Problems while solving the differential equation. $$x^2\frac{d^2y}{dx^2}+x^2\frac{dy}{dx}-2y=0$$
$x=0$ 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.

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) :
$$y=c'_0\left(\frac{1}{x}-\frac{1}{2}\right)$$
