Consider the Legendre equation, $$(1-x^2)y''-2xy'+\alpha (\alpha +1)y=0$$
Let $x=1/t$, then $$\frac{dy}{dx}=-t^2\frac{dy}{dt}$$ $$\frac{d^2y}{dx^2}=t^4\frac{d^2y}{dt^2}+2t^3\frac{dy}{dt}$$
The equations becomes (I write $y_t$ to denote a derrivative w.r.t $t$), $$(1-\frac{1}{t^2})(t^4y_{tt}+2t^3y_t)-2(\frac{1}{t})(-t^2y_t)+\alpha(\alpha+1)y=0$$ or, $$t^2(t^2-1)y_{tt}-2t^3y_t+\alpha(\alpha+1)y=0$$ We see we have a singular point at $t=0\ (x=\infty)$ since, $$b_1(t)=t\frac{2t^3}{t^2(t^2-1)}=\frac{2t^2}{(t^2-1)}$$ $$b_2(t)=t^2\frac{\alpha (\alpha+1)}{t^2(t^2-1)}=\frac{\alpha (\alpha+1)}{(t^2-1)}$$ are both analytic around $t=0$. Thus it is a regular singular point. The indicial polynomial is (I'm not sure if this is correct), $$r(r-1)+b_1(0)r+b_2(0)=0$$ $$r(r-1)+0r-\alpha(\alpha +1)=0$$ which has solutions, $r_1=\alpha+1,\ r_2=-\alpha$.
Now my question is concerning a special case of the Legendre polynomial when $\alpha =1$, or, $$(1-x^2)y''-2xy'+2y=0$$ I am asked to find 2 L.I. solutions of the form, $$\phi(x)=x^{-r}\sum_{k=0}^{\infty}c_kx^{-k}$$ valid for $|x|>1$.
My attempt $$\phi(x)=\sum_{k=0}^{\infty}c_kx^{-k-r}$$ $$\phi'(x)=\sum_{k=0}^{\infty}c_k(-k-r)x^{-k-r-1}$$ $$\phi''(x)=\sum_{k=0}^{\infty}c_k(-k-r)(-k-r-1)x^{-k-r-2}=\sum_{k=0}^{\infty}c_k(k+r)(k+r+1)x^{-k-r-2}$$ Plugging this into the equation I get, $$\sum_{k=0}^{\infty}c_k(k+r)(k+r+1)x^{-k-r-2}-\sum_{k=0}^{\infty}c_k(k+r)(k+r+1)x^{-k-r}-\sum_{k=0}^{\infty}2c_k(-k-r)x^{-k-r}+\sum_{k=0}^{\infty}2c_kx^{-k-r}=0$$
Doing the change of indecies in the first some for $k\to k-2$ and combining the last 3 sums gives, $$\sum_{k=2}^{\infty}c_{k-2}(k+r-2)(k+r-1)x^{-k-r}+\sum_{k=0}^{\infty}c_kx^{-k-r}[-(k+r)(k+r+1)+2(k+r)+2]=0$$ Taking out the first 2 terms of the last sum gives, $$ c_0x^{-r}[-r(r+1)+2r+2]+c_1x^{-r-1}[-(r+1)(r+2)+2(r+1)+2]+\sum_{k=2}^{\infty}x^{-k-r}\{c_{k-2}[(k+r-2)(k+r-1)]+c_k[-(k+r)(k+r+1)+2(k+r)+2]\}=0$$
If $c_0 \neq 0$ and $c_1 \neq 0$, then we get 2 indicial polynomials for the roots of r, $$-r(r+1)+2r+2=0$$ and, $$-(r+1)(r+2)+2(r+1)+2=0$$
which both have different roots, so I'm not sure if I simply made a calculation error somewhere, or if I am missing something.
Furthermore, only the second indicial equation would match up with the one found earlier, with roots $-\alpha$ and $\alpha +1$.