This is going to be tough without LaTeX...

I'm going to define a symbol to try to make things easier. Let S(f(i),i,a,b) be the sum of the function f(i) as i goes from a to b.

The method of Frobenius, for expanding a solution about a non-singular point is that

y(x) = S(a_n x^(n+k),n,0,infinity)

where a_n is a "subscript" n are a series of constants, and k is a constant (I believe it is required to be an integer, but I'm not certain of that.) So basically:

y(x) = a_0 x^k + a_1 x^(k+1) + a_2 x^(k+2) + ...

(Note: Our expansion is in powers of x, and not powers of (x-c) because we are expanding about x = 0.)

Plug this into your differential equation, and reduce the sums so that you can write the equation in terms of a single summation, then set each coefficient of x to zero.

So

y(x) = S(a_n x^(n+k),n,0,infinity)

y'(x) = S(a_n (n+k)x^(n+k-1),n,1,infinity)

y''(x) = S(a_n (n+k)(n+k-1)x^(n+k-2),n,2,infinity)

(Write these out explicitly as a power series if you can't follow the notation. And for the sake of your sanity, rewrite these on paper in terms of summation symbols!)

Thus:

x^2y"-(x+2)y=0 becomes

x^2 S(a_n (n+k)(n+k-1)x^(n+k-2),n,2,infinity) - (x+2) S(a_n x^(n+k),n,0,infinity) = 0

S(a_n (n+k)(n+k-1)x^(n+k),n,2,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2 S(a_n x^(n+k),n,0,infinity) = 0

Now we want to get these three summations to be all one sum. The first step is to note that the first and third summations already have the same power of x listed, but they have a different lower sum limit. (n = 2 for the first, and n = 0 for the last.) So I'm going to rewrite the third summation as:

S(a_n x^(n+k),n,0,infinity) = a_0 x^k + a_1 x^(k+1) + S(a_n x^(n+k),n,2,infinity)

by listing out the first two terms explicity. Thus the equation becomes:

S([a_n (n+k)(n+k-1) - 2a_n]x^(n+k),n,2,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

Now we need to match the coefficients of x in the two summations. (The summation variable is a dummy index, so we can change it to just about whatever we like.) I am going to find a new summation variable n' such that

n + k = n' + k + 1

Thus n = n' + 1 and n' = n - 1.

So replace n with n' + 1 in the function in the summation, and we have a lower limit on the summation of n = 2, so this becomes n' = 2 - 1 = 1. The upper limit, being infinity, stays the same.

Thus:

S([a_(n'+1) (n'+1+k)(n'+1+k-1) - 2a_(n'+1)]x^(n'+1+k),n',1,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

Or, changing the n' back to n for clarity:

S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1)]x^(n+k+1),n,1,infinity) - S(a_n x^(n+k+1),n,0,infinity) - 2a_0 x^k - 2a_1 x^(k+1) = 0

Now we have matching exponents, we need matching summation limits. So write the second summation as:

S(a_n x^(n+k+1),n,0,infinity) = a_0 x^(k+1) + S(a_n x^(n+k+1),n,1,infinity)

And we get:

S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1)]x^(n+k+1),n,1,infinity) - S(a_n x^(n+k+1),n,1,infinity) - 2a_0 x^k - 2a_1 x^(k+1) - a_0 x^(k+1) = 0

Or, collapsing the two summations into one:

S([a_(n+1) (n+k+1)(n+k) - 2a_(n+1) - a_n]x^(n+k+1),n,1,infinity) - 2a_0 x^k - 2a_1 x^(k+1) - a_0 x^(k+1) = 0

Now we start setting powers of x equal to zero.

The lowest power of x in the expansion is x^k. It has a coefficient of:

-2a_0 = 0 Thus a_0 = 0.

The x^(k+1) coefficient gives:

-2a_1 - a_0 = 0. Since a_0 = 0, so a_1 = 0.

The x^(k+2) coefficient gives:

a_2 (1+k+1)(1+k) - 2a_2 - a_1 = 0 ==> a_2[(k+2)(k+1) - 2] = 0 since a_1 = 0.

I am going to assume that a_2 is NOT zero (else we will wind up with a_n = 0 for all n which gives the trivial solution y(x) = 0), because we can set (k+2)(k+1) - 2 = 0. This will give two values of k, representing two linearly independent solutions. So find these values and use each one individually to get a recursion relation relating the rest of the a_n s.

-Dan