Find a Frobenius type solution for the differential equation xy" + y' - y = 0

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-

Explore the Frobenius series solution for the differential equation xy'' + y' - y = 0. This detailed solution includes power series expansions and covers key concepts in differential equations. Suitable for college-level mathematics students.

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