91Ó°ÊÓ

When dealing with homogeneous linear differential equations with constant coefficients, a key step is to derive the characteristic equation. This equation is an algebraic expression derived from the differential equation itself by substituting the differential operator \( \frac{d}{dx} \) with a variable, often denoted as \( m \).

• This transformation simplifies the analysis by converting the differential equation into a polynomial equation.
• The degree of this polynomial reflects the order of the differential equation. For a second-order differential equation, like the one in our exercise, the characteristic equation is a quadratic equation.

In our specific problem, replacing \( \frac{d}{dx} \) with \( m \) in the equation \( \frac{d^2 y}{d x^2} - \frac{d y}{d x} - 6y = 0 \) yields the characteristic equation \( m^2 - m - 6 = 0 \). This characteristic equation is crucial because its roots will help determine the general solution of the differential equation.

Constant coefficients play a vital role in simplifying and solving linear differential equations. These coefficients, as the name suggests, are constant values in front of each derivative term in the equation.

• Equations with constant coefficients are easier to manage as they lead to characteristic equations that are polynomials with constant terms.
• In our exercise, the coefficients are implicit in the equation \( \frac{d^2 y}{d x^2} - \frac{d y}{d x} - 6y = 0 \), where the coefficients for \( \frac{d^2 y}{d x^2} \), \( \frac{d y}{d x} \), and \( y \) are 1, -1, and -6, respectively.

These coefficients remain consistent throughout the problem, allowing us to establish and solve the characteristic equation effectively. Understanding that these constants do not change enables us to focus on how the equation structure dictates the behavior and solution of the differential equation.

The general solution of a differential equation represents a family of all possible solutions. It is derived from the roots of the corresponding characteristic equation.

• There are several cases depending on the nature of these roots: real and distinct, real and repeated, or complex.
• For the given equation, the characteristic roots \( m_1 = 3 \) and \( m_2 = -2 \) are real and distinct.

In the context of real and distinct roots, the general solution takes the form:\[ y(x) = C_1 e^{m_1 x} + C_2 e^{m_2 x} \]Where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions, and \( m_1 \) and \( m_2 \) are the characteristic roots. Thus, for our specific problem, the solution is:\[ y(x) = C_1 e^{3x} + C_2 e^{-2x} \]This general solution signifies that the combination of exponential functions based on the distinct roots fully describes the behavior of the original differential equation under various conditions.