Context:
My friend (The same one who gave me this ODE) originally challenged me to obtain the general solution to this ODE: $$\frac{d^3 y}{dx^3}+\alpha x\frac{dy}{dx}+\beta y=0 \tag{1}$$ Where $\alpha,\beta \in \mathbb{R}$.
I could not figure out any substitution without the use of power series. I thus decided to play around with Wolfram|Alpha, and decided to generalize the order of the ODE, like this:
$$\frac{d^n y}{dx^n}+\alpha x\frac{dy}{dx}+\beta y=0 \tag{2}$$
Upon increasing the value of $n$, I noticed an interesting pattern:
When $n=3$, the solution is: $$\small y(x)=\frac{\sqrt[3]{\alpha}\cdot c_2 x \cdot {_1F_2}\left(\frac{\beta}{3\alpha}+\frac{1}{3};\frac{2}{3},\frac{4}{3}; -\frac{x^3 \alpha}{9}\right)}{3^{2/3}}+c_1\cdot {_1F_2}\left(\frac{\beta}{3\alpha};\frac{1}{3};\frac{2}{3};-\frac{x^3\alpha}{9}\right)+\frac{\alpha^{2/3} c_3 x^2\cdot {_1F_2}\left(\frac{\beta}{3\alpha}+\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{x^3 \alpha}{9}\right)}{\sqrt[3]{3}} \tag{3}$$ When $n=4$, the solution is: $$\small y(x)=\frac{\sqrt[4]{\alpha}c_2 \cdot {_1F_3}\left(\frac{\beta}{4\alpha}+\frac{1}{4};\frac{1}{2},\frac{3}{4},\frac{5}{4};-\frac{x^4\alpha}{64}\right)}{\sqrt{2}}+c_1\cdot {_1 F_3}\left(\frac{\beta}{4\alpha};\frac{1}{4},\frac{1}{2},\frac{3}{4};-\frac{x^4 \alpha}{64}\right)+\frac{\alpha^{3/4} c_4 x^3\cdot {_1F_3}\left(\frac{\beta}{4\alpha}+\frac{3}{4};\frac{5}{4},\frac{3}{2},\frac{7}{4};-\frac{x^4 \alpha}{64}\right)}{\sqrt{2}}+\sqrt{\alpha}c_3 x^2 {_1F_3}\left(\frac{\beta}{4\alpha}+\frac{1}{2};\frac{3}{4},\frac{5}{4},\frac{3}{2};-\frac{x^4 \alpha}{64}\right)$$ When $n=5$, the solution is: $$\small y(x)=\frac{\sqrt[5]{\alpha} c_2 x \cdot {_1F_4}\left(\frac{\beta}{5\alpha}+\frac{1}{5};\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5};-\frac{x^5 \alpha}{625}\right)}{5^{4/5}}+c_1\cdot {_1F_4}\left(\frac{\beta}{5\alpha};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};-\frac{x^5 \alpha}{625}\right)+\frac{\alpha^{4/5}c_5 x^4\cdot {_1F_4}\left(\frac{\beta}{5\alpha}+\frac{4}{5};\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};-\frac{x^5 \alpha}{625}\right)}{\sqrt[5]{5}}+\frac{\alpha^{3/5} c_4 x^3\cdot {_1F_4}\left(\frac{\beta}{5\alpha}+\frac{3}{5};\frac{4}{5},\frac{6}{5},\frac{7}{5},\frac{8}{5};-\frac{x^5 \alpha}{625}\right)}{5^{2/5}}+\frac{\alpha^{2/5} c_3 x^2 {_1F_4}\left(\frac{\beta}{5\alpha}+\frac{2}{5};\frac{3}{5},\frac{4}{5},\frac{6}{5},\frac{7}{5};-\frac{x^5 \alpha}{625}\right)}{5^{3/5}}$$ If anyone is interested, here are the solutions for $n=6$, $n=7$, $n=8$ and $n=9$. I apologize if I mistyped one of the solutions.
As we can see, there is a pattern. I would therefore like to find a reason for this pattern.
