91Ó°ÊÓ

When working with differential equations, particularly the Euler-Cauchy type, finding the general solution is paramount. The Euler-Cauchy equation, like the one given, is a second-order linear differential equation. The general form of these equations is usually expressed as \(x^2y'' + axy' + by = 0\). To find the general solution, we focus on a type of assumed solution.

In our exercise, we assume a solution of the form \(y = x^m\). This substituted form is simple yet effective because it takes advantage of the equation's structure. Once we assume \(y = x^m\), we calculate its first and second derivatives and substitute these into the original equation.

The goal is to simplify the equation into a form that allows easy solving for \(m\), which represents the powers used in our general solution. Finally, with distinct roots derived from the characteristic equation, the solution is expressed as a combination of these roots: \(y(x) = C_1 x^{m_1} + C_2 x^{m_2}\). This form accounts for all potential solutions to the equation and is what makes it 'general.'

The characteristic equation is a pivotal part of solving an Euler-Cauchy differential equation. Once the equation is transformed with the assumed solution of \(y = x^m\), and its derivatives are substituted back, this transformation leads to a polynomial equation, often quadratic in nature.

Our complication derived from this is: \(m(m-1) - 6 = 0\). This type of polynomial arising out of the manipulation always provides a characteristic equation. The characteristic equation here is quadratic, which reflects the nature of the problem: specifically the order of the differential equation.

Solving this characteristic polynomial gives us the values of \(m\), which are critical for defining the general solution. It precisely transforms a second-order equation into an algebraic problem that is more straightforward to solve, emphasizing the power of algebra in tackling differential equations.

The roots of the quadratic equation play a crucial role in finding the solution to the Euler-Cauchy equation. To find roots of the characteristic equation \(m^2 - m - 6 = 0\), we make use of the quadratic formula:

• \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For our characteristic equation, use the coefficients:

• \(a = 1\)
• \(b = -1\)
• \(c = -6\)

Plugging these values into the quadratic formula provides us with:

• \(m = \frac{1 \pm \sqrt{1 + 24}}{2}\)

Simplifying further, we get:

• \(m_1 = 3\)
• \(m_2 = -2\)

These roots \(m_1\) and \(m_2\) are distinct and they help in defining the structure of the general solution. Distinct roots are crucial in linear differential equations, as they ensure the solutions are linearly independent, allowing the combination to truly cover all possible solutions.