Is the Method of Frobenius Appropriate for this DE? If so, how to proceed?

Question

To not bore with motivation, I'll get straight to the point. I have struggled for quite some time finding analytical solutions to the differential equation $$y'' + \frac{a}{x}y' - \frac{b}{x^2\left(1 - \frac{c}{x^{n-2}}\right)}y = 0,\tag{1}$$ where $a,b,c$ are positive constants, and $n \geq 4$ a positive integer. I am interested in the behavior of this equation around $x = x_p \equiv c^{n-2}$. My approach was to use Frobenius' method, and posit the two solutions $$y_1(x) = \sum_{k \ge 0}a_k(x - x_p)^{k+1} \hspace{0.2cm}\text{ and } \hspace{0.2cm}y_2(x) = C\log(x-x_p)y_1(x) + \sum_{k\ge0}b_k(x - x_p)^k,$$ where $a_k,b_k$ are to be determined constants, with $a_0,b_0 \neq 0$, and $C$ a constant that may be zero. [These solutions were obtained by first computing the indices of the associated indicial equation for $x = x_p$ (they turn out to be 0 and 1).] I arrive at some trouble when trying to deduce recurrence relations for $a_k$ and $b_k$. I end up multiplying (1) through by $x^2(1 - \frac{x}{x^{n-2}})$ to simplify the form, but I then obtain the unforgiving expression (for $y_1$) $$x^2\left(1 - \frac{c}{x^{n-2}}\right)\sum_{k\ge 0}a_k(k+1)k(x-x_p)^{k-1} + ax\left(1 - \frac{c}{x^{n-2}}\right)\sum_{k\ge 0}a_k(k+1)(x-x_p)^k - b\sum_{k\ge 0}a_k(x-x_p)^k = 0$$ It is not at all clear to me how to write this solely in terms of $(x-x_p)$ factors, with no stray factors of $x$ or $(x + x_p)$, etc. Perhaps there is a better approach to solving (1)? Or maybe this method is fine, but I'm not seeing how to proceed. Any input is appreciated.

Answer 1

$y''+\dfrac{a}{x}y'-\dfrac{b}{x^2\left(1-\dfrac{c}{x^{n-2}}\right)}y=0$

$x^2\dfrac{d^2y}{dx^2}+ax\dfrac{dy}{dx}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$

With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0216.pdf:

Let $r=x^{n-2}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=(n-2)x^{n-3}\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left((n-2)x^{n-3}\dfrac{dy}{dr}\right)=(n-2)x^{n-3}\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)x^{n-3}\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)x^{n-3}\dfrac{d^2y}{dr^2}(n-2)x^{n-3}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}=(n-2)^2x^{2n-6}\dfrac{d^2y}{dr^2}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}$

$\therefore x^2\left((n-2)^2x^{2n-6}\dfrac{d^2y}{dr^2}+(n-2)(n-3)x^{n-4}\dfrac{dy}{dr}\right)+a(n-2)x^{n-2}\dfrac{dy}{dr}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$

$(n-2)^2x^{2n-4}\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)x^{n-2}\dfrac{dy}{dr}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$

$(n-2)^2r^2\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)r\dfrac{dy}{dr}-\dfrac{br}{r-c}y=0$

$(n-2)^2r(r-c)\dfrac{d^2y}{dr^2}+(n-2)(a+n-3)(r-c)\dfrac{dy}{dr}-by=0$

Which reduces to Gaussian hypergeometric equation.