Here, I attempt to explain it using a power series solution:
We make the ansatz: $$y=\sum_{k=0}^{\infty} A_k x^k$$ Thus, if we substitute this into our ODE on $(2)$, we obtain: $$\sum_{k=n}^{\infty} k(k-1)(k-2)\cdots (k-n+1)A_{k}x^{k-n}+\alpha\cdot \sum_{k=1}^{\infty} k A_k x^{k}+\beta\cdot \sum_{k=0}^{\infty} A_k x^k=0$$ Shifting the first sum, we obtain: $$\sum_{k=0}^{\infty} (k+n)(k+n-1)(k+n-2)\cdots (k+1)A_{k+n}x^{k}+\alpha\cdot \sum_{k=1}^{\infty} k A_k x^{k}+\beta\cdot \sum_{k=0}^{\infty} A_k x^k=0$$ Hence, we have: $$\small n!\cdot A_n+\beta\cdot A_0+\sum_{k=1}^{\infty} (k+n)(k+n-1)(k+n-2)\cdots (k+1)A_{k+n}x^{k}+\alpha\cdot \sum_{k=1}^{\infty} k A_k x^{k}+\beta\cdot \sum_{k=1}^{\infty} A_k x^k=0$$ Thus, we obtain: $$\small n!\cdot A_n+\beta\cdot A_0+\sum_{k=1}^{\infty} \left[(k+n)(k+n-1)(k+n-2)\cdots (k+1)A_{k+n}+\alpha k A_k+\beta\cdot A_k\right]\cdot x^k=0$$ If the series is a solution, then all the coefficients must equal zero: $$\begin{cases} A_n=-\frac{\beta}{n!}\cdot A_0 & k=0\\ A_{k+n}=-\frac{\alpha k+\beta}{(k+n)(k+n-1)(k+n-2)\cdots (k+1)}A_k & k>0, k\in \mathbb{N}\end{cases}$$ The system of equations is redundant, thus one may just solve the following recurrence relation: $$\small A_{k+n}=-\frac{\alpha k+\beta}{(k+n)(k+n-1)(k+n-2)\cdots (k+1)}A_k=-\frac{k!(\alpha k+\beta)}{(k+n)!}A_k=-\frac{\alpha k+\beta}{(k+1)_n} A_k \tag{4}$$ Where $(x)_n$ is the Pochhammer Symbol.
The Hypergeometric function ${_1P_q}(a_1; b_1,\cdots,b_q;x)$ is defined as: $${_1P_q}(a_1; b_1,\cdots,b_q;x)=\sum_{k=0}^{\infty} \frac{(a_1)_k}{(b_1)_k\cdots (b_q)_k}\cdot \frac{x^k}{k!}$$ Given the recurrence, this should follow the pattern for general values of $n\geq 3, n \in \mathbb{Z}^+$.
The problem:
However, I am not satisfied with this. I would prefer if one could use one (or several) substitution(s) which works for general values of $n$ and which does not involve a series solution. My idea was to reduce the differential equation on $(2)$ to Kummer's Equation to obtain these hypergeometric functions on their solutions: $$z\frac{d^2 w}{dz^2}+(b-z)\frac{dw}{dz}-aw=0 \tag{5}$$ To do this, I've attempted to use the same substitution as done on my previous question: $$\ln y=\ln{z}-\frac{(1+x)^2}{4}$$ However, the resulting ODE turns out to be even worse than it was previously (Even for $n=3$).
The question:
Is there any way we can make several substitutions such that we obtain the general solution for general values of $n$ where $n\geq 3, n \in \mathbb{Z^+}$?
If this is too difficult or not possible, is there a way we can use several substitutions for the specific case $n=3$ to obtain the general solution? i.e. The same one, or in a similar form to the one on equation $(3)$.
Thanks in advance.
Define $s=-\frac{\alpha x^n}{n^{n-1}}$ and introduce Euler operators $\vartheta=s\frac{d}{ds}$ and $\delta=x\frac{d}{dx}=n\vartheta$.
Lemma. $x^n\frac{d^n}{dx^n}=\delta(\delta-1)(\delta-2)\ldots (\delta-n+1)$.
Corollary 1. $x^n\frac{d^n}{dx^n}=n^n\vartheta\left(\vartheta-\frac1n\right)\left(\vartheta-\frac2n\right)\ldots \left(\vartheta-\frac{n-1}n\right)$.
Corollary 2. After the change of independent variable $y(x)=f(s)$, the equation (2) becomes $$\left[\vartheta\left(\vartheta-\frac1n\right)\left(\vartheta-\frac2n\right)\ldots \left(\vartheta-\frac{n-1}n\right)-s\left(\vartheta+\frac{\beta}{n\alpha}\right)\right]f(s)=0.$$ This is a particular case of the standard form of the generalized hypergeometric equation - namely, the equation for $_1F_{n-1}\left(\frac{\beta}{n\alpha};\frac1n,\ldots,\frac{n-1}{n};s\right)$. Other $n-1$ linear independent solutions can be chosen as in formula 16.8.6 here.