll 1$, with $\psi = \sum_j \psi_j b^j$. We have \begin{align} \sum_{j=0} b^j (z^2 \psi_j'' + a z \psi_j' + c \psi_j) + b^{j+1} (\psi_j' + z \psi_j) & = 0 \\ z^2 \psi_0'' + a z\psi_0' + c \psi_0 + \sum_{j=1} b^j (z^2 \psi_j'' + a z \psi_j' + c \psi_j + \psi_{j-1}' + z \psi_{j-1}) & = 0 \end{align} Thus, we need to solve the following systems iteratively: \begin{align} & \mathcal O(1): \quad L \psi_0:=z^2 \psi_0'' + a z\psi_0' + c \psi_0 = 0\\ & \mathcal O(b^j): \quad L \psi_j = -(\psi_{j-1}' + z \psi_{j-1}), \quad j \ge 1. \end{align} You can build up your solution with the number of terms desired. If $b$ is not small, then this will not work well.