91Ó°ÊÓ

Understanding a characteristic equation is fundamental to solving some types of differential equations. In the context of the second-order linear homogeneous equation discussed here, it helps identify the behavior of solutions. For the given equation \[ \frac{d^2 y}{d x^2} + 6 \frac{d y}{d x} + 9 y = 0, \] we assume a solution of the form \( y = e^{rx} \). Why? Because exponential functions are easy to differentiate and integrate, making them suitable candidates.When we plug \( y = e^{rx} \) into the differential equation, each term transforms accordingly, resulting in a new equation involving \( r \), called the **characteristic equation**. This equation \[ r^2 + 6r + 9 = 0 \] is quadratic in nature. The solutions of this equation, \( r \), help us define the behavior of the solution to the differential equation.

Differential equations can often be categorized into two types: homogeneous and non-homogeneous. Here, we are dealing with a homogeneous differential equation. This simply means that the right-hand side of the equation is zero: \[ \frac{d^2 y}{d x^2} + 6 \frac{d y}{d x} + 9 y = 0. \]The zero signifies that there's no external force or input acting on the system described by the equation. Homogeneous equations are defined as:

• No terms are independent of the variable \( y \).
• Solutions can be found using characteristic equations.
• They often illustrate natural behaviors of systems, such as natural decay or oscillation.

Such equations are crucial in physics and engineering, where they often describe natural, undisturbed phenomena, like the cooling of an object without external heat, or a pendulum swinging without air resistance.

The quadratic formula is a mathematical principle used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It's especially useful for finding the roots of the characteristic equation derived from a differential equation. For our characteristic equation \[ r^2 + 6r + 9 = 0, \]we can identify:

• \( a = 1 \)
• \( b = 6 \)
• \( c = 9 \)

The quadratic formula is expressed as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]When applied here, it gives:\[ r = \frac{-6 \pm \sqrt{36 - 36}}{2} = \frac{-6 \pm 0}{2} = -3. \]Thus, the roots are real and equal. The quadratic formula doesn't just find roots but also indicates the nature of these roots (real vs. complex, distinct vs. repeated). In this case, equal roots signify a repeated solution in the context of second-order differential equations.