To provide a complete answer, the full equation needs to be included. However, based on the given information, it seems like the differential equation is of the form:

x^2y'' + x^2y = 0

Using the method of Frobenius, we can assume a power series solution of the form:

y(x) = ∑[n=0 to ∞] a_nx^(n + r)

where r is the root of the indicial equation.

The indicial equation is found by substituting y(x) into the differential equation and equating the coefficients of like powers of x to zero.

The differential equation becomes:

x^2y'' + x^2y = ∑[n=0 to ∞] (n + r)(n + r - 1)a_nx^(n + r) + ∑[n=0 to ∞] a_nx^(n + r + 2) = 0

Equating the coefficients of like powers of x to zero, we get:

(n + r)(n + r - 1)a_n + a_n-2 = 0

This equation holds for all values of n, so we have two cases:

Case 1: n is even
In this case, let n = 2m. The equation becomes:

(2m + r)(2m + r - 1)a_2m + a_2m-2 = 0

Simplifying, we get:

[(2m + r)(2m + r - 1)a_2m]/a_2m-2 = -1

This equation holds for all values of m, so we can write:

a_2m = (-1)^(m+1)(r)(r - 1)(r - 2)...(r - 2m + 1)a_0

Case 2: n is odd
In this case, let n = 2m + 1. The equation becomes:

(2m + r + 1)(2m + r)a_2m+1 + a_2m-1 = 0

Simplifying, we get:

[(2m + r + 1)(2m + r)a_2m+1]/a_2m-1 = -1

This equation holds for all values of m, so we can write:

a_2m+1 = (-1)^(m+1)(r + 1)(r)(r - 1)...(r - 2m + 2)a_1

Combining the even and odd terms, we can write the general solution as:

y(x) = a_0x^r [1 + ∑[m=1 to ∞] (-1)^(m+1)(r)(r - 1)...(r - 2m + 1)a_0/(2m)!x^(2m)] + a_1x^(r + 1) [∑[m=0 to ∞] (-1)^(m+1)(r + 1)(r)(r - 1)...(r - 2m + 2)a_1/(2m + 1)!x^(2m + 1)]

Note that the coefficients c1 and c2 mentioned in the initial question are not present in the final solution