91Ó°ÊÓ

To tackle the Cauchy-Euler equation, we leverage a clever transformation. Consider the substitution: \( x = e^t \), which leads to \( t = \ln{x} \), and thus we rewrite the variable x in terms of t. This substitution simplifies the differential equation significantly. The derivatives transform as follows using the chain rule:
We have \(y' = \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{dy}{dt} \frac{1}{x}\) and \(y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}\big(\frac{dy}{dx}\big) \) which ultimately leads to:
\(y'' = \frac{d}{dx}\big( \frac{1}{x} \frac{dy}{dt} \big) = x^{-2}\frac{d^2 y}{d t^2} - x^{-3}\frac{d y}{d t}\). By substituting these transformations back into the original Cauchy-Euler equation, we convert it into a more manageable form: \(a \frac{d^2 y}{dt^2} + (b - a) \frac{d y}{dt} + c y = 0\). This paves the way for further analysis and solution methods.

The characteristic equation is a crucial step in solving differential equations. For the transformed Cauchy-Euler equation \(a \frac{d^2 y}{dt^2} + (b - a) \frac{d y}{dt} + c y = 0\), we set up the characteristic polynomial as \(a r^2 + (b - a) r + c = 0\). Solving this quadratic equation provides the roots that indicate the nature of the solutions to the differential equation. If this quadratic has repeated roots, then the discriminant must be zero:
\((b - a)^2 - 4ac = 0\). Solving for r gives the repeated root as:
\(r = \frac{-(b - a)}{2a}\). This result is essential in determining whether we need additional techniques such as reduction of order to find the second solution.

Reduction of order is a powerful method used to find a second solution when one solution is already known. In the context of our problem, if the solution \(y_1 = x^r\) is found, reduction of order can help find the second independent solution. Assume the second solution takes the form:
\(y_2 = v(x)y_1 = v(x)x^r\). For simplicity, let \(v(x) = \ln{x}\). Differentiating, we get:
\(y_2' = r x^{r - 1} \ln{x} + x^r \frac{1}{x} = r x^{r - 1 } + x^{r - 1} = x^{r - 1}(r + \ln{x})\). Verify that this form satisfies the Cauchy-Euler equation. This process reaffirms the solution's validity and ensures that it is indeed linearly independent from the first solution.

The Wronskian is a determinant used to check if two solutions to a differential equation are linearly independent. For our solutions \(y_1 = x^r\) and \(y_2 = x^r \ln{x}\), we set up the Wronskian as:
\[W(y_1, y_2) = \begin{vmatrix} x^r & x^r \ln {x} \ r x^{r - 1} & x^{r - 1}(r \ln{x} + 1) \end{vmatrix}\].
Simplifying, \[ W = x^{2r - 1}\]. This non-zero Wronskian confirms that the two solutions \(x^r\) and \(x^r \ln{x}\) are linearly independent. Linear independence is crucial because it means we have a complete set of solutions for the differential equation, which we can use to construct the general solution.