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A characteristic equation is fundamental when solving second-order differential equations. It serves as a bridge between differential equations and algebra. To create the characteristic equation, we substitute derivatives in the differential equation with powers of a variable, typically denoted as \( r \). This results in a polynomial equation that is often easier to solve. In the given problem, the differential equation is \( \frac{d^2 y}{d x^2} + 6 \frac{d y}{d x} + 2 y = 0 \). By making substitutions, we convert the problem into an algebraic one:

• Replace \( \frac{d^2 y}{d x^2} \) with \( r^2 \)
• Replace \( \frac{d y}{d x} \) with \( r \)
• Replace \( y \) with 1

These substitutions produce the characteristic equation: \( r^2 + 6r + 2 = 0 \). Solving this equation is crucial as it provides the roots needed to form the general solution of the differential equation.

A homogeneous ordinary differential equation (ODE) is one where every term is a function of the dependent variable and its derivatives with zero as the right-hand side. In essence, it means there is no 'extra' function added to the equation. Our differential equation \( \frac{d^2 y}{d x^2} + 6 \frac{d y}{d x} + 2 y = 0 \) is an example of a linear homogeneous equation because:

• It's linear: Each term (\( y \, \frac{d y}{d x}, \, \frac{d^2 y}{d x^2} \)) can be separated by coefficients.
• It's homogeneous: All terms combined equal zero.

The solutions to homogeneous equations are generally easier to determine because they depend solely on finding the roots of the characteristic equation. In our case, solving the characteristic equation, \( r^2 + 6r + 2 = 0 \), and analyzing its roots will yield the general solution.

The quadratic formula is a timeless mathematical tool used when factoring a quadratic equation directly is impractical. It allows us to find the roots of any quadratic equation \( ar^2 + br + c = 0 \). The formula is written as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This method is crucial in the solution of our characteristic equation \( r^2 + 6r + 2 = 0 \). Using the specific coefficients from our equation \( a = 1, b = 6, c = 2 \), we substitute them into the formula:\[ r = \frac{-6 \pm \sqrt{36 - 8}}{2} = \frac{-6 \pm \sqrt{28}}{2} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7} \]The roots \( r_1 = -3 + \sqrt{7} \) and \( r_2 = -3 - \sqrt{7} \) are then used to construct the general solution of the differential equation, demonstrating the importance and utility of the quadratic formula in solving differential equations.