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The indicial equation plays a crucial role in solving Cauchy-Euler differential equations. First, let's understand what an indicial equation is. It is an auxiliary polynomial equation derived when we assume a solution of the form \( y = x^r \) for a differential equation of the Cauchy-Euler type. This assumption helps transform the Cauchy-Euler equation into a simpler algebraic form.

To find the indicial equation, we substitute \( y = x^r \) into the original differential equation and simplify. For instance, with the equation \( x^2 y'' + 3x y' + (1 - 2x) y = 0 \), substituting the assumed form leads to:
\[r(r-1) + 3r + 1 = 0\]

The roots of the indicial equation tell us much about the potential solutions of the differential equation. Solving the above, we see that it can be rewritten as a perfect square trinomial:
\[(r + 1)^2 = 0\]

This equation has a repeated root \( r = -1 \). Repeated roots will lead to a different form of the solution compared to distinct roots. These roots help in constructing the fundamental set of solutions, which we will explore next.

A fundamental set of solutions is a collection of solutions to a differential equation that can be used to express any solution of that equation. It is akin to a basis in vector spaces, where the solutions in the set are linearly independent. In the context of second-order linear differential equations like Cauchy-Euler equations, finding this set is crucial.

When dealing with the Cauchy-Euler equation \( x^2 y'' + 3x y' + (1 - 2x) y = 0 \), and solving its indicial equation gives repeated roots \( r = -1 \), we obtain the following solutions:

• \( y_1 = x^{-1} = \frac{1}{x} \)
• \( y_2 = x^{-1} \ln(x) = \frac{1}{x} \ln(x) \)

These functions \( y_1 \) and \( y_2 \) form the fundamental set of solutions for our differential equation. They are particularly useful because every solution to the differential equation can be formed by a linear combination of \( y_1 \) and \( y_2 \):
\[y = C_1 y_1 + C_2 y_2\]
where \( C_1 \) and \( C_2 \) are arbitrary constants. Now, let's explore why these solutions are linearly independent.

In mathematics, two functions are said to be linearly independent if neither is a scalar multiple of the other. For a differential equation, having linearly independent solutions ensures that they span the solution space, providing us the most specific representation of any solution.

Consider our two solutions \( y_1 = \frac{1}{x} \) and \( y_2 = \frac{1}{x} \ln(x) \). To confirm their linear independence, one can use the Wronskian test. The Wronskian \( W(y_1, y_2) \) is calculated by the determinant as follows:
\[W(y_1, y_2) = \begin{vmatrix}y_1 & y_2 \y_1' & y_2'\end{vmatrix}\]

Calculating this gives us:
\[W(y_1, y_2) = \begin{vmatrix}\frac{1}{x} & \frac{1}{x} \ln(x) \-\frac{1}{x^2} & -\frac{1}{x^2} \ln(x) + \frac{1}{x^2}\end{vmatrix}\]

Simplifying this determinant results in a non-zero expression, confirming that \( y_1 \) and \( y_2 \) are indeed linearly independent. This linear independence is what makes \( \{y_1, y_2\} \) a robust fundamental set of solutions.