To find a Frobenius series solution for the differential equation

$xy'' + y' - y = 0,$

we will attempt to find a solution in the form of a power series around $x = 0$:

$y(x) = \sum_{n=0}^{\infty} a_n x^{n+r},$

where $r$ is the indicial exponent to be determined, and $a_n$ are the series coefficients.

Step 1: Compute derivatives of the series

First, we calculate the first and second derivatives of the proposed solution.

1. The first derivative $y'(x)$ is:

$y'(x) = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1}.$

2. The second derivative $y''(x)$ is:

$y''(x) = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2}.$

Step 2: Substitute into the differential equation

We now substitute $y(x)$, $y'(x)$, and $y''(x)$ into the original differential equation:

$x y'' + y' - y = 0.$

Substitute each term:

• ( x y'' = x \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-