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$$x^2\frac{d^2y}{dx^2}+x^2\frac{dy}{dx}-2y=0$$

$x=0$ is a regular singular point.

$$y=\sum_{n=0}^\infty c_nx^{n+r}$$

$$\frac{dy}{dx}=(n+r)\sum_{n=0}^\infty c_nx^{n+r-1}$$

$$\frac{d^2y}{dx^2}=(n+r)(n+r-1)\sum_{n=0}^\infty c_nx^{n+r-2}$$

$$(n+r)(n+r-1)\sum_{n=0}^\infty c_nx^{n+r}+(n+r-1)\sum_{n=1}^\infty c_{n-1}x^{n+r}-2\sum_{n=0}^\infty c_nx^{n+r}=0$$

Taking out a few terms

$$(r)(r-1)c_0x^r+-2c_0x^r+(n+r)(n+r-1)\sum_{n=1}^\infty c_nx^{n+r}+ (n+r-1)\sum_{n=1}^\infty c_{n-1}x^{n+r}-2\sum_{n=1}^\infty c_nx^{n+r}=0$$

The incidal equation is

$$r^2-r-2=0$$

The recurrence formula is,

$$(n+r)(n+r-1)c_n+(n+r-1)c_{n-1}-2 c_n=0$$

Bigger roots

Let $r=r_1=2$

$$(n+2)(n+1)c_n+(n+1)c_{n-1}-2c_n=0$$

$$c_n=\frac{-(n+1)c_{n-1}}{n^2+3n}$$

$$c_1=\frac{-c_0}{2},$$

$$c_2=\frac{-c_{1}}{6}$$

Taking the smaller root, $r=r_2=-1$

$$(n-1)(n-2)c_n+(n-2)c_{n-1}-2c_n=0$$

$$c_n=\frac{(2-n)c_{n-1}}{n^2-3n}$$

How shall I continue this any further? Any help would be appreciated. Can someone hint me on this question.

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In your equations they are some terms with $n$ in them.

These terms should be in the $\displaystyle{\sum_n}$ , not outside.

Nevertheless, your result is correct which is well. But one can go further.

Case $r=2$ : $$c_n=\frac{-(n+1)}{n^2+3n}c_{n-1}$$ $$c_n=c_0\prod_{k=1}^n\left(\frac{-(k+1)}{k^2+3k}\right)= (-1)^n c_0\prod_{k=1}^n\left(\frac{k+1}{k(k+3)}\right)$$ Develop and simplify : $$c_n=\frac{6(-1)^n(n+1)}{(n+3)!}c_0$$ $$y=\sum_{n=0}^\infty c_nx^{n+2}=6c_0\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{(n+3)!}x^{n+2} = -6c_0\sum_{n=3}^\infty \frac{(-1)^n(n-2)}{n!}x^{n-1}$$ $$y= -6c_0\sum_{n=3}^\infty \frac{(-1)^n}{(n-1)!}x^{n-1} +12c_0x^{-1}\sum_{n=3}^\infty \frac{(-1)^n}{n!}x^{n}$$ $$y= 6c_0\sum_{n=2}^\infty \frac{(-1)^n}{n!}x^{n} +12c_0x^{-1}\sum_{n=3}^\infty \frac{(-1)^n}{n!}x^{n}$$

$$y= 6c_0\left(e^{-x}-1+x \right)+12c_0x^{-1}\left(e^{-x}-1+x-\frac{x^2}{2} \right)$$ So, the first family of solutions is obtained : $$y= 6c_0\left(e^{-x}+\frac{2e^{-x}}{x} -\frac{2}{x}+1\right)$$

Case $r=-1$ :

Proceed on the same manner to obtain the second family of solutions (It is simpler because they are only two terms in the series) : $$y=c'_0\left(\frac{1}{x}-\frac{1}{2}\right)$$

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