What is the pattern for writing the indicial equation for higher order Euler Ordinary Differential Equations?
For first order Euler ODEs, the indicial equation coefficients mirror the ODE coefficients exactly, i.e., $xy' + 5y = 0$ yields $s + 5 = 0$. Define this pattern as $[0, 0]$, as nothing is added to either term.
For second order Euler ODEs, $1$ must be subtracted from the $xy'$ coefficient, i.e., $x^2 y'' + 9xy' - y = 0$ yields $s^2 + 8s - 1 = 0$. This is the pattern $[0, -1, 0]$, as $-1$ is added to the middle term.
These patterns can be computed by converting a first order constant coefficient linear ODE to the associated Euler ODE via the substitution $x = e^t$. However, checking the patterns by doing this conversion for higher order ODEs would require applying a mess of third/fourth/fifth derivative chain rules and product rules.
What is the pattern for higher order Euler ODEs? My guess is that something related to binomial coefficients ($[0, -1, -1, 0]\ [0, -1, -2, -1, 0]$, perhaps up to some sign oscillation) or another collection of numbers famous in Combinatorics appears.
For the ODE \begin{align} \sum_{n=0}^m\alpha_nx^n\frac{\mathrm d^ny}{\mathrm dx^n}=0, \end{align} assume the solution $y=x^r$, so \begin{align} x^r\sum_{n=0}^m\alpha_n(r)_n=0, \end{align} where $(r)_n$ is the falling factorial. Equating the coefficients to $0$ gives \begin{align} \sum_{n=0}^m\alpha_n(r)_n=0, \end{align} writing out the falling factorial in terms of stirling numbers gives \begin{align} \sum_{n=0}^m\alpha_n \sum_{k=0}^n s(n,k)r^k=0. \end{align} Not sure how to concisely do this, but I manually wrote out and added like terms of $r^k$ to get that the $k^{\text{th}}$ coefficient is \begin{align} c_k=\sum_{n=0}^m\alpha_n s(n,k). \end{align}
To actually answer your question, to find the change from the original $k^{\text{th}}$ constant you could simply compute \begin{align} c_k-\alpha_k=\sum_{n=0}^m\alpha_n s(n,k)-\alpha_k, \end{align} which is indeed shorter than computing a $7^{\text{th}}$ order chain rule. . . .